User simon henry - MathOverflow

On the openness of the map X^I -> X * X.

2013-04-04T10:13:05Z

Hello !

Let $X$ be a locale or a topological spaces. $I$ denote the unit interval of the real numbers, and $X^I$ the space of function from $I$ to $X$ (The locale exponential if $X$ is a locale or the set of function endowed with the open-compact topology if $X$ is a topological space.)

In both case, we consider the map $X^I \rightarrow X \times X$, which is simply the evaluation at the two endpoints.

In "Connected locally connected toposes are path connected" (available here : http://www.ams.org/journals/tran/1986-295-02/S0002-9947-1986-0833712-3/) Moerdijk and Wraith showed that if $X$ is a locale (or actually even a topos) which is connected and locally connected then the map $X^I \rightarrow X \times X$ is an open surjection.

At the very end of the paper, they mentioned without proof that in the case of topological space, the map $X^I \rightarrow X \times X$ is open if and only if $X$ is "semi-locally path connected".

But if I assume that $X^I \rightarrow X \times X$ is an open map, and $U$ is any open set of $X$, then $U^I$ is an open of $X^I$. hence by restriction, the map $U^I \rightarrow U \times U$ is also an open map. In particular its image which is the relation "$x$ is path connected with $y$ in $U$ " is open in $U \times U$.

So if $x$ is any point of $U$ the path connected component of $x$ in $U$ is the set of $y$ path connected to $x$ in $U$, and hence is open by the previous observation. Finally, any open of $X$ has open path connected component, ie $X$ is locally path connected.

I don't exactly know what is meant by "semi-locally path connected" but I though it was weaker than "locally path connected" so there is something weird here !

My questions are :

Is there a mistake in what I just said, a mistake in the paper, or am I misunderstanding the notion "semi-locally path connected " ?
For a general locale, doesn't the openness of the map $X^I \rightarrow X \times X$ actually equivalent to the fact that $X$ is locally connected ?

Thank you !

Answer by Simon Henry for questions of localization of topos

2013-03-30T10:02:04Z

Hello !

$i^*$ is the functor which send an object $X \in T$ to $X \times F$ with the natural projection as map into $F$.

$i_*$ is a little harder to describe, if $p: Y \rightarrow F$ is an object of $T/F$, then $i_*(Y)$ is the sub-object of $[F,Y]$ (the internal hom object) which corresponds to map $f$ from $F$ to $Y$ such that $p \circ f =Id_F$ this can be express as an equaliser or with internal language as you prefers)

You also have a $i_!$ functor (left adjoint to $i^*$ ) which is just the forget functor who send $p:Y \rightarrow F$ to $Y$.

For your second question :

You can chose any generating family $B$ of $T$ (for example the image by the yoneda embeddings of a site of definition of $T$) add $F$ to this familly, the familly $B'$ you obtain seen as a full subcategory of $T$ and endowed with the canonical topology of $T$ is (by the Grothendieck comparison lemma) a site of definition of $T$, and you simply choose $U$ to be $F$.

On the descent homomorphsim of Kasparov equivariant KK theory

2013-02-08T09:52:05Z

Hello,

I have recently read about the construction of the descent map in Kasparov KK theory, which, for a group $G$ and two $G$-equivariant $C^*$ algebra $A$ and $B$ send $KK_i^G(A,B)$ to $KK_i(A \rtimes_{red} G, B \rtimes_{red} G)$.

I havn't checked all the detail, but it seems to me that the same construction should also give a map from $KK_i^G(A,B)$ to $KK_i(A \rtimes_{max} G, B \rtimes_{max} G)$ : One can also construct bimodule over $( . \rtimes_{max} G)$ from a $G$-equivariant bimodule. But as this one is never mentioned anywhere, I guess something goes wrong with it.

My question is "why this second map does not appears in literature ?"

Is the construction of the application impossible, and then what exactly goes wrong ? Or the construction is possible but this map doesn't bring more information than the classical descent homomorphism ? Or the map exists, is interesting but simply wasn't needed in the application as we have no idea of what is the $K$-theory of $A \rtimes_{max} G$ when it is different from $A \rtimes_{red} G$ ?

Note : I'm mostly interested in the case where $G$ is a discrete (countable) group.

Thank you !

Are $\infty$-topoi determined by their localic points ?

2013-01-22T11:53:01Z

Hello !

If $T$ is an infinity topos, then you can consider the infinity category of geometric morphism from $Sh_{\infty}(\mathcal{L})$ to $T$ for any locale $\mathcal{L}$. This associate to $T$ an infinity stacks over the category of all locale (at least for the etale topology, but also probably for some stronger topology).

