TITLE: Is the property of being an inner product space a topological notion?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.