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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 $a_{ij}^k\in F$ such that $$[x_i,x_j]=\sum_{k=1}^{n}a_{ij}^{k}x_k$$ 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 $\forall a,b\in L$ $$[a,b]=\sum_{i,j,t}^{n,n,n^2}c_{t}[x_i,x_j]=\sum_{i,j,t}^{n,n,n^2}c_{t}\Bigl(\sum_{k=1}^{n}a_{ij}^kx_k\Bigr)$$ where $c_t \in F$ ?

TheLast Cipher
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2 Answers2

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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 $$[u,v]=\sum_{i,j}c_id_j[x_i,x_j] =\sum_k\left(\sum_{i,j}a_{i,j}^kc_id_j\right)x_k.$$ This is nothing more than the bi-linearity of the Lie bracket.

Angina Seng
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It means that there is only one Lie algebra structure $[\cdot,\cdot]$ on $L$ for which it is true that$$(\forall i,j\in\{1,2,\ldots,n\}):[x_i,x_j]=\sum_{k=1}^na_{ij}^kx_k.$$This follows from the fact that $\{x_1,\ldots,x_n\}$ is a basis an that the Lie bracket is bilinear.

José Carlos Santos
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  • 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 $F$. – TheLast Cipher May 04 '19 at 08:55