# Definition of the Lie algebra and the Lie bracket for general vector fields

I've started to go deep into the theory of Lie groups to eventually understand their representation theory. I picked up a text online and right on the first chapter something started to bother me. When the author defines a Lie algebra over a vector field, we get the standard definition:

Definition 1.2 A Lie algebra over $$\mathbb K = \mathbb R$$ or $$\mathbb C$$ is a vector space $$V$$ over $$\mathbb K$$ with a skew-symmetric $$\mathbb K$$-bilinear form (the Lie bracket) $$[\,,\,]:V\times V\rightarrow V$$ which satisfies the Jacobi identity $$[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$$ for all $$X,Y,Z\in\mathbb K$$.

However, when the author first goes to use the notion of a Lie bracket to define the Lie algebra of a Lie group, he states that the pushforward of a diffeomorphism $$f:M\rightarrow N$$ is compatible with the Lie bracket, i.e. that

$$f_*[X,Y]=[f_* X,f_* Y].$$

As such, it seems like we completely disregarded the original abstract definition of the bracket on 1.2 and chose to use the Lie derivative because it happens to satisfy these properties and is nice to work with. This immediately seemed very arbitrary to me.

Are there any assumptions that I haven't picked up on to maybe uniquely determine that the Lie algebra of left-invariant vector fields can only have the Lie derivative as a consistent bracket (up to a scalar multiple, of course)?

• This is a really weird question. It's like asking, "why is the group operation on $GL_n(\mathbb{R})$ matrix multiplication, when the definition of a group doesn't say that the group operation must be matrix multiplication?". – Eric Wofsey Jan 11 at 3:13
• Oh, looking at the text you're reading, it looks like it's not your fault and the text just does a terrible job of explaining what's going on. – Eric Wofsey Jan 11 at 3:18

Suppose that there is a basis $$v_1,v_2,\cdots,v_n$$, you can define a consistent bracket just through the commutation relation which is completely determined by some constants.