**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 $$\varepsilon\_{a\_1\cdots a\_n} \varepsilon\_{b\_1\cdots b\_n} = \text{(antisymmetrization over $a\_1\cdots a\_n$ of)} ~ h\_{a\_1 b\_1} \cdots h\_{a\_n b\_n}.$$



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> **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 $L\_n$?
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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.



Smooth and $GL(n)$-equivariant implies algebraic?
If I understand you correctly, the answer is 'no'. 


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.


