Why is the group action on the vector space of polynomials naturally a left action? When seeking irreducible representations of a group (for example $\text{SL}(2,\mathbb{C})$ or $\text{SU}(2)$), one meets the following construction. Let $V$ be the space of polynomials in two variables. Define a group action on $V$ by $g\cdot P(z) = P(zg)$ (in some texts the convention is $P(g^{-1}z)$). Here $z=(z_1 \; z_2)$ is a row vector and matrix multiplication is implied.
So my question is, why is this a left action? In both conventions (multiplication on the right, multiplication on the left by inverse) it naively appears to be a right action. I check that $g_1\cdot(g_2\cdot P(z))=g_1\cdot P(zg_2)=P(zg_2g_1)=(g_2g_1)\cdot P(z)$, which certainly looks like a right action. What am I doing wrong here?
And how can I understand this issue on a more fundamental level? Is there some contravariant functor lurking around which converts the action? I found this discussion by Michael Joyce, which seemed relevant. The space of (homogeneous) polynomials can actually be realized as $\operatorname{Sym}^k(V^*).$ Maybe that dual space functor explains something?
 A: The action defined here is on functions, i.e. $P$, not vectors $z$. Write $R=g_2\cdot P$.
Then $(g_1\cdot R)(z)=R(zg_1)$. Combine this with the fact that $R(x)=P(xg_2)$ for any vector $x$. Here $x=zg_1$, so we get that
$$
(g_1\cdot(g_2\cdot P))(z)=(g_1\cdot R)(z)=R(zg_1)=(g_2\cdot P)(zg_1)=P((zg_1)g_2).
$$
Another way of seeing this is to realize that $g\cdot P$ is the function gotten by composing $\rho_g:z\mapsto zg$ with $P$ from the right (i.e. ahead of $P$). So
$$
g\cdot P = P \circ \rho_g.
$$
Therefore
$$
g_1\cdot(g_2\cdot P)=g_1\cdot(P\circ \rho_{g_2})=(P\circ \rho_{g_2})\circ\rho_{g_1}.
$$
By associativity of composition this latter composition is then
$$
P\circ(\rho_{g_2}\circ\rho_{g_1})=P\circ\rho_{g_1g_2}=(g_1g_2)\cdot P.
$$
Here $\rho_{g_2}\circ\rho_{g_1}=\rho_{g_1g_2}$ because 
$$(\rho_{g_2}\circ\rho_{g_1})(z)=\rho_{g_2}(\rho_{g_1}(z))=\rho_{g_2}(zg_1)=(zg_1)g_2=z(g_1g_2)=\rho_{g_1g_2}(z).
$$
A: First, you have a left action on vectors, $g \cdot z = zg^{-1}$.  (Here $g$ is an invertible matrix, $z$ is a row vector and $g \cdot z = zg^{-1}$ is the result of multiliplying $z$ times $g^{-1}$.  You need to multiply on the right by $g^{-1}$ to make it a left action.)  Then $(g \cdot P)(z) = P(g^{-1} \cdot z) = P(zg)$.
Let's check that this is a left action, i.e. $g_1 \cdot (g_2 \cdot P) = (g_1 g_2) \cdot P$:
$$
(g_1 \cdot (g_2 \cdot P) )(z) = (g_2 \cdot P)(z g_1) = P(z g_1 g_2) = ((g_1 g_2) \cdot P)(z).
$$
The reason that the action is given by $(g \cdot P)(z) = P(g^{-1} \cdot z)$ is because the polynomials on $V$ are the elements of the symmetric algebra of $V^*=\text{Hom}(V,\mathbb{C})$.  Given a left $G$-action on $V$ (i.e. a representation), there is a natural way to construct a left $G$-action on $V^*$ -- the contragredient representation -- by letting $(g \cdot l)(v) := l(g^{-1} \cdot v)$.  The argument for why this defines an action on $V^*$ follows from the calculation above.  (Indeed, just let $P = l$ be linear.)  Similarly, there is also a natural way to get a $G$-representaion of $Sym^{\bullet}(V^*)$ from a $G$-representation of $V$, which yields the action on the polynomial ring that you are interested in.
