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Section 18.3-19.1. Today we will discuss finite-dimensional associative algebras and their representations. Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element a ∈ A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if all of its elements are. Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con- versely, if I ⊂ A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is A. Example. The algebra of all strictly lower- (or upper-) triangular n×n matrices is nilpotent. Proposition 3. If the algebra A is nilpotent, then An = 0 for some n ∈ Z+, that is the product of any n elements if the algebra A equals 0. Proof. Let B ⊂ A be the maximal subspace for which there exists n ∈ Z+ such that Bn = 0. Note that B is closed under multiplication, i.e. B ⊃ Bk for any k ∈ Z+. Assume B 6= A and choose an element a ∈ A\B. Since aBn = 0, there exists k ∈ Z+ such that aBk 6⊂ B. Replacing a with a non-zero element in aBk we obtain aB ⊂ B. Recall that there exists m ∈ Z+ so that am = 0. Now, let us set C = B ⊕〈a〉. Then we have Cmn = 0 which contradicts the definition of the subspace B. � Unless A is commutative, the set of all nilpotent elements of A does not have to be an ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A, then so is I + J = {x + y |x ∈ I,y ∈ J} . Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal of A. Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A). The algebra A is semisimple if rad(A) = 0. If char(F) = 0, there exists an alternative description of semisimple algebras. Consider the regular representation of the algebra A ρ: A → L(A), ρ(a)(b) = ab, and define a “scalar product” on A via (a,b) = tr(ρ(ab)) = tr(ρ(a)ρ(b)). One can see that (·, ·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc). Definition 5. For any ideal I ⊂ A, its orthogonal complement I⊥ is defined as I⊥ = {a ∈ A |(a,i) = 0 for all i ∈ I} . Proposition 6. For any ideal I ⊂ A, its orthogonal complement I⊥ ⊂ A is also an ideal. Proof. For any a ∈ A, x ∈ I⊥, and y ∈ I we have (xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0. � 1 2 Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a ∈ A orthogonal to all of its powers is nilpotent. Proof. Let a ∈ A be such that (a,an) = tr ρ(a)n+1 = 0 for all n ∈ Z+. Let L ⊃ F be the splitting field of the characteristic polynomial f(x) of the operator ρ(a). Then, over L we have f(x) = tk0 s∏ i=1 (t−λi)ki, where λi are distinct for i = 1, . . . ,s, and tr ρ(a)n+1 = s∑ i=1 kiλ n+1 i = 0. Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous linear equations in variables k1, . . . ,ks. The determinant of this system equals λ21 . . .λ 2 nV (λ1, . . . ,λn), where V (λ1.

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- Section 18.3-19.1. Today we will discuss finite-dimensional associative algebras and their representations. Definition 1. Let A be a finite-dimensional associative algebra over a field F . An element a ∈ A is nilpotent if an = 0 for some positive integer n. An algebra is said to be nilpotent if all of its elements are. Exercise 2. Every subalgebra and factor algebra of a nilpotent algebra are nilpotent. Con- versely, if I ⊂ A is a nilpotent ideal, and the quotient algebra A/I is nilpotent, then so is A. Example. The algebra of all strictly lower- (or upper-) triangular n×n matrices is nilpotent. Proposition 3. If the algebra A is nilpotent, then An = 0 for some n ∈ Z+, that is the product of any n elements if the algebra A equals 0. Proof. Let B ⊂ A be the maximal subspace for which there exists n ∈ Z+ such that Bn = 0. Note that B is closed under multiplication, i.e. B ⊃ Bk for any k ∈ Z+. Assume B 6= A and choose an element a ∈ AB. Since aBn = 0, there exists k ∈ Z+ such that aBk 6⊂ B. Replacing a with a non-zero element in aBk we obtain aB ⊂ B. Recall that there exists
- m ∈ Z+ so that am = 0. Now, let us set C = B ⊕〈a〉. Then we have Cmn = 0 which contradicts the definition of the subspace B. � Unless A is commutative, the set of all nilpotent elements of A does not have to be an ideal (in general, not even a subspace). On the other hand, if I,J are nilpotent ideals in A, then so is I + J = {x + y |x ∈ I,y ∈ J} . Therefore, there exists the maximal nilpotent ideal which contains every other nilpotent ideal of A. Definition 4. The radical of A is the maximal nilpotent ideal of A it is denoted rad(A). The algebra A is semisimple if rad(A) = 0. If char(F) = 0, there exists an alternative description of semisimple algebras. Consider the regular representation of the algebra A ρ: A → L(A), ρ(a)(b) = ab, and define a “scalar product” on A via (a,b) = tr(ρ(ab)) = tr(ρ(a)ρ(b)). One can see that (·, ·) is a symmetric bilinear function on A satisfying (ab,c) = (a,bc). Definition 5. For any ideal I ⊂ A, its orthogonal complement I⊥ is defined as
- I⊥ = {a ∈ A |(a,i) = 0 for all i ∈ I} . Proposition 6. For any ideal I ⊂ A, its orthogonal complement I⊥ ⊂ A is also an ideal. Proof. For any a ∈ A, x ∈ I⊥ , and y ∈ I we have (xa,y) = (x,ay) = 0 and (ax,y) = (y,ax) = (ya,x) = 0. � 1 2 Proposition 7. If A is an algebra over a field F with char(F) = 0, then every element a ∈ A orthogonal to all of its powers is nilpotent. Proof. Let a ∈ A be such that (a,an) = tr ρ(a)n+1 = 0 for all n ∈ Z+. Let L ⊃ F be the splitting field of the characteristic polynomial f(x) of the operator ρ(a). Then, over L we have f(x) = tk0 s∏ i=1 (t−λi)ki, where λi are distinct for i = 1, . . . ,s, and tr ρ(a)n+1 =
- s∑ i=1 kiλ n+1 i = 0. Letting n run through the set 1, . . . ,s, the above equation yields a system of s homogeneous linear equations in variables k1, . . . ,ks. The determinant of this system equals λ21 . . .λ 2 nV (λ1, . . . ,λn), where V (λ1, . . . ,λn) is the Vandermonde determinant. Therefore, the system is non-degenerate and we derive that kj = 0 in F for all j = 1, . . . ,s. If char(F) = 0 the latter is equivalent to f(x) = tk0, which implies that the operator ρ(a) is nilpotent. Therefore, for some m ∈ Z+ we have am+1 = T(a)m(a) = 0, and we conclude that a ∈ A is nilpotent. � Definition 8. The scalar product (·, ·) is non-degenerate if for any a ∈ A there exists a′ ∈ A such that (a,a′) 6= 0. Equivalently, it is nilpotent if A⊥ = 0. Theorem 9.
- (1) If the scalar product (a,b) = tr(ρ(ab)) is non-degenerate, then the algebra A is semisimple. (2) If the algebra A is semisimple and char(F) = 0, then the scalar product is non- degenerate. Proof. (1) Let I ⊂ A be a nilpotent ideal. Then for any x ∈ I and any a ∈ A, their product xa is nilpotent and hence (a,x) = tr ρ(ax) = 0. Therefore, I ⊂ A⊥ = 0. (2) Conversely, if char(F) = 0 the previous two propositions show that A⊥ is a nilpotent ideal, therefore A⊥ = 0. � Definition 10. An algebra is simple if it has no proper ideals. Note that a simple algebra is semi-simple unless it is nilpotent. However, in that case A2 ⊂ A is a proper ideal of A, hence A2 = 0. Therefore, a simple algebra is semi-simple, unless A is 1-dimensional with 0 multiplication. 3 Theorem 11. Any semisimple associative algebra A can be decomposed into a direct sum of simple algebras
- A = A1 ⊕···⊕As, moreover, any ideal of A is a sum of some direct summands in the above decomposition. Proof. We will prove this theorem in the assumption char(F) = 0. If A is simple, we have s = 1 and the statement is trivial. Otherwise, let A1 ⊂ A be a minimal ideal of A. Then either A = A1 ⊕A⊥ 1 , or A1 ⊂ A⊥ 1 . In the second case, A1 has to be nilpotent, which would contradict the semisim- plicity of A. Then, indeed A is a direct sum of A1 and A ⊥ 1 , in which case every ideal of A1 or A⊥ 1 is also an ideal of A. This allows us to conclude that A1 is simple and A ⊥ 1 is semisimple. We know apply the same reasoning to the algebra A⊥ 1 . Let now I ⊂ A be any ideal. Denote by πk : A → Ak the canonical projections, and set Ik = πk(I). Then Ik is an ideal of the algebra Ak. If Ik 6= 0 we have Ik = Ak which yields Ak = A 2 k = AkIk = AkI ⊂ I, which means that I is the sum of several summands Aj. �
- Theorem 12. Any non-trivial simple associative algebra A over an algebraically closed field F is isomorphic to L(V ) for some vector space V over F . Moreover, any non-trivial irre- ducible representation of A is isomorphic to the tautological representation of L(V ). Proof. Let ρ: A → L(A) be the regular representation, and V be a minimal invariant subspace, equivalently a minimal left ideal in A. Denote φ = ρ|V : A → L(V ). Using the fact that ker(φ) ⊂ A is an ideal, and one of the results from the previous lecture, we get either A ' φ(A) = L(V ) or dim(V ) = 1 and φ(A) = 0. In the latter case, we see that AV = 0 and therefore the subspace A0 = {x ∈ A |Ax = 0} is nonzero. Since A0 ⊂ A is an ideal, we must have A = A0 which would imply that A is trivial. Therefore, φ(A) = L(V ) for some vector space V , and φ is a tautological representation of A in V . In that case we have previously shown that the regular representation ρ: A → L(A) is isomorphic to the sum of φ with itself n times: ρ = nφ. Now, let τ : A → L(U) be any irreducible representation of A, and u0 ∈ U be any nonzero
- vector. Then the map T : A → U, T(a) = τ(a)(u0) is a morphism between the regular representation (A,ρ) and the representation (U,τ), indeed for any x ∈ A we have T(ρ(a)(x)) = T(ax) = τ(ax)(u0) = τ(a)τ(x)(u0) = τ(a)T(x). If the representation (U,τ) is non-trivial, we have im(T) = U, therefore U is a factor repre- sentation of the regular representation ρ. Since U is irreducible, we get (U,τ) ' (V,φ). � 4 Definition 13. Subalgebra Z(A) ⊂ A of an associative algebra A defined as Z(A) = {z ∈ A |za = az ∀ a ∈ A} is called the center of A. Corollary 14. Every semisimple associative algebra over an algebraically closed field F is isomorphic to the algebra A = L(V1) ⊕···⊕L(Vs) where s = dim Z(A) for some vector spaces V1, . . . ,Vs over F . If dim(Vi) = ni for i = 1, . . . ,s, we get dim(A) = n21 + · · · + n 2
- s. Proof. The only statement that have not been yet proven is that the number of direct summands is equal to dim Z(A). It follows from the fact that the center of each summand is 1-dimensional and consists of scalar matrices. The latter statement is left as an exercise. � Corollary 15. Let ρj for j = 1, . . . ,s be the irreducible representation of the algebra A ' L(V1) ⊕···⊕L(Vs) in the space Vj obtained via projection πj : A → L(Vj). Then any non-trivial irreducible representation of A is isomorphic to one of the representations ρj. Remark 16. Note that representations ρj are pairwise non- isomorphic since their kernels are distinct. Let us now get back to the representation theory of finite groups. Theorem 17. Let G be a finite group of order n. Then the algebra FG is semisimple provided char(F) does not divide n. Proof. Let ρ: FG → FG be the regular representation. Then it is easy to see that tr ρ(g) = {
- n, g = e, 0, g 6= e. Therefore, for any g,h ∈ G we have (g,h) = { n, gh = e, 0, gh 6= e. If char(F) does not divide n, this scalar product is non0degenerate, therefore FG is semisim- ple. � In what follows we will assume that F is an algebraically closed field whose characteristic does not divide |G|. By the results of the previous lecture, FG is isomorphic to a direct sum of matrix algebras L(Vi). Theorem 18. For any finite group G with |G| = n, there are only finitely many non- isomorphic irreducible representations of G over the field F . Dimensions n1, . . . ,ns of these representations satisfy n21 + · · · + n 2 s = n, and s is the number of conjugacy classes in G.
