Mixing
Lie algebra generated by elements of the symmetric group $S_N$
For an arbitrary group $G$, the group algebra $mathbb{C}(G)$ is defined as the set of all formal linear combinations of the elements of $G$: $mathbb{C}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$. Multiplication in $mathbb{C}(G)$ is induced by the group multiplication of $G$ and extends to $mathbb{C}(G)$ by linearity. In a similar fashion, we can define a "group Lie algebra" $mathfrak{L}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$, which is the same as $mathbb{C}(G)$ as a linear vector space, but in $mathfrak{L}(G)$ only the Lie bracket $[,]:mathfrak{L}(G)times mathfrak{L}(G)to mathfrak{L}(G)$ is defined, to be the commutator in $mathbb{C}(G)$, i.e. $[l_1,l_2]=l_1cdot l_2-l_2cdot l_1$, where "$cdot$" is the multiplication in $mathbb{C}(G)$. My problem is to determine the structure of the Lie algebra $mathfrak{L}(G)$ for the symmetric group $G=S_N$.
Let us take $S_3$ as an example. Since $|S_3|=6$, the group algebra $mathbb{C}(S_3)$ of $S_3$ is 6-dimensional, spanned by ${1,P_{12},P_{23},P_{13},P_{12}P_{23},P_{23}P_{12}}$. It is easy to see that ${1,I,I^2}$ are central elements (i.e. commute with everything in $mathbb{C}(S_3)$), where $I=P_{12}+P_{23}+P_{13}$. The 3-dimensional subspace orthogonal to these central elements is spanned by ${P_{12}-P_{23},P_{12}-P_{13},[P_{12},P_{23}]}$, which forms an $mathfrak{su}(2)$ Lie algebra, their relation to the spin generators are
$$P_{12}-P_{23}=sqrt{6}(s_x-s_y),~~P_{12}-P_{13}=sqrt{6}(s_x-s_z),~~[P_{12},P_{23}]=2i(s_x+s_y+s_z),$$
where $[s_i,s_j]=iepsilon_{ijk}s_k$. Therefore we conclude that $mathfrak{L}(S_3)=mathfrak{u}(1)^3oplus mathfrak{su}(2)$.
But when trying to determine the structure of $mathfrak{L}(S_N)$ for arbitrary $N$, this kind of brute force calculation seems hopeless. I don't even know the structure of $mathfrak{L}(S_4)$. For large $N$, I want to know this: if $mathfrak{L}(S_N)$ is a direct sum of simple Lie algebras, let $d_N$ be the dimension of its largest irreducible(simple) component (e.g. $d_3=3$). Does $d_N$ grow algebraically fast or exponentially fast with $N$?
Any hints, suggestions, or relevant references will be welcomed. Thanks.
abstract-algebra permutations lie-algebras symmetric-groups
asked Nov 20 '18 at 3:06
Lagrenge
1208
add a comment |
For an arbitrary group $G$, the group algebra $mathbb{C}(G)$ is defined as the set of all formal linear combinations of the elements of $G$: $mathbb{C}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$. Multiplication in $mathbb{C}(G)$ is induced by the group multiplication of $G$ and extends to $mathbb{C}(G)$ by linearity. In a similar fashion, we can define a "group Lie algebra" $mathfrak{L}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$, which is the same as $mathbb{C}(G)$ as a linear vector space, but in $mathfrak{L}(G)$ only the Lie bracket $[,]:mathfrak{L}(G)times mathfrak{L}(G)to mathfrak{L}(G)$ is defined, to be the commutator in $mathbb{C}(G)$, i.e. $[l_1,l_2]=l_1cdot l_2-l_2cdot l_1$, where "$cdot$" is the multiplication in $mathbb{C}(G)$. My problem is to determine the structure of the Lie algebra $mathfrak{L}(G)$ for the symmetric group $G=S_N$.
