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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?2012-04-26
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    @JasonDeVito indeed it is...2012-04-26
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    Great. What about derivatives? Do you know the chain rule for maps between manifolds?2012-04-26
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    @JasonDeVito Here's where I'm failing...I've no idea how to compute that derivative...2012-04-26
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    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."?2012-04-26
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    @JasonDeVito Yeah, we defined it entry-wise...identifying it with an ordinary curve in R^(n^2) methinks.2012-04-26
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    Then I'm afraid I don't see of a slick way of doing it short of either deriving the general theory (as Eric suggested), or going a lengthy and annoying calculation. If I come up with something clever to reduce the annoying calculation, I'll let you know. Sorry I wasn't more of a help!2012-04-26
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    @JasonDeVito is there a better defined way of taking the derivative that makes it simpler? It's probably more likely that I'm wrong in stating what we did...2012-04-26

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