A standard result in the invariant theory of the orthogonal group states the following.
Theorem Let $(E, \langle .,. \rangle)$ be an n-dimensional euclidean vector space, let $f : E^m \rightarrow {\bf R}$ a polynomial function satisfying $f(g(v_1), ... g(v_m)) = f(v_1,...,v_m)$ for all isometries $g$ of $E$ and $v_1$,..., $v_m \in E$. Then such a function is a polynomial function in the quantities $\{\langle{v_i}{v_j}\rangle\}_{i,j = 1...m}$.
Does the theorem holds in the topological setting, namely when polynomial is replaced by continuous ?
My guess is that it should be true and the proof should be simpler than its algebraic counterpart, maybe a short computation using SVD. All references I know present the algebraic proof though. Same question in the differential setting.