Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I do not understand the topology of a Lie group clearly. Let $G$ be a Lie group and $T_eG$ be its tangent space at the identity $e \in G$. Why $Aut(T_eG)$ is an open subset of the vector space of endomorphisms of $T_eG$ (i.e. $End(T_eG)$)? What does "open" mean?

share|cite|improve this question
up vote 4 down vote accepted

First, this has nothing to do with Lie groups or Lie algebras. The only important part is that $T_eG$ is a vector space.

For any vector space $V$ (say, finite dimensional over the reals), the set $End(V)$ is naturally a finite dimensional vector space. Hence, $End(V)$ is isomorphic to $\mathbb{R}^N$ for some $N$ (in fact, $N = $(dim$V)^2$). Use any choice of isomorphism to topologize $End(V)$. This choice of isomorphism is equivalent to choosing a basis of $V$.

Now, one has the determinant $det:End(V)\rightarrow\mathbb{R}$ which is given as a polynomials in the entries of the matrices in $End(V)$ (they are matrices after choosing a basis), and hence is continuous.

Since $det$ is continuous, $det^{-1}(\mathbb{R}-\{0\})$ is an open subset of $End(V)$. But this subset is precisely $Aut(V)$, the invertible transformations from $V$ to $V$.

share|cite|improve this answer
I don't agree 100% with "The only important part is that $T_eG$ is a vector space." In general vector spaces can be topologized in more than one way, so there should be some reason from the context why the choice you gave is natural. For instance, inducing the topology on $T_eG$ via an isomorphism with $\mathbb{R}^n$ is consistent with how the topology on the tangent bundle of $G$ is defined. Once you have the vector space $V$ topologized, you can then correspondingly define the topology on $End(V)$, for instance via the topology of uniform convergence on bounded sets. – Jonas Meyer Jan 2 '11 at 6:11
I'm just nitpicking of course. There's only one Hausdorff topological vector space structure on a finite dimensional real vector space, with many different ways of describing it. But I think the reasons for topologizing in certain ways in certain contexts are important, too. – Jonas Meyer Jan 2 '11 at 6:14
Thank you very much. – user Jan 2 '11 at 14:46
Hi Jason, thank you very much. – user Jan 2 '11 at 17:06
Hi Jonas, thank you very much. – user Jan 2 '11 at 17:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.