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So I'm given the following definition:

$h(g)p(z)=p(g^{-1}z)$ where g is an element of $SL(3,\mathbb{C})$, $p$ is in the vector space of homogenous complex polynomials of $3$ variables and $z$ is in $\mathbb{C}^3$.

What I'm having trouble showing is that mapping $g$ to $h(g)$ is a group homomorphism. Namely, I know that $h(ab)p(z)=p(b^{-1}a^{-1}z)$, but I can't seem to make sense that $h(a)(h(b)(p(z))$ is not $p(a^{-1}b^{-1}z)$.

This is probably trivial, so condescending replies are welcome!

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You should really write $h(g) p(z)$ as $(h(g) p)(z)$. In other words, $p$ is a polynomial function on $\mathbb{C}^3$ and so is $h(g) p$, with its value at $z$ being given by your formula, i.e. $(h(g) p)(z) = p(g^{-1} z)$.

Now recompute $(h(a) h(b) p)(z)$, giving $(h(a) h(b) p)(z) = (h(b) p)(a^{-1} z) = p(b^{-1} a^{-1} z)$. For what it is worth, this may well be classified as "trivial", but I found keeping straight what acts on what and how was a significant stumbling block for me when I first studied representation theory and invariant theory. It's easy to get actions wrong and have calculations fall apart due to these "trivial" issues, so they are most definitely worth straightening out.

You can see also this thread for further explanation in a slightly more abstract setting.

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    Ah thanks! It makes sense that the action is on P and not on the value, but the order is still tripping me up. Why does b act on P 'before' a does?2012-04-25
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    Yeah, like I said, it's confusing for a while. But the story is that $b$ acts on $P$ (producing $h(b) P$) and then $a$ acts on $h(b) P$ (producing $h(a) h(b) P$). Now, if you want to actually compute the value $(h(a) h(b) P)(z)$, you have to work backwards. By the definition of $h(a)$, applied to the function $h(b) P$, you get $(h(a) (h(b) P))(z) = (h(b) P)(a^{-1} z)$. And then, evaluating $h(b) P$ at the point $a^{-1} z$ gives $P(b^{-1} (a^{-1} z))$. Hope that helps! (If it seems unhelpful, my best suggestion is to follow the parenthetization carefully.)2012-04-25
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    Just clicked...you're parenthetically gifted sir.2012-04-25
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    http://math.stackexchange.com/questions/137020/lie-algebra-homomorphism-question In case you feel like digging me out of another rut...2012-04-26