My question is : is there anything know about this functor ? is it fully faithful ? or does it has some kind of "conservativity" properties that could allow to give an answer to the question in the tittle ? Or in the contrary is there example of non trivial infinity topos with no (or not enough) morphism from non trivial locale ?

thank you !

Answer by Simon Henry for Compactification of topological spaces

2012-11-12T17:59:03Z

Note : In order to define the alexandrov compactification of $X$ you have to take the algebra of function converging at infinity (not just those converging to $0$)

If by a compactification of $X$ you mean a compact Hausdorff topological space $Y$ equiped a continuous map $ X \rightarrow Y$ having a dense image, then the answer is true :

If you consider instead the algebra of complex function then it is immediate from the gelfand duality :

A such space is just a compact Hausdorff topological space equipped with a surjective continuous map form $\overline{X}^{sc}$. But the gelfand duality assert that compact Hausdorff topological space corresponds to commutative $C^ * $ algebra. And it is easy to see, that a continuous map between compact space is surjective if and only if the induce map between the $C^ * $ algebra is injective (and hence isometric by classical result on $C^ * $ algebra).

So, compactification of $X$ corresponds to sub-$C^*$-algebra of $C^b(X)$. ie sub-algebra of $C^b(X)$ which are closed under complexe conjugaition, closed for the normt topology, and which contain $1$.

One can then easily see that such algebra also corresponds to sub-algebra of the algbera of real bounded function, which contain $1$ and are closed for the norm topology.

NB : local compactness is actually useless for all this to work.

Non perfect Type one C^* Algebra, and a lemma in fourier analysis.

2012-10-25T14:37:22Z

Hello !

I Would like to know if the following is true :

Let $\mathcal{H}$ be the complex hilbert space $L^2([0,1])$ for the Lebesgue measure. Let $q$ be the orthogonal projection on the subspace of $\mathcal{H}$ spammed by the $(exp(\pm 2 i \pi 2^k x ))_{k \in \mathbb{N}}$.

Let $f_n$ be a sequence of elements of $\mathcal{H}$ such that :

$\Vert f_n \Vert_2 =1$
$\displaystyle \forall \epsilon >0, \lim_{n \rightarrow \infty} \int_0^\epsilon |f_n|^2 = 1$.

Then :

$\lim_{n \rightarrow \infty} \Vert q f_n \Vert_2 = 0$

In the first place I thought it will be false, but I haven't been able to found a counterexemple and in second thought It is reasonable because functions in the image of $q$ have quasi-periodicity property, and hence it shouldn't be possible to produce of sequence with the desired properties inside the image of $q$.

The motivation for the question lie in the paper of F.W. Shultz "Pure states as a dual object for C-algebras". In the end of this paper, the author give an exemple of a non perfect type one $C^ * $ algebra : the $C^* $ algebra generated in $B(\mathcal{H})$ by the compact operator and $\mathcal{C}([0,1])$ acting on $\mathcal{H}$ by multiplication. The pure state of this algebra are :

The vectors state from the action on $\mathcal{H}$
the character of $\mathcal{C}([0,1])$ extended to $C$ by sending all the compact operator to $0$.

Hence the atomic part of the enveloping algebra is $B(\mathcal{H}) \oplus l^{\infty}([0,1])$ (where $l^2$ is for the counting measure on the discret set.) and (if I'm understanding well) the author affirm without proof that $(q,0)$, is continuous on the stat space. I'm not convinced by this fact, and I would like to understand this point better, and my question is (if I'm not mistaken) equivalent to this continuity at the character $ev_0$ (the sequence $f_n$ is exactly a sequence of vector state converging weakly to the character $ev_0$, and $\Vert f_n q \Vert_2$ is the evaluation of $q$ on the state corresponding to $f_n$. )

Thanks !

$\infty$-topos and localic $\infty$-groupoids ?

2012-04-08T20:23:10Z

Hello !

It's known that every classical (Grothendieck) topos is equivalent to the topos of sheaves on a localic groupoid (a groupoid in the category of locales).

For the record, this is proved by, starting form a topos $T$, constructing a locale $L$ and a surjection $L \rightarrow T$ 'nice enough' (like a proper surjection, or an open surjection depending on the proof). Then $(L, L \times_T L, L \times_T L \times_T L)$ is a truncated simplicial locale, which can be seen as a localic groupoid. There is a canonical geometric morphism from the topos of sheaves on this groupoid to $T$, and if the surjection $L \rightarrow T$ was nice enough it's an isomorphism.

