Smooth and $GL(n)$-equivariant implies algebraic?

Context: Let $B_n$ be the space of symmetric bilinear forms on $\mathbb{R}^n$ and $L_n\subset B_n$ be the subset of non-degenerate forms of Lorentzian signature $(-,+,\ldots,+)$. Let $T$ be a finite dimensional real vector space. Both $B_n$ and $T$ carry linear representations of $GL^+(n,\mathbb{R})$, where $+$ means the connected component of the identity, with the action on $B_n$ restricting to $L_n$. We assume that, for $g\in GL^+(n,\mathbb{R})$, the corresponding linear maps acting on $B_n$ and $T$ are polynomial in the components of $g$ up to a multiple of $\left|\det g\right|$ to a rational power. For $h\in L_n$, consider its components $h_{ab}$ (using some fixed basis on $\mathbb{R}^n$, say) as functions on $L_n$. I can also define another set of functions which are the components $\varepsilon_{a_1\cdots a_n}$ of the Levi-Civita tensor, defined as a solution to the equation $$\varepsilon_{a_1\cdots a_n} \varepsilon_{b_1\cdots b_n} = \text{(antisymmetrization over a_1\cdots a_n of)} ~ h_{a_1 b_1} \cdots h_{a_n b_n}.$$

Question: If I have a (non-linear) smooth equivariant map $A\colon L_n \to T$, can I conclude that $A$ is actually polynomial in the functions $h_{ab}$ and $\varepsilon_{a_1\cdots a_n}$ (up to a multiple of a rational power of $\left|\det h_{ab}\right|$) on $L_n$?

I believe that the answer is Yes. Unfortunately, I don't know which (probably well-known) result implies that. In the absence of a direct answer, some hints about how to find one in the vast literature on Invariant Theory would also be appreciated.

Note: Most of the literature in Invariant Theory seems to already make the assumption that all maps involved are polynomial, rational or algebraic. On the other hand, my question is about reducing to one of the contexts starting with smoothness and some other assumptions.

Because the open set $L_n\subset B_n$ is an orbit of $\mathrm{GL}^+(n,\mathbb{R})$ under the natural representation of $\mathrm{GL}^+(n,\mathbb{R})$ on $B_n$, it follows that, if $\rho:\mathrm{GL}^+(n,\mathbb{R})\to \mathrm{GL}(T)$ is the representation that defines the $\mathrm{GL}^+(n,\mathbb{R})$-action on $T$, then any $\mathrm{GL}^+(n,\mathbb{R})$-equivariant mapping $A:L_n\to T$ must satisfy $$A(g\cdot b) = \rho(g)\bigl(A(b)\bigr)$$ for all $g\in\mathrm{GL}^+(n,\mathbb{R})$ and $b\in L_n$. In particular, since $\rho$ is smooth, it follows that $\mathrm{GL}^+(n,\mathbb{R})$-equivariance already implies that $A$ is smooth. Conversely, given any such representation $\rho$ and any fixed $b_0\in L_n$, any choice of an element $A(b_0)\in T$ defines a $\mathrm{GL}^+(n,\mathbb{R})$-equivariant mapping $A:L_n\to T$ by the above formula. (Note that, if one regards $L_n$ as the space of $n$-by-$n$ symmetric matrices of index $n{-}2$, then $g\cdot b = g b g^T$.)
Thus, this is really a question about the representation $\rho$. Consider the representation $\rho:\mathrm{GL}^+(n,\mathbb{R})\to \mathrm{SL}(2,\mathbb{R})$ defined by $$\rho(g) = \begin{pmatrix} 1 & \log\bigl(\det(g)\bigr)\\0&1\end{pmatrix}.$$ Then, for, say, $b_0 = -{x_1}^2 + {x_2}^2 + \cdots + {x_n}^2$ and $A(b_0) = \begin{pmatrix}0\\1\end{pmatrix}$, the above formula gives the smooth equivariant map $$A(b) = \begin{pmatrix} \log\bigl(-\det(b)\bigr)\\1\end{pmatrix},$$ for $b\in L_n$, and this mapping is not polynomial in the functions that you name in your question.
• Hi, Robert. Thanks, that's a good counter example to keep in mind. However, it is excluded in my case, since the components of $\rho(g)$ are polynomial up to an overall multiple of a power of $\det(g)$ (hmm... yes, I did put that into the question). So, with that constraint the answer is positive. Your idea to use the first displayed equation essentially solves the problem. Since I'm not used to thinking in terms of equivariant maps, it took a few days after I posted the question for the same idea to occur to me. Then it was clear! – Igor Khavkine Oct 5 '14 at 22:50
• Actually, your question was not clear. You seemed to be asserting that the representation on $T$ is always polynomial up to a power of $\det(g)$, not assuming it. I was providing a counterexample to that assertion. – Robert Bryant Oct 5 '14 at 22:58