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I've been studying some category theory lately and in particular, I became acquainted with the notions of products and coproducts, which led me to ponder the following:

Consider the category of all complex Hilbert spaces (the morphisms being linear isometries). This category has coproducts, due to the direct sum construction: if $X_{\alpha} , {\alpha\in\Lambda}$ is family of Hilbert spaces, define $X := \bigoplus_{\alpha\in\Lambda}X_{\alpha}$ as the set of all "$\Lambda$-tuples" $(x_\alpha)_{\alpha \in \Lambda}$ such that:

$x_\alpha \in X_\alpha \\ \forall \alpha$ and $\sum_{\alpha \in \Lambda} \|x_{\alpha}\|^2 < \infty$

Then one can define addition, scalar multiplication and an inner product on $X$ in an obvious way, and we have the canonical inclusion maps.

However, I don't see any way to make this construction into a product, though maybe there is another construction I don't know of.

I'm sorry if this question is elementary for category theorists, but to me it's not so obvious.

EDIT: Thanks for the replies. As it was pointed out, this category doesn't even admit finite products with morphisms being linear isometries. As I don't see any more natural choice for morphisms, I suppose there isn't any good answer to my question (other than "no" :)).

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If you insist that morphisms are linear isometries (clearly not supposed surjective) then even the category of finite-dimensional Hilbert spaces won't have products. All morphisms are injective linear maps; so what could the product of a 1-dimensional and a 2-dimensional Hilbert space be? –  Robin Chapman Jul 31 '10 at 21:09
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Are you sure that your coproduct construction has the universal property? Suppose I map the $X_\alpha$'s into some Hilbert space $Y$ by linear isometries whose ranges are not orthogonal; how would those give a morphism from $X$ to $Y$? For the simplest case, consider the sum of two copies of the same $X$ and try to form the co-diagonal map to $X$; it won't be an isometry (or even injective). –  Andreas Blass Jul 31 '10 at 22:26
    
Just as the natural concept of a "product" of vector spaces is the tensor product, there is a notion of completed tensor product for Hilbert spaces. See the Wikipedia page en.wikipedia.org/wiki/Tensor_product_of_Hilbert_spaces. However, this is not a categorical product: there are not natural maps from a tensor product of two spaces to the two spaces separately, etc. Nevertheless, I think this might be something you should look at. –  KConrad Jul 31 '10 at 23:19
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KConrad - I'm not a categorist but I'm not convinced that the tensor product should be thought of as the natural concept of a "product" for vector spaces. If anything, on some occasions it has more of a coproduct feel. –  Yemon Choi Aug 1 '10 at 0:54
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Typically, the category of Hilbert spaces is taken to have as arrows bounded linear maps, rather than linear isometries. In such a case, finite direct sums are biproducts. That is, they are simultaneously products and coproducts, where for each projection p_j and inject i_j, p_j o i_j = 1. Yemon's comment seems to show why arbitrary direct sums fail to be biproducts. –  Aleks Kissinger Aug 4 '10 at 0:50

1 Answer 1

up vote 2 down vote accepted

The category you specified does not have products, because it doesn't have a product of zero objects. The product of zero objects is an final object, if it exists, but final objects need to have unique maps (in this case, isometries) from all other objects. Such a final Hilbert space would need to be at least as large as any other Hilbert space (hence nonzero), but it cannot have nonidentity automorphisms (such as minus one).

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