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Given two Hilbert Spaces $H$ and $K$, a natural inner product on $H\otimes K$(= vector space tensor product of $H$ and $K$) is given by

$\hspace{.5in}\langle h_1\otimes k_1,h_2\otimes k_2\rangle=\langle h_1,h_2\rangle_H$ $\langle k_1,k_2\rangle_K $ for $h_1,h_2\in H$ and $k_1,k_2\in K$.

Now , suppose we are given another inner product $\langle\cdot,\cdot\rangle_1$ on $H\otimes K$ which satisfies the following:

$\hspace{.5in }$$\langle h\otimes k,h\otimes k\rangle_1=\langle h,h\rangle_H$ $\langle k,k\rangle _K$ for each $h\in H$ and $k\in K.$

Does this imply that the new inner product $\langle\cdot,\cdot\rangle_1$ is same as the natural inner product i.e. does the following hold?

$\hspace{.5in} \langle h_1\otimes k_1,h_2\otimes k_2\rangle_1=\langle h_1\otimes k_1,h_2\otimes k_2\rangle=\langle h_1,h_2\rangle_H$ $\langle k_1,k_2\rangle_K $

for $h_1,h_2\in H$ and $k_1,k_2\in K$.

I am trying to apply universal property of tensor product of vector spaces to show the mentioned but I can't solve. Or, is there any different universal property for tensor product of Hilbert Spaces or tensor product of inner product spaces which makes the natural inner product unique?

If the above is not true, then can you provide a different inner product than the natural one?

P.S. The question may be thought just for inner product spaces instead of Hilbert Spaces.

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    $\begingroup$ Isn't this just the polarisation identity? $\endgroup$ – Johannes Hahn Nov 7 '17 at 11:28
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    $\begingroup$ @JohannesHahn The assumption for the new inner product is only for simple tensors $h\otimes k$ and not for all elements of tensor product. But in polarization expansion, we shall get elements which are not simple tensors. $\endgroup$ – Manish Kumar Nov 7 '17 at 11:45
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    $\begingroup$ Use polarization twice: first with fixed $k$, you obtain that $\langle h_1 \otimes k,h_2\otimes k\rangle_1= \langle h_1,h_2\rangle \langle k,k\rangle$. Then with fixed $h_1,h_2$ you obtain $\langle h_1 \otimes k_1,h_2\otimes k_2\rangle_1= \langle h_1,h_2\rangle \langle k_1,k_2\rangle$. $\endgroup$ – Mikael de la Salle Nov 7 '17 at 14:09

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