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1. UtilitasMathematica
ISSN 0315-3681 Volume 120, 2023
84
Action of Representation for Lie Groups
Zinah Kadhim1, Taghreed Majeed2
1
Mathematics Department - College of Education, Mustansiriyah University,
zenah73@uomustansiriyah.edu.iq
2
Mathematics Department - College of Education, Mustansiriyah University,
taghreedmajeed@uomustansiriyah.edu.iq
Abstract
The primary purpose of this research is to work out a new action of Lie group through dual
representation. In our paper we mention the basic definitions, the tensor product of two
representations, Weβll discuss the study of action for Lie group on Hom-spase using equivalence
relationship between Hom and tensor product. Weβll their action to study on a structure
consisting of five vector spaces. In the end we obtain new generalizations using dual action of
representation for Lie group Η€.
Keywords: Lie group, representation of Lie group, dual representation of Lie group, tensor
product for representation of Lie group.
1. Introduction
A Lie group Η€ is finite dimensional smooth manifold, together with a group structure on Η€, such
that the multiplication Η€ΓΗ€βΗ€ and attaching of an inverse Η€βΗ€βΆ Η₯ β Η₯β1
being smooth maps
[1]. Lie group homomorphism from Η€ to β is linear map such that Η€ and β represent Lie
groups , Π: Η€ β β represents the group homomorphism , and Π represent πΆβ
βmap on β [6].
The direct sum of ππ of Lie group Η€ acting on the vector spaces ππ over the field M is:
π1(π )π1, β¦ , ππ(π )ππ [4]. π»ππ(π1, π2) is linear map from vector space π1 to vector space π2
such that: π»ππ(π1, π2) β π1
β
β π2[2].
And, in this paper, We will symbolize for representations quadrilaterals and pentagons by (πππ΄)
and (πππ΄) respectively. see [7] and [3].
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2. Dual representation of Lie Groups:
Definition (2.1): [5]
Let Lie group Η€ , be a finite dimensional real or complex representation of Η€ being a Lie group
homomorphism π: Η€ β Η€πΏ(π, β) = (π β₯ 1). In general, a Lie group homomorphism is π: Η€ β
Η€πΏ(π) where π being a finite dimensional real or complex vector space with a dim π β₯ 1.
Example (2.2):
The 1-dimensional complex vector space (β΅). For any Lie group Η€ , one can describe the trivial
representation of Η€, π: Η€ β Η€πΏ(1, β΅), via the formular:
π(π ) = 1, for all π β Η€.
Definition (2.3): [1]
Let Η€ be a Lie group and ππ, π = 1,2, β¦ , π be representation of Η€ affects the vector spaces
ππ, π = 1,2, β¦ , π, then the direct sum of ππ, bring the representation defined by: {π1 β π2 β β¦ β
ππ(π )}(π1, π2, β¦ , π
π) = π1(π )π1, π2(π )π2, β¦ , ππ(π )π
π for all (π β Η€), π1, π2, β¦ , π
π β
π1, π2 Γ β¦ Γ π
π.
Example (2.4):
Let π1: β β Η€πΏ(2, β), such that π1(π ) = (
1 π
0 1
) for all π β β.
And π2: β β Η€πΏ(2, β), such that π2(π‘) = (
β1 0
π‘ 1
) for all π β β.
{π1 + π2(π )}(π1, π2) = (π1(π )π1, π2(π )π2) = ((
1 π
0 1
) π1 , (
β1 0
π‘ 1
) π2 ).
Definition (2.5): [9]
Let both Η€ and β be the Lie groups , let π1 be a representation of Η€ affects the space π1 and let
π2 be representation of β affects the space π2, then the tensor product of π1and π2 being the
representation on (π1 β π2)(π , π‘) = π(π ) β π2(π‘).
for all π β Η€ and π‘ β β.
Example (2.6):
Let π1: β β Η€πΏ(2, β), such that π1(π ) = (
1 π
0 1
) for all π β β.
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Let π2: β β Η€πΏ(2, β), such that π2(π‘) = (
1 0
βπ‘ β1
) for all π β β.
