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I've been trying to check the claim that the vector space direct sum $L \oplus D$ is a Lie algebra, and I'm having a lot of trouble with verifying the Jacobi identity. It's defined where $L$ is a Lie algebra and $D$ is a subalgebra of Der(L) with bracket $[x_1 + d_1, x_2 + d_2] = [x_1,x_2] + d_1(x_2) - d_2(x_1) + [d_1,d_2]$.

I started with $[x,[y,z]] + [y,[z,x]] + [z,[x,y]]$ for $x = x_1 + d_1, y = x_2 + d_2, z = x_3 + d_3$, with the aim of showing that this simplifies to 0 by bilinearity and definition of the bracket. I simplified all the inner brackets first, but then get confused when trying to break down further. I end up getting what seem like strange compositions of elements of L and derivations.

Andreas Caranti
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Phdetermined
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1 Answers1

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Linearity of the Lie product in each component, and the cyclic format of the Jacobi identify, allows you to deal with four cases.

  1. $x, y, z \in L$, that's Jacobi in $L$.

  2. $x, y, z \in D$, that's Jacobi in $D$.

  3. $x, y\in L$, and $z \in D$, that's the fact that the elements of $D$ are derivations of $L$.

  4. $x\in L$, and $y, z \in D$, that's the fact that $D$ is a Lie subalgebra of $\operatorname{Der}(L) $.

Andreas Caranti
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  • Can you help me to understand why we can deal with 4 cases?? We never assume that the bracket operation being defined is bilnear to make the simplification, right? Rather, we use that $(s_1,x_1) = (s_1,0)+(0,x_1)$... and how do using the cyclic format of the Jacobi identity help to allow us to make this simplification? Thank you for your time – FireFenix777 Sep 20 '20 at 18:48
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    @FireFenix777, sorry, I am short of time, still, without Jacobi I would have to consider also the cases when $x, z \in L$, and $y \in D$, etc. – Andreas Caranti Sep 21 '20 at 13:44