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.



Continuous version of the fundamental theorem of invariant theory for the orthogonal group
Yes. It suffices to show that if one has a sequence $\vec v^{(n)} = (v^{(n)}\_1,\dots,v^{(n)}\_m) \in E^m$ whose Gram matrix $(\langle v^{(n)}\_i, v^{(n)}\_j \rangle)\_{i,j=1,\dots,m}$ converges to a Gram matrix $(\langle v\_i, v\_j \rangle)\_{i,j=1,\dots,m}$ of a tuple $\vec v = (v\_1,\dots,v\_m) \in E^m$, then after applying linear isometries to each of the $\vec v^{(n)}$, that $\vec v^{(n)}$ converges to $\vec v$.


By permuting indices we may assume that $v\_1,\dots,v\_k$ are linearly independent, and $v\_{k+1},\dots,v\_m$ are in the span of $v\_1,\dots,v\_k$, for some $0 \leq k \leq m$. Applying the Gram-Schmidt process to $v\_1,\dots,v\_k$ and transforming $v^{(n)}\_1,\dots,v^{(n)}\_k$ appropriately we may assume that $v\_1,\dots,v\_k$ are orthonormal. It is not difficult to inductively apply isometries to the $\vec v^{(n)}$ so that $\vec v^{(n)}\_i$ converges to $v\_i$ for $i=1,\dots,k$. If we write each $v\_j, j=k+1,\dots,m$ as a linear combination $v\_j = a\_{j1} v\_1 + \dots + a\_{jk} v\_k$ of $v\_1,\dots,v\_k$, then from the Gram matrix convergence we see that $\| v\_j^{(n)} - (a\_{j1} v^{(n)}\_1 + \dots + a\_{jk} v^{(n)}\_k) \|^2$ converges to zero, hence $v\_j^{(n)}$ converges to $v\_j$ by the triangle inequality, and the claim follows.


