For two inner product spaces $(\mathcal{V}, (\cdot,\cdot)\_V)$ and $(\mathcal{W}, (\cdot,\cdot)\_W)$, we can put an inner product on their tensor product in the obvious way:
$$
(1) ~~~~ \langle v \otimes w, v' \otimes w'\rangle := \langle v,v'\rangle\_V \langle w,w'\rangle\_W.
$$
This then implies that 
$$
(2) ~~~~ \|v \otimes w\| = \|v\|\_V \|w\|\_W.
$$
Is there an example of an inner product on $\mathcal{V} \otimes \mathcal{W}$ such that (2) holds, but the inner product is not of the form (1)? Are such things of interest? Do they have a name?



Non-standard tensor products of inner product spaces
Assuming complex scalars, no, any inner product which satisfies (2) for all $v \in \mathcal{V}$ and $w \in \mathcal{W}$ has the given form. To see this, let $[\cdot,\cdot]$ be any inner product which satisfies (2). So right away we know that $[v\otimes w,v\otimes w] = \langle v\otimes w, v\otimes w\rangle$ for all $v$ and $w$. Next, for $v,v' \in \mathcal{V}$ and $w \in \mathcal{W}$, we know that $$[(v + v')\otimes w, (v + v')\otimes w] = \langle (v + v')\otimes w, (v + v')\otimes w\rangle.$$ Expanding this out and applying $[v\otimes w,v\otimes w] = \langle v\otimes w,v\otimes w\rangle$ plus the same for $v'\otimes w$ yields $2{\rm Re}[v\otimes w, v'\otimes w] = 2{\rm Re}\langle v\otimes w, v'\otimes w\rangle$. Replacing $v$ with $iv$ yields equality of the imaginary parts too. So now we know that $[v\otimes w,v'\otimes w] = \langle v\otimes w,v'\otimes w\rangle$ for all $v$, $v'$, and $w$.


Finally, for any $v,v' \in \mathcal{V}$ and $w,w' \in \mathcal{W}$ we have $$[(v + v')\otimes (w + w'), (v + v')\otimes (w + w')] = \langle (v + v')\otimes (w + w'), (v + v')\otimes (w + w')\rangle.$$ Expanding this out and cancelling all the equalities we already have then leaves only $$2{\rm Re}([v\otimes w, v'\otimes w'] + [v\otimes w', v'\otimes w]) = 2{\rm Re}(\langle v\otimes w, v'\otimes w'\rangle + \langle v\otimes w', v'\otimes w\rangle).$$ Replacing $v$ with $iv$ yields equality of the imaginary parts too, so that $$[v\otimes w, v'\otimes w'] + [v\otimes w', v'\otimes w] = \langle v\otimes w, v'\otimes w'\rangle + \langle v\otimes w', v'\otimes w\rangle.$$ Then replacing $w'$ with $iw'$ yields $$[v\otimes w, v'\otimes w'] - [v\otimes w', v'\otimes w] = \langle v\otimes w, v'\otimes w'\rangle - \langle v\otimes w', v'\otimes w\rangle,$$ so that $[v\otimes w, v'\otimes w'] = \langle v\otimes w,v'\otimes w'\rangle$. As every element of the algebraic tensor product $\mathcal{V}\otimes\mathcal{W}$ is a linear combination of elementary tensors, this shows that $[\cdot,\cdot] = \langle\cdot,\cdot\rangle$.


I feel there ought to be a one-line proof of this, but I don't quite see it.


