Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
I have added dollar signs \$...\$ to your formulas so that they display properly. – Andreas Caranti Feb 11 '13 at 13:11
Thank you. Looks much better. – Phdetermined Feb 11 '13 at 21:30

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) $.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.