Given two representations $\pi_1$ and $\pi_2$ of a group $G$ (let's say it's a compact Lie group), a natural thing to study are linear equivariant maps A between them: $$ A \pi_1 = \pi_2 A $$ I'm wondering if it is possible to study non-linear (e.g. polynomial) equivariant maps beween representations.

To give a concrete example, consider the following representations of $\operatorname{SO}(2)$: $$ \pi_1(\theta) = \begin{bmatrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \\ \pi_2(\theta) = \begin{bmatrix}\cos(2\theta) & -\sin(2\theta) \\ \sin(2\theta) & \cos(2\theta) \end{bmatrix} \\ $$

These are inequivalent irreducible representations (over the reals), so there are no non-trivial linear equivariant maps A. But are there non-linear maps $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ that are equivariant? Is there a name for this type of thing? In what field (if any) are these maps studied? When (if ever) do they exist?

I'm guessing that if we assume that $G$ is an algebraic group, there can be polynomial equivariant maps, but I may be wrong.

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    $\begingroup$ Sure. You can identify $SO(2)$ with the unit circle and the mapping $(x,y)\mapsto (x^2-y^2,2xy)$, i.e. squaring of complex numbers, should do it. I somehow doubt that such mappings are very interesting. After all, winding numbers (of the unit circle) are all over the place. Also the theory of weights (central to rep theory of algebraic groups and/or Lie groups) needs to make the distinction between $\pi_1$ and $\pi_2$. $\endgroup$ – Jyrki Lahtonen Aug 11 '15 at 12:43
  • $\begingroup$ Thanks! You're right that forgetting about the distinction between inequivalent representations (considering them as equivalent) would destroy almost all structure, but I'm still wondering when exactly it is possible to find such non-linear equivariant maps. Does it work for any algebraic group? Does it work for representations that have different dimensionality? Is there a procedure for explicitly determining these maps given two representations? $\endgroup$ – John von N. Aug 11 '15 at 14:20
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    $\begingroup$ Perhaps an equivariant (degree $d$) polynomial map $A : V \to W$ is equivalent (or similar?) to an equivariant linear map $Sym^d(V) \to W$? $\endgroup$ – Jake Levinson Aug 11 '15 at 16:22

The ring of polynomial functions $V\to K$ is called the coordinate ring and is denoted $K[V]$ (although personally, I think it could've been denoted $K[V^*]$ instead, because polynomials are sums of products of functionals $V\to K$ no?) It is so-called because if $x_1,\cdots,x_n$ are coordinate projections $V\to K$ in some basis then $K[V]=K[x_1,\cdots,x_n]$.

If $V$ is a representation of $G$, then $G$ acts by automorphisms on the ring $K[V]$. As an algebra and as a representation, if $K$ has characteristic $0$ then the coordinate ring $K[V]$ is isomorphic to the full symmetric algebra ${\rm Sym}(V^*)$ which decomposes as an infinite product $\bigoplus_d {\rm Sym}^d(V^*)$.

The ring of polynomial functions $V\to W$ is then $K[V]\otimes W$. Then the $G$-equivariant polynomial functions $V\to W$ are the $G$-invariants of $K[V]\otimes W$, so it reduces to computing the invariants of the graded components $({\rm Sym}^d(V^*)\otimes W)^G$, which is a task for linear representation theory; if one has the character table of $V$ and $W$ (and $G$ is finite) then for any given $d$ one can reduce finding its dimension to a finite numerical calculation. Indeed, given ordered bases for $V$ and $W$ and explicit matrices for elements of $G$, one can compute a basis for the $G$-invariants of the $d$th component.

Maybe look into invariant theory to see if it covers more. Dunno about nonpolynomial stuff.

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    $\begingroup$ One small comment to add to your answer. If $W$ is irreducible, computing invariants of $\mathrm{Sym}^{d}(V^{*}) \otimes W$ is equivalent to computing the $W^{*}$-isotypic component of $\mathrm{Sym}^{d}(V^{*})$. $\endgroup$ – Siddharth Venkatesh Aug 12 '15 at 23:46

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