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Following the paper by Kubok (Infinite-dimensional Lie algebras with null Jacobson radical), we have the following definitions. Let $L$ be a Lie algebra. If every finite subset of $L$ is contained in a finite-dimensional subalgebra (resp. a solvable subalgebra) of $L$ then $L$ is called locally finite (resp. locally solvable). For a locally finite Lie algebra $L$, $\sigma(L)$ is the maximal locally solvable ideal of $L$ and $L$ is said to be semisimple if $\sigma(L)=0$.

Suppose that $L$ is an arbitrary Lie algebra over a field of characteristic zero and that $L=\sum_{i\in\mathcal I}L_i$, where each $L_i$ is a finite-dimensional semisimple ideal of $L$.

If $\mathcal I$ is finite then $L$ is semisimple. In order to show this, it is enough to verify the case $L=L_1+L_2$ and apply induction. As $L_1$ is semisimple the Killing form $\kappa(\cdot,\cdot)$ of $L$ restrict $L_1$ is non-degenerated and thus $L=L_1\oplus L_1^\perp$, where $L_1^\perp=\{x\in L\mid \kappa(x,y)=0, \mbox{ for } y\in L_1 \}$. Now, since $L=L_1+L_2$ it follows that $$L_1^\perp \cong L/L_1=(L_1+L_2)/L_1\cong L_2/(L_1\cap L_2)$$is semisimple. Hence $L=L_1\oplus L_1^\perp$ is semisimple.

But, what happens if $\mathcal I$ is infinite? Is $L$ semisimple? I believe that the answer is no, but I can't find a counterexample. Is there some condition on $L$ or $L_i$ that guarantees the semisimplicity of $L$?

I appreciate any help.

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    $\begingroup$ What is your definition of "semisimple" for infinite-dimensional Lie algebras? $\endgroup$ Nov 30 '20 at 15:16
  • $\begingroup$ In fact, the case of infinite dimension is more delicate to define. I added the definition of semi-simplicity for infinite dimension. $\endgroup$ Nov 30 '20 at 20:02
  • $\begingroup$ You only have defined it now for a locally finite Lie algebra, right? $\endgroup$ Nov 30 '20 at 20:04
  • $\begingroup$ yeah! the definition is only for locally finite Lie algebras. $\endgroup$ Nov 30 '20 at 20:08
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Each $L_i$, being finite-dimensional semisimple, is a finite direct sum of simple ideals, say

$$L_i = \bigoplus_{j \in J(i)} S_j$$

where for each $i \in \mathcal I$, $J(i)$ is some finite set.

Now if $A, B$ are simple ideals of any Lie algebra, then either $A \cap B = \{0\}$ or $A=B$.

Consequently, we can w.l.o.g. assume that $S_j \cap S_k =\{0\}$ if $j \in J(i_1)$ and $k \in J(i_2)$ for $i_1 \neq i_2 \in \mathcal I$, consequently $L_{i_1} \cap L_{i_2} = \{0\}$ for all $i_1 \neq i_2 \in \mathcal I$ and we have

$$\displaystyle L = \sum L_i = \bigoplus L_i= \bigoplus_{i \in I} \bigoplus_{j \in J(i)} S_j = \bigoplus_{j\in \coprod_{i\in \mathcal I} J(i)} S_j$$

with all the $S_j$ being simple (and finite-dimensional) ideals and the direct sum being one of Lie algebras. Frome here it should be easy to describe all ideals of $L$ and in particular see that the only solvable ideal of $L$, hence $\sigma(L)$, is zero.

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  • $\begingroup$ Thanks, very useful! $\endgroup$ Dec 1 '20 at 1:08
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If $L=L_1\oplus L_2$ is a direct sum of simple Lie algebras, then $L$ is semisimple. However, for the sum $L_1+L_2$, an arbitrary sum, this need not be true in general - see our Example $4.10$ here: $$ L=\mathfrak{sl}_n(\Bbb C)\rtimes V(n)=\mathfrak{sl}_n(\Bbb C)+\mathfrak{sl}_n(\Bbb C) $$ is not semisimple. So $\sum_i L_i$ need not be semisimple, even for finitely many summands.

Edit: An infinite direct sum $L$ of simple Lie algebras will be infinite-dimensional. What is your definition of "semisimple" for infinite-dimensional Lie algebras?

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  • $\begingroup$ It is right @Dietrich Burde. But, I believe that in this case, $\mathfrak{sl}_n(\mathbb{C})$ is not a semisimple ideal of $L$. $\endgroup$ Nov 30 '20 at 14:38
  • $\begingroup$ Yes, you are right. If we have ideals then the sum is of course direct. I was not sure because of your title "arbitrary" sum. In the end the sum then is not so arbitrary but just a direct sum. $\endgroup$ Nov 30 '20 at 14:58
  • $\begingroup$ Even better (or worse), one can write any Lie algebra $L$ (of dimension $\ge 2$) as a nontrivial sum of subalgebras, namely, choosing a basis $(x_j)_{j \in J}$, just write $L = \sum_{j \in J} \langle x_j \rangle$. This is a direct sum of vector spaces, but of course not of Lie algebras unless $L$ happens to be abelian. I agree that OP should clarify what definition of semisimplicty they use in the infinite-dimensional case. $\endgroup$ Nov 30 '20 at 18:09

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