(π1 β π2)(π , π‘) = π1(π ) β π2(π‘) = (
1 π
0 1
) β (
1 0
βπ‘ β1
) = (
(
1 0
βπ‘ β1
) (
π 0
βπ‘π βπ
)
0 (
1 0
βπ‘ β1
)
) =
(
1 0
βπ‘ β1
π 0
βπ‘π βπ
0 0
0 0
1 0
βπ‘ β1
)
4Γ4
, the matrix in Η€πΏ(4, β).
Proposition (2.7): [7]
Let π be a representation for Lie group Η€ affects the finite dimensional vector space π, then
representation of dual of π is representation of Η€ on πβ
given by:
πβ(π ) = [π(π β1)]π‘π
, the dual representation is also called contragredient representation.
Example (2.8):
Let π: π β1
β ππ(2, β΅), where π 1{(cos β , sin β), 0 β€ββ€ 2π}, and
π 1
= ππβ
= cos β + π sin β
ππ(2, β) = {(
cos β β sin β
sin β cos β
) , 0 β€ββ€ 2π} such that π(cos β , sin β) =
(
cos β β sin β
sin β cos β
) π(ππβ
) = (
cos β β sin β
sin β cos β
) , 0 β€ββ€ 2π.
π is a representation for Lie group π1
, let π = ππβ
, π(ππβ
) = (
cos β sin β
βsin β cos β
)
π1
= cos β β π sin β , π(π )β1
= (
cos β β sin β
sin β cos β
)
{π(π )β1}π‘π
= (
cos β β sin β
sin β cos β
).
3. The action of representation for Lie groups
Through the lemma Schur's is the action idea upon the tensor product of two Lie algebra's
representation where it is :
Assume that πβ²1 and πβ²2 are representation of Lie algebra (Η₯) affects the finite-dimensional
spaces π1 as well as π2, correspondingly. defined as an action of (Η₯) on π»πππ(π2, π1), where
π be field, Π: Η₯ β Η₯πΏ(π»πππ(π2, π1)). By for all π β Η₯, β β π»πππ(π2, π1).
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Π(π ) = πβ²1(π )β = βπβ²2(π ). And π»πππ(π2, π1) β π2
β
β π1 as equivalence of representations,
see [2] .
Proposition (3.1): [4]
Let ππ: Η€ β Η€πΏ(ππ) be representations of Lie group Η€ on π β finite dimensional vector space
(ππ), for π = 1,2, correspondingly, and ππ
β
: Η€ β Η€πΏ(ππ
β
) the dual representation on (ππ
β
),for π =
1,2, which is give by ππ
β
(π ) = βπ β ππ(π ) , for all π β Η€, where βπ: ππ β π. See the diagram:
Proposition (3.2):
Let ππ, π = 1,2,3, πππ 4 be for representations of Lie group affects the vector spaces ππ, π =
1,2,3, πππ 4 respectively, and let
π»πππ(π4
β
, Hom (π3
β
, π2), π»ππ(π2, π1)), be vector space of the whole linear mapping from
π2 and from Hom (π3
β
, π2) to Hom (π3
β
, π1) as well as from (π4, Hom (π3
β
, π2) to
Hom(π3, π1)), then the (πππ΄) of Lie group on
π»πππ(π4
β
, Hom (π3
β
, π2), π»ππ(π2, π1)).
Proof:
Define Π
Μ : Η€ β Η€πΏ(π»πππ(π4
β
, Hom (π3
β
, π2), π»ππ(π2, π1)). Such that:
Π
Μ (π )β
Μ [(π1(π ) β β
Μ1 β π2(π )) β (π2(π ) β β
Μ2π3(π )β1
)] β β
Μ3 β π4(π )β1
,
for all (π β Η€) and βπ: ππ β π.
The following diagram shows the use acting of group Η€ into
Η€πΏ(π»πππ(π4
β
, Hom (π3
β
, π2), π»ππ(π2, π1)).
