Lie bracket, trace and an interesting exact sequence.

 

I see an interesting question on 知乎：全体XY-YX型矩阵可以构成一个线性空间吗？ - 知乎 (zhihu.com)。

Considering a vector space $U$$U$ over a field $\mathbb{F}$$\mathbb{F}$.

Does the set of operators on $U$$U$ such that $S:=\{XY-YX:X,Y\in {\mathrm{End}}_{\mathbb{F}-\mathrm{Vect}}(U)\}$$S := \{XY - YX : X, Y \in \mathrm{End}_{\mathrm{\mathbb{F}-Vect}}(U)\}$ form a vector space?

The link offers a method to prove that $S$$S$ is a vector space when $\dim U<\mathrm{\infty }$$\dim U < \infty$. The author uses a lot of matrix theory to prove that

$$\begin{array}{}\text{(1)}& S=\mathrm{Ker}(\mathrm{Tr})\end{array}$$

Where $\mathrm{Tr}$$\mathrm{Tr}$ refer to trace.

Here is a generalization to arbitrary vector spaces $U$$U$.

For convenience, denote ${\mathrm{End}}_{\mathbb{F}-\mathrm{Vect}}(U)$$\mathrm{End}_{\mathrm{\mathbb{F}-Vect}}(U)$ as $V$$V$.

Observe that the Lie bracket $[X,Y]=XY-YX$$[X,Y] = XY - YX$ is a bilinear map. By the universal property of the tensor product, we have a unique linear map $\mathrm{\Phi }$$\Phi$ such that:

$$\begin{array}{}\text{(2)}& \mathrm{\Phi }:V\otimes V\to V,\mathrm{\Phi }(X\otimes Y)=[X,Y]\end{array}$$

Hence, $S$$S$ is the image of $\mathrm{\Phi }$$\Phi$, which implies that $S$$S$ is a subspace of $V$$V$. $◻$$\square$

Then we obtain an interesting exact sequence when $\dim U<\mathrm{\infty }$$\dim U < \infty$:

$$\begin{array}{}\text{(3)}& V\otimes V\stackrel{\mathrm{\Phi }}{\to }V\stackrel{\mathrm{Tr}}{\to }\mathbb{F}\stackrel{}{\to }0\end{array}$$