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Given Lie algebras $S$ and $I$ and a Lie homomorphism $\theta \colon S\to \operatorname{Der} I$, we have the semidirect product to be the vector space $S\oplus I$ with operation $$ (s_{1},x_{1})(s_{2}x_{2}) := ([s_{1},s_{2}],[x_{1},x_{2}]+\theta(s_{1})x_{2}-\theta(s_{2})x_{1}). $$ Show that this is a Lie algebra.

So I can easily verify the skew-symmetric but I can't seem to work out a nice way of proving the Jacobi identity. Am I missing a simple trick or must you perform the tedious calculation to show this? Thanks.

Jendrik Stelzner
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user320140
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  • You can at least simplify the calculation and break it into parts by proving the Jacobi identity for a basis (or generating system) of $S\oplus I$. – Berci Mar 05 '16 at 15:42
  • I'm not entirely sure how I could simplify the calculation in this way since $S$ and $I$ are arbitrary Lie algebras so we know nothing of their basis structure? – user320140 Mar 05 '16 at 16:20
  • I only wanted to mean that you can check it only on special form entities like $(s_1,0)$ or $(0,x_1)$. It simplifies the calculation but introduces many cases. I guess though this is not the ultimate answer you're after. – Berci Mar 05 '16 at 16:33
  • Actually I do think this makes the calculation SLIGHTLY more bearable since the Lie bracket is bilinear so we can write $$[(s_{1},x_{1}),[(s_{2},x_{2}),(s_{3},x_{3})]]=[(s_{1},0),[(s_{2},x_{2}),(s_{3},x_{3})]]+[(0,x_{1}),[(s_{2},x_{2}),(s_{3},x_{3})]]$$ Thanks for your input, I will take all the help I can get! – user320140 Mar 05 '16 at 19:04

1 Answers1

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The calculation is no longer tedious if you split it up into four cases. Since the Jacobi identity is trilinear we only need to check it for one of the following cases: $(s_1,0),(s_2,0),(s_3,0)$, or $(s_1,0),(s_2,0),(0,x_3)$, or $(s_1,0),(0,x_2),(0,x_3)$ or $(0,x_1),(0,x_2),(0,x_3)$. The cases themselves are immediate, because they follow from the facts that either $S$ is a Lie algebra, or that $I$ is a Lie algebra, or that the $\theta(s_i)$ are derivations, or that $\theta$ is a Lie algebra homomorphism.

Dietrich Burde
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  • I see that makes a lot of sense, thanks! – user320140 Mar 05 '16 at 19:07
  • You are welcome. – Dietrich Burde Mar 05 '16 at 19:16
  • I'm trying to understand why we can break this down into these 4 cases. In breaking it down to four cases, we never assume that $[-,-]$ is bilnear.. – FireFenix777 Sep 20 '20 at 18:46
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    The Lie bracket is bilinear, so the "Jacobiator" is trilinear. So we are assuming it. – Dietrich Burde Sep 20 '20 at 18:50
  • But the question is asking to verify that $[-,-]$ is indeed a well defined lie bracket operation, which means it is bilinear, antisymmetric and satisfies the Jacobi identity. Why don't we have to prove that this interesting candidate for a product of a lie algebra is bilinear?

    I get that $[-,-]$ is bilinear on $S$ and $I$, but when we are trying to extend $[-,-]$ in a nonobvious way to $S \oplus I$ it seems to be that we would have to verify the bilinearity, which I am having a hard time doing

     – FireFenix777 Sep 21 '20 at 15:11
  • Infact, it seems like it's not bilinear... – FireFenix777 Sep 21 '20 at 15:27
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    It is certainly bilinear. The OP made a typo in the formula. – Dietrich Burde Sep 21 '20 at 18:30