Let us take $S_3$ as an example. Since $|S_3|=6$, the group algebra $mathbb{C}(S_3)$ of $S_3$ is 6-dimensional, spanned by ${1,P_{12},P_{23},P_{13},P_{12}P_{23},P_{23}P_{12}}$. It is easy to see that ${1,I,I^2}$ are central elements (i.e. commute with everything in $mathbb{C}(S_3)$), where $I=P_{12}+P_{23}+P_{13}$. The 3-dimensional subspace orthogonal to these central elements is spanned by ${P_{12}-P_{23},P_{12}-P_{13},[P_{12},P_{23}]}$, which forms an $mathfrak{su}(2)$ Lie algebra, their relation to the spin generators are
$$P_{12}-P_{23}=sqrt{6}(s_x-s_y),~~P_{12}-P_{13}=sqrt{6}(s_x-s_z),~~[P_{12},P_{23}]=2i(s_x+s_y+s_z),$$
where $[s_i,s_j]=iepsilon_{ijk}s_k$. Therefore we conclude that $mathfrak{L}(S_3)=mathfrak{u}(1)^3oplus mathfrak{su}(2)$.
But when trying to determine the structure of $mathfrak{L}(S_N)$ for arbitrary $N$, this kind of brute force calculation seems hopeless. I don't even know the structure of $mathfrak{L}(S_4)$. For large $N$, I want to know this: if $mathfrak{L}(S_N)$ is a direct sum of simple Lie algebras, let $d_N$ be the dimension of its largest irreducible(simple) component (e.g. $d_3=3$). Does $d_N$ grow algebraically fast or exponentially fast with $N$?
Any hints, suggestions, or relevant references will be welcomed. Thanks.
abstract-algebra permutations lie-algebras symmetric-groups
asked Nov 20 '18 at 3:06
Lagrenge
1208
The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components.
– Sheel Stueber
Nov 20 '18 at 3:14
add a comment |
For an arbitrary group $G$, the group algebra $mathbb{C}(G)$ is defined as the set of all formal linear combinations of the elements of $G$: $mathbb{C}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$. Multiplication in $mathbb{C}(G)$ is induced by the group multiplication of $G$ and extends to $mathbb{C}(G)$ by linearity. In a similar fashion, we can define a "group Lie algebra" $mathfrak{L}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$, which is the same as $mathbb{C}(G)$ as a linear vector space, but in $mathfrak{L}(G)$ only the Lie bracket $[,]:mathfrak{L}(G)times mathfrak{L}(G)to mathfrak{L}(G)$ is defined, to be the commutator in $mathbb{C}(G)$, i.e. $[l_1,l_2]=l_1cdot l_2-l_2cdot l_1$, where "$cdot$" is the multiplication in $mathbb{C}(G)$. My problem is to determine the structure of the Lie algebra $mathfrak{L}(G)$ for the symmetric group $G=S_N$.
Let us take $S_3$ as an example. Since $|S_3|=6$, the group algebra $mathbb{C}(S_3)$ of $S_3$ is 6-dimensional, spanned by ${1,P_{12},P_{23},P_{13},P_{12}P_{23},P_{23}P_{12}}$. It is easy to see that ${1,I,I^2}$ are central elements (i.e. commute with everything in $mathbb{C}(S_3)$), where $I=P_{12}+P_{23}+P_{13}$. The 3-dimensional subspace orthogonal to these central elements is spanned by ${P_{12}-P_{23},P_{12}-P_{13},[P_{12},P_{23}]}$, which forms an $mathfrak{su}(2)$ Lie algebra, their relation to the spin generators are
$$P_{12}-P_{23}=sqrt{6}(s_x-s_y),~~P_{12}-P_{13}=sqrt{6}(s_x-s_z),~~[P_{12},P_{23}]=2i(s_x+s_y+s_z),$$
where $[s_i,s_j]=iepsilon_{ijk}s_k$. Therefore we conclude that $mathfrak{L}(S_3)=mathfrak{u}(1)^3oplus mathfrak{su}(2)$.
