There are two important theorems about Lie group and Lie algebra:

First, consider a Lie group homomorphism $\rho: G \rightarrow H$.

  1. $d\rho(Ad(g)(v)) = Ad(\rho(g))(d\rho(v))$
  2. $d\rho_e(ad(X)(Y)) = ad(d\rho_e(X))(d\rho_e(Y))$, $ad$ is differential of $Ad$

I don't quite get the reason to prove these two theorems. It seems if Lie group is finite dimension, we can just use matrix $A$ for Lie algebra $T_e G$, $e^A$ for Lie group G, $[X, Y]$ for $ad(X)(Y)$. So proof should just be simple computation of matrices and each step is very clear.

  1. Why do we need to bother using a lot of stuffs in representation theorem?
  2. What are the most nontrivial parts about proving these two theorems?

1 Answer 1


Well for a start we cannot write every element of $G$ as a matrix nor as an exponential. Lie algebras can always be written as matrices by Ado's theorem but that is not true for Lie groups. Of course we should really strive for a proof without matrices anyway since writing something in terms of matrices requires us to not only pick a representation but pick a basis of that representation. We would really need to show that our proof did not depend on these choices and even then a more direct proof is cleaner. Then the second point. Even if we restrict ourselves to only considering connected Lie groups the exponential map (which depends on which Lie group of our Lie algebra we are considering and is not in general just the matrix exponential) will, in general, fail to be surjective unless, for example, the group is compact or nilpotent. Instead the image of the exponential generates the group (or more generally the connected component of the identity).

The results you are referring to are simply showing the derivative of a Lie group homomorphism is a Lie algebra homomorphism $\mathfrak{g}\to \mathfrak{h}$ (number 2) and that this is equivariant with respect to the natural adjoint actions on $\mathfrak{g}$ and $\mathfrak{h}$ (number 1). So to answer your first question: we are not really, except for the adjoint representation. Using matrices would be invoking representation theory. For your second, I'm not sure. These are somewhat basic facts of the Lie algebra-Lie Group correspondence. I don't think they rely on any overly technical parts of Lie theory as long as you are comfortable dealing with Lie Groups/Algebras in the abstract.

  • $\begingroup$ Thank you, very explicit answer. I am still struggling with various definition of Lie algebra, like differential operators, matrices, free associative tensor algebra, derivation algebra. $\endgroup$ Aug 20, 2022 at 11:27
  • $\begingroup$ I think the simplest way is just as a vector space with a Lie bracket. Matrices are useful but we should be careful thinking that they are the same as the Lie algebra themselves. That will lead into trouble, especially if you want consider more than one representation. $\endgroup$
    – Callum
    Aug 20, 2022 at 19:55

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