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As an answer to this post Additive category that is not abelian

it was said that the additive category of Hilbert spaces is not abelian.

Why is that?

Also what category of Hilbert spaces is this? In particular what are the morphisms.

Is there another choice of morphisms that will make the category abelian?

Don Stanley
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2 Answers2

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First, the morphisms are continuous(=bounded) linear maps.

One can see easily that every morphism $T: H\to H'$ has kernel and cokernel: The kernel is the closed subspace $K=\{x\in H: T(x)=0\}$ and cokernel is the quotient of $H'$ by the closure of $T(H)$. But in this category "image" = kernel of cokernel is equal to the closure of $T(H)$ whereas "coimage" = cokernel of kernel is equal to $H/K$. And the natural map from $H/K$ to the closure of $T(H)$ surjects to $T(H)$ and is not always isomorphism. So the last axiom "image = coimage" is not satisfied.

(Maybe it is easier to check another equivalent axiom: "Every monomorphism is a kernel".)

I have no candidate for an abelian category on Hilbert spaces.

Mostafa - Free Palestine
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Recall that in an abelian category $\mathcal A$ a morphism $f$ is an isomorphism if and only if $f$ is both monic and epic.

Now, the category of Hilbert spaces is the category $\mathsf{Hilb}$ whose objects are Hilbert spaces and whose morphisms are continuous linear maps. Proving that $\mathsf{Hilb}$ is additive is quite straightforward.

To see that $\mathsf{Hilb}$ is not abelian, let $f:X\to Y$ be an injective but nonsurjective morphism of Hilbert spaces whose image is dense in $Y$. One easily shows that $f$ is both monic and epic but not an isomorphism.

Exercise: Can you construct such a morphism?

Brian Fitzpatrick
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  • You can take a self map of $l^2(\mathbb N)$, sending the $n$-th basis vector to $1/2^n$ times itself. Thanks! – Don Stanley Mar 12 '15 at 20:28
  • I was thinking of a slightly more restrictive category than Hilb, where images and kernels of maps had to be closed, but I'm not sure if that works. – Don Stanley Mar 12 '15 at 20:30