But when trying to determine the structure of $mathfrak{L}(S_N)$ for arbitrary $N$, this kind of brute force calculation seems hopeless. I don't even know the structure of $mathfrak{L}(S_4)$. For large $N$, I want to know this: if $mathfrak{L}(S_N)$ is a direct sum of simple Lie algebras, let $d_N$ be the dimension of its largest irreducible(simple) component (e.g. $d_3=3$). Does $d_N$ grow algebraically fast or exponentially fast with $N$?
Any hints, suggestions, or relevant references will be welcomed. Thanks.
abstract-algebra permutations lie-algebras symmetric-groups
asked Nov 20 '18 at 3:06
Lagrenge
1208
For an arbitrary group $G$, the group algebra $mathbb{C}(G)$ is defined as the set of all formal linear combinations of the elements of $G$: $mathbb{C}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$. Multiplication in $mathbb{C}(G)$ is induced by the group multiplication of $G$ and extends to $mathbb{C}(G)$ by linearity. In a similar fashion, we can define a "group Lie algebra" $mathfrak{L}(G)={c_1g_1+c_2g_2+ldots+c_n g_n|c_iin mathbb{C},g_iin G}$, which is the same as $mathbb{C}(G)$ as a linear vector space, but in $mathfrak{L}(G)$ only the Lie bracket $[,]:mathfrak{L}(G)times mathfrak{L}(G)to mathfrak{L}(G)$ is defined, to be the commutator in $mathbb{C}(G)$, i.e. $[l_1,l_2]=l_1cdot l_2-l_2cdot l_1$, where "$cdot$" is the multiplication in $mathbb{C}(G)$. My problem is to determine the structure of the Lie algebra $mathfrak{L}(G)$ for the symmetric group $G=S_N$.
Let us take $S_3$ as an example. Since $|S_3|=6$, the group algebra $mathbb{C}(S_3)$ of $S_3$ is 6-dimensional, spanned by ${1,P_{12},P_{23},P_{13},P_{12}P_{23},P_{23}P_{12}}$. It is easy to see that ${1,I,I^2}$ are central elements (i.e. commute with everything in $mathbb{C}(S_3)$), where $I=P_{12}+P_{23}+P_{13}$. The 3-dimensional subspace orthogonal to these central elements is spanned by ${P_{12}-P_{23},P_{12}-P_{13},[P_{12},P_{23}]}$, which forms an $mathfrak{su}(2)$ Lie algebra, their relation to the spin generators are
$$P_{12}-P_{23}=sqrt{6}(s_x-s_y),~~P_{12}-P_{13}=sqrt{6}(s_x-s_z),~~[P_{12},P_{23}]=2i(s_x+s_y+s_z),$$
where $[s_i,s_j]=iepsilon_{ijk}s_k$. Therefore we conclude that $mathfrak{L}(S_3)=mathfrak{u}(1)^3oplus mathfrak{su}(2)$.
But when trying to determine the structure of $mathfrak{L}(S_N)$ for arbitrary $N$, this kind of brute force calculation seems hopeless. I don't even know the structure of $mathfrak{L}(S_4)$. For large $N$, I want to know this: if $mathfrak{L}(S_N)$ is a direct sum of simple Lie algebras, let $d_N$ be the dimension of its largest irreducible(simple) component (e.g. $d_3=3$). Does $d_N$ grow algebraically fast or exponentially fast with $N$?
Any hints, suggestions, or relevant references will be welcomed. Thanks.
abstract-algebra permutations lie-algebras symmetric-groups
abstract-algebra permutations lie-algebras symmetric-groups
asked Nov 20 '18 at 3:06
Lagrenge
1208
asked Nov 20 '18 at 3:06
Lagrenge
1208
asked Nov 20 '18 at 3:06
Lagrenge
1208
asked Nov 20 '18 at 3:06
Lagrenge
1208
asked Nov 20 '18 at 3:06
Lagrenge
1208
1208
The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components.
– Sheel Stueber
Nov 20 '18 at 3:14
add a comment |
The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components.
– Sheel Stueber
Nov 20 '18 at 3:14
The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components.
– Sheel Stueber
Nov 20 '18 at 3:14
The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components.