My question is : Can we hope for a similar result for $\infty$-toposes ? for example by replacing localic groupoids by localic $\infty$-groupoids (I'm not sure of how to define it in a way to be able to construct an $\infty$-topos from it...)

Thank you !

Edit (13/10/2012) :

I returned to this question a few days ago, and I have some new idea about it :

One might define a localic $\infty$-groupoid simply as a simplicial locale $\mathcal{L}_n$ (I mean a simplicial object of the category of locale). And then associate an $\infty$-topos to it simply by seeing $\mathcal{L}_n$ as a diagram of localic $\infty$-topos and taking its colimite in the $\infty$-category of $\infty$-topos.

Maybe one will need to assume some "kan complexe" hypothesis on $\mathcal{L}_n$ or maybe every simplical locale is going to be equivalent to one which have enough "kan complexe" properties. But I Think we can deduce (or at least try to ^^ ) a simpliciale model category on simplicale locale from this construction.

The important point is that I have thought about a few example, and it's seems to me that (maybe) if we consider two $\infty$-Topos $\mathcal{A}$,$\mathcal{B}$ which both come from simpliciales locales them $Hom(\mathcal{A},\mathcal{B})$ is always small. Which would show that $\infty$-Groupoide form a strict but really interesting sub-category of the category of $\infty$-topos.

unfortunately, I'm still not comfortable enough with infinity category and model category to answer those question at this time...

Answer by Simon Henry for Imaginary part of a spectrum

2012-10-12T10:39:30Z

There is probably an elementary proof, but that's an immediate consequence of the continuous functional calculus :

$ \frac{A-A^*}{2i} = f (A) $

where $f$ is the imaginary part function on $\mathbb{C}$. And when you apply a continuous function $f$ to a normal operator $A$ you have : $Spec(f(A)) = f(Spec(A))$ (you can see that by restricting to the abelian sub-$C^*$-algebra generated by $A$).

Answer by Simon Henry for Simplification in Semi-continuous real ?

2012-09-27T12:27:41Z

Ok I think I finally found an internaly valid proof by my self, so I explain it briefly here in case someone is interested some day :

If $U \in \Omega$ is a subterminal object, you can define the element $1_U \in R$ as :

$q \in 1_U$ when $q < 0 \cup (U \cap q<1)$.

(this correspond to the indicator function of an open set... )

It's easy to show that if $x+1_U \leqslant y + 1_U$ then $x \leqslant y$ :

if $q < x$, then $q < x+1_U$, hence, $\exists u,v$ such that $q=u+v, u < y$ and $v < 0 \cup (v<1 \cap U)$

First case : $v<0$ then, $q < u < y$, we have $q < y$.
Second case : we have 'U' then $1_U = 1$ so $q+1 < x+1_U$,

$q+1 < y +1$

$q < y$

In both case, $q < y$ so $x \leqslant y$.

Now, if $x$ is any bounded positive element, let $N$ be an integer such that $x < N$.

on can define :

$\displaystyle h = \frac{1}{n} \sum_{i=1}^{N.n } 1_{x> (i/n)} $

and one can show that (but it's a little longer) : $h \leqslant x \leqslant h+\frac{1}{n}$.

It's now easy to conclude if $x$ is positive and bounded and $a + x \leqslant x + b$ then if $q < a$, take $q'$ another rational such that $q < q' < a$, let $n$ an integer such that $q'-q < \frac{1}{n}$. and let $h$ be as above :

$ a + h \leqslant a +x \leqslant b+x \leqslant b + h + 1/n $

but by a simple recurence, h can be simplified so :

$ a \leqslant b+1/n $

so $q' < b+1/n$, and finaly $q < q'-1/n < b $

If $x$ is not positive, just take some $q < x$, and $x-q$ will be positive

Simplification in Semi-continuous real ?

2012-09-25T08:38:44Z

Hi !

I'm considering in a general topos $T$ the object $R$ of lower semi-continuous real (one sided lower non-empty Dedekind cuts, as for exemple in http://ncatlab.org/nlab/show/one-sided+real+number ).

I want to know if, even if substraction is not possible, there is (internally) some sort of simplification rules for addition like :

If $x$ is 'bounded' (there exist a rational q such that $x \leqslant q$ ) then $x+a=x+b$ imply $a=b$.

It's seem true to me, but only because of an argument involving a covering of $T$ by a boolean topos, if it's possible I would prefer a completely internal argument, and I can't find it.

