Let $V$ be a finite dimensional euclidean vector space (that is, a vector space with an inner product $\langle,\rangle$. I read somewhere that for such vector spaces, there is a canonical isomorphism between a vector space and its dual, namely $x \mapsto \langle x,-\rangle$. This makes sense to me, but I am left wondering, does this make $V$ natually isomorphic to $V^{*}$? That is, in the category of finite dimensional euclidean vector spaces, is the identity functor isomorphic to the dualization functor?

The dualization functor is of course a contravariant functor, so we would need the notion of a dinatural transformation. This would be a collection of isomorphisms $m_V$ with a corresponding naturality condition. To follow through with my inutition of what the natural isomorphism would ve, I have set $m_V: x \mapsto \langle x,- \rangle$

For all $f: V \to W$, the following diagram commutes

$$\require{AMScd}\begin{CD} V @>m_V>> V^{*}\\ @VfVV @AAf^{*}A\\ W @>>m_{W}> W^{*} \end{CD}$$

Here, as usual, $f^{*}: W^{*} \to V^{*}$ is defined by $\phi \mapsto \phi \circ f$. Then, this diagram reads:

For all $v \in V$, we have $(f^{*} \circ m_W \circ f) (v) = m_V (v)$.

Directly evaluating gives:

$$ (f^{*} \circ m_W \circ f) (v) = f^{*} \circ m_V (f(v))$$ $$ = f^{*} (\langle f(v),- \rangle)$$ $$ \langle f(v), f(-) \rangle$$

Thus, the naturality condiditon reads that for all $v \in V$, we have $\langle v, - \rangle = \langle f(v), f(-) \rangle$, where both the left and right hand side of the equation are viewed as elements of $V^{*}$. This however, is only true when $f$ is orthogonal, right?

Furthermore, setting $f=0$ says that $\langle v,- \rangle =0$ for all $v$, which is certainly not true, so this definition of naturality doesn't seem to be the right one. My question then is, how can we make precise the naturality of the map $x \mapsto \langle x,-\rangle$?


What you've proven is precisely that the naturality is with respect to orthogonal (/unitary) maps. These are precisely the maps that preserve the inner product, so this is a natural choice of morphisms for the category of inner product spaces.

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    $\begingroup$ Ah of course! That makes quite a lot of sense. I hadn't given much thought to what the morphisms of this category would be; I just assumed it would make sense to stick with linear maps. That begs the question though, is there anything interesting to be said about the category with objects euclidean vector spaces and morphisms just all linear maps? The map $x \mapsto \langle x, - \rangle$ still seems like it ought to be natural, since it essentially just means that we have prespecified an agreed-upon basis. Is there anything to be said about the naturality of such a map within that category? $\endgroup$ – ASKASK Mar 8 '17 at 4:37
  • $\begingroup$ @ASKASK: yes, it has the structure of what's called a dagger category (en.wikipedia.org/wiki/Dagger_category), where the dagger is given by taking adjoints. In a dagger category it makes sense to talk about unitary as well as self-adjoint maps, but category theory for dagger categories is very weird and I can't say I understand it very well. The dagger structure can be used to recover the inner product (exercise), but in any case you can't really expect a construction making use of the dagger structure to be natural with respect to arbitrary morphisms, since those don't refer to the... $\endgroup$ – Qiaochu Yuan Mar 8 '17 at 5:09
  • $\begingroup$ ...dagger structure at all. The most naturality it's reasonable to hope for is with respect to unitaries. But maybe there is some notion of "dagger naturality" relevant here? I don't know. Dagger categories are really weird. $\endgroup$ – Qiaochu Yuan Mar 8 '17 at 5:09

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