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Proposition (3.3):
Let ππ, π = 1,2,3, πππ 4 be for representations of Lie group affects the vector spaces ππ, π =
1,2,3, πππ 4 respectively, and let
π»πππ(π4, π»ππ(π3, π2), π»ππ(π2
β
, π1
β)), be M-vector space of the whole linear mapping
from π4 to π»ππ(π3, π2), π»ππ(π2
β
, π1
β). Then the (πππ΄) of Lie group on
π»πππ(π4, π»ππ(π3, π2), π»ππ(π2
β
, π1
β)).
Proof:
Define Π
Μ : Η€ β Η€πΏπ»πππ(π4, π»ππ(π3, π2), π»ππ(π2
β
, π1
β)), such that
Π
Μ (π )β
Μ [(π1(π )β1
β β
Μ1 β π2(π )β1
) β (π2(π ) β β
Μ2 β π3(π ))] β β
Μ3 β π4(π ) ,
for all π β Η€ and βπ: ππ β π.
The following diagram shows the acting of group Η€ is into
Η€πΏ(π»πππ(π4, π»ππ(π3, π2), π»ππ(π2
β
, π1
β))).
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Proposition (3.4):
Let ππ, π = 1,2,3, πππ 4 be for representations of Lie group affects the vector spaces ππ, π =
1,2,3, πππ 4 respectively, and let
π»πππ(π4, π»ππ(π3, π2
β)), π1
β
), be M-vector space of the whole linear mapping from π3 to
π2
β
and from π»ππ(π3, π2
β))into π1
β
as well as π4 into
π»ππ(π»ππ(π3, π2
β)), π1
β
).Then the (πππ΄) of Lie group is on
π»πππ(π4, π»ππ(π3, π2
β)), π1
β
)).
Proof:
Define π: Η€ β Η€πΏ(π»πππ(π4, π»ππ(π3, π2
β)), π1
β
)). Such that
Π
Μ (π )β
Μ [(π1(π )β1
β β
Μ1 β (π2(π )β1
β β
Μ2 β π3(π ))] β β
Μ3 β π4(π ) ,
for all π β Η€ , βπ: ππ β π.
Proposition (3.5):
Let ππ, π = 1,2,3, πππ 4 be for representations of Lie group affects the vector spaces ππ, π =
1,2,3, πππ 4 respectively, and let
π»πππ(π4, π»ππ(π3
β
, π2)), π1
β
), be M-vector space of the whole linear mapping from π3 to π2
and from π»ππ(π3
β
, π2))into π1 as well as π4 into
π»ππ(π»ππ(π3
β
, π2)).Then the (πππ΄) of Lie group is on
π»πππ(π4, π»ππ(π3
β
, π2)), π1
β
)).
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Proof:
Define Π
Μ : Η€ β Η€πΏπ»πππ(π4, π»ππ(π3
β
, π2), π1
β
), by proposition (3.2) such that: Π
Μ (π )β
Μ =
[(π1(π )β1
β β
Μ1 β (π2(π ) β β
Μ2 β π3(π )β1
)] β β
Μ3 β π4(π ).
for all π β Η€ and βπ: ππ β π.
Proposition (3.6):
Let ππ, π = 1,2,3, πππ 4 be four representations of Lie group affects the vector spaces ππ, π =
1,2,3, πππ 4 respectively, and let
π»πππ(π4
β
, π»ππ(π3
β
, π2)), π1), be M-vector space of the whole linear mapping from π3
β
to
π2 and π»ππ(π3
β
, π2))into π1 as well as π4
β
into
π»ππ(π»ππ(π3
β
, π2)).Then the (πππ΄) of Lie group is on
π»πππ(π4
β
, π»ππ(π3
β
, π2)), π1)).
Proof:
Define Π
Μ : Η€ β Η€πΏπ»πππ(π4
β
, π»ππ(π3
β
, π2), π1), by proposition (3.4) such that: Π
Μ (π )β
Μ =
[(π1(π ) β β
Μ1 β (π2(π ) β β
Μ2 β π3(π )β1
)] β β
Μ3 β π4(π )β1
.
for all (π β Η€) and βπ: ππ β π.