- 5 Proof. The first statement and the equality n21 + · · · + n 2 s = n follow from the fact that FG is a semisimple associative algebra, and a correspondence between representations of FG and G. We also know that s = dim Z(FG). An element a = ∑ g∈ G agg belongs to the center Z(FG) if and only if hah−1 = ∑ g∈ G aghgh −1 = ∑ g∈ G ah−1ghg = a for all h ∈ G. The latter is equivalent to the statement that the coefficients ag are constant on conjugacy classes of G: ag = ahgh−1
- for all g ∈ G. Therefore, Z(FG) is a linear span of elements of the form ∑ g∈ C g, where C ⊂ G is a conjugacy class, and dim Z(FG) equals the number of conjugacy classes of G. � Example. (1) Any irreducible representation of an abelian group is 1-dimensional, there- fore a finite abelian group of order n has precisely n non- isomorphic irreducible rep- resentations. (2) The permutation group S3 has a trivial representation, a 1- dimensional sign repre- sentation, where σ 7→ sgn(σ), and a 2-dimensional irreducible representation, where S3 acts by permuting vertices of a regular triangle. Up to isomorphism, these are all irreducible representations of S3 because 6 = 12 + 12 + 22. (3) In a similar fashion, one can find the following non- isomorphic irreducible represen- tations of S4: • the trivial one; • sign representation; • a composition of the projection S4/K4 ' S3 with the irreducible 2-dimensional representation of S3 considered above; • an isomorphism with the group of rotations of a cube (composed with its natural
- 3-dimensional representation); • an isomorphism with the permutation group of the vertices of a regular tetrahe- dron composed (composed with its natural 3-dimensional representation). Checking dimensions we get 24 = 12 + 12 + 22 + 32 + 32, and conclude that these are all irreducible representations of S4 up to isomorphism. Let us summarize, what we have proved so far. Let G be a finite group, and ρi : V → Vi, i = 1, . . . ,s be all distinct (up to isomorphism) irreducible representations of G over an algebraically closed field F , whose characteristic does not divide |G|. Then the group algebra FG can be written as FG ' L(V1) ⊕···⊕L(Vs), and ρi is simply a projection onto the i-th summand. Then the subspaces L(Vi) are the isotypic components of the regular representation ρ of the algebra FG, and the restriction of ρ onto L(Vi) is isomorphic to niρi where ni = dim(Vi). Therefore, for any a,b ∈ FG, we have (a,b) = s∑ i=1 ni tr ρi(a)ρi(b).
- 6 Now, let us consider the (finite-dimensional) space F [G] of F- valued functions on G. Every function f ∈ F[G] can be uniquely continued to a linear function on FG by f( ∑ g∈ G agg) = ∑ g∈ G agf(g), which allows us to identify F[G] with the space of linear functionals on FG, dual to FG. On the other hand, the non-degenerate scalar product (·, ·) : FG×FG −→ F provides a bijection between FG and its dual via∑ g∈ G agg 7→ ∑ g∈ G agfg where fg(h) = (g,h). Since
- (g,h) = { n, gh = e, 0, gh 6= e, we see that fg = nδg−1, where δg−1 is the Kronecker delta. Above isomorphisms allow us to drag the scalar product (·, ·) from FG onto F[G]. In terms of δ-functions we obtain (δg,δh) = 1 n2 (g−1,h−1) = { 1 n , gh = e, 0, gh 6= e, and for any functions f1,f2 ∈ F[G] we have (f1,f2) = 1 n
- ∑ g∈ G f1(g)f2(g −1). Now, recall that FG = L(V1) ⊕···⊕L(Vs). Let us now choose a basis 〈ei〉, i = 1, . . . ,ni in the representation Vi, and denote by fijk ∈ F[G] the (j,k) matrix coefficient in the representation Vi, that is fijk(g) = 〈ej,ρi(g)ek〉 . Then one can see that (fijk,f i′ j′k′) = 1 ni δii′δjk′δj′k. Definition 19. A character χφ ∈ F [G] of the representation φ: G → GL(V ) is defined by χ(g) = tr φ(g) for any g ∈ G. It is clear that he character of a sum of two representations equals the sum of the characters: χφ+ψ = χφ + χψ,
- and that a character is a class function: χ(g) = χ(hgh−1) for any g,h ∈ G. The following theorem is a direct corollary of above calculations: 7 Theorem 20. Let χi, i = 1, . . . ,s be the characters of irreducible representations ρi of the group G. Then (χi,χj) = δij, equivalently, characters χi form an orthonormal basis in the subspace of class functions in F[G]. Corollary 21. Let χ be the character of a representation φ: G → GL(V ). Then, V ' s∑ i=1 (χ,χi)Vi. In other words, (χ,χi) is the multiplicity of the irreducible representation Vi in the decompo- sition of the representation V . Proof. Representation V can be written as
- V = s∑ i=1 kiVi for some ki ∈ Z>0. Therefore χ = s∑ i=1 kiχi and (χ,χi) = ki. � Corollary 22. A representation φ: G → GL(V ) with character χ is irreducible if and only if (χ,χ) = 1. Proof. By above, (χ,χ) = s∑ i=1 k2i which is equal to 1 if and only if one of the ki is 1, and all the other are 0. � Remark 23. Above corollaries show that a character, which is clearly less data than a representation, since it only knows about traces of operators
- rather than operators themselves, is enough to recover the whole representation. Finally, if F = C it is more convenient for computations to replace the bilinear form (·, ·) with the following Hermitian scalar product: (f1 |f2) = 1 n ∑ g∈ G f1(g)f2(g), where x denotes the complex conjugate to x. If now in each of the spaces Vi, one chooses a basis orthonormal with respect to the Hermitian inner product, they would get fikj(g −1) = fijk(g), which yields (fijk |fi ′ j′k′) = (f i jk,f i′ k′j′) =
- 1 ni δii′δjj′δkk′. Example. 8 (1) A character of a 1-dimensional representation coincides with the representation. If G = 〈g |gn = e〉 is a cyclic group of order n, its irreducible representations are defined by ρk(g) = ω k−1, where ω = e2πi/n. Therefore, the table of characters of G takes the following form χ0 χ1 . . . χn−1 e 1 1 . . . 1 g 1 ω . . . ωn−1 g2 1 ω2 . . . ω2(n−1) ... ... ... . . .
- ... gn−1 1 ωn−1 . . . ω(n−1) 2 (2) Using the description of irreducible representations of the group S4, we can write out its table of characters as well χ1 χ ′ 1 χ2 χ3 χ ′ 3 e 1 1 2 3 3 1 (12) 1 -1 0 -1 1 6 (12)(34) 1 1 2 -1 -1 3 (123) 1 1 -1 0 0 8 (1234) 1 -1 0 1 -1 6 where the first column shows representatives of conjugacy classes in S4, and the last column shows number of elements in the corresponding conjugacy class. For example one gets (χ2,χ3) = (χ2 |χ3) = 1 24 (2 · 3 · 1 + 0 · (−1) · 6 + 2 · (−1) · 3 + (−1) · 0 · 8 + 0 · 1 · 6) = 0.
- (3) Let V be a 6-dimensional space of functions on the set of faces of a cube. An iso- morphism between S4 and the rotation group of a cube defines a representation of φ: S4 → GL(V ), let χ be its character. Every element g ∈ S4 permutes the faces of a cube, and therefore permutes δ-functions of these faces accordingly. Therefore, χ(g) = tr φ(g) equals the number of faces fixed by g. This allows us to compute the character χ: e (12) (12)(34) (123) (1234) χ 6 0 2 0 2 Computing scalar products of χ with those of irreducible representations of S4 we find (χ |χ1) = 1, (χ |χ′1) = 0, (χ |χ2) = 1, (χ |χ3) = 1, (χ |χ ′ 3) = 0. Therefore, φ ' ρ1 ⊕ρ2 ⊕ρ3. Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] The Project Gutenberg EBook of Phaedra, by Jean Baptiste Racine
- This eBook is for the use of anyone anywhere at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online at www.gutenberg.org Title: Phaedra Author: Jean Baptiste Racine Translator: Robert Bruce Boswell Release Date: October 30, 2008 [EBook #1977] Last Updated: February 7, 2013 Language: English Character set encoding: ASCII *** START OF THIS PROJECT GUTENBERG EBOOK PHAEDRA *** Produced by Dagny, John Bickers, and David Widger PHAEDRA By Jean Baptiste Racine Translated by Robert Bruce Boswell Contents INTRODUCTORY NOTE
- PHAEDRA ACT I ACT II Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] ACT III ACT IV ACT V INTRODUCTORY NOTE JEAN BAPTISTE RACINE, the younger contemporary of Corneille, and his rival for supremacy in French classical tragedy, was born at Ferte-Milon, December 21, 1639. He was educated at the College of Beauvais, at the great Jansenist school at Port Royal, and at the College d'Harcourt. He attracted notice by an ode written for the marriage of Louis XIV in 1660, and made his first really great dramatic success with his "Andromaque." His tragic masterpieces include "Britannicus," "Berenice," "Bajazet," "Mithridate," "Iphigenie," and "Phaedre," all written between 1669 and 1677. Then for some years he gave up dramatic composition, disgusted by the intrigues of enemies who sought to injure his career by exalting above him an
- unworthy rival. In 1689 he resumed his work under the persuasion of Mme. de Maintenon, and produced "Esther" and "Athalie," the latter ranking among his finest productions, although it did not receive public recognition until some time after his death in 1699. Besides his tragedies, Racine wrote one comedy, "Les Plaideurs," four hymns of great beauty, and a history of Port Royal. The external conventions of classical tragedy which had been established by Corneille, Racine did not attempt to modify. His study of the Greek tragedians and his own taste led him to submit willingly to the rigor and simplicity of form which were the fundamental marks of the classical ideal. It was in his treatment of character that he differed most from his predecessor; for whereas, as we have seen, Corneille represented his leading figures as heroically subduing passion by force of will, Racine represents his as driven by almost uncontrollable passion. Thus his creations appeal to the modern reader as more warmly human; their speech, if less exalted, is simpler and more natural; and he succeeds more brilliantly with his portraits of women than with those of men. All these characteristics are exemplified in "Phaedre," the tragedy of Racine which has made an appeal to the widest audience. To the legend as treated by Euripides, Racine added the love of Hippolytus for Aricia, and thus supplied a motive for Phaedra's jealousy, and at the same time he made the nurse instead of Phaedra the calumniator of his son to Theseus. PHAEDRA CHARACTERS
- THESEUS, son of Aegeus and King of Athens. PHAEDRA, wife of Theseus and Daughter of Minos and Pasiphae. HIPPOLYTUS, son of Theseus and Antiope, Queen of the Amazons. ARICIA, Princess of the Blood Royal of Athens. OENONE, nurse of Phaedra. THERAMENES, tutor of Hippolytus. ISMENE, bosom friend of Aricia. PANOPE, waiting-woman of Phaedra. GUARDS. The scene is laid at Troezen, a town of the Peloponnesus. Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] ACT I SCENE I HIPPOLYTUS, THERAMENES HIPPOLYTUS My mind is settled, dear Theramenes, And I can stay no more in lovely Troezen. In doubt that racks my soul with mortal anguish, I grow ashamed of such long idleness. Six months and more my father has been gone, And what may have befallen one so dear I know not, nor what corner of the earth Hides him. THERAMENES
- And where, prince, will you look for him? Already, to content your just alarm, Have I not cross'd the seas on either side Of Corinth, ask'd if aught were known of Theseus Where Acheron is lost among the Shades, Visited Elis, doubled Toenarus, And sail'd into the sea that saw the fall Of Icarus? Inspired with what new hope, Under what favour'd skies think you to trace His footsteps? Who knows if the King, your father, Wishes the secret of his absence known? Perchance, while we are trembling for his life, The hero calmly plots some fresh intrigue, And only waits till the deluded fair— HIPPOLYTUS Cease, dear Theramenes, respect the name Of Theseus. Youthful errors have been left Behind, and no unworthy obstacle Detains him. Phaedra long has fix'd a heart Inconstant once, nor need she fear a rival. In seeking him I shall but do my duty, And leave a place I dare no longer see. THERAMENES Indeed! When, prince, did you begin to dread These peaceful haunts, so dear to happy childhood, Where I have seen you oft prefer to stay, Rather than meet the tumult and the pomp Of Athens and the court? What danger shun you, Or shall I say what grief? HIPPOLYTUS That happy time Is gone, and all is changed, since to these shores The gods sent Phaedra.
