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Let $\text{FinVec}$ denote the category of finite dimensional vector spaces over some field $k$, and let $F:\text{FinVec}\to \text{FinVec}$ be a contravariant functor such that $F^2$ is naturally isomorphic to the identity. Is $F$ naturally isomorphic to the canonical duality functor $V\mapsto V^*=\text{Hom}(V,k)$?

I suspect this question has been asked before on MO, but I couldn't find it.

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    $\begingroup$ $F = \mathrm{id}$ satisfies those requirements, but probably isn't what you're looking for. Also, over $\mathbb{C}$, there's the complex conjugate functor, right? $\endgroup$
    – Arun Debray
    Commented May 24, 2017 at 13:17
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    $\begingroup$ @ArunDebray those functors are covariant, not contravariant. $\endgroup$
    – Chris Schommer-Pries
    Commented May 24, 2017 at 13:20
  • $\begingroup$ Ah, I missed that assumption.Thank you. $\endgroup$
    – Arun Debray
    Commented May 24, 2017 at 13:47
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    $\begingroup$ There is a sense in which the duality is not canonical: the space of functors $F$ equipped with an isomorphism $F^2 \cong \mathrm{id}$ is not contractible. Indeed, any such functor admits a nontrivial natural automorphism, namely multiplication by $-1$. $\endgroup$
    – Theo Johnson-Freyd
    Commented May 24, 2017 at 16:33

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$FinVec$ and its opposite are enriched in finite dimensional $k$-vector spaces. Assume that $F$ is an enriched functor. Then consider the covariant functor:

$$F(-)^*: FinVec \to FinVec$$

It is a $Vec$-enriched functor. Finite direct sums are absolute limits and so are preserved by this functor (see here). Since every finite dimensional vector space is isomorphic to a finite direct sum of copies of $k$ it follows that this functor is completely determined by the single vector space $L=F(k)^*$, and there is a natural isomorphism:

$$F(V)^* \cong V \otimes L$$

Thus $F(V) \cong V^* \otimes L^*$ as functors. Your condition that $F^2 \cong id$ implies that $L$ must be one-dimensional (hence isomorphic to $k$) and so $F(-) \cong (-)^*$ is the standard duality functor.

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    $\begingroup$ I think that any equivalence $G\colon FinVec\to FinVec$ is automatically enriched. Indeed, it gives a ring map $Z\text{End}(V)\to Z\text{End}(GV)$, and both source and target are $\mathbb{R}$ (provided that $V\neq 0$), and it is not hard to show that any ring endomorphism of $\mathbb{R}$ is the identity. $\endgroup$
    – Neil Strickland
    Commented May 24, 2017 at 14:10
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    $\begingroup$ @NeilStrickland It depends on your field $k$, which the OP left unspecified. For example the complex conjugation functor mentioned by Arun Debray is an equivalence, but not $\mathbb{C}$-linear, hence not enriched. I agree that over $\mathbb{R}$ the enrichment is automatic, but that is basically equivalent to the fact that the Galois group of $\mathbb{R}$ over $\mathbb{Q}$ is trivial. So I think in general $F$ has to be a Galois twist of the standard duality functor. $\endgroup$
    – Chris Schommer-Pries
    Commented May 24, 2017 at 15:06

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