MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.