- THERAMENES I perceive the cause Of your distress. It is the queen whose sight Offends you. With a step-dame's spite she schemed Your exile soon as she set eyes on you. But if her hatred is not wholly vanish'd, It has at least taken a milder aspect. Besides, what danger can a dying woman, One too who longs for death, bring on your head? Can Phaedra, sick'ning of a dire disease Of which she will not speak, weary of life And of herself, form any plots against you? HIPPOLYTUS It is not her vain enmity I fear, Another foe alarms Hippolytus. I fly, it must be own'd, from young Aricia, The sole survivor of an impious race. THERAMENES What! You become her persecutor too! The gentle sister of the cruel sons Of Pallas shared not in their perfidy; Why should you hate such charming innocence? HIPPOLYTUS I should not need to fly, if it were hatred. THERAMENES May I, then, learn the meaning of your flight? Is this the proud Hippolytus I see, Than whom there breathed no fiercer foe to love And to that yoke which Theseus has so oft Endured? And can it be that Venus, scorn'd So long, will justify your sire at last?
- Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] Has she, then, setting you with other mortals, Forced e'en Hippolytus to offer incense Before her? Can you love? HIPPOLYTUS Friend, ask me not. You, who have known my heart from infancy And all its feelings of disdainful pride, Spare me the shame of disavowing all That I profess'd. Born of an Amazon, The wildness that you wonder at I suck'd With mother's milk. When come to riper age, Reason approved what Nature had implanted. Sincerely bound to me by zealous service, You told me then the story of my sire, And know how oft, attentive to your voice, I kindled when I heard his noble acts, As you described him bringing consolation To mortals for the absence of Alcides, The highways clear'd of monsters and of robbers, Procrustes, Cercyon, Sciro, Sinnis slain, The Epidaurian giant's bones dispersed, Crete reeking with the blood of Minotaur. But when you told me of less glorious deeds, Troth plighted here and there and everywhere, Young Helen stolen from her home at Sparta, And Periboea's tears in Salamis, With many another trusting heart deceived
- Whose very names have 'scaped his memory, Forsaken Ariadne to the rocks Complaining, last this Phaedra, bound to him By better ties,—you know with what regret I heard and urged you to cut short the tale, Happy had I been able to erase From my remembrance that unworthy part Of such a splendid record. I, in turn, Am I too made the slave of love, and brought To stoop so low? The more contemptible That no renown is mine such as exalts The name of Theseus, that no monsters quell'd Have given me a right to share his weakness. And if my pride of heart must needs be humbled, Aricia should have been the last to tame it. Was I beside myself to have forgotten Eternal barriers of separation Between us? By my father's stern command Her brethren's blood must ne'er be reinforced By sons of hers; he dreads a single shoot From stock so guilty, and would fain with her Bury their name, that, even to the tomb Content to be his ward, for her no torch Of Hymen may be lit. Shall I espouse Her rights against my sire, rashly provoke His wrath, and launch upon a mad career— THERAMENES The gods, dear prince, if once your hour is come, Care little for the reasons that should guide us. Wishing to shut your eyes, Theseus unseals them; His hatred, stirring a rebellious flame Within you, lends his enemy new charms. And, after all, why should a guiltless passion Alarm you? Dare you not essay its sweetness, But follow rather a fastidious scruple?
- Fear you to stray where Hercules has wander'd? What heart so stout that Venus has not vanquish'd? Where would you be yourself, so long her foe, Had your own mother, constant in her scorn Of love, ne'er glowed with tenderness for Theseus? What boots it to affect a pride you feel not? Confess it, all is changed; for some time past You have been seldom seen with wild delight Urging the rapid car along the strand, Or, skilful in the art that Neptune taught, Making th' unbroken steed obey the bit; Less often have the woods return'd our shouts; A secret burden on your spirits cast Has dimm'd your eye. How can I doubt you love? Vainly would you conceal the fatal wound. Has not the fair Aricia touch'd your heart? HIPPOLYTUS Theramenes, I go to find my father. THERAMENES Will you not see the queen before you start, My prince? HIPPOLYTUS That is my purpose: you can tell her. Yes, I will see her; duty bids me do it. But what new ill vexes her dear Oenone? SCENE II HIPPOLYTUS, OENONE, THERAMENES OENONE
- Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] Alas, my lord, what grief was e'er like mine? The queen has almost touch'd the gates of death. Vainly close watch I keep by day and night, E'en in my arms a secret malady Slays her, and all her senses are disorder'd. Weary yet restless from her couch she rises, Pants for the outer air, but bids me see That no one on her misery intrudes. She comes. HIPPOLYTUS Enough. She shall not be disturb'd, Nor be confronted with a face she hates. SCENE III PHAEDRA, OENONE PHAEDRA We have gone far enough. Stay, dear Oenone; Strength fails me, and I needs must rest awhile. My eyes are dazzled with this glaring light So long unseen, my trembling knees refuse Support. Ah me! OENONE Would Heaven that our tears Might bring relief! PHAEDRA Ah, how these cumbrous gauds, These veils oppress me! What officious hand
- Has tied these knots, and gather'd o'er my brow These clustering coils? How all conspires to add To my distress! OENONE What is one moment wish'd, The next, is irksome. Did you not just now, Sick of inaction, bid us deck you out, And, with your former energy recall'd, Desire to go abroad, and see the light Of day once more? You see it, and would fain Be hidden from the sunshine that you sought. PHAEDRA Thou glorious author of a hapless race, Whose daughter 'twas my mother's boast to be, Who well may'st blush to see me in such plight, For the last time I come to look on thee, O Sun! OENONE What! Still are you in love with death? Shall I ne'er see you, reconciled to life, Forego these cruel accents of despair? PHAEDRA Would I were seated in the forest's shade! When may I follow with delighted eye, Thro' glorious dust flying in full career, A chariot— OENONE Madam? PHAEDRA Have I lost my senses?
- What said I? and where am I? Whither stray Vain wishes? Ah! The gods have made me mad. I blush, Oenone, and confusion covers My face, for I have let you see too clearly The shame of grief that, in my own despite, O'erflows these eyes of mine. OENONE If you must blush, Blush at a silence that inflames your woes. Resisting all my care, deaf to my voice, Will you have no compassion on yourself, But let your life be ended in mid course? What evil spell has drain'd its fountain dry? Thrice have the shades of night obscured the heav'ns Since sleep has enter'd thro' your eyes, and thrice The dawn has chased the darkness thence, since food Pass'd your wan lips, and you are faint and languid. To what dread purpose is your heart inclined? How dare you make attempts upon your life, And so offend the gods who gave it you, Prove false to Theseus and your marriage vows, Ay, and betray your most unhappy children, Bending their necks yourself beneath the yoke? That day, be sure, which robs them of their mother, Will give high hopes back to the stranger's son, To that proud enemy of you and yours, To whom an Amazon gave birth, I mean Hippolytus— Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM]
- PHAEDRA Ye gods! OENONE Ah, this reproach Moves you! PHAEDRA Unhappy woman, to what name Gave your mouth utterance? OENONE Your wrath is just. 'Tis well that that ill-omen'd name can rouse Such rage. Then live. Let love and duty urge Their claims. Live, suffer not this son of Scythia, Crushing your children 'neath his odious sway, To rule the noble offspring of the gods, The purest blood of Greece. Make no delay; Each moment threatens death; quickly restore Your shatter'd strength, while yet the torch of life Holds out, and can be fann'd into a flame. PHAEDRA Too long have I endured its guilt and shame! OENONE Why? What remorse gnaws at your heart? What crime Can have disturb'd you thus? Your hands are not Polluted with the blood of innocence? PHAEDRA Thanks be to Heav'n, my hands are free from stain. Would that my soul were innocent as they!
- OENONE What awful project have you then conceived, Whereat your conscience should be still alarm'd? PHAEDRA Have I not said enough? Spare me the rest. I die to save myself a full confession. OENONE Die then, and keep a silence so inhuman; But seek some other hand to close your eyes. Tho' but a spark of life remains within you, My soul shall go before you to the Shades. A thousand roads are always open thither; Pain'd at your want of confidence, I'll choose The shortest. Cruel one, when has my faith Deceived you! Think how in my arms you lay New born. For you, my country and my children I have forsaken. Do you thus repay My faithful service? PHAEDRA What do you expect From words so bitter? Were I to break silence Horror would freeze your blood. OENONE What can you say To horrify me more than to behold You die before my eyes? PHAEDRA When you shall know My crime, my death will follow none the less, But with the added stain of guilt.
- OENONE Dear Madam, By all the tears that I have shed for you, By these weak knees I clasp, relieve my mind From torturing doubt. PHAEDRA It is your wish. Then rise. OENONE I hear you. Speak. PHAEDRA Heav'ns! How shall I begin? OENONE Dismiss vain fears, you wound me with distrust. PHAEDRA O fatal animosity of Venus! Into what wild distractions did she cast My mother! OENONE Be they blotted from remembrance, And for all time to come buried in silence. Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] PHAEDRA My sister Ariadne, by what love
- Were you betray'd to death, on lonely shores Forsaken! OENONE Madam, what deep-seated pain Prompts these reproaches against all your kin? PHAEDRA It is the will of Venus, and I perish, Last, most unhappy of a family Where all were wretched. OENONE Do you love? PHAEDRA I feel All its mad fever. OENONE Ah! For whom? PHAEDRA Hear now The crowning horror. Yes, I love—my lips Tremble to say his name. OENONE Whom? PHAEDRA Know you him, Son of the Amazon, whom I've oppress'd So long? OENONE
- Hippolytus? Great gods! PHAEDRA 'Tis you Have named him. OENONE All my blood within my veins Seems frozen. O despair! O cursed race! Ill-omen'd journey! Land of misery! Why did we ever reach thy dangerous shores? PHAEDRA My wound is not so recent. Scarcely had I Been bound to Theseus by the marriage yoke, And happiness and peace seem'd well secured, When Athens show'd me my proud enemy. I look'd, alternately turn'd pale and blush'd To see him, and my soul grew all distraught; A mist obscured my vision, and my voice Falter'd, my blood ran cold, then burn'd like fire; Venus I felt in all my fever'd frame, Whose fury had so many of my race Pursued. With fervent vows I sought to shun Her torments, built and deck'd for her a shrine, And there, 'mid countless victims did I seek The reason I had lost; but all for naught, No remedy could cure the wounds of love! In vain I offer'd incense on her altars; When I invoked her name my heart adored Hippolytus, before me constantly; And when I made her altars smoke with victims, 'Twas for a god whose name I dared not utter. I fled his presence everywhere, but found him— O crowning horror!—in his father's features. Against myself, at last, I raised revolt,
- And stirr'd my courage up to persecute The enemy I loved. To banish him I wore a step—dame's harsh and jealous carriage, With ceaseless cries I clamour'd for his exile, Till I had torn him from his father's arms. I breathed once more, Oenone; in his absence My days flow'd on less troubled than before, And innocent. Submissive to my husband, I hid my grief, and of our fatal marriage Cherish'd the fruits. Vain caution! Cruel Fate! Brought hither by my spouse himself, I saw Again the enemy whom I had banish'd, And the old wound too quickly bled afresh. No longer is it love hid in my heart, But Venus in her might seizing her prey. I have conceived just terror for my crime; I hate my life, and hold my love in horror. Dying I wish'd to keep my fame unsullied, And bury in the grave a guilty passion; But I have been unable to withstand Tears and entreaties, I have told you all; Content, if only, as my end draws near, Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] You do not vex me with unjust reproaches, Nor with vain efforts seek to snatch from death The last faint lingering sparks of vital breath. SCENE IV PHAEDRA, OENONE, PANOPE
- PANOPE Fain would I hide from you tidings so sad, But 'tis my duty, Madam, to reveal them. The hand of death has seized your peerless husband, And you are last to hear of this disaster. OENONE What say you, Panope? PANOPE The queen, deceived By a vain trust in Heav'n, begs safe return For Theseus, while Hippolytus his son Learns of his death from vessels that are now In port. PHAEDRA Ye gods! PANOPE Divided counsels sway The choice of Athens; some would have the prince, Your child, for master; others, disregarding The laws, dare to support the stranger's son. 'Tis even said that a presumptuous faction Would crown Aricia and the house of Pallas. I deem'd it right to warn you of this danger. Hippolytus already is prepared To start, and should he show himself at Athens, 'Tis to be fear'd the fickle crowd will all Follow his lead. OENONE Enough. The queen, who hears you, By no means will neglect this timely warning.
