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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.

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What do you mean by $v^2$ ? –  Belgi Oct 23 '12 at 11:31
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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 $g$.

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I'll need some time to understand it, but I suspected the answer would be something like that. If somebody knows a reference to a detailed derivation, you are welcome to share. –  Yrogirg Oct 23 '12 at 17:27
    
"Now if you have any invariant, its level sets must be unions of orbits" Could you please elaborate this part? Is that a theorem or what? –  Yrogirg Oct 25 '12 at 13:04
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Being invariant means precisely being constant on orbits. Therefore the set where an invariant takes a specific value is necessarily a union of (zero or more) orbits. –  Marc van Leeuwen Oct 25 '12 at 13:34
    
"The orbits of that group are easily shown to be the sets of the $v$ with $\langle v,v\rangle=c$ for some constant $c$." How do you do it? And how would you do it for cases when the group acts not just on $V$, but for example on $V \otimes V^*$ or something of that kind? Oh should I ask it as a separate question? –  Yrogirg Oct 29 '12 at 8:28
    
@Yrogirg: See the added paragraph. The case where the group of isometries acts on $V\otimes V^*$ is another kettle of fish; the elements of that space are not vectors of $V$ so there is no such condition as $\langle v,v\rangle=c$ to begin with. Elements $V\otimes V^*$ of can be identified with linear maps $V\to V$ (endomorphisms), and even for the action of all of $GL_n(\Bbb R)$, the orbits are not described by polynomial invariants alone in this case. –  Marc van Leeuwen Oct 30 '12 at 9:45
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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, $v \mapsto u$.


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 $v$.

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It's not the answer to the question. Why there is no other function $h : V \to \mathbb R$ so that if $h(v) = h(u) = L$ there exists an isometry $u \mapsto v$? For a simplicity I took real inner product spaces. –  Yrogirg Oct 23 '12 at 17:26
    
As you say, there are lots of such functions: $\sqrt{v^2}$, $e^{v^2}$, etc. But they are all functions of $v^2$, as this argument shows. See if the last paragraph I just added helps. –  David Speyer Oct 23 '12 at 23:42
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