[Edited by Andrej Bauer] A lower Dedekind cut is a subset $L \subseteq \mathbb{Q}$ which is

rounded: $q \in L \iff \exists r \in L . q < r$
inhabited $\exists q \in \mathbb{Q} . q \in L$
bounded: $\exists q \in \mathbb{Q} \forall r \in L . r < q$

Answer by Simon Henry for Characterization of Stone-Cech compactifications

2012-09-23T07:54:11Z

I confirme my comment :

$X$ is the stone-cech compactification of a discrete space if and only if $X$ is compact, haussdorf, extremally disconected, and has a dense set of open points.

here is a sketches of the proof :

If X is a stone-chech compactification of a discret set Y, then it is clear that X is compact, hausdorf, the point of Y form a dense set of open points, and it is well know that X is extremally disconected.

Asume now that $X$ is a topological space satisfying all those hypothesis.

Let $Y$ be the set of open point of $X$.

It's a routine to check to see that the map which to a subset $P$ of $Y$ associate it's closure in $X$, and the map which to a clopen of $X$ associate it's intersection with $Y$, are reciprocal bijection between the parts of $Y$ and the clopen set of $X$.

considere now a point $x \in X$, then {x} is the intersection of clopen set containing $X$, and the set of clopen of $X$ containing $x$ correspond through the previous bijection to an ultrafilter on $Y$.

After that, consider an ultrafilter $\mathcal{F}$ on $Y$, you can see that $\displaystyle \bigcap_{P \in \mathcal{F}} \overline{P} $ is a singleton (it contains a point because it is an intersection of non-empty compact, and it can't contain two point because of the properties ultrafilter).

those two application will induce an homeomorphism between $X$ and the space of ultrafilter of $Y$.

Intersection of open sublocale of a compact regular locale ?

2012-09-15T16:23:38Z

Hello !

It's well know that any sublocale of regular locale is the intersection of a familly of open sublocale. Hence if $X$ is a regular locale, the map which to a sublocal $Y \subset X$ associate $ \lbrace o \in \mathcal{O}(X), y \subset o \rbrace $ is injective. my question is : do we know it's image, at least when $X$ is compact and regular ?

If I'm not mistaken, a subset $I$ of $\mathcal{O}(X)$ correspond to a sublocale of $X$ if and only if :

$u\in I, v \geqslant u \Rightarrow v \in I $
if $\forall i, u_i \in I$ and $u = \bigcap u_i $ as sublocale, then $ u \in I$.

So I'm wondering if there is a way to make the last condition more explicit... (or equivalently to detect if $\cap u_i = \emptyset$ as sublocale ).

Thank you !

surjection of localic infinity toposes?

2012-06-14T15:07:30Z

Hello!

Is there a simple 'topological' condition to detect whenever a morphism of locales $f : X \rightarrow Y$ induces a surjection of infinity-toposes $f : \mathrm{Sh}_{\infty}(X) \rightarrow \mathrm{Sh}_{\infty}(Y)$ (i.e. such that $f^*$ is conservative)?

It's not enough to assume that f is a surjection of locales: indeed, if we take a topological space $X$ such that $\mathrm{Sh}_{\infty}(X)$ is not hypercomplete, and $X^{\mathrm{disc}}$ is its space of points endowed with the discrete topology, then $\mathrm{Sh}_{\infty}(X^{\mathrm{disc}}) \rightarrow \mathrm{Sh}_\infty (X)$ can't be a surjection, because the pullback of an $\infty$-connected map in $\mathrm{Sh}_\infty (X)$ is a weak equivalence in $\mathrm{Sh}_{\infty}(X^{\mathrm{disc}})$ ...

Thank you!

Topos Without point, from the point of view of logic.

2012-06-03T17:35:46Z

Hello !

I am a little troubled by the following "paradox" :

Let $X$ be a non trivial (Grothendieck) Topos without point.

We want to look this situation from the point of view of logic, $X$ classify some geometric theory $T$. The assumption on $X$ means that $T$ is consistent but have no model in Set. This is not in contradiction with Godel theorem because the theory $T$ might not be a "finitary first order theory".

But as we have been able to proove that $X$ doesn't have point, hence we have a proof (using boolean logic and possibly the axiome of choice) that the theory $T$ doesn't have any model.

So let now $Y$ a boolean topos with internal axiom of choice. It should be possible to apply the previous proof in the internal logic of $Y$, and then prevent $T$ to have any model in $Y$... But this is not the case : The Barr covering theorem imply that there is a such topos $Y$ that cover $X$ and hence which have a $T$-model.