- SCENE V PHAEDRA, OENONE OENONE Dear lady, I had almost ceased to urge The wish that you should live, thinking to follow My mistress to the tomb, from which my voice Had fail'd to turn you; but this new misfortune Alters the aspect of affairs, and prompts Fresh measures. Madam, Theseus is no more, You must supply his place. He leaves a son, A slave, if you should die, but, if you live, A King. On whom has he to lean but you? No hand but yours will dry his tears. Then live For him, or else the tears of innocence Will move the gods, his ancestors, to wrath Against his mother. Live, your guilt is gone, No blame attaches to your passion now. The King's decease has freed you from the bonds That made the crime and horror of your love. Hippolytus no longer need be dreaded, Him you may see henceforth without reproach. It may be, that, convinced of your aversion, He means to head the rebels. Undeceive him, Soften his callous heart, and bend his pride. King of this fertile land, in Troezen here His portion lies; but as he knows, the laws Give to your son the ramparts that Minerva Built and protects. A common enemy Threatens you both, unite them to oppose Aricia. PHAEDRA To your counsel I consent. Yes, I will live, if life can be restored,
- If my affection for a son has pow'r To rouse my sinking heart at such a dangerous hour. ACT II SCENE I ARICIA, ISMENE Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] ARICIA Hippolytus request to see me here! Hippolytus desire to bid farewell! Is't true, Ismene? Are you not deceived? ISMENE This is the first result of Theseus' death. Prepare yourself to see from every side. Hearts turn towards you that were kept away By Theseus. Mistress of her lot at last, Aricia soon shall find all Greece fall low, To do her homage. ARICIA 'Tis not then, Ismene, An idle tale? Am I no more a slave? Have I no enemies? ISMENE The gods oppose Your peace no longer, and the soul of Theseus Is with your brothers.
- ARICIA Does the voice of fame Tell how he died? ISMENE Rumours incredible Are spread. Some say that, seizing a new bride, The faithless husband by the waves was swallow'd. Others affirm, and this report prevails, That with Pirithous to the world below He went, and saw the shores of dark Cocytus, Showing himself alive to the pale ghosts; But that he could not leave those gloomy realms, Which whoso enters there abides for ever. ARICIA Shall I believe that ere his destined hour A mortal may descend into the gulf Of Hades? What attraction could o'ercome Its terrors? ISMENE He is dead, and you alone Doubt it. The men of Athens mourn his loss. Troezen already hails Hippolytus As King. And Phaedra, fearing for her son, Asks counsel of the friends who share her trouble, Here in this palace. ARICIA Will Hippolytus, Think you, prove kinder than his sire, make light My chains, and pity my misfortunes? ISMENE
- Yes, I think so, Madam. ARICIA Ah, you know him not Or you would never deem so hard a heart Can pity feel, or me alone except From the contempt in which he holds our sex. Has he not long avoided every spot Where we resort? ISMENE I know what tales are told Of proud Hippolytus, but I have seen Him near you, and have watch'd with curious eye How one esteem'd so cold would bear himself. Little did his behavior correspond With what I look'd for; in his face confusion Appear'd at your first glance, he could not turn His languid eyes away, but gazed on you. Love is a word that may offend his pride, But what the tongue disowns, looks can betray. ARICIA How eagerly my heart hears what you say, Tho' it may be delusion, dear Ismene! Did it seem possible to you, who know me, That I, sad sport of a relentless Fate, Fed upon bitter tears by night and day, Could ever taste the maddening draught of love? The last frail offspring of a royal race, Children of Earth, I only have survived War's fury. Cut off in the flow'r of youth, Mown by the sword, six brothers have I …
- Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] The Project Gutenberg EBook of Phaedra, by Jean Baptiste Racine This eBook is for the use of anyone anywhere at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this eBook or online at www.gutenberg.org Title: Phaedra Author: Jean Baptiste Racine Translator: Robert Bruce Boswell Release Date: October 30, 2008 [EBook #1977] Last Updated: February 7, 2013 Language: English Character set encoding: ASCII *** START OF THIS PROJECT GUTENBERG EBOOK PHAEDRA *** Produced by Dagny, John Bickers, and David Widger PHAEDRA
- By Jean Baptiste Racine Translated by Robert Bruce Boswell Contents INTRODUCTORY NOTE PHAEDRA ACT I ACT II Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] ACT III ACT IV ACT V INTRODUCTORY NOTE JEAN BAPTISTE RACINE, the younger contemporary of Corneille, and his rival for supremacy in French classical tragedy, was born at Ferte-Milon, December 21, 1639. He was educated at the College of Beauvais, at the great Jansenist school at Port Royal, and at the College
- d'Harcourt. He attracted notice by an ode written for the marriage of Louis XIV in 1660, and made his first really great dramatic success with his "Andromaque." His tragic masterpieces include "Britannicus," "Berenice," "Bajazet," "Mithridate," "Iphigenie," and "Phaedre," all written between 1669 and 1677. Then for some years he gave up dramatic composition, disgusted by the intrigues of enemies who sought to injure his career by exalting above him an unworthy rival. In 1689 he resumed his work under the persuasion of Mme. de Maintenon, and produced "Esther" and "Athalie," the latter ranking among his finest productions, although it did not receive public recognition until some time after his death in 1699. Besides his tragedies, Racine wrote one comedy, "Les Plaideurs," four hymns of great beauty, and a history of Port Royal. The external conventions of classical tragedy which had been established by Corneille, Racine did not attempt to modify. His study of the Greek tragedians and his own taste led him to submit willingly to the rigor and simplicity of form which were the fundamental marks of the classical ideal. It was in his treatment of character that he differed most from his predecessor; for whereas, as we have seen, Corneille represented his leading figures as heroically subduing passion by force of will, Racine represents his as driven by almost uncontrollable passion. Thus his creations appeal to the modern reader as more warmly human; their speech, if less exalted, is simpler and more natural; and he succeeds more brilliantly with his portraits of women than with those of men. All these characteristics are exemplified in "Phaedre," the tragedy of Racine which has made an appeal to
- the widest audience. To the legend as treated by Euripides, Racine added the love of Hippolytus for Aricia, and thus supplied a motive for Phaedra's jealousy, and at the same time he made the nurse instead of Phaedra the calumniator of his son to Theseus. PHAEDRA CHARACTERS THESEUS, son of Aegeus and King of Athens. PHAEDRA, wife of Theseus and Daughter of Minos and Pasiphae. HIPPOLYTUS, son of Theseus and Antiope, Queen of the Amazons. ARICIA, Princess of the Blood Royal of Athens. OENONE, nurse of Phaedra. THERAMENES, tutor of Hippolytus. ISMENE, bosom friend of Aricia. PANOPE, waiting-woman of Phaedra. GUARDS. The scene is laid at Troezen, a town of the Peloponnesus. Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] ACT I SCENE I HIPPOLYTUS, THERAMENES HIPPOLYTUS My mind is settled, dear Theramenes,
- And I can stay no more in lovely Troezen. In doubt that racks my soul with mortal anguish, I grow ashamed of such long idleness. Six months and more my father has been gone, And what may have befallen one so dear I know not, nor what corner of the earth Hides him. THERAMENES And where, prince, will you look for him? Already, to content your just alarm, Have I not cross'd the seas on either side Of Corinth, ask'd if aught were known of Theseus Where Acheron is lost among the Shades, Visited Elis, doubled Toenarus, And sail'd into the sea that saw the fall Of Icarus? Inspired with what new hope, Under what favour'd skies think you to trace His footsteps? Who knows if the King, your father, Wishes the secret of his absence known? Perchance, while we are trembling for his life, The hero calmly plots some fresh intrigue, And only waits till the deluded fair— HIPPOLYTUS Cease, dear Theramenes, respect the name Of Theseus. Youthful errors have been left Behind, and no unworthy obstacle Detains him. Phaedra long has fix'd a heart Inconstant once, nor need she fear a rival. In seeking him I shall but do my duty, And leave a place I dare no longer see. THERAMENES Indeed! When, prince, did you begin to dread These peaceful haunts, so dear to happy childhood,
- Where I have seen you oft prefer to stay, Rather than meet the tumult and the pomp Of Athens and the court? What danger shun you, Or shall I say what grief? HIPPOLYTUS That happy time Is gone, and all is changed, since to these shores The gods sent Phaedra. THERAMENES I perceive the cause Of your distress. It is the queen whose sight Offends you. With a step-dame's spite she schemed Your exile soon as she set eyes on you. But if her hatred is not wholly vanish'd, It has at least taken a milder aspect. Besides, what danger can a dying woman, One too who longs for death, bring on your head? Can Phaedra, sick'ning of a dire disease Of which she will not speak, weary of life And of herself, form any plots against you? HIPPOLYTUS It is not her vain enmity I fear, Another foe alarms Hippolytus. I fly, it must be own'd, from young Aricia, The sole survivor of an impious race. THERAMENES What! You become her persecutor too! The gentle sister of the cruel sons Of Pallas shared not in their perfidy; Why should you hate such charming innocence? HIPPOLYTUS
- I should not need to fly, if it were hatred. THERAMENES May I, then, learn the meaning of your flight? Is this the proud Hippolytus I see, Than whom there breathed no fiercer foe to love And to that yoke which Theseus has so oft Endured? And can it be that Venus, scorn'd So long, will justify your sire at last? Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] Has she, then, setting you with other mortals, Forced e'en Hippolytus to offer incense Before her? Can you love? HIPPOLYTUS Friend, ask me not. You, who have known my heart from infancy And all its feelings of disdainful pride, Spare me the shame of disavowing all That I profess'd. Born of an Amazon, The wildness that you wonder at I suck'd With mother's milk. When come to riper age, Reason approved what Nature had implanted. Sincerely bound to me by zealous service, You told me then the story of my sire, And know how oft, attentive to your voice, I kindled when I heard his noble acts, As you described him bringing consolation To mortals for the absence of Alcides,
- The highways clear'd of monsters and of robbers, Procrustes, Cercyon, Sciro, Sinnis slain, The Epidaurian giant's bones dispersed, Crete reeking with the blood of Minotaur. But when you told me of less glorious deeds, Troth plighted here and there and everywhere, Young Helen stolen from her home at Sparta, And Periboea's tears in Salamis, With many another trusting heart deceived Whose very names have 'scaped his memory, Forsaken Ariadne to the rocks Complaining, last this Phaedra, bound to him By better ties,—you know with what regret I heard and urged you to cut short the tale, Happy had I been able to erase From my remembrance that unworthy part Of such a splendid record. I, in turn, Am I too made the slave of love, and brought To stoop so low? The more contemptible That no renown is mine such as exalts The name of Theseus, that no monsters quell'd Have given me a right to share his weakness. And if my pride of heart must needs be humbled, Aricia should have been the last to tame it. Was I beside myself to have forgotten Eternal barriers of separation Between us? By my father's stern command Her brethren's blood must ne'er be reinforced By sons of hers; he dreads a single shoot From stock so guilty, and would fain with her Bury their name, that, even to the tomb Content to be his ward, for her no torch Of Hymen may be lit. Shall I espouse Her rights against my sire, rashly provoke His wrath, and launch upon a mad career—
- THERAMENES The gods, dear prince, if once your hour is come, Care little for the reasons that should guide us. Wishing to shut your eyes, Theseus unseals them; His hatred, stirring a rebellious flame Within you, lends his enemy new charms. And, after all, why should a guiltless passion Alarm you? Dare you not essay its sweetness, But follow rather a fastidious scruple? Fear you to stray where Hercules has wander'd? What heart so stout that Venus has not vanquish'd? Where would you be yourself, so long her foe, Had your own mother, constant in her scorn Of love, ne'er glowed with tenderness for Theseus? What boots it to affect a pride you feel not? Confess it, all is changed; for some time past You have been seldom seen with wild delight Urging the rapid car along the strand, Or, skilful in the art that Neptune taught, Making th' unbroken steed obey the bit; Less often have the woods return'd our shouts; A secret burden on your spirits cast Has dimm'd your eye. How can I doubt you love? Vainly would you conceal the fatal wound. Has not the fair Aricia touch'd your heart? HIPPOLYTUS Theramenes, I go to find my father. THERAMENES Will you not see the queen before you start, My prince? HIPPOLYTUS That is my purpose: you can tell her. Yes, I will see her; duty bids me do it.
