# Lie Algebra Homomorphism Question

So this is a bit of a follow-up to my recent question. I don't mean to inundate the feed with my quandaries, but as I move through the theory I keep hitting stumbling blocks (which y'all so kindly help me through).

As done previously, the group action defined by $h(g)p(z)=p(g^{-1}z)$ where $g\in\rm{SL}(3,\Bbb C)$, $p$ is from the vector space of polynomials of degree $\le2$ in three variables, and $z\in \Bbb C^3$.

I'm now introduced to a new function, $dh(X)=\left.\frac{d}{dt}h(e^{XT}]\right|_{t=0}$, where $X$ resides in the lie algebra of $\rm SL(3,\Bbb C)$ [ie $\mathfrak{sl}(3,\Bbb C)$]. The goal is: show that this is a lie algebra homomorphism $\mathfrak{sl}(3,\Bbb C)\to \mathfrak{gl}(6,\Bbb C)$.

Our basis in our space of polynomials is the standard one. Namely, the degree 2 terms in their various permutations.

I've been looking into the complexification of $\mathfrak{su}(3,\Bbb C)$ as a way of making sense of the polynomial when acted on by h, but I can't seem to get a handle on how to understand the derivative. I have a hunch this probably isn't even remotely the right course of action.

Any and all help is much appreciated. Feel free to assume I know the bare minimum.

Edit: While the induced homomorphism approach is awesome, a direct proof would help me get a better feel for how the matrix exponential is affecting the polynomial. Also, I'm too "machinery illiterate" to understand some of the more general formalisms at this point.

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What is your definition of $e^X$? Is it just a power series? –  Jason DeVito Apr 26 '12 at 3:24
@JasonDeVito indeed it is... –  AsinglePANCAKE Apr 26 '12 at 3:24
Great. What about derivatives? Do you know the chain rule for maps between manifolds? –  Jason DeVito Apr 26 '12 at 3:26
@JasonDeVito Here's where I'm failing...I've no idea how to compute that derivative... –  AsinglePANCAKE Apr 26 '12 at 3:28
In whatever notes/book you're studying, how is the derivative defined? Does it mean "look at $h(e^{tX})$ as a matrix with $t$s in it and take the usual derivative of that."? –  Jason DeVito Apr 26 '12 at 3:33
$h: SL(3,\mathbb C) \to GL(6,\mathbb C)$ is a homomorphism of Lie groups. Then your map $dh$ is the induced homomorphism of Lie algebras $\mathfrak{sl}(3,\mathbb C) \to \mathfrak{gl}(6,\mathbb C)$. So this is really just the general (and very important/fundamental fact) that the derivative at the identity of a Lie group homomorphism is a Lie algebra homomorphism. Proving this in generality would be a good exercise.
EDIT: Here's a brief outline (you can refer to the book I mentioned in the comments for more details if you need them though if you're familiar with some of these concepts they aren't that hard to prove). Let $\phi : G \to H$ be a Lie group homomorphism. Let $X_e, Y_e \in T_e G$ and denote by $X,Y$ the corresponding left invariant vector fields on $G$. Then one shows that the left-invariant vector field on $H$ that is equal to $\phi_*\vert_e X_e$ at the identity is $\phi$ related to $X$, i.e. for all $g \in G$ we have $\phi_*\vert_g X_g = {L_{\phi(g)}}_* \phi_*\vert_e X_e$. So call this vector field $\phi_* X$ (this is only a slight abuse of notation). Now by properties of Lie brackets, $[\phi_* X, \phi_* Y] = \phi_* [X,Y]$ so that $\phi_*\vert_e$ is a Lie algebra homomorphism.