– Sheel Stueber
Nov 20 '18 at 3:14
add a comment |
1 Answer
1
active
oldest
votes
$newcommand{fru}{mathfrak{u}}
newcommand{frsu}{mathfrak{su}}
newcommand{frgl}{mathfrak{gl}}
newcommand{frsl}{mathfrak{sl}}
newcommand{frL}{mathfrak{L}}
newcommand{End}{operatorname{End}}
newcommand{KG}{Kleft[Gright]}
newcommand{CC}{mathbb{C}}
newcommand{LG}{mathfrak{L}left(Gright)}$
Yes. For each $N geq 0$, we have $frLleft(S_Nright) cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right)$, where the direct sum is over all partitions $lambda$ of $N$, and where $V_lambda$ denotes the Specht module corresponding to the partition $lambda$.
But this isn't really specific to the symmetric groups. More generally, we can answer this question for any finite group $G$ if we know its representation theory.
So let $G$ be a finite group, and let $K$ be a field of characteristic $0$ over which $G$ splits. (It is sufficient, but not necessary, to take $K = CC$.)
For each $K$-algebra $A$, we let $A^-$ be the Lie algebra on the $K$-vector space $A$ whose Lie bracket is the commutator of $A$. Thus, what you call $frLleft(Gright)$ is just $left(CCleft[Gright]right)^-$.
Since the field $K$ has characteristic $0$, we have
begin{equation}
frglleft(mright) cong K^- oplus frslleft(mright)
label{darij1.eq.gl-sl}
tag{1}
end{equation}
(an isomorphism of $K$-Lie algebras) for each positive integer $m$.
Let $V_i$ (with $i$ running over some finite indexing set $I$) be all irreducible representations of $G$ over $K$ (up to isomorphism, without repetitions). Theorem 4.1.1 (ii) in Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina, Introduction to representation theory, Student Mathematical Library #59, AMS 2011 shows that $KG cong bigopluslimits_{i in I} End left(V_iright)$ as $K$-algebras. Thus,
begin{align}
left(KGright)^-
&cong left(bigopluslimits_{i in I} End left(V_iright)right)^-
cong bigopluslimits_{i in I} underbrace{left(End left(V_iright)right)^-}_{= frglleft(V_iright) cong frglleft(dim V_iright) } \
&cong bigopluslimits_{i in I} underbrace{frglleft(dim V_iright) }_{substack{cong K^- oplus frslleft(dim V_i right) \ text{(by eqref{darij1.eq.gl-sl})}}}
cong bigopluslimits_{i in I} left( K^- oplus frslleft(dim V_i right) right) .
label{darij1.eq.gen-form}
tag{2}
end{align}
When $G$ is the symmetric group $S_N$, we can take $I$ to be the set of all partitions $lambda$ of $N$ (this is a well-known fact from the representation theory of symmetric group; see, e.g., §5.12 in op. cit.), and the corresponding irreducible representations $V_lambda$ are the so-called Specht modules. For each partition $lambda$ of $N$, the dimension $dim V_lambda$ has a nice expression called the hook-length formula (§5.17 in op. cit.), and there is a basis of $V_lambda$ indexed by the standard tableaux of shape $lambda$ (see, e.g., Mark Wildon, Representation theory of the symmetric group, 2014).
In light of this, eqref{darij1.eq.gen-form} (applied to $G = S_N$) becomes
begin{align}
left(Kleft[S_Nright] right)^-
cong bigopluslimits_{lambda vdash N} left( K^- oplus frslleft(dim V_lambda right) right) .
end{align}
When $K = CC$, this further rewrites as
begin{align}
left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( underbrace{CC^-}_{cong fruleft(1right)} oplus underbrace{frslleft(dim V_lambda right)}_{cong frsuleft(dim V_lambdaright)} right)
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right) .
end{align}
Thus,
begin{align}
frLleft(S_Nright)
= left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right)
cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right) .
end{align}
This is exactly my claim.
