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This might seem a very basic question, but I can't manage to find a proper proof in the books I have on my desk (or simply cannot see that it's "just that"). So be sure of what we talk about, let $G$ a Lie group, $V$ a (finite dimentional) representation of $G$ and $X$ an element of the Lie algebra $LG$ of $G$. Let also $\exp : LG \to G$ denote the usual exponential map. Then define $$ L_X(v):=\lim_{t\to 0}\frac{1}{t}(\exp(tX)v-v) $$ It is clear to me that it is linear in $v$, and that $L_{sX}v = sL_Xv$, but I'm having trouble proving that $L_{X+Y}v = L_Xv + L_Yv$.

I tried the following :

a) Rewrite $L_Xv$ as $$ L_Xv = \left.\frac{d}{dt}\right|_{t=0}\exp(tX)v $$ then use the fact that $\exp$ is the unique map such that $\left.\frac{d}{dt}\right|_{t=0}\exp(tX) = X$ to get $L_{X+Y}v = \left.\frac{d}{dt}\right|_{t=0}\exp(t(X+Y))v = (X+Y)v = Xv + Yv = L_Xv + L_Yv$ but I'm not sure if I can do this. So if yomeone could tell me if I can, and why, that would be great !

b) In matrix form, $\frac{1}{t}(\exp(tX) - Id) = \sum_{k=1}^\infty \frac{t^{k-1}}{k!}X^k $ so $$ \lim_{t\to 0}\frac{1}{t}(\exp(tX)v-v) = Xv $$ and then conclude as in a). But here again, I don't know if I can do that in general.

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I guess the gap in your proof is that $\rho(\exp(tX))=\exp(t\rho_*(X))$. Since $\exp(tX)$ is the unique one-parameter subgroup of $G$ with tangent vector $X$ at the identity of $G$, $\rho(\exp(tX))=\exp(t\rho_*(X))$ by the definition of $\rho_*$ as the tangent map of $\rho$ at the identities. –  Yuchen Liu Jun 11 '13 at 16:04
    
Yeah, exactly ! It's a great way to put it, thanks ! –  Samuel T Jun 12 '13 at 11:45

1 Answer 1

up vote 1 down vote accepted

If $\rho: G \to \mathrm{GL}(V)$ is a sufficiently nice, say smooth, real/complex representation, you may think of it as a smooth map between manifolds. The push-forward/derivative of $\rho$ is the linear map $\rho_{*}: TG \to T\mathrm{GL}(V)$. It is exactly what you compute there, $\rho_{*}(X)v= \frac{d}{dt}\big\vert_{t=0} \rho(\exp(tX))v$.

Personally I would call that map the induced representation of the Lie algebra and not the Lie derivative.

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Well, yes, but how the to prove that this is linear ? And it is the Lie derivative with respect to the vector field generated by $X$ and the action of $G$. –  Samuel T Jun 11 '13 at 11:26
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The push-forward is fibre-wise linear by definition. Or am I not getting your question? –  Dominik Jun 11 '13 at 11:52
    
Oh, yes, I failed to see this. >_< Thank you ! =) –  Samuel T Jun 11 '13 at 15:53

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