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Proposition (3.7):
Let ππ, π = 1,2,3,4, πππ 5 be five representation of Lie group affects the vector spaces ππ, π =
1,2,3,4, πππ 5 respectively, and let
π»πππ(π5, π»ππ(π4, π3
β), π2
β
β π1
β
)), be M-vector space of the whole linear mapping from
π4 to π3
β
and π»ππ(π4, π3
β))into π2
β
β π1
β
as well as π5 into
π»ππ(π»ππ(π4, π3
β), π2
β
β π1
β
).Then the (πππ΄) of Lie group is on
π»πππ(π5, π»ππ(π»ππ(π4, π3), π2
β
β π1
β
)).
Proof:
Define β
Μ : Η€ β Η€πΏ(π»πππ(π5, π»ππ(π»ππ(π4, π3
β), π2
β
β π1
β
))), by proposition (3.4) such
that:
β
Μ (π )β
Μ = [(π1(π )β1
β (π2(π )β1
) β β
Μ1 β (π3(π )β1
β β
Μ2 β π4(π ))] β β
Μ3 β π5(π ).
for all π β Η€ and βπ: ππ β π.
Proposition (3.8):
Let ππ, π = 1,2,3,4, πππ 5 be five representation of Lie group affects the vector spaces ππ, π =
1,2,3,4, πππ 5 respectively, and let
π»πππ(π5, π»ππ(π4
β
, π3
β), π»ππ(π2, π1
β
), be M-vector space of the whole linear mapping
from π2 to π1
β
and π»ππ(π2
β
, π3
β))into π»ππ(π2, π1
β
) as well as π5 into (π»ππ(π4
β
β
π3
β), π»ππ(π2, π1
β)). Then the (πππ΄) of Lie group is on
π»πππ(π5, π»ππ(π4
β
, π3
β)π»ππ(π2, π1
β)).
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Proof:
Define πΜ : Η€ β Η€πΏ(π»πππ(π5, (π4
β
β π3
β)π»ππ(π2, π1
β))),by proposition (3.4) such that:
πΜ (π )β
Μ = [(π1(π )β1
β β
Μ1 β π2(π )) β β
Μ2 β (π3(π )β1
ββ π4(π )β1
)] β β
Μ3 β π5(π ).
for all π β Η€ and βπ: ππ β π.
Proposition (3.9):
Let ππ, π = 1,2,3,4,5 be five representation of Lie group affects the vector spaces ππ, π = 1,2,3,4,5
respectively, and let
π»πππ(π»ππ(π5, π4
β), π»ππ(π3, π2 β π1
β)) be M-vector space of all linear mapping from π5
to π4
β
and from π3 into (π2 β π1
β)) as well as from π»ππ(π5, π4
β) into π»ππ(π3, π2 β π1
β)
Then the (πππ΄) of Lie group on
π»πππ(π»ππ(π5, π4
β), π»ππ(π3, π2 β π1
β).
Proof:
Define πΜ : Η€ β Η€πΏ(π»πππ(π»ππ(π5, π4
β), π»ππ(π3, π2 β π1
β))),by proposition (3.4) such
that:
πΜ (π )β
Μ = [(π1(π )β1
β π2(π )) β β
Μ1 β π3(π ))] β β
Μ2 β π4(π )β1
β β
Μ3 β π5(π ).
for all π β Η€ and βπ: ππ β π.
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4. The dual action of representation for Lie Group
Proposition (4.1): [6]
Let π1and π2 are finite dimensional vector space, and (π) be a natural no, then the action of Lie
group action on tensor product is :
1. π
2
ββ,..β
β
π
β π1 β π2 β π1, if π is an even number.
2. π
2
ββ,..β
β
π
β π1 β π2
β
β π1, if π is an odd number.
3. π2 β π
2
ββ,..β
β
π
β π2 β π1, if π is an even number.
4. π2 β π
2
ββ,..β
β
π
β π2 β π1
β
, if π is an odd number.