- But what new ill vexes her dear Oenone? SCENE II HIPPOLYTUS, OENONE, THERAMENES OENONE Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] Alas, my lord, what grief was e'er like mine? The queen has almost touch'd the gates of death. Vainly close watch I keep by day and night, E'en in my arms a secret malady Slays her, and all her senses are disorder'd. Weary yet restless from her couch she rises, Pants for the outer air, but bids me see That no one on her misery intrudes. She comes. HIPPOLYTUS Enough. She shall not be disturb'd, Nor be confronted with a face she hates. SCENE III PHAEDRA, OENONE PHAEDRA We have gone far enough. Stay, dear Oenone; Strength fails me, and I needs must rest awhile. My eyes are dazzled with this glaring light So long unseen, my trembling knees refuse
- Support. Ah me! OENONE Would Heaven that our tears Might bring relief! PHAEDRA Ah, how these cumbrous gauds, These veils oppress me! What officious hand Has tied these knots, and gather'd o'er my brow These clustering coils? How all conspires to add To my distress! OENONE What is one moment wish'd, The next, is irksome. Did you not just now, Sick of inaction, bid us deck you out, And, with your former energy recall'd, Desire to go abroad, and see the light Of day once more? You see it, and would fain Be hidden from the sunshine that you sought. PHAEDRA Thou glorious author of a hapless race, Whose daughter 'twas my mother's boast to be, Who well may'st blush to see me in such plight, For the last time I come to look on thee, O Sun! OENONE What! Still are you in love with death? Shall I ne'er see you, reconciled to life, Forego these cruel accents of despair? PHAEDRA Would I were seated in the forest's shade!
- When may I follow with delighted eye, Thro' glorious dust flying in full career, A chariot— OENONE Madam? PHAEDRA Have I lost my senses? What said I? and where am I? Whither stray Vain wishes? Ah! The gods have made me mad. I blush, Oenone, and confusion covers My face, for I have let you see too clearly The shame of grief that, in my own despite, O'erflows these eyes of mine. OENONE If you must blush, Blush at a silence that inflames your woes. Resisting all my care, deaf to my voice, Will you have no compassion on yourself, But let your life be ended in mid course? What evil spell has drain'd its fountain dry? Thrice have the shades of night obscured the heav'ns Since sleep has enter'd thro' your eyes, and thrice The dawn has chased the darkness thence, since food Pass'd your wan lips, and you are faint and languid. To what dread purpose is your heart inclined? How dare you make attempts upon your life, And so offend the gods who gave it you, Prove false to Theseus and your marriage vows, Ay, and betray your most unhappy children, Bending their necks yourself beneath the yoke? That day, be sure, which robs them of their mother, Will give high hopes back to the stranger's son, To that proud enemy of you and yours,
- To whom an Amazon gave birth, I mean Hippolytus— Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] PHAEDRA Ye gods! OENONE Ah, this reproach Moves you! PHAEDRA Unhappy woman, to what name Gave your mouth utterance? OENONE Your wrath is just. 'Tis well that that ill-omen'd name can rouse Such rage. Then live. Let love and duty urge Their claims. Live, suffer not this son of Scythia, Crushing your children 'neath his odious sway, To rule the noble offspring of the gods, The purest blood of Greece. Make no delay; Each moment threatens death; quickly restore Your shatter'd strength, while yet the torch of life Holds out, and can be fann'd into a flame. PHAEDRA Too long have I endured its guilt and shame!
- OENONE Why? What remorse gnaws at your heart? What crime Can have disturb'd you thus? Your hands are not Polluted with the blood of innocence? PHAEDRA Thanks be to Heav'n, my hands are free from stain. Would that my soul were innocent as they! OENONE What awful project have you then conceived, Whereat your conscience should be still alarm'd? PHAEDRA Have I not said enough? Spare me the rest. I die to save myself a full confession. OENONE Die then, and keep a silence so inhuman; But seek some other hand to close your eyes. Tho' but a spark of life remains within you, My soul shall go before you to the Shades. A thousand roads are always open thither; Pain'd at your want of confidence, I'll choose The shortest. Cruel one, when has my faith Deceived you! Think how in my arms you lay New born. For you, my country and my children I have forsaken. Do you thus repay My faithful service? PHAEDRA What do you expect From words so bitter? Were I to break silence Horror would freeze your blood. OENONE
- What can you say To horrify me more than to behold You die before my eyes? PHAEDRA When you shall know My crime, my death will follow none the less, But with the added stain of guilt. OENONE Dear Madam, By all the tears that I have shed for you, By these weak knees I clasp, relieve my mind From torturing doubt. PHAEDRA It is your wish. Then rise. OENONE I hear you. Speak. PHAEDRA Heav'ns! How shall I begin? OENONE Dismiss vain fears, you wound me with distrust. PHAEDRA O fatal animosity of Venus! Into what wild distractions did she cast My mother! OENONE Be they blotted from remembrance, And for all time to come buried in silence.
- Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] PHAEDRA My sister Ariadne, by what love Were you betray'd to death, on lonely shores Forsaken! OENONE Madam, what deep-seated pain Prompts these reproaches against all your kin? PHAEDRA It is the will of Venus, and I perish, Last, most unhappy of a family Where all were wretched. OENONE Do you love? PHAEDRA I feel All its mad fever. OENONE Ah! For whom? PHAEDRA Hear now The crowning horror. Yes, I love—my lips Tremble to say his name.
- OENONE Whom? PHAEDRA Know you him, Son of the Amazon, whom I've oppress'd So long? OENONE Hippolytus? Great gods! PHAEDRA 'Tis you Have named him. OENONE All my blood within my veins Seems frozen. O despair! O cursed race! Ill-omen'd journey! Land of misery! Why did we ever reach thy dangerous shores? PHAEDRA My wound is not so recent. Scarcely had I Been bound to Theseus by the marriage yoke, And happiness and peace seem'd well secured, When Athens show'd me my proud enemy. I look'd, alternately turn'd pale and blush'd To see him, and my soul grew all distraught; A mist obscured my vision, and my voice Falter'd, my blood ran cold, then burn'd like fire; Venus I felt in all my fever'd frame, Whose fury had so many of my race Pursued. With fervent vows I sought to shun Her torments, built and deck'd for her a shrine, And there, 'mid countless victims did I seek The reason I had lost; but all for naught,
- No remedy could cure the wounds of love! In vain I offer'd incense on her altars; When I invoked her name my heart adored Hippolytus, before me constantly; And when I made her altars smoke with victims, 'Twas for a god whose name I dared not utter. I fled his presence everywhere, but found him— O crowning horror!—in his father's features. Against myself, at last, I raised revolt, And stirr'd my courage up to persecute The enemy I loved. To banish him I wore a step—dame's harsh and jealous carriage, With ceaseless cries I clamour'd for his exile, Till I had torn him from his father's arms. I breathed once more, Oenone; in his absence My days flow'd on less troubled than before, And innocent. Submissive to my husband, I hid my grief, and of our fatal marriage Cherish'd the fruits. Vain caution! Cruel Fate! Brought hither by my spouse himself, I saw Again the enemy whom I had banish'd, And the old wound too quickly bled afresh. No longer is it love hid in my heart, But Venus in her might seizing her prey. I have conceived just terror for my crime; I hate my life, and hold my love in horror. Dying I wish'd to keep my fame unsullied, And bury in the grave a guilty passion; But I have been unable to withstand Tears and entreaties, I have told you all; Content, if only, as my end draws near, Phaedra, by Jean Baptiste Racine
- https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] You do not vex me with unjust reproaches, Nor with vain efforts seek to snatch from death The last faint lingering sparks of vital breath. SCENE IV PHAEDRA, OENONE, PANOPE PANOPE Fain would I hide from you tidings so sad, But 'tis my duty, Madam, to reveal them. The hand of death has seized your peerless husband, And you are last to hear of this disaster. OENONE What say you, Panope? PANOPE The queen, deceived By a vain trust in Heav'n, begs safe return For Theseus, while Hippolytus his son Learns of his death from vessels that are now In port. PHAEDRA Ye gods! PANOPE Divided counsels sway The choice of Athens; some would have the prince, Your child, for master; others, disregarding The laws, dare to support the stranger's son. 'Tis even said that a presumptuous faction Would crown Aricia and the house of Pallas.
- I deem'd it right to warn you of this danger. Hippolytus already is prepared To start, and should he show himself at Athens, 'Tis to be fear'd the fickle crowd will all Follow his lead. OENONE Enough. The queen, who hears you, By no means will neglect this timely warning. SCENE V PHAEDRA, OENONE OENONE Dear lady, I had almost ceased to urge The wish that you should live, thinking to follow My mistress to the tomb, from which my voice Had fail'd to turn you; but this new misfortune Alters the aspect of affairs, and prompts Fresh measures. Madam, Theseus is no more, You must supply his place. He leaves a son, A slave, if you should die, but, if you live, A King. On whom has he to lean but you? No hand but yours will dry his tears. Then live For him, or else the tears of innocence Will move the gods, his ancestors, to wrath Against his mother. Live, your guilt is gone, No blame attaches to your passion now. The King's decease has freed you from the bonds That made the crime and horror of your love. Hippolytus no longer need be dreaded, Him you may see henceforth without reproach. It may be, that, convinced of your aversion, He means to head the rebels. Undeceive him, Soften his callous heart, and bend his pride. King of this fertile land, in Troezen here
- His portion lies; but as he knows, the laws Give to your son the ramparts that Minerva Built and protects. A common enemy Threatens you both, unite them to oppose Aricia. PHAEDRA To your counsel I consent. Yes, I will live, if life can be restored, If my affection for a son has pow'r To rouse my sinking heart at such a dangerous hour. ACT II SCENE I ARICIA, ISMENE Phaedra, by Jean Baptiste Racine https://www.gutenberg.org/files/1977/1977-h/1977- h.htm[10/4/2019 1:44:51 PM] ARICIA Hippolytus request to see me here! Hippolytus desire to bid farewell! Is't true, Ismene? Are you not deceived? ISMENE This is the first result of Theseus' death. Prepare yourself to see from every side. Hearts turn towards you that were kept away By Theseus. Mistress of her lot at last, Aricia soon shall find all Greece fall low, To do her homage.
- ARICIA 'Tis not then, Ismene, An idle tale? Am I no more a slave? Have I no enemies? ISMENE The gods oppose Your peace no longer, and the soul of Theseus Is with your brothers. ARICIA Does the voice of fame Tell how he died? ISMENE Rumours incredible Are spread. Some say that, seizing a new bride, The faithless husband by the waves was swallow'd. Others affirm, and this report prevails, That with Pirithous to the world below He went, and saw the shores of dark Cocytus, Showing himself alive to the pale ghosts; But that he could not leave those gloomy realms, Which whoso enters there abides for ever. ARICIA Shall I believe that ere his destined hour A mortal may descend into the gulf Of Hades? What attraction could o'ercome Its terrors? ISMENE He is dead, and you alone Doubt it. The men of Athens mourn his loss. Troezen already hails Hippolytus As King. And Phaedra, fearing for her son,
- Asks counsel of the friends who share her trouble, Here in this palace. ARICIA Will Hippolytus, Think you, prove kinder than his sire, make light My chains, and pity my misfortunes? ISMENE Yes, I think so, Madam. ARICIA Ah, you know him not Or you would never deem so hard a heart Can pity feel, or me alone except From the contempt in which he holds our sex. Has he not long avoided every spot Where we resort? ISMENE I know what tales are told Of proud Hippolytus, but I have seen Him near you, and have watch'd with curious eye How one esteem'd so cold would bear himself. Little did his behavior correspond With what I look'd for; in his face confusion Appear'd at your first glance, he could not turn His languid eyes away, but gazed on you. Love is a word that may offend his pride, But what the tongue disowns, looks can betray. ARICIA How eagerly my heart hears what you say, Tho' it may be delusion, dear Ismene! Did it seem possible to you, who know me,
- That I, sad sport of a relentless Fate, Fed upon bitter tears by night and day, Could ever taste the maddening draught of love? The last frail offspring of a royal race, Children of Earth, I only have survived War's fury. Cut off in the flow'r of youth, Mown by the sword, six brothers have I … Section 18.1-2. In the next 2-3 lectures we will have a lightning introduction to representations of finite groups. For any vector space V over a field F, denote by L(V ) the algebra of all linear operators on V , and by GL(V ) the group of invertible linear operators. Note that if dim(V ) = n, then L(V ) = Matn(F) is isomorphic to the algebra of n×n matrices over F , and GL(V ) = GLn(F) to its multiplicative group. Definition 1. A linear representation of a set X in a vector space V is a map φ: X → L(V ), where L(V ) is the set of linear operators on V , the space V is called the space of the rep- resentation. We will often denote the representation of X by (V,φ) or simply by V if the homomorphism φ is clear from the context. If X has any additional structures, we require that the map φ is a homomorphism. For example, a linear representation of a group G is a homomorphism φ: G → GL(V ).
