# $SL(3,\mathbb{C})$ acting on Complex Polynomials of $3$ variables of degree $2$

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!

#### Solutions Collecting From Web of "$SL(3,\mathbb{C})$ acting on Complex Polynomials of $3$ variables of degree $2$"

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.