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For example, by viewing Hilbert spaces as enriched categories in some fashion? (I suppose the same idea of considering the inner product of a Hilbert space as a generalized Hom-set has also been pondered in connection with the similarity of adjoint transformations and adjoint functors, though I'm not aware of any more formal correspondence there either)

Edit for clarification: The full similarity I see is this:

For any Hilbert space $H$, we have its inner product, a continuous linear map from $H^{op} \otimes H$ to $\mathbb{C}$ [where $H^{op}$ is $H$ with its inner product's argument order flipped]; currying this gives a continuous linear map from $H^{op}$ into the Hilbert space of continuous linear maps from $H$ to $\mathbb{C}$. The Riesz representation theorem says this is an isomorphism.

Similarly, for any category $H$, we have its Hom functor, a continuous functor from $H^{op} \times H$ to $Set$ [where $H^{op}$ is $H$ with its Hom functor's argument order flipped]; currying this gives a continuous functor from $H^{op}$ into the category of continuous functors from $H$ to $Set$. The Yoneda embedding lemma says this is an embedding; furthermore, under suitable conditions (e.g., if $H^{op}$ is equivalent to a presheaf category), this is an equivalence.

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What exactly is the similarity here? – Qiaochu Yuan Jun 13 '11 at 20:47
Here's the similarity: ⓵ If you know $\langle -, \xi \rangle : H\to \mathbb C$ then you know the vector $\xi\in H$. ⓶ If you know $Hom(-,X): \mathcal C\to Set$, then you know the object $X\in \mathcal C$. – André Henriques Jun 13 '11 at 21:35
Okay. Then see also… . – Qiaochu Yuan Jun 13 '11 at 21:41
If $\mathcal C$ is a semisimple $\mathbb C$-linear dagger-category, then X,Y ↦ dim(Hom(X,Y)) extends to an inner product on $K_0(\mathcal C)$. This inner product declares the basis of simple objects to be an orthonormal basis. – André Henriques Jun 13 '11 at 21:51
Perhaps the similarity should be state for pre-Hilbert-spaces and categories; see André's comment above. In these cases, we have an embedding. Then in some "completion", we actually get an equivalence. – Martin Brandenburg Jun 14 '11 at 9:34

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