# Does the category of Hilbert spaces possess a product?

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
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
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
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