Continuous functions on products

Stream: maths

[David Loeffler]

Let $X, Y$ be topological spaces and $A$ a topological ring. What are some reasonable conditions implying that the natural map $C(X, A) \otimes C(Y, A) \to C(X \times Y, A)$, sending $(f, g)$ to $(x, y) \mapsto f(x) g(y)$, has dense image? (Here $C(X \times Y, A)$ has the compact-open topology.)

I have an open PR which proves this when $X, Y$ are profinite (actually it suffices that both are compact Hausdorff and one of them is profinite). But I'm wondering if there's a "natural" degree of generality for this statement. (I could only find references online assuming $A = \mathbb{R}, \mathbb{C}$ and I don't want to make that restriction; the case I actually care about is $A = \mathbb{Q}_p$.)

[Anatole Dedecker]

You mean total range rather than dense range right? If so, then for $\mathbb{R}$ and $\mathbb{C}$ you only need compact Hausdorff thanks to the Stone-Weierstrass theorem, but I don’t know about the p-adic case.

[David Loeffler]

My question indeed had a typo, thanks Anatole for pointing this out – I meant that the span of the image of $C(X) \times C(Y)$ should be dense, or equivalently the image of the tensor product should be dense. Now corrected.

[Anatole Dedecker]

Aha, I think you can use one of my favourite facts of topology: any compact Hausdorff space is the quotient of a profinite set (namely Ultrafilter X, or in other words the Stone-Cech compactification of X seen as a discrete space). Furthermore quotients and products do commute for compact Hausdorff spaces, so I expect the general case to follow from your work in the profinite setting. I haven't checked it carefully though.

[Anatole Dedecker]

Actually I'm not sure at all that this works. I'll have to stop thinking about this for now, but consider this more of a random thought than a claim of proof.

[David Loeffler]

(I was just about to post saying I don't understand the argument you're proposing.)

[Anatole Dedecker]

Yes sorry, I was thinking while eating and should have thought a bit more before posting...

[David Loeffler]

The proof for profinite $X$ is quite easy, so I'd be really surprised if one could deduce the general compHaus case from that, it would feel like getting something for nothing.

[Anatole Dedecker]

Well the "nothing" would go through Stone-Cech compactification, so it's not that free. The reason I thought about this is that it reminded be of a very nice proof of the Riesz-Markov for measures in the compact metrizable case, where you first prove the theorem (quite easily) on the Cantor set and then use that every compact metrizable space is a quotient of the Cantor set to get the full result """for free""".

[Johan Commelin]

Won't you need some sort of condition on $A$? If it isn't a field, don't you expect some Tor group as cokernel, in some sort of Kunneth-formula?

[David Loeffler]

Johan Commelin said:

Won't you need some sort of condition on $A$? If it isn't a field, don't you expect some Tor group as cokernel, in some sort of Kunneth-formula?

I think that’s a bit of a different situation, there are no derived functors involved here.