Corollary (4.2):
If (π)1and (π2) are finite- dimensional vector spaces, and (π) be a natural number, then the Lie
group affects the tensor product and π»πππ(π2, π1):
1. π2 β π1
β
β π»πππ(π2, π1
β)
2. π2 β π3 β π1
β
β π»πππ(π1, π2 β π3)
} if π is an odd number.
Proposition (4.3):
Let π1: Η€ β Η€πΏ(ππ), π π
β
: Η€ β Η€πΏ(ππ
β
) for π = 1,2, and the Lie group Η€ affects the π(π )β =
π1(π ) β β β π2(π ), for every π β Η€, β β π»ππ(π2, π1).
Then the Lie group Η€ affects (π»ππ(π2, π1))β
being also given provided a representation πβ
,
where: πβ(π ) = π(π )β1
β ββ
β π1(π ), for all π β Η€ and ββ
β (π»ππ(π2, π1))β
.
Proof:
Let action of Lie group Η€ affects π»ππ(π2, π1) being induced via the representation
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π: Η€ β Η€πΏ(π»ππ(π2, π1)), where π(π ) = π1(π ) β β β π2(π )β1
, for π β Η€, and
β β π»ππ(π2, π1). Such that πβ(π ) = π(π )β1
β ββ
β π1(π ) representation for all π β Η€ and ββ
β
(π»ππ(π2, π1))β
.
Since: β β(π ) = (π1(π ) β β β π2(π )β1
, )β
= π2
β(π )β1
β ββ
β π1
β(π ) for all (π β Η€) and ββ
: π1
β
β
π2
β
, we have
πβ(π π‘) = (π(π π‘))β
= (π(π‘) β β β π(π ))β
= πβ
(π ) β ββ
β πβ
(π‘)
Thus, πβ
is a representation from Η€(πβ
. π ππππ’π βπππππππβππ π ππ Η€)
Corollary (4.4):
Let π1: Η€ β Η€πΏ(π, π), Η€πΏ(π1) β Η€πΏ(π, π)
And π2: Η€ β Η€πΏ(β, π), Η€πΏ(π2) β Η€πΏ(β, π).
Where π1 and π2 have matrix representation, if Lie group Η€ affects π»ππ(π2, π1) being a
representation π: Η€ β Η€πΏ(π»ππ(π2, π1)), where
π(π ) = π1(π ) β β β π2(π )β1
, for π β Η€. Then Lie group Η€ affects π»ππ(π2, π1) is a
representation πβ
: Η€ β Η€πΏ(π»πππ(π2, π1))β
.
Where: πβ(π ) = (π2
β(π ))
π‘π
β ββ
β (π1(π )β1)π‘π
, for all π β Η€.
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Proof:
Since (π ) = π1(π ) β β β π2(π )β1
, πβ(π ) = π(π )β1
β ββ
β π1
β(π ) = (π2
β(π ))
π‘π
β ββ
β (π1(π )β1)π‘π
And πβ
is a representation a matrix representation of dimension πβ, then
πβ(π π‘) = (π(π π‘)β1)π‘π
= (π(π‘)β1)π‘π
β (π(π )β1)π‘π
= πβ(π‘) β πβ(π )
Example (4.5):
Let π1 = π1
β ππ(2) β Η€πΏ(2, β΅) and
π2 = π1
β ππ(3) β Η€πΏ(3, β΅)
Where Η€ = π1(π = 2, β = 3) and (π1) is the β΅ β vector space of dimensional 2, and ( π2) being
the β΅ β vector space of dimensional 3, then by collar we have
πβ
(ππβ
) = π2
β
(ππβ
) β ββ
β π1
β
(ππβ
)
= (π2
β
(ππβ
))
π‘π
β ββ
β (π1(ππβ
))
π‘π
= (
1 0 0
0 cos β β sin β
0 sin β cos β
)
π‘π
β ββ
β (
cos β sin β
βsin β cos β
)
π‘π
= (
1 0 0
0 cos β β sin β
0 sin β cos β
)
π‘π
β ββ
β (
cos β β sin β
sin β cos β
)
π‘π
Let π΄ = (
1 0 0
0 cos β β sin β
0 sin β cos β
)
Thus πβ
(ππβ
) = (
(cos β)π΄ (β sin β)π΄
(sin β)π΄ (cos β)π΄
)
6Γ6
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=
(
cos β 0 0
0 cos β2
cos β sin β
0 β cos β sin β cos β2
β sin β 0 0
0 β sin β cos β β sin β2
0 sin β2
sin β cos β
sin β 0 0
0 sin β cos β sin β2
0 β sin β2
sin β cos β
cos β2
0 0
0 cos β2
cos β sin β
0 β cos β sin β cos β2 )
Is the representation of Η€πΏ(6, β΅) acting on π»πππ(π2, π1) of dimensional 6.
