Mathematics

robjohn

12:00 AM

@tb I escaped the vampires :-)

t.b.

@robjohn I'm very glad :)

Did you bring enough garlic?

Zhen Lin

Interesting self-enrichment occurs when a category $\mathcal{C}$ has a monoidal closed structure. Then, every morphism $X \to Y$ corresponds to a unique morphism $I \otimes X \to Y$, and so to a unique morphism $I \to [X, Y]$. So then one can factor $\textrm{Hom} : \mathcal{C}^\textrm{op} \times \mathcal{C} \to \textbf{Set}$ through the functor $\textrm{Hom}(I, -) : \mathcal{C} \to \textbf{Set}$.

robjohn

@tb of course! and standing by mirrors helps :-)

Zhen Lin

There's also non-interesting self-enrichment like assigning to every pair of objects the unit object $I$ as the internal hom...

robjohn

I guess waiting for the downvoter will be just as fruitful as waiting for an acceptance yesterday.

t.b.

12:05 AM

@ZhenLin okay, I think I got that. But now Qiaochu wants to replace Set by Top, doesn't he?

Zhen Lin

Yes. But he's also avoiding the problem that $\textbf{Top}$ is not self-enriched by changing the target category to some category enriched in $\textbf{Top}$.

t.b.

But something must be wrong...

Here's the category I know best: Ban. The internal hom is the space of linear maps equipped with the operator norm. It is simply not true that a continuous map $G \times V \to V$ corresponds to a continuous map $G \to [V,V]$.

(action I mean).

Zhen Lin

Really? That's disturbing. Is there a correspondence in at least one direction?

t.b.

Yes, a continuous map $G \to [V,V]$ gives a continuous map $G \times V \to V$.

Zhen Lin

Hmmm. Then perhaps one might argue that a continuous map $G \to [V, V]$ is the right notion of a continuous representation...

t.b.

12:12 AM

That's tempting but uninteresting: If $G$ is a compact group then a continuous map $G \to [V,V]$ is a direct sum of a finite-dimensional representation and a Banach space with trivial action.

So you exclude natural and important stuff like $L^p(G)$ even for compact groups.

Zhen Lin

Ah.

Hm. I'm not sure what to think. Maybe there ought to be two notions of group action.

t.b.

I think the problem is that $[V,V]$ ought to carry the compact-open topology for this to work but the operator norm topology is much stronger.

Zhen Lin

Well, could we change the enrichment of $\textbf{Ban}$ to make it work?

t.b.

Well, I think it should work when considering Ban as enriched over locally convex spaces.

This actually gives a looser notion of representations than the one people are actually interested in, but at least a reasonable one.

@robjohn I suspect that it might be the other one who posted an answer. One reason might be that you didn't post code but actually a nice FSA...

Zhen Lin

I'm confused. Are you saying that when we take $\textbf{Ban}$ as enriched over locally convex spaces, then we get continuous maps $G \to [V, V]$ which don't give continuous maps $G \times V \to V$?

t.b.

12:26 AM

@ZhenLin No, the bijection is fine. People usually don't want to consider mere continuous actions but rather continuous actions by isometries. I don't know how to implement that nicely in terms of enrichments.

That's what I meant.

Zhen Lin

Ah, I see. That must be the unitary representation thing I keep hearing about.

t.b.

Exactly if you take Hilbert spaces. That's certainly by far the most important and riches case.

Zhen Lin

We could again change the target category by removing all the non-isometries... but this is really getting quite contrived.

t.b.

That's essentially what I'm saying. It's getting quite complicated and messy quite quickly. You deal with at least 3 if not four different categories at the same time and exploit their interactions.

Zhen Lin

"Dans la $2$-catégorie $\mathfrak{Top}$ les $2$-produits sont représentables."

I'm not sure why the Grothendieck school decided that that "répresentables" should also mean "[limits] exist"...

t.b.

12:32 AM

Wild guess: SAFT?

Zhen Lin

Well, without having read much of their work, I'd guess that it's because the limit of a diagram $F : \mathcal{J} \to \mathcal{C}$ exists if and only if the functor $\varprojlim \mathcal{C}(-, F) : \mathcal{C} \to \textbf{Set}$ is representable.

t.b.

Oh, right.

Zhen Lin

I suppose it's possible that our modern notion of limit via universal cones post-dates the definition via representable functors.

On the other hand it seems incredibly prescient: no matter what limits $\mathcal{C}$ has, the Yoneda embedding $\mathcal{C} \to [\mathcal{C}^\textrm{op}, \textbf{Set}]$ will preserve them, so we can always take limits in $ [\mathcal{C}^\textrm{op}, \textbf{Set}]$ and regard them as surrogates for the limits which may or may not exist in $\mathcal{C}$...

t.b.

It might be that many things were developed in parallel: limits via cones certainly are in Kan's adjoint functors. But limits are also in Tohoku (I don't remember how G. defined them) and I thought the importance of representables became only apparent in a Séminaire Cartan by Grothendieck (where he also proves or rather mentionsthe Yoneda lemma).

I just checked: Tohoku also uses the cone definition for limits.

Zhen Lin

Interesting. Hmmm.

Blah, the theorem I want is stated but the proof is hidden somewhere. In French.

t.b.

12:49 AM

Is it long?

Zhen Lin

Well, I need to find the proof first...

t.b.

Oh :)

Zhen Lin

Ah, it seems that the entire section is dedicated to the proof of the theorem.

Or "paragraphe", as she calls it...

"Le but de ce paragraphe est de montrer qu'à tout $\mathscr{U}$-topos annelé $(\mathsf{X}, A)$, on peut associer de façon universelle un couple formé d'un $\mathscr{U}$-topos annelé en anneaux locaux [...]"

t.b.

Oy!

I posted a comment to Qiaochu's answer

Zhen Lin

Good find!

t.b.

12:58 AM

This must be a duplicate

Zhen Lin

Huh, it's 1am already. I haven't made any significant progress on my essay in the last 3 days because I keep getting sidetracked by topos theory. Hm. Maybe I should just abandon this little point and come back to it later when I've written the more important things...

t.b.

This sounds like a reasonable decision.

Zhen Lin

I wonder if this sounds as awkward in French as the obvious English translation would be... "Soit $(\mathsf{X}, A)$ un $\mathscr{U}$-topos et soit $\mathsf{E} = (\mathsf{S}, J)$, où $\mathsf{S}$ est une $\mathscr{U}$-catégorie et où $J$ est une $\mathscr{U}$-topologie sur $\mathsf{S}$, un $\mathscr{U}$-site standard de définition de $\mathsf{X}$."

t.b.

Not particularly elegant in my opinion...

Especially the end looks murky

Zhen Lin

Indeed. I got confused reading it the first time.

t.b.

1:11 AM

Okay, I have to go now. See you around!

Zhen Lin

Bye!

t.b.

Bye!