How to prove that $\langle v, v \rangle$ is the only scalar invariant for a vector under the isometry?

Question

Consider an inner product space $V$ and $v \in V$.

It seems that the only scalar invariants a vector can have under the isometry come from $\langle v, v \rangle$. For example $\langle v, v \rangle$, $2\langle v, v \rangle^2 - 3$ or $\sin(\langle v, v \rangle)$ for real spaces etc. But how can it be proved?

I'm interested not just in a proof as is, but in insights behind it. Various proofs and/or references would be nice.

Answer 1

I suppose that by "the isometry" you mean the orthogonal group of all isometries. The orbits of that group are easily shown to be the sets of the $v$ with $\langle v,v\rangle=c$ for some constant $c$. Now if you have any invariant, its level sets must be unions of orbits, which for a polynomial invariant can only be a finite union for reasons of dimension. It can be seen algebraically that any polynomial function of the coordinates whose level sets are unions of sets of the form $\langle v,v\rangle=c$ is itself a polynomial in $\langle v,v\rangle$.

Added, since apparently not so obvious, an explanation why orbits are sets of the $v$ with $\langle v,v\rangle=c$ for some constant $c$. On one hand, since any isometry preserves the scalar product, it is clear that the value of $\langle v,v\rangle$ cannot vary along an orbit: $\langle v',v'\rangle=\langle g(v),g(v)\rangle=\langle v,v\rangle$ whenever there is some isometry $g$ mapping $v$ to $v'$. On the other hand if $\langle v,v\rangle=\langle v',v'\rangle=c>0$ we must find an isometry $g$ mapping $v$ to $v'$: note that $v/\sqrt c$ and $v'/\sqrt c$ are vectors of unit length, so each one can be (individually) extended to an orthonormal basis; then the linear map sending the orthonomal basis extended from $v$ to the one extended from $v'$ is such an isometry $g$.

Answer 2

Let $u$ and $v$ be two vectors with $u^2 = v^2=L$. I will show that there is an isometry taking $u$ to $v$. Specifically, I will reflect in the hyperplane orthogonal to $u-v$. Here is the computation:

Let $w = u - v$. If $w=0$, then $u=v$ and we are done, so we may assume $w \neq 0$. Reflection in $w^{\perp}$ is given by the formula $$x \mapsto x - 2 \frac{\langle w,x \rangle}{\langle w,w \rangle} w.$$ Since $w \neq 0$, the denominator is not $0$.

We have $\langle u,w \rangle = \langle u,u \rangle-\langle u,v \rangle = L - \langle u,v \rangle$ and $\langle w,w \rangle - \langle u,u \rangle - 2 \langle u,v \rangle + \langle v,v \rangle = 2L - 2 \langle u, v \rangle$. So $\langle u,w \rangle/ \langle w,w \rangle = 1/2$. (Challenge: Explain this formula in terms of Euclidean geometry!) Plugging into the formula for the reflection, $$u \mapsto u-2 (1/2) w=u-(u-v) = v.$$ Similarly, $v \mapsto u$.

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Let's spell out why this means that every invariant function must be of the form $f(\langle v, v \rangle)$ for some function $f$. Let $g: \mathbb{R}^n \to \mathbb{R}$ be invariant under isometries. Define $f(x) = g(\sqrt{L}, 0,0,0, \ldots, 0)$. Then for any vector $v$ with $v^2=L$, there is an isometry taking $v$ to $(\sqrt{L},0,0,\ldots,0)$ (by the above). So $g(v) = g(\sqrt{L},0,0,0\ldots,0) = f(L)$. In short, for any invariant $g$, there is a function $f:\mathbb{R}_{+} \to \mathbb{R}$ so that $g(v) = f(v^2)$ for all $v$.