Can someone explain me why this is not working ? or give an example where T and Y are explicit ?

Thank you !

Counterexemple to Urysohn's lemma in a topos without denombrable choice ?

2012-04-26T14:03:13Z

Hello !

The Urysohn's Lemma assert that in every topological spaces which is normal two closed subset may be separated by a real valued function. It's proof use axiom of countable choice (but not the law of excluded middle).

I would like to find a counterexample to this theorem in the internal logic of a topos in which the axiom of countable choice does not hold (for exemple, the topos of smooth action of some non discrete locally pro-finite group, or the topos of sheaf on [0,1].)

I need a counterexample which is compact, but If you have an example involving not a topological space but a local (an example of compact regular local which does not have enough functions with value in the Dedekind real) it's perfectly fine for me.

Thank you !

Link between internal groupoids and stacks on a topos ?

2012-04-13T11:03:12Z

Hello !

If I a have a grothendieck Site (C,J), I can consider :

The Stacks on (C,J) : category fibered in groupoid over C which statisfy suitable descent condition with respect to the covering sieve of J...
Internal Groupoid in the topos Sh(C,J).

Clearly those two notions are very close. but not exactly equivalent. I found a lot of book/article/thesis who define stacks but none of them were in term of the internal logic of the topos and it appears to me that some properties would be a lot more natural if they were stated in terms of an internal groupoid and the internal logic.

My feeling is that stacks corresponds to internal groupoids "up to weak categorical equivalences" (we invert the functor between groupoid which are internally fully faithfull and essentially surjective). But that this equivalence rely on the (external) axiom of choice. Am I right ? (and if I am, how can we make this more explicit ? )

Oh, and What about Higher Stacks ?

Thank you !

Relation between topos and $\infty$-topos

2012-03-28T13:18:19Z

Hello,

I'm currently reading The book of Jacob Lurie, 'Higher Topos Theory', and I'm a little confused by the relation between classical topos and $\infty$-topos :

to an $\infty$-topos I can attach the ordinary topos of it's $0$-truncatured object. And to a classical topos I have several way to associate $\infty$-topos 'above' it.

Jacob Lurie (in his book, section 6.4) present this relation as similar to the relation between a classical topos and it's local of sub-terminal objects. In this situation, I know I can have plenty of topos (a proprer class) that are associated to the same local, even if this local is juste a point. But i have no idea of what happen in the case of $\infty$-Topos : I have seen that in some case there might be several non equivalent $\infty$-topos above a same ordinary topos, but I see them more like "different way of doing homotopy theory in the internal logic of $X$ because some classical result of homotopy theory (like Whitehead's theorem) may fail in the internal logic" than completely different objects that just share a small property" (like the class of topoi whose local of subterminal object is reduce to a point is just the class of topos whose internal logic is two-valued)

So for example : Is there several (non equivalent) $\infty$-topoi, whose topos of $0$-truncatured object is the topos of set ? if it's the case, can I have example ? are we able to 'classify' them ?

Etale topos as a classifyng topos ?

2012-03-23T14:37:29Z

Hello !

If $X$ is a scheme, we can consider the etale topos of $X$ whose object are etale scheme above $X$ with the etale topology.

My question is : is there a know way to express this topos as the classifying topos of some geometric theory ? Of course it is possible, just because it's a grothendieck topos, but I'm looking for an explicit theory at least on some particular case (like when $X$ is affine, or when $X$ is the spectrum of the ring of integer of a number field, or when $X$ is a projective curve over a finite field... )

For example, if $A$ is a ring, then the Zariski topos of $Spec A$ (topos of finite presentation scheme above $Spec A$ with the Zariski topology) is the classifying topos of the theory of local $A$ algebra. (the universal local $A$ algebra being the structural sheaf).

Answer by Simon Henry for What is the smallest variety of algebras containing all fields?

2012-03-22T09:24:19Z

If I'm not mistaken, your answer is 'yes' : Let $M(A)$ be the set of maximal ideal of your commutative inverse ring $A$. Then you have a map :

$$A \rightarrow \prod_{\rho \in M(A) } A / \rho $$.

Each projection is surjective. the kernel of this map si the jacobson radical $R$ of $A$

so let $x$ be in $R$ then $(1-xy)$ is invertible for all in $y$, in particular for $y$ the multiplicative inverse of $x$. but as $y(1-xy) =0$, then automaticaly $y=0$, and hence $x=0$.

