Regarding basis vectors of a Lie algebra.Are there finite-dimensional Lie algebras which are not defined over the integers?Direct sum decomposition of weight spaces and relation to Tensor products.Structure constants for and the adjoint representation and meaning in $sl(2,F)$Is a formal deformation of a Lie algebra an example of a formal group law?Basis of Lie Algebra $mathfraksu(3)$Regarding lower central series of a Lie algebra LRegarding Levi's decomposition of a Lie algebraRepresentation of $mathfrakgl(n,R)$ on $R[X_1,dots,X_n]$ over arbitrary ringWhat does the Jacobi identity impose on structure constants?Exercise 7.7.3 in Weibel (computation of $H^3(mathfraksl_2,k)$ via Chevalley-Eilenberg complex)
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Regarding basis vectors of a Lie algebra.
Are there finite-dimensional Lie algebras which are not defined over the integers?Direct sum decomposition of weight spaces and relation to Tensor products.Structure constants for and the adjoint representation and meaning in $sl(2,F)$Is a formal deformation of a Lie algebra an example of a formal group law?Basis of Lie Algebra $mathfraksu(3)$Regarding lower central series of a Lie algebra LRegarding Levi's decomposition of a Lie algebraRepresentation of $mathfrakgl(n,R)$ on $R[X_1,dots,X_n]$ over arbitrary ringWhat does the Jacobi identity impose on structure constants?Exercise 7.7.3 in Weibel (computation of $H^3(mathfraksl_2,k)$ via Chevalley-Eilenberg complex)
$begingroup$
From the book "Introduction to Lie Algebras" by Erdmann & Wildon:
If $L$ is a Lie algbra over a field $F$ with basis $(x_1,cdots, x_n)$, then $[-,-]$ is complete determined by the products $[x_i,x_j]$. We define scalars $a_ij^kin F$ such that $$[x_i,x_j]=sum_k=1^na_ij^kx_k$$ The $a_ij^k$ are the structure constants of $L$ with respect to this basis.
What does "completey determined by the products $[x_i,x_j]$" mean? Does it mean that $forall a,bin L$ $$[a,b]=sum_i,j,t^n,n,n^2c_t[x_i,x_j]=sum_i,j,t^n,n,n^2c_tBigl(sum_k=1^na_ij^kx_kBigr)$$ where $c_t in F$ ?
lie-algebras
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
$endgroup$
add a comment |
$begingroup$
From the book "Introduction to Lie Algebras" by Erdmann & Wildon:
If $L$ is a Lie algbra over a field $F$ with basis $(x_1,cdots, x_n)$, then $[-,-]$ is complete determined by the products $[x_i,x_j]$. We define scalars $a_ij^kin F$ such that $$[x_i,x_j]=sum_k=1^na_ij^kx_k$$ The $a_ij^k$ are the structure constants of $L$ with respect to this basis.
What does "completey determined by the products $[x_i,x_j]$" mean? Does it mean that $forall a,bin L$ $$[a,b]=sum_i,j,t^n,n,n^2c_t[x_i,x_j]=sum_i,j,t^n,n,n^2c_tBigl(sum_k=1^na_ij^kx_kBigr)$$ where $c_t in F$ ?
lie-algebras
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
$endgroup$
$begingroup$
What are $a$ and $b$?
$endgroup$
– Lord Shark the Unknown
May 4 at 8:40
$begingroup$
elements of $L$.
$endgroup$
– TheLast Cipher
May 4 at 8:41
$begingroup$
There's nothing corresponding to either $a$ or $b$ on the right side of your equation.
$endgroup$
– Lord Shark the Unknown
May 4 at 8:44
add a comment |
$begingroup$
From the book "Introduction to Lie Algebras" by Erdmann & Wildon:
If $L$ is a Lie algbra over a field $F$ with basis $(x_1,cdots, x_n)$, then $[-,-]$ is complete determined by the products $[x_i,x_j]$. We define scalars $a_ij^kin F$ such that $$[x_i,x_j]=sum_k=1^na_ij^kx_k$$ The $a_ij^k$ are the structure constants of $L$ with respect to this basis.
What does "completey determined by the products $[x_i,x_j]$" mean? Does it mean that $forall a,bin L$ $$[a,b]=sum_i,j,t^n,n,n^2c_t[x_i,x_j]=sum_i,j,t^n,n,n^2c_tBigl(sum_k=1^na_ij^kx_kBigr)$$ where $c_t in F$ ?
lie-algebras
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
$endgroup$
From the book "Introduction to Lie Algebras" by Erdmann & Wildon:
If $L$ is a Lie algbra over a field $F$ with basis $(x_1,cdots, x_n)$, then $[-,-]$ is complete determined by the products $[x_i,x_j]$. We define scalars $a_ij^kin F$ such that $$[x_i,x_j]=sum_k=1^na_ij^kx_k$$ The $a_ij^k$ are the structure constants of $L$ with respect to this basis.
What does "completey determined by the products $[x_i,x_j]$" mean? Does it mean that $forall a,bin L$ $$[a,b]=sum_i,j,t^n,n,n^2c_t[x_i,x_j]=sum_i,j,t^n,n,n^2c_tBigl(sum_k=1^na_ij^kx_kBigr)$$ where $c_t in F$ ?
lie-algebras
lie-algebras
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
asked May 4 at 8:26
TheLast CipherTheLast Cipher
789715
789715
$begingroup$
What are $a$ and $b$?
$endgroup$
– Lord Shark the Unknown
May 4 at 8:40
$begingroup$
elements of $L$.
$endgroup$
– TheLast Cipher
May 4 at 8:41
$begingroup$
There's nothing corresponding to either $a$ or $b$ on the right side of your equation.