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
add a comment |
Your Answer
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1 Answer
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$newcommand{fru}{mathfrak{u}}
newcommand{frsu}{mathfrak{su}}
newcommand{frgl}{mathfrak{gl}}
newcommand{frsl}{mathfrak{sl}}
newcommand{frL}{mathfrak{L}}
newcommand{End}{operatorname{End}}
newcommand{KG}{Kleft[Gright]}
newcommand{CC}{mathbb{C}}
newcommand{LG}{mathfrak{L}left(Gright)}$
Yes. For each $N geq 0$, we have $frLleft(S_Nright) cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right)$, where the direct sum is over all partitions $lambda$ of $N$, and where $V_lambda$ denotes the Specht module corresponding to the partition $lambda$.
But this isn't really specific to the symmetric groups. More generally, we can answer this question for any finite group $G$ if we know its representation theory.
So let $G$ be a finite group, and let $K$ be a field of characteristic $0$ over which $G$ splits. (It is sufficient, but not necessary, to take $K = CC$.)
For each $K$-algebra $A$, we let $A^-$ be the Lie algebra on the $K$-vector space $A$ whose Lie bracket is the commutator of $A$. Thus, what you call $frLleft(Gright)$ is just $left(CCleft[Gright]right)^-$.
Since the field $K$ has characteristic $0$, we have
begin{equation}
frglleft(mright) cong K^- oplus frslleft(mright)
label{darij1.eq.gl-sl}
tag{1}
end{equation}
(an isomorphism of $K$-Lie algebras) for each positive integer $m$.
Let $V_i$ (with $i$ running over some finite indexing set $I$) be all irreducible representations of $G$ over $K$ (up to isomorphism, without repetitions). Theorem 4.1.1 (ii) in Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina, Introduction to representation theory, Student Mathematical Library #59, AMS 2011 shows that $KG cong bigopluslimits_{i in I} End left(V_iright)$ as $K$-algebras. Thus,
begin{align}
left(KGright)^-
&cong left(bigopluslimits_{i in I} End left(V_iright)right)^-
cong bigopluslimits_{i in I} underbrace{left(End left(V_iright)right)^-}_{= frglleft(V_iright) cong frglleft(dim V_iright) } \
&cong bigopluslimits_{i in I} underbrace{frglleft(dim V_iright) }_{substack{cong K^- oplus frslleft(dim V_i right) \ text{(by eqref{darij1.eq.gl-sl})}}}
cong bigopluslimits_{i in I} left( K^- oplus frslleft(dim V_i right) right) .
label{darij1.eq.gen-form}
tag{2}
end{align}
When $G$ is the symmetric group $S_N$, we can take $I$ to be the set of all partitions $lambda$ of $N$ (this is a well-known fact from the representation theory of symmetric group; see, e.g., §5.12 in op. cit.), and the corresponding irreducible representations $V_lambda$ are the so-called Specht modules. For each partition $lambda$ of $N$, the dimension $dim V_lambda$ has a nice expression called the hook-length formula (§5.17 in op. cit.), and there is a basis of $V_lambda$ indexed by the standard tableaux of shape $lambda$ (see, e.g., Mark Wildon, Representation theory of the symmetric group, 2014).
In light of this, eqref{darij1.eq.gen-form} (applied to $G = S_N$) becomes
begin{align}
left(Kleft[S_Nright] right)^-
cong bigopluslimits_{lambda vdash N} left( K^- oplus frslleft(dim V_lambda right) right) .
end{align}
When $K = CC$, this further rewrites as
begin{align}
left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( underbrace{CC^-}_{cong fruleft(1right)} oplus underbrace{frslleft(dim V_lambda right)}_{cong frsuleft(dim V_lambdaright)} right)
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right) .
end{align}
Thus,
begin{align}
frLleft(S_Nright)
= left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right)
cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right) .
end{align}
This is exactly my claim.
