$\newcommand{\<}{\langle} \newcommand{\>}{\rangle} $Let $||\cdot||$ be a norm on a finite dimensional real vector space $V$.

Does there always exist some inner product $\<,\>$ on $V$ such that $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ ?


As pointed by Qiaochu Yuan the answer is positive.

This raises the question of uniqueness of the inner product $\<,\>$ which satisfies $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$.

Is it unique (up to scalar multiple)?


1) Determining $\<,\>$ (up to scalar multiple) is equivalent to determining $\text{ISO}(\<,\>)$.

Clearly if we know the inner product we know all its isometries. The other direction follows as a corollary from an argument given here which shows which inner products are preserved by a given automorphism.

2) Since there are "rigid" norms (whose only isometries are $\pm Id$ ) the uniqueness certainly doesn't hold in general.

One could hope for that in the case of "rich enough norms" (norms with many isometries, see this question) the subset $\text{ISO}(|| \cdot ||)\subseteq \text{ISO}(\<,\>)$ will be large enough to determine $\text{ISO}(\<,\>)$.

(which by remark 1) determines $(\<,\>)$).


2 Answers 2


Yes. This is because an isometry group is always compact (with respect to the topology on $\text{End}(V)$ induced by the operator norm: this is a consequence of the Heine-Borel theorem). Hence you can average an inner product over it with respect to Haar measure.

  • 4
    $\begingroup$ @Asaf: that is not the suggestion. The suggestion is the following: pick an arbitrary inner product $\langle v, w \rangle$. Construct the new inner product $\langle v, w \rangle_G = \int_G \langle gv, gw \rangle \, dg$. (There is some work necessary to prove that this is still an inner product but it's not so bad.) By construction, this new inner product is $G$-invariant, so every $g \in G$ is an isometry with respect to the corresponding norm. This is a standard maneuver in representation theory called "Weyl's unitary trick"; you can try to google that term for references. $\endgroup$ Jun 14, 2015 at 19:38

$\newcommand{\<}{\langle} \newcommand{\>}{\rangle} $

For completeness, I am writing more detailes of the solution suggested by Qiaochu:

Denote by $G$ the isometry group of $(V,\|\cdot\|)$. $G\subseteq \text{End}(V)$. On $\text{End}(V)$ we have the operator norm $\| \|_{op}$ (w.r.t the given norm $\|\cdot\|$), which induces a topology on $\text{End}(V)$.

Lemma 1: $G$ is compact in $\text{End}(V)$

Proof: $\text{End}(V)$ is a finite dimensional normed space, hence it is linearly homeomorphic to $\mathbb{R}^n$ (This is in fact true for every finite dimensioanl real topological vector space). So, the Heine-Borel theorem aplies. (Every closed and bounded subset is compact).

$G$ is bounded since for every isometry $g\in G, \|g\|_{op}=1$, hence $G$ is contained in the unit sphere of $(\text{End}(V),\| \|_{op})$ .

$G$ is closed: Assume $g_n \rightarrow g,g_n\in G$. Fix some $v\in V$. $\|g_n(v)-g(v)\|_V\leq \|g_n-g\|_{op}\cdot\|v\|_V \xrightarrow{n\to\infty} 0$. So $g_n(v)\xrightarrow{n\to\infty}g(v)$. Now by the continuity of the norm $\| \cdot \|$ (w.r.t to the topology it induces on $V$) we get that: $\|v\| \stackrel{g_n isometry}{=} \|g_n(v)\|\xrightarrow{n\to\infty}\|g(v)\|$. This forces $g$ to be an isometry.

Corollary1: $G$ is locally compact Hausdorff topological group.

Proof: $\text{End}(V)$ is Hausdorff (Any metric space is...), and every subspace of Hausdorff is also Hausdorff. It is sdandard fact that $GL(V)$ is a topological group (t.g), and any subgroup of a t.g is a t.g.

Now, there exists a left-invariant measure $\mu$ on the Borel $\sigma$-algebra of $G$ such that $\mu(G)>0$. (This is the Haar measure which can be constructed on any locally compact Hausdorff topological group).

Now take any inner product $\<,\>$ on $V$. Fix $v,w \in V$. Define $f_{v,w}:G\rightarrow \mathbb{R},f_{v,w}(g)=\<gv,gw\>$.

Lemma 2: $f_{v,w}$ is continuous

Proof: Since $G$ is a metric space (a subspace of the normed space $\text{End}(V)$) it is enough to check sequential continuity. Take
$g_n \rightarrow g,g_n\in G$. We already showed this implies $g_n(v)\xrightarrow{n\to\infty}g(v)$ so by the continuity of the inner product $f_{v,w}(g_n)=\<g_nv,g_nw\> \xrightarrow{n\to\infty} f_{v,w}(g) $.

In particular $f_{v,w}$ is measurable, so we can integrate it. (Compactness of $G$ implies $f_{v,w}$ is bounded, and $G$ being a finite measure space guarantees the integral will be finite).

So we define: $\<v, w \>' = \int_G f_{v,w} \, d\mu = \int_G \< gv, gw \> \, d\mu$.

Now all is left is to show $\<, \>'$ is an inner product on $V$ that is presrved by each $h\in G$. (since this means $G=\text{ISO}(V,\|\cdot\|) \subseteq \text{ISO}(V,\<, \>')$ as required.

Lemma 3: $\<,\>'$ is an inner product.

The only non-trivial thing is positive-definiteness. (The rest follows from the linearity of $g\in G$ and the integral, and the bilinearity of $\<,\>$). But this follows from standard measure theory:

Fix $v\neq 0$. $f_{v,v} > 0$ on $G$ (since each $g\in G$ is injective and the original inner prodcut is positive). But this forces $\<v,v\>'>0$ as required.

Lemma 4: $\<, \>'$ is $G$-invariant. But this follows from another standard proposition in measure theory: (See "Real Analaysis" by H.L.Royden) chapter 22 pg 488 proposition 10).

  • 2
    $\begingroup$ Still not enough langles and rangles... $\endgroup$ Jun 16, 2015 at 3:19
  • $\begingroup$ You are right... I thought I already fixed it but I was wrong. I hope now its OK. $\endgroup$ Jun 16, 2015 at 7:10
  • 3
    $\begingroup$ I really like that you accepted Qiaochu's answer and then added the details in your own. Nice! $\endgroup$
    – Joachim
    Jun 21, 2015 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.