$endgroup$
– Lord Shark the Unknown
May 4 at 8:44
add a comment |
$begingroup$
What are $a$ and $b$?
$endgroup$
– Lord Shark the Unknown
May 4 at 8:40
$begingroup$
elements of $L$.
$endgroup$
– TheLast Cipher
May 4 at 8:41
$begingroup$
There's nothing corresponding to either $a$ or $b$ on the right side of your equation.
$endgroup$
– Lord Shark the Unknown
May 4 at 8:44
$begingroup$
What are $a$ and $b$?
$endgroup$
– Lord Shark the Unknown
May 4 at 8:40
$begingroup$
What are $a$ and $b$?
$endgroup$
– Lord Shark the Unknown
May 4 at 8:40
$begingroup$
elements of $L$.
$endgroup$
– TheLast Cipher
May 4 at 8:41
$begingroup$
elements of $L$.
$endgroup$
– TheLast Cipher
May 4 at 8:41
$begingroup$
There's nothing corresponding to either $a$ or $b$ on the right side of your equation.
$endgroup$
– Lord Shark the Unknown
May 4 at 8:44
$begingroup$
There's nothing corresponding to either $a$ or $b$ on the right side of your equation.
$endgroup$
– Lord Shark the Unknown
May 4 at 8:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A typical element of $L$ is $u=sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=sum_j d_j x_j$. Then
$$[u,v]=sum_i,jc_id_j[x_i,x_j]
=sum_kleft(sum_i,ja_i,j^kc_id_jright)x_k.$$
This is nothing more than the bi-linearity of the Lie bracket.
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
$endgroup$
add a comment |
$begingroup$
It means that there is only one Lie algebra structure $[cdot,cdot]$ on $L$ for which it is true that$$(forall i,jin1,2,ldots,n):[x_i,x_j]=sum_k=1^na_ij^kx_k.$$This follows from the fact that $x_1,ldots,x_n$ is a basis an that the Lie bracket is bilinear.
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
$endgroup$
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A typical element of $L$ is $u=sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=sum_j d_j x_j$. Then
$$[u,v]=sum_i,jc_id_j[x_i,x_j]
=sum_kleft(sum_i,ja_i,j^kc_id_jright)x_k.$$
This is nothing more than the bi-linearity of the Lie bracket.
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
$endgroup$
add a comment |
$begingroup$
A typical element of $L$ is $u=sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=sum_j d_j x_j$. Then
$$[u,v]=sum_i,jc_id_j[x_i,x_j]
=sum_kleft(sum_i,ja_i,j^kc_id_jright)x_k.$$
This is nothing more than the bi-linearity of the Lie bracket.
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
$endgroup$
add a comment |
$begingroup$
A typical element of $L$ is $u=sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=sum_j d_j x_j$. Then
$$[u,v]=sum_i,jc_id_j[x_i,x_j]
=sum_kleft(sum_i,ja_i,j^kc_id_jright)x_k.$$
This is nothing more than the bi-linearity of the Lie bracket.
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
$endgroup$
A typical element of $L$ is $u=sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=sum_j d_j x_j$. Then
$$[u,v]=sum_i,jc_id_j[x_i,x_j]
=sum_kleft(sum_i,ja_i,j^kc_id_jright)x_k.$$
This is nothing more than the bi-linearity of the Lie bracket.
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
answered May 4 at 8:44
Lord Shark the UnknownLord Shark the Unknown
111k1163138
111k1163138
add a comment |
add a comment |
$begingroup$
It means that there is only one Lie algebra structure $[cdot,cdot]$ on $L$ for which it is true that$$(forall i,jin1,2,ldots,n):[x_i,x_j]=sum_k=1^na_ij^kx_k.$$This follows from the fact that $x_1,ldots,x_n$ is a basis an that the Lie bracket is bilinear.
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
$endgroup$
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
add a comment |
$begingroup$
It means that there is only one Lie algebra structure $[cdot,cdot]$ on $L$ for which it is true that$$(forall i,jin1,2,ldots,n):[x_i,x_j]=sum_k=1^na_ij^kx_k.$$This follows from the fact that $x_1,ldots,x_n$ is a basis an that the Lie bracket is bilinear.
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
$endgroup$
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
add a comment |
$begingroup$
It means that there is only one Lie algebra structure $[cdot,cdot]$ on $L$ for which it is true that$$(forall i,jin1,2,ldots,n):[x_i,x_j]=sum_k=1^na_ij^kx_k.$$This follows from the fact that $x_1,ldots,x_n$ is a basis an that the Lie bracket is bilinear.
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
$endgroup$
It means that there is only one Lie algebra structure $[cdot,cdot]$ on $L$ for which it is true that$$(forall i,jin1,2,ldots,n):[x_i,x_j]=sum_k=1^na_ij^kx_k.$$This follows from the fact that $x_1,ldots,x_n$ is a basis an that the Lie bracket is bilinear.
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
answered May 4 at 8:47
José Carlos SantosJosé Carlos Santos
183k24143257
183k24143257
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
add a comment |
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
$begingroup$
Thanks! I always get confused when my text says "the lie algebra $L$" and wonder if $L=(V,[-,-])$ where $V$ is some vector space. Where in fact, it mean there is a lie algebra on $L$ over $F$.
$endgroup$
– TheLast Cipher
May 4 at 8:55
add a comment |
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$begingroup$
What are $a$ and $b$?
$endgroup$
– Lord Shark the Unknown
May 4 at 8:40
$begingroup$
elements of $L$.
$endgroup$
– TheLast Cipher
May 4 at 8:41
$begingroup$
There's nothing corresponding to either $a$ or $b$ on the right side of your equation.
$endgroup$
– Lord Shark the Unknown
May 4 at 8:44