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Let $(E,\lVert\cdot\rVert)$ denote a normed vector space. Recall that an inner product space $E$ is a NVS with an additional gadget, namely an inner product that induces the norm. But, a NVS space $E$ is an IPS if and only if the norm satisfies the parallelogram law, i.e. if for all $x,y\in E$, $$\lVert x+y\rVert^2 + \lVert x-y\rVert^2 = 2\lVert x\rVert^2 + 2\lVert y\rVert^2$$ which allows us to define an inner product by letting $$\langle x,y\rangle = \frac12\left(\lVert x+y\rVert^2-\lVert x\rVert^2 - \lVert y\rVert^2\right).$$ Now, this tells us that the inner product is uniquely determined by the NVS structure.

Can we take this further? Can we say that the IPS structure on $E$ is uniquely defined by the structure of $E$ as a topological vector space? Or do there exist non isomorphic IPS's that are isomorphic as topological vector spaces?

Dominic Wynter
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    My suspicion is that the answer is no, because I suspect there are multiple distinct NVS's with the same topology. This is probably related the fact that, in general, metrizable topologies have many distinct metrics that are compatible with the topology in question. – Justin Benfield Apr 27 '17 at 22:05
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    When do you consider two inner products equivalent? – user251257 Apr 27 '17 at 23:32
  • When we have an isometry between the two. – Dominic Wynter Apr 28 '17 at 21:18

2 Answers2

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It depends on the sense of equivalence you care about. There is a notion of equivalent norms which says that $\|x\|_1 \equiv \|x\|_2 \iff \exists a,b \,\mathrm{where}\, a\|x\|_1 \leq \|x\|_2\leq b\|x\|_1$ which is to say that they induce the same topology.

If this is all you care about, then all finite dimensional normed vector spaces are equivalent. In this case, the norm is very much a topological notion, and it indeed induces a topology unique up to equivalence. However, there are of course different inner products that induce the same norm. To me, the inner product is not a topological notion, but rather geometric in the stricter sense that it cares about angles, orthogonality, etc. which are not at all topological notions.

Andres Mejia
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  • Exactly. Any inner product on $\mathbb C^2$ gives the (unique!) topological vector space topology on it, but, obviously, we can make unequal (not even by scaling) distinct inner products in two or more dimensions. – paul garrett Apr 27 '17 at 22:52
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It seems my comment is correct, the norm on a vector space induces a metric, and through that, a topology for the vector space. You can have many different metrics that induce the same topology on a metrizable topological space, hence there are multiple NVS's with the same topology.

See https://en.wikipedia.org/wiki/Normed_vector_space, and https://en.wikipedia.org/wiki/Metric_(mathematics)

The theorem characterizing those topological spaces that are metrizable is https://en.wikipedia.org/wiki/Nagata%E2%80%93Smirnov_metrization_theorem

Justin Benfield
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  • Do you have any specific examples? – Dominic Wynter Apr 27 '17 at 22:14
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    An example of 2 different metrics that induce the same topology for the plane, $\mathbb{R}^2$ are the standard euclidean metric: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the 'taxicab' metric, $d=|x_2-x_1|+|y_2-y_1|$, which both induce the standard topology on $\mathbb{R}$. – Justin Benfield Apr 27 '17 at 22:34
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    Only one of those has an associated inner product though. My question is whether any topological vector space can admit two non-isometric inner products. – Dominic Wynter Apr 27 '17 at 22:36
  • I suspect the answer is still yes, I was merely giving an example of two different metrics can induce the same topology. Since inner products define an associated norm, the only way that you can have non- isometric inner products is if the norms are also different, but induce the same topology (which is at least plausible). – Justin Benfield Apr 27 '17 at 23:04
  • @MonstrousMoonshine Any two separable Hilbert spaces are isomorphic as Hilbert spaces – 3-in-441 Apr 28 '17 at 00:59