# 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?

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@Arturo: Sorry; I first read the first part of your comment as "it is a left action because you chose to write it on the left". I see now the confusion is because the OP made a mistake in $g_1\cdot P(zg_2)=P(zg_2g_1)$ -- replacing $z$ in $P(zg_2)$ with $zg_1$ should have yielded $P(zg_1g_2)$, consistent with being a left action. –  Henning Makholm Mar 1 '12 at 20:51
@Henning: Right; I didn't make that clear, but Jyrki has. –  Arturo Magidin Mar 1 '12 at 20:53

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 $$(\rho_{g_2}\circ\rho_{g_1})(z)=\rho_{g_2}(\rho_{g_1}(z))=\rho_{g_2}(zg_1)=(zg_1)g_2=z(g_1g_2)=\rho_{g_1g_2}(z).$$

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Looking at it with your second point of view, we could say that we get a left action because there are two "contravariant" operations (which reverse the order of group multiplication): one is let multiplication act on the right, the other is to let $\rho_g$ compose with $P$ on the right. And I guess this is quite general? Given any left group action on any set, then we have a left group action on functions on that set, by the above construction? –  Joe Hannon Mar 1 '12 at 22:14
@ziggurism: A key difference is whether the action on functions affects the functions inputs or outputs. Acting on inputs reverses the order. Here the group originally acted on vectors, and a function mapped a vector to a number, so the only possibility is for the group to act on the inputs. –  Jyrki Lahtonen Mar 2 '12 at 4:18

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.

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I think I do understand why the dual space representation carries that $g^{-1}$; it makes sense that a contravariant functor acting on a rep would switch the order of the group multiplication. But so the mistake I made, I failed to see where we switched the order a second time. It was in the expression $(g_1\cdot(g_2\cdot P(z))$. I wanted to apply the innermost operation first, whereas you apply the outermost operation first. This upheaval of the traditional order of operations is disturbing to me. Is there another way I can make sense of that? –  Joe Hannon Mar 1 '12 at 22:24
I guess it's just a normal feature of things like pullbacks which take functions as arguments and precompose them, huh? –  Joe Hannon Mar 1 '12 at 22:29
The main thing is that $g_1 \cdot (g_2 \cdot P(z))$ doesn't make sense ... you're asking $g_2$ to act on the value $P(z)$. (Or maybe you mean $g_1 \cdot ((g_2 \cdot P)(z))$, but then you're asking $g_1$ to act on a value.) Instead, you always act on the function $P$, i.e. you're looking at $g_1 \cdot (g_2 \cdot P)$ and then you want to compute its value at $z$, which gives $(g_1 \cdot (g_2 \cdot P))(z)$, and not the expression you wrote. –  Michael Joyce Mar 1 '12 at 22:49
I was easily confused by such matters when I first learned representation theory. The best thing to do, as you're doing with this question, is to focus on what's acting on what. –  Michael Joyce Mar 1 '12 at 22:50
@ziggurism: I think that you are getting there. The innermost operation is done first (as it should), but the innermost operation amounts to precomposing $P$ with the action on its input. So the outermost operation, when it is done afterwards, tags on another precomposition, and the inputs get hit with the outermost operation first. If the operation were a postcomposition the reverse would hold. Contra/covariant. –  Jyrki Lahtonen Mar 2 '12 at 4:14