1
$\begingroup$

Let $\mathsf{Hilb}_1$ (resp. $\mathsf{Hilb}_{\leq 1}$) denote the category of Hilbert spaces with linear isometries (resp. short linear maps).

Does $\mathsf{Hilb}_1$ (resp. $\mathsf{Hilb}_{\leq 1}$) have binary coproducts?

Notice what is usually called the direct sum of Hilbert spaces is not a coproduct, since here the summands are orthogonal to each other, which is a special relation not suitable for a coproduct. But perhaps there is another construction which turns out to be the coproduct.

$\endgroup$
2
$\begingroup$

This is an answer for $\mathsf{Hilb}_1$. Let $X$ be any nonzero Hilbert space. I claim the coproduct $X \vee X$ doesn't exist in $\mathsf{Hilb}_1$.

Assume that it did, call it $X \xrightarrow{i_0} X \vee X \xleftarrow{i_1} X$. Then by the universal property of the coproduct, we get an isometry $f : X \vee X \xrightarrow{(\operatorname{id}, \operatorname{id})} X$ such that $f \circ i_0 = \operatorname{id}_X$ and $f \circ i_1 = \operatorname{id}_X$. Since $f$ is an isometry and hence injective, it follows that $i_0 = i_1$.

Now consider the isometry $\theta : X \to X$ given by $x \mapsto -x$. Again by the universal property there is an isometry $g : X \vee X \xrightarrow{(\operatorname{id}, \theta)} X$ such that $g \circ i_0 = \operatorname{id}_X$ and $g \circ i_1 = \theta$. But this is absurd, since $\theta$ isn't the identity but $i_0 = i_1$.

$\endgroup$
  • $\begingroup$ Thank you! Perhaps something similar works for short linear maps. $\endgroup$ – Martin Brandenburg Sep 9 '15 at 9:46
  • 2
    $\begingroup$ What you have shown is this: if you have a category in which every morphism is monic and coproducts exist, then the category is a preorder. $\endgroup$ – Zhen Lin Sep 9 '15 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.