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
add a comment |
$newcommand{fru}{mathfrak{u}}
newcommand{frsu}{mathfrak{su}}
newcommand{frgl}{mathfrak{gl}}
newcommand{frsl}{mathfrak{sl}}
newcommand{frL}{mathfrak{L}}
newcommand{End}{operatorname{End}}
newcommand{KG}{Kleft[Gright]}
newcommand{CC}{mathbb{C}}
newcommand{LG}{mathfrak{L}left(Gright)}$
Yes. For each $N geq 0$, we have $frLleft(S_Nright) cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right)$, where the direct sum is over all partitions $lambda$ of $N$, and where $V_lambda$ denotes the Specht module corresponding to the partition $lambda$.
But this isn't really specific to the symmetric groups. More generally, we can answer this question for any finite group $G$ if we know its representation theory.
So let $G$ be a finite group, and let $K$ be a field of characteristic $0$ over which $G$ splits. (It is sufficient, but not necessary, to take $K = CC$.)
For each $K$-algebra $A$, we let $A^-$ be the Lie algebra on the $K$-vector space $A$ whose Lie bracket is the commutator of $A$. Thus, what you call $frLleft(Gright)$ is just $left(CCleft[Gright]right)^-$.
Since the field $K$ has characteristic $0$, we have
begin{equation}
frglleft(mright) cong K^- oplus frslleft(mright)
label{darij1.eq.gl-sl}
tag{1}
end{equation}
(an isomorphism of $K$-Lie algebras) for each positive integer $m$.
Let $V_i$ (with $i$ running over some finite indexing set $I$) be all irreducible representations of $G$ over $K$ (up to isomorphism, without repetitions). Theorem 4.1.1 (ii) in Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina, Introduction to representation theory, Student Mathematical Library #59, AMS 2011 shows that $KG cong bigopluslimits_{i in I} End left(V_iright)$ as $K$-algebras. Thus,
begin{align}
left(KGright)^-
&cong left(bigopluslimits_{i in I} End left(V_iright)right)^-
cong bigopluslimits_{i in I} underbrace{left(End left(V_iright)right)^-}_{= frglleft(V_iright) cong frglleft(dim V_iright) } \
&cong bigopluslimits_{i in I} underbrace{frglleft(dim V_iright) }_{substack{cong K^- oplus frslleft(dim V_i right) \ text{(by eqref{darij1.eq.gl-sl})}}}
cong bigopluslimits_{i in I} left( K^- oplus frslleft(dim V_i right) right) .
label{darij1.eq.gen-form}
tag{2}
end{align}
When $G$ is the symmetric group $S_N$, we can take $I$ to be the set of all partitions $lambda$ of $N$ (this is a well-known fact from the representation theory of symmetric group; see, e.g., §5.12 in op. cit.), and the corresponding irreducible representations $V_lambda$ are the so-called Specht modules. For each partition $lambda$ of $N$, the dimension $dim V_lambda$ has a nice expression called the hook-length formula (§5.17 in op. cit.), and there is a basis of $V_lambda$ indexed by the standard tableaux of shape $lambda$ (see, e.g., Mark Wildon, Representation theory of the symmetric group, 2014).
In light of this, eqref{darij1.eq.gen-form} (applied to $G = S_N$) becomes
begin{align}
left(Kleft[S_Nright] right)^-
cong bigopluslimits_{lambda vdash N} left( K^- oplus frslleft(dim V_lambda right) right) .
end{align}
When $K = CC$, this further rewrites as
begin{align}
left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( underbrace{CC^-}_{cong fruleft(1right)} oplus underbrace{frslleft(dim V_lambda right)}_{cong frsuleft(dim V_lambdaright)} right)
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right) .
end{align}
Thus,
begin{align}
frLleft(S_Nright)
= left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right)
cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right) .
end{align}
This is exactly my claim.
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
add a comment |
$newcommand{fru}{mathfrak{u}}
newcommand{frsu}{mathfrak{su}}
newcommand{frgl}{mathfrak{gl}}
newcommand{frsl}{mathfrak{sl}}
newcommand{frL}{mathfrak{L}}
newcommand{End}{operatorname{End}}
newcommand{KG}{Kleft[Gright]}
newcommand{CC}{mathbb{C}}
newcommand{LG}{mathfrak{L}left(Gright)}$
Yes. For each $N geq 0$, we have $frLleft(S_Nright) cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right)$, where the direct sum is over all partitions $lambda$ of $N$, and where $V_lambda$ denotes the Specht module corresponding to the partition $lambda$.