- Definition 2. A morphism of representations φ: X → L(V ) and ψ : X → L(U) is a linear map T : V → U, such that ψ(x) ◦T = T ◦φ(x) for all x ∈ X. In other words, T makes the following diagram commutative V φ(x) // T �� V T �� U ψ(x) // U An invertible morphism of two representation is called an isomorphism, and two representa- tions are called isomorphic (or equivalent ) if there exists an isomorphism between them. Example. (1) A representation of a one-element set in a vector space V is simply a linear operator
- on V . Two such representations are isomorphic if two operators have the same Jordan Normal Form. (2) For any vector space, identity maps id : GL(V ) → GL(V ) and id : L(V ) → L(V ) give tautological representations of the group GL(V ) and the algebra L(V ) respectively. (3) For each representation φ: X → L(V ) one can define its dual representation φ∗ : X → L(V ∗ ), φ∗ (x)(α)(v) = α(φ(x)(v)) for all x ∈ X, v ∈ V , and α ∈ V ∗ . (4) The trivial representation of a group G is a homomorphism φ: G → GL(V ), where V is a 1-dimensional vector space over the field F, and φ(g) = 1 ∈ GL(V ) = F× for all g ∈ G. (5) Recall the dihedral group Dn = ⟨ r,s |rn = s2 = 1,rs = sr−1 ⟩ . The following map r 7→ ( cos (2π/n) −sin (2π/n) sin (2π/n) cos (2π/n) )
- , s 7→ ( 0 1 1 0 ) defines a 2-dimensional representation φ: Dn → GL(2,R). 1 2 (6) The following map gives a representation of a symmetric group S4: (12) 7→ 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 0
- 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 (7) There is a natural representations of Sn on the space of polynomials in n variables: (σf)(x1, . . . ,xn) = f(xσ(1), . . . ,xσ(n)). Note, that symmetric polynomials stay invariant under this action. (8) Any Galois group Gal(K/F) comes with its representation in K over the field F. Definition 3. An invariant subspace of a representation φ: X → L(V ) is a subspace W ⊂ V invariant under φ(x) for all x ∈ X. An invariant subspace gives rise to a subrepresentation φW : X → L(W), φW (x) = φ(x)|W , and a factor representation φV/W : X → L(V/W), φV/W (x)(v + W) = φ(x)(v).
- Remark 4. If one chooses a basis of the space V of a representation φ: X → L(V ) in such a way that the first k vectors span an invariant subspace W ⊂ V , then any operator φ(x) is written in matrix form as φ(x) = ( φW (x) ∗ 0 φV/W (x) ) Example. (1) Let Sn act on F n by permuting basis vectors, as in example (4) above. Then the subspace {a,a, . . . ,a) |a ∈ F} ⊂ Fn is a 1-dimensional invariant subspace. There is a complementary invariant subspace of dimension n− 1 defined by {(a1, . . . ,an) |a1 + · · · + an = 0} . Note that Fn is isomorphic to the direct sum of these two subspaces. (2) For any positive integer d, the subspace Poldn of polynomials in n variables of degree d in the space Poln of all polynomials of n variables gives a subrepresentation of Sn. Definition 5.
- (1) A representation is irreducible (or simple) if it (or rather the underlying vector space) has no invariant subspaces other than 0 and itself, otherwise it is reducible. (2) A representation is indecomposable if it can not be written as a direct sum of two invariant subrepresentations. (3) A representation is completely reducible if it can be written as a direct sum of irre- ducible representations. Example. The representation of a 1-element set given by a 2- dimensional space V = F2 and an operator A = ( a b 0 c ) is reducible, since the subspace 〈e1〉 ⊂ V is invariant, however it is indecomposable unless b = 0. If b = 0, the representation V is completly reducible: V = 〈e1〉⊕ 〈e2〉. Exercise 6. If T : V → U is a morphism of representations of X, then ker(T) ⊂ V and im(T) ⊂ U are invariant subspaces. 3
- Corollary 7. If V and U are irreducible representations of X, any morphism T : V → U is either 0 or an isomorphism. Proof. Since V is irreducible, the subrepresentation ker(T) is either 0, in which case T is injective, or coincides with the whole V , in which case T = 0. Since U is irreducible, we must now have im(T) = U unless T = 0, which forces T to be an isomorphism. � Corollary 8 (Schur’s lemma). Any endomorphism T : V → V of an irreducible represen- tation V of X over an algebraically closed field is a scalar, that is T = λ · Id for some λ ∈ F . Proof. Let T ∈ EndX(V ) = HomX(V,V ) be an endomorphism of V . Then for any λ ∈ F, so is T−λ·Id. Choosing λ to be an eigenvalue of T (which we can always do over an algebraically closed field) we see that ker(T −λ · Id) is a subrepresentation of X in V . Therefore, since V is irreducible, we must have T = λ · Id. � Corollary 9. Let φ: X → L(V ) and ψ : X → L(U) be a pair of irreducible representations over an algebraically closed field. Then any morphisms T,S : V → U differ by a scalar factor. Proof. If one of the two morphisms is zero, the statement is obvious. Otherwise, both mor- phisms are isomorphisms, hence ST−1 is an endomorphism of an irreducible representation U which implies ST−1 = λ Id. � Corollary 10. Any irreducible representation of an abelian group over an algebraically closed field is 1-dimensional.
- Proof. If G is abelian and φ: G → GL(V ) is a representation of G, operators φ(g) commute for all g ∈ G, hence any of them can be viewed as an endomorphism of V . If V is irre- ducible, By Schur’s lemma, all of them must be scalar, therefore any subspace is invariant and representation V can be irreducible if and only if dim(V ) = 1. � Example. The 2-dimensional plane R2 is an irreducible representation of the abelian group SO(2,R) of plane rotations G = {( cos θ −sin θ sin θ cos θ )∣ ∣ ∣ ∣ θ ∈ R } . However, over the field of complex numbers, we see that( cos θ −sin θ sin θ cos θ ) ∼ ( eiθ 0 0 e−iθ ) .
- Proposition 11. Every subrepresentation and factor representation of a completely reducible representation is completely reducible. Proof. Let V be a completely reducible representation, and U be its subrepresentation. Then for any U1 ⊂ U, there exists an invarinat subspace V ′ ⊂ V such that V ' U1 ⊕ V ′. Setting U2 = V ′ ∩U we see that U2 is invariant and U ' U1 ⊕U2. Now, let π : V → V/W be a canonical projection to a factor representation, and U1 ⊂ V/W be the subrepresentation. Then V1 = π −1(U1) is an invariant subspace of V , which con- tains W . Now, we have V = V1 ⊕ V2 for some invariant subspace V2, which implies that V/W ' U1 ⊕U2 where U2 = π(V2) is an invariant subspace in V/W . � Proposition 12. If a representation φ: X → L(V ) is completely reducible, then V can be written as a direct sum of minimal (nonzero) irreducible subspaces. Conversely, if V can be written as a sum of minimal invariant subspaces V = V1 + · · · + Vn, then V is completely reducible. 4 Proof. The first statement is immediate: choose any minimal invariant subspace V1 ⊂ V
- and write V = V1⊕V2, continue the same process for V2. Now, let V = V1 +· · ·+Vn for some minimal invariant subspaces Vi ⊂ V , and U ⊂ V be any invariant subspace. Then U ∩Vj is either 0 or coincides with Vj for any j = 1, . . . ,n. Setting U ′ = ∑ j:U∩Vj=∅ Vj we get V = U ⊕U ′. � Definition 13. A sum of representations φ: X → L(V ) and ψ : X → L(U) is a representa- tion φ + ψ : X → V ⊕U, (φ + ψ)(x) = (φ(x),ψ(x)). Remark 14. In matrix form we have (φ + ψ)(x) = ( φ(x) 0 0 ψ(x) ) Corollary 15. (1) A representation is completely reducible if and only if it is isomorphic to a sum of irreducible subrepresentations: V = V1 ⊕···⊕Vn, φ = (φ1, . . . ,φn).
- (2) In that case, any subrepresentation and factor representation of V is isomorphic to a sum of some of the representations Vj. Proof. Left as an exercise. � Definition 16. Note that in the above corollary, some of the irreducible summands Vj might be isomorphic. (1) Let S be an irreducible subrepresentation of a completely reducible representation V , then the S-isotypic component of V is the subspace VS ⊂ V defined as VS = ⊕ j |Vj'S Vj. (2) A representation V is isotypic if it has single isotypic component. (3) We will say that the representation V has simple spectrum if it can be written as a direct sum of pairwise non-isomorphic irreducible representations, equivalently if all of its isotypic components are irreducible. Corollary 17. Let φ: X → L(V ) be a completely reducible representation with simple spec- trum, and V1, . . . ,Vn be its irreducible subrepresentations. Then the following holds.
- (1) The decomposition of V into a sum of irreducible subrepresentations Vj is unique. (2) If the base field F is algebraically closed, then every endomorphism of V takes the form T(x) = λjx, for any x ∈ Vj. Proof. Left as an exercise. � It is convenient to describe isotypic representations in the following way. Let φ: X → L(V ) be a representation, and Z be any vector space (over the same field). Define a representation φ̃: L(V ⊗Z), φ̃(x)(v ⊗z) = φ(x)(v) ⊗z. 5 If {z1, . . .zn} is a basis of Z we get V ⊗Z ' (V ⊗z1) ⊕···⊕ (V ⊗zn), which provides a decomposition of an isotypic representation V ⊗Z into irreducible subrep- resentations. Proposition 18. Let φ: X → L(U) be an irreducible representation of X over an alge- braically closed field F , Z be a vector space over F , and ψ : X → L(U ⊗ Z) be the isotypic representation of X discussed above. Then every X-invariant
- subspace of U ⊗ Z is of the form U ⊗Z0, for some subspace Z0 ⊂ Z. Proof. It is enough to prove the statement for a minimal invariant subspace W ⊂ U ⊗ Z. Any element w ∈ W can be written as w = S1(w) ⊗z1 + · · · + Sn(w) ⊗zn, where Sj : W → U are certain morphisms between representations (W,ψ|W ) and (U,φ). We know that any such isomorphisms have to differ by scalar multiples, hence Sj = λjS for some morphism S : W → U and λ1, . . . ,λn ∈ F . This way we obtain w = S(w) ⊗ (λ1z1 + · · · + λnzn), and therefore W = U ⊗ (λ1z1 + · · · + λnzn). � Theorem 19. Let φ: X → L(V ) be an irreducible representation of X over an algebraically closed field F . Then the subalgebra of L(V ) generated by operators φ(x), x ∈ X coincides with L(V ) unless dim(V ) = 1 and φ = 0. Proof. Recall the following isomorphism of vector spaces: V ⊗V ∗ ' L(V ), v ⊗α 7−→ lα,v where `α,v(u) = α(u)v for all α ∈ V ∗ and u,v ∈ V . Note that under this isomorphism, for any operator T ∈ L(V ) we have T`α,v = `α,T(v) = `T∗ (α),v, (∗ )
- where T∗ ∈ L(V ∗ ) defined by T∗ (α)(v) = α(Tv). Due to a canonical bijection between subspaces of V and V ∗ , which sends each subspace U ⊂ V to its annihilator Ann(U) ⊂ V ∗ , we see that the dual representation φ∗ : X → L(V ∗ ) is also irreducible. Let us now define representations Tl and Tr of X on the space L(V ), precisely Tl(x)(A) = φ(x)A and Tr(x)A = Aφ(x). Thanks to the formula (∗ ), representations Tl and Tr are isotypic (we can think that they act on only one tensor factor in V ⊗V ∗ ). Note now that the subspace φ(X) ⊂ L(V ) is invariant under both Tl and Tr. Using the previous Proposition, we conclude that φ(X) can be written simultaneously as V ⊗W0 and as V0 ⊗ V ∗ , where V0 and W0 are subspaces of V and V ∗ respectively. The latter can only happen if φ(X) = L(V ) or φ(X) = 0 in which case one has to assume dim(V ) = 1 since otherwise V would not be irreducible. � 6 Theorem 20. Any representation of a finite group G over a field F such that char(F) does not divide |G| is completely reducible.
