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I am beginning to learn some materials about representation of Lie algebra. Here I define $\mathfrak{sl}_2$ triple as: $$ \{H,X,Y\in \text{End}_{\mathbb{C}}(V)|H,X,Y \text{are nonzero},\quad [H, X]=2 X,\quad[H, Y]=-2 Y,\quad[X, Y]=H\} $$ where $V$ is a finite dimensional vector space over $\mathbb{C}$ (or equivalently, finite dimensional representation of $\mathfrak{sl}_2\mathbb{C}$). Here $[A, B]=A B-B A, \quad \forall A, B \in \operatorname{End} V$.

I suspect from the definition of $\mathfrak{sl}_2$ triple it is enough to imply $H$ is semisimple (since we are working with $\mathbb{C}$, this is equivalent to say $H$ is diagonalizable).

I try to prove it using induction on dimension of $V$, when $\text{dim}_{\mathbb{C}}V=2$, after fixing a basis, I wonder is it enough to deduce from commutation relation that $$ H=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right], \quad X=\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right], \quad Y=\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right] $$ So in particular, $H$ is semisimple. It is a little bit painful to write out the coefficients in terms of basis and solve the linear equations, I want to ask is there any clever way to prove that?

For induction hypothesis, I assume $H$ is semisimple when $\text{dim}_{\mathbb{C}}V\leq n-1$. Let $\lambda_1,\lambda_2,\cdots, \lambda_k$ be eigenvalues of $H$, and let $$ W=V_{\lambda_{1}} \oplus \cdots \oplus V_{\lambda_{k}} $$ where $V_{\lambda_{k}}$ is eigenspace of $H$ with respect to eigenvalue $\lambda_{k}$. I can show $W$ is an invariant space for $H,X,Y$, thus we can consider the quotient representation $$H,X,Y \in \text{End}_{\mathbb{C}}(V/W)$$ After using induction hypothesis, $H$ is semisimple in $V/W$. Does $H$ is semisimple in both $W$ and $V/W$ implies $H$ is semisimple in $V$? I am not sure about this...

I guess there are more ways to prove this proposition. Any insight is welcomed and thank you in advance!

Edward Z. Miao
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  • Can you recall what is the definition of an element being semisimple? – Arctic Char May 23 '20 at 02:20
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    @ArcticChar Of course! In my textbook, $\mathcal{A} \in \text{End}_{\mathbb{F}}(V)$ is called semisimple if for all $\mathcal{A}$-invariant subspace $V_1$, its complement is also a $\mathcal{A}$-invariant subspace. One can check this definition is equvalent with $\mathcal{A}$ is diagonalizable when $\mathbb{F}=\mathbb{C}$. Hope this helps. – Edward Z. Miao May 23 '20 at 02:25

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In your set-up $H$, $X$ and $Y$ are a basis of a Lie algebra $H\cong\mathfrak{sl}_2$ which happens to be a Lie subalgebra of $\text{End}(V)$ where $V$ is a finite-dimensional vector spaces. From the theory of representations of $\mathfrak{sl}_2$ (weight spaces), $H$ acts semisimply on the $L$-module $V$.

Your argument seems to be mimicking the usual weight space theory. You can show the sum of eigenspaces of $H$ (these are the weight spaces) is an $L$-submodule $W$ of $V$. Inductively you have $H$ acting semisimply on $V/W$. The problem is that although a matrix acts semisimply on $V/W$ and $W$ it does not imply that it acts semisimply on $V$. Consider a typical nilpotent matrix $\pmatrix{0&1\\0&0}$.

The extra ingredient one need here is complete reducibility: that $V\cong W\oplus (V/W)$ as $L$-modules. This is usually proved by considering the action of the Casimir element of the universal enveloping algebra of $\mathfrak{sl}_2$.

Angina Seng
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  • One reality check: does your $L$ stands for Lie algebra $\cong \mathfrak{s l}_{2}$? Besides, in second paragraph, I guess there is a minor typo? It should be "The problem is that although a matrix acts semisimply on $V/W$ and $W$ it does not imply that it acts semisimply on $V$ ". – Edward Z. Miao May 23 '20 at 05:37
  • Do you mean the fact that representation of semisimple Lie algebra $\mathfrak{sl_2}(\mathbb{C})$ is complete reducible can rule out the nilpotent case such as $\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right)$? – Edward Z. Miao May 23 '20 at 05:45