But this isn't really specific to the symmetric groups. More generally, we can answer this question for any finite group $G$ if we know its representation theory.
So let $G$ be a finite group, and let $K$ be a field of characteristic $0$ over which $G$ splits. (It is sufficient, but not necessary, to take $K = CC$.)
For each $K$-algebra $A$, we let $A^-$ be the Lie algebra on the $K$-vector space $A$ whose Lie bracket is the commutator of $A$. Thus, what you call $frLleft(Gright)$ is just $left(CCleft[Gright]right)^-$.
Since the field $K$ has characteristic $0$, we have
begin{equation}
frglleft(mright) cong K^- oplus frslleft(mright)
label{darij1.eq.gl-sl}
tag{1}
end{equation}
(an isomorphism of $K$-Lie algebras) for each positive integer $m$.
Let $V_i$ (with $i$ running over some finite indexing set $I$) be all irreducible representations of $G$ over $K$ (up to isomorphism, without repetitions). Theorem 4.1.1 (ii) in Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina, Introduction to representation theory, Student Mathematical Library #59, AMS 2011 shows that $KG cong bigopluslimits_{i in I} End left(V_iright)$ as $K$-algebras. Thus,
begin{align}
left(KGright)^-
&cong left(bigopluslimits_{i in I} End left(V_iright)right)^-
cong bigopluslimits_{i in I} underbrace{left(End left(V_iright)right)^-}_{= frglleft(V_iright) cong frglleft(dim V_iright) } \
&cong bigopluslimits_{i in I} underbrace{frglleft(dim V_iright) }_{substack{cong K^- oplus frslleft(dim V_i right) \ text{(by eqref{darij1.eq.gl-sl})}}}
cong bigopluslimits_{i in I} left( K^- oplus frslleft(dim V_i right) right) .
label{darij1.eq.gen-form}
tag{2}
end{align}
When $G$ is the symmetric group $S_N$, we can take $I$ to be the set of all partitions $lambda$ of $N$ (this is a well-known fact from the representation theory of symmetric group; see, e.g., §5.12 in op. cit.), and the corresponding irreducible representations $V_lambda$ are the so-called Specht modules. For each partition $lambda$ of $N$, the dimension $dim V_lambda$ has a nice expression called the hook-length formula (§5.17 in op. cit.), and there is a basis of $V_lambda$ indexed by the standard tableaux of shape $lambda$ (see, e.g., Mark Wildon, Representation theory of the symmetric group, 2014).
In light of this, eqref{darij1.eq.gen-form} (applied to $G = S_N$) becomes
begin{align}
left(Kleft[S_Nright] right)^-
cong bigopluslimits_{lambda vdash N} left( K^- oplus frslleft(dim V_lambda right) right) .
end{align}
When $K = CC$, this further rewrites as
begin{align}
left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( underbrace{CC^-}_{cong fruleft(1right)} oplus underbrace{frslleft(dim V_lambda right)}_{cong frsuleft(dim V_lambdaright)} right)
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right) .
end{align}
Thus,
begin{align}
frLleft(S_Nright)
= left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right)
cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right) .
end{align}
This is exactly my claim.
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
$newcommand{fru}{mathfrak{u}}
newcommand{frsu}{mathfrak{su}}
newcommand{frgl}{mathfrak{gl}}
newcommand{frsl}{mathfrak{sl}}
newcommand{frL}{mathfrak{L}}
newcommand{End}{operatorname{End}}
newcommand{KG}{Kleft[Gright]}
newcommand{CC}{mathbb{C}}
newcommand{LG}{mathfrak{L}left(Gright)}$
Yes. For each $N geq 0$, we have $frLleft(S_Nright) cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right)$, where the direct sum is over all partitions $lambda$ of $N$, and where $V_lambda$ denotes the Specht module corresponding to the partition $lambda$.
