Consider Napkin Problem 13D (a):

Let $V$ and $W$ be finite-dimensional inner product spaces over $k$, where $k$ is either $\mathbb{R}$ or $\mathbb{C}$. Find a canonical way to make $V \otimes_k W$ into an inner product space too.

I approached this by working on special cases. Let $e_1, \dots , e_n$ and $f_1, \dots , f_m$ are orthonormal bases for $V$ and $W$, respectively.

I think we want $\langle e_1 \otimes f_1, e_1 \otimes f_1 \rangle_{V \otimes W} = 1$, since it feels like $\|e_1\| = \|f_1\| = 1$ should imply that $\| e_1 \otimes f_1 \|$. We also want $\langle e_1 \otimes f_1, e_2 \otimes f_2 \rangle_{V \otimes W} = 0$, since it feels like the fact that $e_1$ and $e_2$ are orthogonal and $f_1$ and $f_2$ are orthogonal should imply that $e_1 \otimes f_1$ and $e_2 \otimes f_2$ are orthogonal.

So for any $v_1, v_2 \in V$ and $w_1, w_2 \in W$, the inner form $\langle v_1 \otimes w_1, v_2 \otimes w_2 \rangle_{V \otimes W} := \langle v_1, v_2 \rangle_V \langle w_1, w_2 \rangle_W$ seems like a good choice, and it ended up being the intended inner form.

But I'm still very uncomfortable with the intuition behind this; is there a more solid way to understand why we take the product of the individual inner forms to get the inner form of the tensor product?

  • $\begingroup$ Are you familiar with the tensor product of bilinear maps? The inner product here is just the tensor product of the two inner products for the two spaces, so it is a natural choice. $\endgroup$
    – blargoner
    Commented Dec 4, 2022 at 22:40
  • $\begingroup$ Change all your "it feels like"s to "we are free to choose". There is not just one inner product on $V\times_k W$ extending the inner products on $V$ and $W$. So what inner product we get depends on arbitrary choices we make. What is the least arbitrary choice to make? Well, the simplest thing is to just make $V$ and $W$ orthogonal to each other, and treat their unit vectors as being the same size. Which leads exactly to your choices. $\endgroup$ Commented Dec 5, 2022 at 12:30
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    $\begingroup$ I think ronno’s answer is very good, but at a very naive level, my first intuition was that both inner products and tensor products are like a product of vectors, so $\langle v_1\otimes w_1,v_2\otimes w_2\rangle$ is like a product $v_1w_1v_2w_2$, so we just rearrange the only way that makes sense (i.e. to $v_1v_2w_1w_2$ in my extremely informal notation). All this is very informal of course, but the conclusion for me was that I’d be surprised if this was not the answer $\endgroup$
    – Milten
    Commented Mar 4, 2023 at 2:46
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    $\begingroup$ @Milten I think my answer agrees with what you said but at a more formal level. That inner products are "products" is precisely the statement that they are defined on $V \otimes V$, which is the universal place where you can multiply $v, v' \in V$. The rearrangement step is the first isomorphism. $\endgroup$
    – ronno
    Commented Mar 4, 2023 at 19:13
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    $\begingroup$ @Milten added to my answer a rephrasing that I think is fairly close to your comment. $\endgroup$
    – ronno
    Commented Mar 4, 2023 at 19:28

1 Answer 1


Canonical is an imprecise word, but here is a justification. An inner product on $V$ is a map $V \otimes V \to k$ with certain properties. So having inner products on $V$ and $W$ gives a map $$(V \otimes W) \otimes (V \otimes W) \cong (V \otimes V) \otimes (W \otimes W) \to k \otimes k \xrightarrow{(x \otimes y) \mapsto xy} k$$ and I would argue that the first isomorphism and the third map, the multiplication on $k$, are the canonical maps with those domains and targets. This composition is exactly the same as your formula, but all it says is that given bilinear forms on $V$ and $W$ there is a "canonical" bilinear form on $V \otimes W$. You still have to check that it's an inner product if you start with inner products.

To add some intuition in line with @Milten's comment, an inner product (or more generally a bilinear form) on $V$ is a product on $V$ taking values in $k$, so is a function on $V \otimes V$ since that's the universal target for a product on $V$. To have a way to multiply $v_1 \otimes w_1, v_2 \otimes w_2 \in V \otimes W$ (*) it should suffice to know how to:

  1. rearrange to $(v_1 \otimes v_2) \otimes (w_1 \otimes w_2)$;
  2. multiply $v_1$ with $v_2$ and $w_1$ with $w_2$ individually to get elements of $k$;
  3. multiply in $k$.

These are precisely the three maps that are being composed; 1 and 3 are "canonical" and 2 is the input.

(*) I'm ignoring that not all elements are pure tensors, but a "product" should satisfy distributive law so it's enough to know what it does to a generating set anyway.


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