Why Lie derivative provides a representations of a Lie algebra

Question

I am reading a book about general relativity. After introducing the Lie derivative and Lie bracket, the author claims that the Lie derivative provides a representation of the Lie algebra of vector fields from the relation $$[L_V,L_V]=L_{[V,W]}$$ I can not understand this. What is the Lie algebra? and what is algebra relation? For me Lie algebra is defined by some algebra relation for example $$ [T^a,T^b]=if_{abc}T^c$$ and the representaion of Lie algebra is something that satisfies the same algebra relation as Lie algebra components.

Answer 1

A Lie algebra is simply a vector space with an alternating bilinear product (written as Lie brackets) that satisfies the Jacobi identity. I'm not sure what you mean by "relations" here, maybe you are thinking in terms of generators and relations? Since we did not define the Lie algebra that way, don't think of it in that terms.

For the quote, you are looking at the Lie algebra of all (globally defined, smooth) vector fields $\mathcal{X}(M)$ on some smooth manifold $M$, with the usual Lie bracket as the product, i.e, $$ \mathcal{X}(M)\otimes_\mathbb{R}\mathcal{X}(M)\to\mathcal{X}(M) $$ is the usual Lie bracket given locally as $$ X^i\partial_i\otimes Y^j\partial_j\mapsto (X^j(\partial_jY^i)-Y^j(\partial_jX^i))\partial_i. $$ Then the Lie derivative $$ L\colon\mathcal{X}(M)\otimes_\mathbb{R}\Gamma(E)\to \Gamma(E) $$ where $E$ is, say, a smooth vector bundle over $M$, provides a representation of $\mathcal{X}(M)$, i.e., $$ L\colon\mathcal{X}(M)\to\mathrm{End}(\Gamma(E)) $$ is a Lie algebra homomorphism. In other words, $L$ is a linear map and $L$ of a Lie bracket of vector fields is the same as the corresponding Lie bracket of (continuous) endomorphisms of the space of sections of $E$. That is what $[L_V,L_W]=L_{[V,W]}$ tells you.