- Proof. Let G be a finite group, and φ: G → GL(V ) be its representation over a field F with char(F) not dividing |G|. It is enough to show that for any invariant subspace U ⊂ V there exists a complementary invariant subspace W ⊂ V such that V ' U ⊕ W . The latter is equivalent to finding a projector π from V onto U, which is a morphism of representations of G. Indeed, recall that a projector is a linear operator P : V → U such that P(V ) ⊂ U and P(u) = u for any u ∈ U. Then setting W = ker(P) we obtain an invariant subspace W ⊂ V satisfying V ' U ⊕W. Note that the space of all projectors P : V → U forms a linear subspace S in the space L(V ) of all linear operators on V . Consider an action of G on L(V ) by conjugation, that is g ◦A = φ(g)Aφ(g)−1 for all g ∈ G and A ∈ L(V ). Since, U is a subrepresentation of G we see that the subspace S ⊂ L(V ) is preserved under this action. Let us choose any projector P0 ∈ S and define P to be the “centre of mass” of the orbit of P0 under the action of G: P = 1 |G| ∑ g∈ G (g ◦P0).
- Then the projector P ∈ S is invariant under the adjoint action of G, or equivalently, commutes with all operators φ(g) ∈ L(V ). � Finally, let us discuss representations of the group algebra FG of a finite group G, where F is a field. Recall that an algebra A over a ring R is an R- module, which is a ring itself, in particular if R = F is a field, then an algebra A over F is a vector space over F, with a ring structure. As a vector space, algebra FG consists of linear combinations∑ g∈ G agg where ag ∈ F, addition on FG is defined component-wise, and multiplication has the form agg ·ahh = agah(gh). In other words, FG is an algebra over the field F with basis vectors labelled by elements of the finite group G, and with multiplication defined from that of G. As we discussed in the beginning of this lecture, a representation of an algebra A (over the ring R) is a homomorphism φ: A → L(V ) which respects the operations on A. In other words, a representation of A is simply an A-module V (which then forced to be an R-module itself). Now, it is easy to see that there is a bijection between representations of the group G and its group algebra FG, moreover this bijection respects subrepresentations,
- factor representations, etc. Indeed, given a representation φ: G → GL(V ) of the group G, we define (and denote by the same symbol) a representation φ: FG → L(V ), φ( ∑ g∈ G agg) = ∑ g∈ G agφ(g). Conversely, every representation of FG yields a representation of G when restricted to the basis vectors of FG. 7 Example. (1) The trivial representation of G corresponds to a 1- dimensional represen- tation of FG where (agg)(v) = agv for any ag ∈ F, g ∈ G, v ∈ V. (2) Considering FG as a (left) module over itself, we obtain a (left) regular representation of G, which is a representation of dimension |G| defined by g( ∑ h∈ G
- ahh) = ∑ h∈ G ah(gh) for all ah ∈ F, g,h ∈ G. The bijection described above allows us to study representations of finite groups by study- ing those of finite-dimensional associative algebras. Section 14.7. Today we discuss the insolvability of a general quantic in radicals. Definition 1. An extension K/F is cyclic if it is Galois with cyclic Galois group. Proposition 2. Let F be a field of characteristic not dividing n, which contains a primitive n-th root of unity. Then for any a ∈ F , the extension F( n √ a)/F is a cyclic extension of degree dividing n, here n √ a denotes any root of the polynomial xn −a. Proof. Since all roots of xn−a can be obtained by multiplying n √ a by n-th roots of unity all
- of which lie in F , we see that F( n √ a)/F is the splitting field for xn −a and hence is Galois. Since the Galois group G = Gal(F( n √ a)/F) permutes the roots of xn −a, for any σ ∈ G we have σ( n √ a) = ζσ n √ a for certain n-th root of unity ζσ. Therefore, we get a map ψ : G −→ Z/nZ, σ 7−→ ζσ, and we leave it as an exercise to show that ψ is a homomorphism. The kernel of ψ consists of those elements which fix n √ a and is therefore trivial. This shows that ψ is an embedding, and hence the order of G divides n. � Definition 3. A character χ of a group G with values in a field L is a homomorphism χ: G −→ L×. Definition 4. Character χ1, . . . ,χn of a group G with values in L are linearly dependent if there exist elements a1, . . . ,an ∈ L such that a1χ1(g) + · · · + anχn(g) = 0 for all g ∈ G,
- and linearly independent otherwise. Proposition 5. Distinct characters of a group G with values in a field L are linearly inde- pendent. Proof. Suppose that characters χ1, . . . ,χm are linearly dependent. Choose the depen- dence relation with the minimal number n of coefficients ai. Let g0 ∈ G be such that χ1(g0) 6= χn(g0), it exists since χ1 6= χn. Then we have n∑ j=1 ajχj(g0g) = n∑ j=1 ajχj(g0)χj(g) = 0. Subtracting expression χn(g0) n∑ j=1 ajχj(g) = n∑ j=1 ajχn(g0)χj(g) = 0 from the left hand side of the previous equality, we get n−1∑
- j=1 aj (χj(g0) −χn(g0)) χj(g) = 0 for all g ∈ G, which contradicts our assumption that n was the minimal number of non-zero coefficients in a dependence relation for characters χ1, . . . ,χm. � Proposition 6. Let K/F be a cyclic extension of degree n, where the field F contains a primitive n-th root of unity and char(F) does not divide n. Then K = F( n √ a) for some a ∈ F . 1 2 Proof. Let σ ∈ G = Gal(K/F) be the generator of the cyclic Galois group. For any element α ∈ K and any n-th root of unity ζ, consider the Lagrange resolvent (α,ζ) = n−1∑ k=0 ζkσk(α) = α + ζσ(α) + · · · + ζn−1σn−1(α). Since ζ ∈ F and therefore σ(ζ) = ζ, we have
- σ(α,ζ) = n−1∑ k=0 ζkσk+1(α) = ζ−1 n−1∑ k=0 ζk+1σk+1(α) = ζ−1 n∑ k=1 ζkσk(α) = ζ−1(α,ζ), where the last equality follows from the fact that ζn = 1 ∈ F and σn = 1 ∈ G. Now, we see that σ ((α,ζ)n) = ( ζ−1 )n (α,ζ)n = (α,ζ)n, which implies (α,ζ)n ∈ F for any α ∈ K. Let us now choose such an α ∈ K that (α,ζ) 6= 0, which exists because the characters 1,σ, . . . ,σn−1 are independent. Let us also pick ζ to be a primitive root of unity. Then σk(α,ζ) = ζ−k(α,ζ), which means that (α,ζ) ∈ K is not fixed by any nontrivial
- element of G, and therefore does not lie in any proper subfield of K. We now conclude that K = F ((α,ζ)), and since (α,ζ)n ∈ F , we can write K = F( n √ a) for a = (α,ζ)n. � In what follows, for simplicity we will assume char(F) = 0. Nonetheless, all the results will be valid for the fields whose characteristic does not divide orders of the roots taken. Definition 7. (1) Let K = F( n √ a) for some element a ∈ F. Then K/F is a simple radical extension. (2) A field K is a root extension of a field F if there exists a chain of subfields F = K0 ⊂ K1 ⊂ ···⊂ Ks = K, such that Ki/Ki−1 is s simple radical extension for all 1 = 1, . . . ,s. (3) An element α, which is algebraic over F , can be expressed by radicals if α ∈ K for some root extension K/F. (4) A polynomial f(x) ∈ F[x] can be solved by radicals if all of its roots can be expressed by radicals.
