Say we have lie algebras $I$ and $S$ and a Lie algebra morphism $\theta:S\to Der I$. Lets try to turn $S \oplus I$ into a lie algebra by giving it the lie bracket operation:
edit: $$ [(s_{1},x_{1}),(s_{2}x_{2})]:=([s_{1},s_{2}],[x_{1},x_{2}]+\theta(s_{1})x_{2}-\theta(s_{2})x_{1})$$
This construction is known as forming a semi-direct product.
So, I want to show that this is antisymmetric and satisfies the Jacobi identity. Now, this has been done on math exchange before, see How to verify the Jacobi identity for the semidirect product Lie algebra and look at Andreas answer.
Looks great, I agree. However, I am trying to more thoroughly understand why we can simplify the calculation to those 4 mentioned cases. I started with one of the 3 summands in the Jacobi and expanded it out and starting using the fact we are working in the direct product of vector spaces so it is linear in each tuple. I end up with terms like $$([[s_1,s_2],s_3],0)$$ $$(0,[[x_1,x_2],x_3])$$
So i see where those two cases can be satisified. However, I get left with a final term of
$$(0 , [\theta(s_1)x_2-\theta(s_2)x_1,x_3]+\theta([s_1,s_2])x_3-\theta(s_3)([x_1,x_2]+\theta(s_1)x_2-\theta(s_2)x_1)$$
Which I can't break up into the two other simplifications that need to made... Perhaps I made an error in calculations, or perhaps understanding why we can make these simplifications is much simpler than I'm making it. Anyway, I'd like to be shown why these simplifications can be made.
Thank you for your time, I really appreciate it!