Proposition (4.6):
Let (π1)and (π2) be two n- dimensional vector spaces and π is a natural number, then
π»πππ(π»ππ(π3, π2, π1))
βββ¦β
β
π = {
π»πππ(π»ππ(π3, π2, π1))ππ π ππ ππ£ππ
π»πππ(π1
β
, π»ππ(π2
β
, π2)ππ π ππ πππ
β¦
Proof:
We can prove this proposition by mathematical induction.
If π = 1, then (π»πππ(π2, π1))β
= (π»πππ(π1
β
, π2
β).
And hence the action of group Η€ on π»πππ(π2, π1) is:
πβ(π ) = π1
β(π ) β ββ
β π2
β(π )β1
(see proposition (3.2).
Suppose that (3.2) is true when π = π it mean
(π»πππ(π2, π1))
βββ¦β
β
π = {
(π»πππ(π2, π1)), ππ π ππ ππ£ππ β¦ (1)
(π»πππ(π1
β
, π2), ππ π ππ πππ β¦ (2)
If π is even then the acting of Lie group Η€ on π»πππ(π2, π1) is:
π(π ) = π2(π )β1
β β β π1(π ).
And if π is odd then the action of Lie group Η€ on π»πππ(π2, π1) is:
πβ(π ) = π1
β(π )β1
β ββ
β π2
β(π )
We will prove that (3.2) it is true, when π = π + 1 ,so it will be proven
π»πππ(π2, π1)
βββ¦β
β
π+1 = {
π»πππΊ(π2, π1))ππ π ππ πππ
π»πππ(π2, π1))ππ π ππ ππ ππ£ππ
(π(π ))
βββ¦β
β
π+1 = (π1
β(π ) β ββ
π2(π ))
βββ¦β
β
π+1 (π1
β(π ) β ββ
β π2
β(π ))
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ISSN 0315-3681 Volume 120, 2023
97
When π is odd we have π + 1 is even, then
π»πππ(π»ππ(π3, π2), π1))
βββ¦β
β
π+1 = π»πππ(π»ππ(π3, π2), π1)) by(1),
Thus the Lie group Η€ affects π»πππ(π2, π1)
βββ¦β
β
π+1 is:
(π(π ))
βββ¦β
β
π+1 = [(π1(π )β1
β β1 β (π2(π ) β β2 β π3
β1(π ))]
And when π is even we have π + 1 is odd.
Then π»πππ(π»ππ(π3, π2), π1))
βββ¦β
β
π+1 = π»πππ(π1
β
, π»ππ(π2
β
, π3)) by (2),
Thus the affects of Lie group Η€ on π»πππ(π2, π1)
βββ¦β
β
π+1 is:
(π(π ))
βββ¦β
β
π+1 = [(π3(π ) β ββ
1 β π2(π )β1
) β ββ
2 β π1
β1(π )β1]
Then the proposition is true for all β πβ
.
5. Conclusion:
We concerned of Lie group, Lie algebra, we representation of Lie group, representation of Lie
algebra, tensor product of representation of Lie group. We obtain new propositions by using four
and five- representations structure as manifested, which was the starting point for the Schur's
lemma.
6. Acknowledgement:
The authors (Zinah Makki Kadhim and Taghreed Hur Majeed) would be grateful to thank
Mustansiriyah University in Baghdad, Iraq. (www.mustansiriyah.ed.iq) for Collaboration and
Support in the present work.
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