But this isn't really specific to the symmetric groups. More generally, we can answer this question for any finite group $G$ if we know its representation theory.
So let $G$ be a finite group, and let $K$ be a field of characteristic $0$ over which $G$ splits. (It is sufficient, but not necessary, to take $K = CC$.)
For each $K$-algebra $A$, we let $A^-$ be the Lie algebra on the $K$-vector space $A$ whose Lie bracket is the commutator of $A$. Thus, what you call $frLleft(Gright)$ is just $left(CCleft[Gright]right)^-$.
Since the field $K$ has characteristic $0$, we have
begin{equation}
frglleft(mright) cong K^- oplus frslleft(mright)
label{darij1.eq.gl-sl}
tag{1}
end{equation}
(an isomorphism of $K$-Lie algebras) for each positive integer $m$.
Let $V_i$ (with $i$ running over some finite indexing set $I$) be all irreducible representations of $G$ over $K$ (up to isomorphism, without repetitions). Theorem 4.1.1 (ii) in Pavel Etingof, Oleg Golberg, Sebastian Hensel, Tiankai Liu, Alex Schwendner, Dmitry Vaintrob, and Elena Yudovina, Introduction to representation theory, Student Mathematical Library #59, AMS 2011 shows that $KG cong bigopluslimits_{i in I} End left(V_iright)$ as $K$-algebras. Thus,
begin{align}
left(KGright)^-
&cong left(bigopluslimits_{i in I} End left(V_iright)right)^-
cong bigopluslimits_{i in I} underbrace{left(End left(V_iright)right)^-}_{= frglleft(V_iright) cong frglleft(dim V_iright) } \
&cong bigopluslimits_{i in I} underbrace{frglleft(dim V_iright) }_{substack{cong K^- oplus frslleft(dim V_i right) \ text{(by eqref{darij1.eq.gl-sl})}}}
cong bigopluslimits_{i in I} left( K^- oplus frslleft(dim V_i right) right) .
label{darij1.eq.gen-form}
tag{2}
end{align}
When $G$ is the symmetric group $S_N$, we can take $I$ to be the set of all partitions $lambda$ of $N$ (this is a well-known fact from the representation theory of symmetric group; see, e.g., §5.12 in op. cit.), and the corresponding irreducible representations $V_lambda$ are the so-called Specht modules. For each partition $lambda$ of $N$, the dimension $dim V_lambda$ has a nice expression called the hook-length formula (§5.17 in op. cit.), and there is a basis of $V_lambda$ indexed by the standard tableaux of shape $lambda$ (see, e.g., Mark Wildon, Representation theory of the symmetric group, 2014).
In light of this, eqref{darij1.eq.gen-form} (applied to $G = S_N$) becomes
begin{align}
left(Kleft[S_Nright] right)^-
cong bigopluslimits_{lambda vdash N} left( K^- oplus frslleft(dim V_lambda right) right) .
end{align}
When $K = CC$, this further rewrites as
begin{align}
left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( underbrace{CC^-}_{cong fruleft(1right)} oplus underbrace{frslleft(dim V_lambda right)}_{cong frsuleft(dim V_lambdaright)} right)
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right) .
end{align}
Thus,
begin{align}
frLleft(S_Nright)
= left(CC left[S_Nright]right)^-
cong bigopluslimits_{lambda vdash N} left( fruleft(1right) oplus frsuleft(dim V_lambdaright) right)
cong bigopluslimits_{lambda vdash N} left( frsuleft(dim V_lambdaright) oplus fruleft(1right) right) .
end{align}
This is exactly my claim.
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
answered Nov 20 '18 at 3:35
darij grinberg
10.2k33062
10.2k33062
add a comment |
add a comment |
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The center of the group algebra has dimension equal to the number of conjugacy classes of the group. This might be helpful to compute one component but I do not know a general method to compute the rest of the components.
– Sheel Stueber
Nov 20 '18 at 3:14