Isometry group of a norm is always contained in some Isometry group of an inner product?


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

Update:

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)?

Remarks:

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 $(\<,\>)$).

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.

• @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. – Qiaochu Yuan Jun 14 '15 at 19:38


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).

• Still not enough langles and rangles... – Qiaochu Yuan Jun 16 '15 at 3:19
• You are right... I thought I already fixed it but I was wrong. I hope now its OK. – Asaf Shachar Jun 16 '15 at 7:10
• I really like that you accepted Qiaochu's answer and then added the details in your own. Nice! – Joachim Jun 21 '15 at 21:00