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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?

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    $\begingroup$ However, there are various different norms on the tensor products of Hilbert spaces which do not come from an inner product (i.e. they do not satisfy the parallelogram law). $\endgroup$ – Matthias Ludewig May 31 at 3:59
  • $\begingroup$ There is a construction called the diamond norm that I think goes close to it but maybe not entirely what you want. $\endgroup$ – lcv May 31 at 4:22
  • $\begingroup$ @lcv It doesn't come from an inner product, and the OP is asking about inner products $\endgroup$ – Yemon Choi May 31 at 12:03
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    $\begingroup$ To be clear - as Yemon Choi says - I am assuming that the norm in (2) is induced by an inner product. $\endgroup$ – Pierre Dubois May 31 at 12:39
  • $\begingroup$ @PierreDubois no problem yes now it's clearer I can remove my answer. $\endgroup$ – lcv May 31 at 15:14
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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.

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  • $\begingroup$ Maybe proving that an orthonormal base for $\[ \cdot , \cdot \] $ is an orthonormal base for $<\cdot , \cdot >$ $\endgroup$ – Alessio Ranallo Jun 11 at 22:50
  • $\begingroup$ @AlessioRanallo: can you explain where you think my argument fails? $\endgroup$ – Nik Weaver Jun 11 at 22:59
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Let $V, W$ be 2-dimensional with orthonormal bases $\{v_0, v_1\}$ and $\{w_0, w_1\}$. If you expand out your condition (2), you obtain the following conditions: $$ \langle v_i \otimes w_j, v_i \otimes w_j \rangle = 1 \\ \langle v_i \otimes w_j, v_i \otimes w_k \rangle = \langle v_i \otimes w_j, v_k \otimes w_j \rangle = 0 \\ \langle v_0 \otimes w_0, v_1 \otimes w_1 \rangle + \langle v_1 \otimes w_0, v_0 \otimes w_1 \rangle = 0 $$ Clearly you can pick any value for $\langle v_0 \otimes w_0, v_1 \otimes w_1 \rangle$. Any nonzero choice will not satisfy your condition (1), since the RHS will be 0. So it is certainly possible. I'm not sure if they're of interest.

EDIT: Since there's disagreement in the comments, I'm posting my calculation. Consider general vectors $a_0 v_0 + a_1 v_1 \in V$, $b_0 w_0 + b_1 w_1 \in W$. Let's mangle Einstein notation by putting $\langle v_i \otimes w_j, v_k \otimes w_l\rangle = g_{ij}^{kl}$.

(2) says that $ \langle a_0 v_0 + a_1 v_1, a_0 v_0 + a_1 v_1 \rangle = (a_0^2 + a_1^2)(b_0^2 + b_1^2)$. Expanding out the LHS gives

$$ a_0^2 b_0^2 g_{00}^{00} + a_1^2 b_0^2 g_{10}^{10} + a_0^2 b_1^2 g_{01}^{01} + a_1^2 b_1^2 g_{11}^{11} + \\ 2\left(a_1 a_0 b_0^2 g_{10}^{00} + a_0^2 b_1 b_0 g_{01}^{00} + a_1^2 b_1 b_0 g^{10}_{11} + a_1 a_0 b_1^2 g^{01}_{11}\right) + \\ 4a_1 a_0 b_1 b_0 \left(g^{00}_{11} + g^{10}_{01} \right).$$

By choosing unit vectors, it is clear that $g^{ij}_{ij} = 1$. So the first line cancels with the RHS. By choosing a sum of two unit vectors, it is clear that $g^{ij}_{ik} = g^{ij}_{kj} = 0$, so the second line vanishes. We are left with $4a_1 a_0 b_1 b_0 (g^{00}_{11} + g^{10}_{01})$ = 0. If we choose our $g$ so that $g^{00}_{11} + g^{10}_{01} = 0$, this quantity will always vanish, and so (2) will hold for any pair of vectors.

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  • $\begingroup$ This isn't correct; you've only assumed (2) for basis vectors, not all $v \in \mathcal{V}$ and $w \in \mathcal{W}$. $\endgroup$ – Nik Weaver May 31 at 3:06
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    $\begingroup$ Hi @NikWeaver -- I've posted my calculation, let me know if there's a mistake. Thanks! $\endgroup$ – Kevin Casto May 31 at 3:31
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    $\begingroup$ At first glance it looks right (assuming real scalars), but I also think my answer is right ... going to have to think about this for a minute. $\endgroup$ – Nik Weaver May 31 at 3:32
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    $\begingroup$ @NikWeaver Heuristic answer: $Sym^2(V^* \otimes W^*) = Sym^2(V^*) \otimes Sym^2(W^*) + \Lambda^2(V^*) \otimes \Lambda^2(W^*)$. This is why, when you fix the inner product on $V$ and $W$, you get something antisymmetric in $V$ and in $W$ (as in my answer). I'm trying to make this precise in terms of kernels vs projections, but in any case counting constants should tell you that (2) doesn't pin down what the total inner product is. $\endgroup$ – Kevin Casto May 31 at 3:40
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    $\begingroup$ Actually, the scalars might be the issue --- my argument for uniqueness uses complex numbers. So we get different answers depending on whether we use real or complex scalars? $\endgroup$ – Nik Weaver May 31 at 3:41

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