Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Essentially, I'm trying to prove that when computing the tangent space for a group that there's nothing special about considering it at only the identity. Namely, there is an isomorphism of vector spaces from the tangent space at the identity to the tangent space at any other element of the group.

The method that I'm being guided along is the following:

First, I'm supposed to prove that if I have any smooth curve through an element g, it is of the form gA(t) where A(t) is a smooth curve through the identity. (Proving this is where I'm failing.) Naturally, once I've done this, the isomorphism is there for the isomorphing...

Apologies for any ambiguities in the wording, I'm not exactly firing on all cylinders...

share|cite|improve this question
What's confusing about the hint? A tangent vector is an equivalence class of smooth curves, and you've just been told how to identity smooth curves through $g$ with smooth curves through the identity. – Zhen Lin Mar 29 '12 at 1:50
Would it help to say: for $C(t)$ a curve through $g$, $g^{-1}C(t)$ is a curve through the identity. I'm not exactly clear as to where you are having trouble . . . – B R Mar 29 '12 at 1:50
@ZhenLin I'm having trouble proving that every curve through g is of the form gA(t). Namely, for a given curve B(t) through g, why is B(t) always of the form gA(t)? – AsinglePANCAKE Mar 29 '12 at 1:54
Because I can multiply by $g^{-1}$ pointwise. – Zhen Lin Mar 29 '12 at 1:55
To spell out my argument more fully, take a curve $B(t)$ through $g$. Then $g^{-1}B(t)$ goes through the identity. Set $A(t)=g^{-1}B(t)$. Then $A(t)$ is a curve through the identity and $gA(t)=B(t)$. – B R Mar 29 '12 at 2:18

It may be useful for you to see the following construction. Take some smooth curve $g(t)$ in Lie group G. Then we have, since G is a group $$ g(t + \epsilon) = h(\epsilon)g(t) $$ for some smooth $h(\epsilon)$ and with $h(0) = e$ since $g(t) = g(t)$.

Then $h(\epsilon) = g^{-1}(t)g(t+\epsilon)$ so for small $\epsilon$ $$ e + \dot{h}(0)\epsilon + O(\epsilon^2) = (e + \dot{g}(t)\epsilon + O(\epsilon^2))g^{-1}(t) $$ so by comparing coefficients we see that $$ \dot{h}(0) = \dot{g}(t) ~ g^{-1}(t) $$ is an element of the Lie Algebra of G for all t. So we may relate some tangent space at $g(t)$ to the tangent space at the identity.

Note the similarity with the pullback/pushforward of functions/tangent spaces.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.