# Lie algebra and Lie Group correspondence

I'm confused as to the relationship between Lie Groups and Lie algebras. From what I know, the Lie algebra of a Lie group is the space of tangent vectors which pass through the identity, together with Lie bracket defined by $[X,Y]=ad(X)(Y)$.

My question is, what actually is $ad(X)(Y)$ and what does it equal? I have found the following link: https://www.maths.gla.ac.uk/~gbellamy/lie.pdf and at the top of page 18, the author shows how $[X,X]=ad(X)(X)=0$, but I don't see how you can't use an analogous method to show $[X,Y]=ad(X)(Y)=0$.

Also, I have seen a lot on left-invariant vector fields, but what is the purpose if the Lie algebra of a Lie group is just the tangent space definition? Is this just an equivalent method of find a Lie algebra from a Lie group?

## 2 Answers

Actually ${\rm ad}(x)(y)=[x,y]$ is the Lie bracket of ${\rm Lie}(G)$, as you said. Let $G$ be the Heisenberg group, and $(x,y,z)$ a basis of $\mathbb{R}^3$, then we obtain $[x,y]=z$ and $[x,z]=[y,z]=0$. Hence the Lie bracket is not $0$. Of course, $[x,x]=0$ for all $x$, and thus $[x,y]=-[y,x]$, because the Lie bracket is bilinear.

For matrix groups, and matrix Lie algebras we can write down the maps $\exp: {\rm Lie}(G)\rightarrow G$ and $\log:G\rightarrow {\rm Lie}(G)$ explicitly. Have a look at the related questions here at MSE.

Another way of defining the Lie bracket is the following: first prove that the Lie bracket of two left-invariant vector field is again a left- invariant vector field. Then, show there is a bijection betwen left- invariant vector fields Ana the tangent space at the identity of the group. The Lie bracket on $$Lie(G)=T_1G$$ is defined from the above identification.