# Infinite sum of semisimple Ideals of a Lie algebra

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.

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

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.

• Thanks, very useful! Dec 1 '20 at 1:08

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?

• It is right @Dietrich Burde. But, I believe that in this case, $\mathfrak{sl}_n(\mathbb{C})$ is not a semisimple ideal of $L$. Nov 30 '20 at 14:38
• 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. Nov 30 '20 at 14:58
• 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. Nov 30 '20 at 18:09