- Remark 8. Simply speaking, a polynomial can be solved by radicals over a field F, if all of its roots can be obtained by applying successive operations of addition, subtraction, multiplication, division, and extraction of n-th roots to elements of F . Lemma 9. For every root extension K/F , there exists a field E ⊃ K, such that E/F is a Galois root extension, and each simple radical extension Ei/Ei−1 is cyclic. Proof. Let L be the Galois closure of K. Then for any σ ∈ Gal(L/F), there exists a chain of subfields F = σ(K0) ⊂ σ(K1) ⊂ ···⊂ σ(Ks) = σ(K), where σ(Ki/Ki−1) is again a simple radical extension: Ki = Ki−1 ( ni √ ai) implies σ(Ki) = σ(Ki−1) ( ni √ σ(ai) ) . 3
- Note also, that a composite of two root extensions is also a root extension. Indeed, if K/F and K′/F are root extensions with intermediate extensions Ki and K ′ j respectively, one can consider a chain F = K0 ⊂ ···⊂ Ks = K = KK′0 ⊂ KK ′ 1 ⊂ KK ′ r = KK ′. It follows that the composite of all fields of the form σ(K), σ ∈ Gal(L/F), is again a root extension, but the latter composite is precisely L. Now we have a Galois root extension L/F with a chain of subfields F = L0 ⊂ L1 ⊂ ···⊂ Lk = L, where Li = Li−1( ni √ ai) for all i = 1, . . . ,k. Choose a number n such that ni |n for all 1 6 i 6 k, and set F ′ = F(ζn), where ζn is a primitive n-th root of unity. Then the extension F ′/F is a simple radical extension. Finally, consider the composition E = LF ′, which is
- Galois over F since both L and F ′ are. Moreover, we have the chain of subfields F ⊂ F ′ = F ′K0 ⊂ F ′K1 ⊂ ···⊂ F ′Ks = E, and each extension F ′Ki/F ′Ki−1 is cyclic by the first proposition in this lecture. � Recall that a group G is solvable if there exists a chain of normal subgroups 1 = Gs CGs−1 C · · ·CG1 CG0 = G, with cyclic quotient groups Gi−1/Gi for all i = 1, . . . ,s. Recall also that subgroups and quotient groups of a solvable group are themselves solvable, and that if both H ⊂ G and G/H are solvable, then so is G. Theorem 10. A polynomial can be solved in radicals if and only if its Galois group is solvable. Proof. Suppose that f(x) ∈ F [x] can be solved in radicals. Then each root of f(x) is con- tained in an extension as in the lemma above. Consider the composite L of these extensions for all roots of f(x). Then L/F is again a Galois root extension with intermediate cyclic extensions Li/Li−1. Now, let G = Gal(L/F) and Gi be the subgroups of G corresponding to the subfields Li. Since Gal(Li/Li−1) = Gi−1/Gi, we see that G is solvable. Moreover, since L contains the splitting field of the polynomial f(x), the Galois group H of f(x) is a subgroup of G and therefore
- solvable. Suppose now that the Galois group H of f(x) is solvable, and let K be the splitting field of f(x). Consider the chain of subgroups 1 = Hs CHs−1 C · · ·CH1 CH0 = H, with cyclic quotient groups Hi−1/Hi, and let Ki = K Hi be the fixed fields of these subgroups. Then we get a chain of cyclic extensions F = K0 ⊂ K1 ⊂ ···⊂ Ks = K. Considering an extension F ′/F as in the preceding lemma, we have a chain of cyclic extensions F ⊂ F ′ = F ′K0 ⊂ F ′K1 ⊂ ···⊂ F ′Ks = F ′K. Now however, each extension F ′Ki/F ′Ki−1 is a simple radical extension by one of the Propo- sitions we proved today, since F ′Ki−1 contains the appropriate roots of unity. Finally, since all roots of f(x) are contained in F ′K, we see that f(x) can be solved in radicals. � Corollary 11. The general equation of degree > 5 is not solvable in radicals. 4 Proof. The Galois group of a general polynomial of degree n is isomorphic to Sn, and Sn is not solvable for n > 5. To show that Sn is not solvable, it is enough to notice that the
- subgroup An ⊂ Sn is simple for n > 5. The latter follows from the following easy to shoe facts: An is generated by 3-cycles for n > 5, any normal subgroup of An contains a 3-cycle, hence contains all 3-cycles. � Section 14.6. Today we discuss Galois groups of polynomials. Let us briefly recall what we know already. If L/F is a Galois extension, then L is the splitting field of a separable polynomial f(x) ∈ F[x]. If f(x) factors into a product of irreducible polynomials f(x) = f1(x) . . .fk(x) with degrees deg(fj) = nj, the Galois group Gal(L/F) embeds into the following product of symmetric groups: Gal(L/F) ⊂ Sn1 ×···×Snk ⊂ Sn, where n = n1 + · · · + nk. Moreover, Gal(L/F) acts transitively on the set of roots of each irreducible factor. The following two examples are also well- known to us by now. The Galois group of (x2 − 2)(x2 − 3) is isomorphic to the Klein 4-group K4 ⊂ S4 and the Galois group of x3 − 2 is isomorphic to S6. In what follows we shall describe Galois groups of polynomials of degrees less than or equal to 4, and prove that the Galois group of a “general” polynomial of degree n is isomorphic to
- Sn. Definition 1. Let x1, . . . ,xn be indeterminates. Define k-th elementary symmetric function by ek = ek(x1, . . . ,xn) = ∑ 16i1<···<ik6n xi1 . . .xin. For example, we have e0(x1,x2,x3) = 1, e1(x1,x2,x3) = x1 + x2 + xn, e2(x1,x2,x3) = x1x2 + x1x3 + x2x3, e3(x1,x2,x3) = x1x2x3. Definition 2. Again, let x1, . . . ,xn be indeterminates. Then the general polynomial of degree n is defined as the product (x−x1) . . . (x−xn). It is easy to see that (x−x1) . . . (x−xn) = n∑ k=0 (−1)kek(x1, . . . ,xn)xn−k With the above definition, we see that for any field F, the
- extension F(x1, . . . ,xn) is the splitting field for the general polynomial of degree n. We now make the permutation group Sn act on F(x1, . . . ,xn) by permuting the variables. It is clear that Sn preserves F(e1, . . . ,en), which shows that Sn is a subgroup of the automorphism group of F(x1, . . . ,xn) over F(e1, . . . ,en). On the other hand, the named automorphism group is itself a subgroup of Sn, therefore we have Aut(F(x1, . . . ,xn)/F(e1, . . . ,en)) ' Sn. Now, if the general polynomial was non-separable, it would force the order of the auto- morphism group Aut(F(x1, . . . ,xn)/F(e1, . . . ,en)) to be strictly less than n!. This allows us to conclude that the F(x1, . . . ,xn) is the splitting field of a separable polynomial with coefficients in F(e1, . . . ,en), and hence the extension F(x1, . . . ,xn)/F(e1, . . . ,en) is Galois. 1 2 Corollary 3. Any symmetric rational function f(x1, . . . ,xn) is a rational function in vari- ables e1, . . . ,en. Proof. By the Fundamental Theorem of Galois Theory, F(e1, . . . ,en) is precisely the Sn-fixed subfield of F(x1, . . . ,xn). � Remark 4. We have just proved that F(x1, . . . ,xn)
- Sn = F(e1, . . . ,en). However, a stronger statement is true: F[x1, . . . ,xn] Sn = F [e1, . . . ,en]. Exercises in this chapter outline a boring proof of this statement. A more conceptual proof uses the notion of an integral extension. Check it out if you are curious! Now, let f(x) ∈ F [x] be a polynomial of degree d with roots x1, . . . ,xd. We will say that f(x) is generic if its roots are algebraically independent, that is there is no such polynomial g(t1, . . . , td) that g(x1, . . . ,xd) = 0. We can see that the Galois group of a generic polynomial f(x) is Sn, because there are no algebraic obstructions for realizing a permutation σ ∈ Sn as an element of the Galois group. Let us now make the following two remarks. First, the Galois group of a polynomial over a finite field is cyclic (since every finite extension is cyclic), therefore, there are no generic polynomials over finite groups. Second, the condition that roots of f(x) are algebraically dependent is equivalent to the condition that its coefficients are algebraically dependent. We will prove one direction and leave the other as an exercise. Assume that the roots x1, . . . ,xd are dependent, that is there exists a polynomial g(t1, . . . , td) such that g(x1, . . . ,xd) = 0. Define the polynomial g̃(t1, . . . , td) = ∏ σ∈ Sd g(tσ(1), . . . , tσ(d)).
- Now one can see that the condition g̃(x1, . . . ,xd) = 0 provides an algebraic dependence of the coefficients of f(x) which are nothing else but the elementary symmetric functions of the roots x1, . . . ,xd. Therefore, we can conclude that the general polynomial is separable over F(e1, . . . ,en). Recall that [Sn : An] = 2, and for n > 5 An is the unique normal subgroup of Sn. Therefore, for n > 5 there is a unique field K such that F(e1, . . .en) ⊂ K ⊂ F(x1, . . .xn) and K/F(e1, . . .en) is Galois. Moreover, K is an extension of degree 2. Definition 5. Define the discriminant of x1, . . . ,xn by D = ∏ i<j (xi −xj)2. The discriminant of a polynomial is defined to be the discriminant of its roots. Note the following properties of the discriminant which will be handy in what follows: (1) D ∈ F [e1, . . . ,en], and if D is viewed as a discriminant of the polynomial f(x) ∈ F(e1, . . .en)[x], then √
- D = ∏ i<j(xi −xj) is contained in the splitting field of f. (2) A permutation σ fixes √ D if and only if σ ∈ An. Therefore, given a field F with char(F) 6= 2, the element √ D generates the fixed field of An, which is a quadratic extension of F(e1, . . . ,en). (3) Let D be the discriminant of a polynomial f(x) ∈ F(e1, . . .en)[x]. Then f is separable if and only if D 6= 0. Therefore, over a perfect field f(x) is reducible if and only if D = 0. 3 (4) Assume that f(x) ∈ F[x] is separable. Reformulating property (2) above we see that the Galois group of f(x) is a subgroup of An if and only if √ D ∈ F . Now, let us finally discuss Galois groups of (the splitting fields of) polynomials of degrees 2, 3, and 4.
- Polynomials of degree 2. Consider f(x) = x2 + ax + b with roots α and β. Then D = (α−β)2 = (α + β)2 − 4αβ = e21 − 4e2 = (−a) 2 − 4b = a2 − 4b. Clearly, f(x) is separable if and only if D 6= 0. Moreover, the Galois group is isomorphic to Z/2Z if D 6= 0 and is trivial otherwise. Polynomials of degree 3. Consider f(x) = x3 + ax2 + bx + c. Making a substitution x = y −a/3, the polynomial becomes g(y) = y3 + py + q where p = 3b−a2 3 and q = 2a3 − 9ab + 27c 27 . The above substitution does not change the splitting field or the discriminant, so we shall only work with g(y) from now on. Let α,β,γ be its roots, then D = ( (α−β)(α−γ)(β −γ)
- )2 . As before, we can write out D in terms of the roots in a straightforward manner, and express the result in terms of coefficients of g(y). However, we are going to use the following trick, which will simplify the work a little: observe that D = −g′(α)g′(β)g′(γ). Using the fact that g′(y) = 3y2 + p, we can derive D = −4p3 − 27q2. Now, if g(y) is reducible over F , it either splits into three linear factors, in which case the Galois group is trivial, or into a linear and an irreducible quadratic factor. In the latter case, the Galois group is isomorphic to Z/2Z. If g(y) is irreducible then a root of g(y) generates an extension of degree 3, and therefore the degree of the splitting field is divisible by 3. Since the Galois group G is a subgroup of S3 we only have two possibilities: G ' A3 ' Z/3Z or G ' S3. The former is realized if and only if √ D ∈ F . Polynomials of degree 4. As before, we consider a polynomial
- f(x) = x4+ax3+bx2+cx+d, and make a substitution x = y −a/4. Then f(x) turns into g(y) = y4 + py2 + qy + r where p = −3a2 + 8b 8 , q = a3 − 4ab + 8c 8 , and r = −3a4 + 16a2b− 64ac + 256d 256 . Let the roots of g(y) be α1, . . . ,α4, and let G be the Galois group of the splitting field of g(y). If g(y) is reducible it either splits into a linear and a cubic, in which case G coincides with the Galois group of the cubic which we already considered, or into a pair of quadratics. 4 In the latter case the splitting field is the extension F(
- √ D1, √ D2), where D1 and D2 are the discriminants of the corresponding quadratics. Then G is either isomorphic to the Klein 4-group if √ D1/D2 6∈ F , or to Z/2Z otherwise. Now, we only need to consider the case when g(y) is irreducible. In that case, the Galois group G ⊂ S4 acts transitively on the roots. Inspecting the subgroups of S4 we conclude that the only subgroups with transitive action on the set of 4 elements are (1) S4 (2) A4 (3) Any of the 3 Sylow 2-subgroups, each isomorphic to D8 (4) K4, the Klein 4-group (5) any of the cyclic groups Z/4Z generated by the cycle of length 4 In order to express the discriminant in terms of the coefficients of g(y) we once again use a trick. Namely, we consider elements θ1 = (α1 + α2)(α3 + α4), θ1 = (α1 + α3)(α2 + α4), θ1 = (α1 + α4)(α2 + α3). Each of them is fixed by a separate Sylow 2-subgroup, and all 3 of them are fixed by the Klein 4-group, which is precisely the intersection of all Sylow
- 2-subgroups in S4. It is also clear that S4 permutes elements θj, therefore h(x) = (x−θ1)(x−θ2)(x−θ3) ∈ F [x]. Moreover, it is rather easy to compute that h(x) = x3 − 2px2 + (p2 − 4r)x + q2. This polynomial is called the resolvent cubic for the quartic g(y). It is also easy to see that the discriminants of g(y) and h(x) coincide, which allows us to compute D = 16p4r − 4p3q2 − 128p2r2 + 144pq2r − 27q4 + 256r3. Let us also notice that the splitting field of the resolvent cubic is a subfield in that of g(y), so the Galois group of the former is a quotient of G. We are finally ready to explore the Galois group G. Suppose first that h(x) is irreducible. If√ D 6∈ F, then G 6⊂ A4, and the Galois group of h(x) is S3, which implies that |G| is divisible by 6. This leaves us with the only option G ' S4. If the resolvent cubic is irreducible but√ D ∈ F, then G ⊂ A4 and |G| is divisible by 3, the order of the Galois group of h(x). Therefore G = A4. Now, it is only left to consider cases when the cubic is reducible. If it splits completely, then G fixes all of θj which implies G ⊂ K4 and therefore G = K4. Otherwise, if the cubic
- splits into a linear and a quadratic, then precisely one of θj lies in F , hence G ⊂ D8 and G 6⊂ K4. This leaves us with 2 possibilities: G ' D8 or G ' Z/4Z. To distinguish between the 2, note that D8∩A4 ' K4 while Z/4Z∩A4 ' Z/2Z, the former group is transitive on the roots of g(y) and the latter is not. Recall also that the fixed field of A4 is F( √ D). Therefore, we can determine G completely by factoring g(y) over F( √ D). Fundamental theorem of algebra. We finish this lecture by proving the Fundamental Theorem of Algebra. Theorem 6. Every polynomial f(x) ∈ C[x] of degree n has precisely n roots in C (counted with multiplicities). Equivalently, C is algebraically closed. Proof. We shall use the following two statements: 5 (1) Any polynomial g(x) ∈ R[x] of odd degree has a root in R. Equivalently, there are no non-trivial extensions of R of odd degree. (2) Any polynomial h(x) ∈ C[x] of degree 2 has roots in C. Equivalently, there are no

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