So the previous map is injective, and $A$ is a subdirect product.

Isomorphism class of locally trivial object classified by some $H^1$ ?

2012-03-14T23:28:33Z

Hello,

I have noticed some fact about cohomologie : when i have some kind of strucutre in a topos (for example the $G$-object for $G$ a group object in the topos) and a particular object $X$ model of this strucutre such that the sheaf of automorphism of $X$ is a sheaf of commutative group, Then there is (at least on a lot of examples) a bijection between isomorphism class of model of this structure which are locally isomorphic to $X$ and $H^1(T,G)$.

Examples I have in mind are the representation of dimension 1 of a group $G$ over a field $k$ which correspond in one hand to one dimensional $k$-vector space in the topos of $G$-set and on the other hand to the cohomology group $H^1(G-set,k^*) $. Or the principal bundle over a topological space $X$ which corresponds to some $H^1(X,G)$ too.

Is there a "general explication" to those facts ? I mean by that a result valid on an arbitrary topos who gave a bijection between a $H^1(T,G)$ and isomorphism class of objects internally isomorph.

And Is there "higher dimensional" generalization ? ( I am working on an example which seem to involve a 2-category of object inside a topos $T$ and where "equivalence class" of objects "localy equivalent" seem to be classified by some $H^2$ group in a way that i don't understand yet... )

Thank you !

Answer by Simon Henry for Dirichlet's theorem on prime density

2012-03-14T10:34:43Z

Try 'Introduction to analytic number theory' by T.Apostol (Chapter 7)

Or the Selberg article "An elementary proof of Dirichlet's theorem about primes in an arithmetic progression"

There is also several course note available on the web, just search for "Introduction to analytic number theory" or "Dirichlet's theorem" on google...

Comment by Simon Henry

2013-04-14T21:24:18Z

One problem with this construction is that : start with X being any $\Gamma$-Set, you can construct a groupoid with $A = X \times X$, this groupoid is suppose to be seen as a trivial one, hence we expect the associated stacks to be also trivial. Be if $X$ doesn't have any $\Gamma$ fixed point this is not the case (for example, $\mathcal{G}({*})$ won't have any point).

Comment by Simon Henry

2013-04-14T14:34:42Z

If you don't give more precision on what you want, it seems to me that the simplest things work. (taking $\mathbb{G}(S)$ to be the groupoid whose object are $\Gamma$-equivariant map from $S$ to $X$ and whose arrow are $\Gamma$-equivariant map from $S$ to $A$, the five strucural map being simply composition, and the functoriality being also given by composition... But this is generally not what we want to do.

Comment by Simon Henry

2013-04-04T09:48:08Z

In Ab, the object $\mathbb{Z}$ can be characterized in purely categorical term : for example it is the only projective object $Z$ such that any other projective can be written as a coproduct of a familly of copies of $Z$. I don't know if $Set[\mathbb{O}]$ can be characterized similarly in a purely categorical way. (or equivalently, does every self-equivalence of the 2-category of toposes preserve $Set[\mathbb{O}]$ up to equivalence). This question is actually equivalent to the the initial questions.

Comment by Simon Henry

2013-04-04T08:01:46Z

I mean't you need to assume that you know which object of $\mathcal{T}$ is equivalent to $Set[\mathbb{O}]$. I don't think there is an actual 'universal property' defining him...

Comment by Simon Henry

2013-04-04T07:45:37Z

Yes but you need to know $\phi(E)$ in order to compute $Geom(\phi(E),Set[\mathbb{O}])$, no ?

Comment by Simon Henry

2013-01-25T10:23:45Z

My motivation for this question is related to the relation between the geometric objects considered in topos theory/stacks theory and those considered in Non commutative geometry. In the first case (with possibly the exception of infinity topos depending of the answer to this question) every object is completely characterize by its morphism from 'commutative' object. but this is not true in non commutative geometry, for example there exist C^* alegebra of type one, which are not characterize by the stacks of their irreducible representation over all locale. (I'm working in Paris).

Comment by Simon Henry

2013-01-25T10:16:23Z

Thank you for your paper, even if i already knew that this was true for classical topos it give really interesting precision on the nature of this embeddings.

Comment by Simon Henry

2012-12-14T17:58:28Z

I'm 'interested in this kind of thing', but, i don't get what you mean... what do you mean by "weak" groupoid. what is for you a 'Morita category', the whole 2-category of C^* algebra - bimodules - map between bimodule ?

Comment by Simon Henry

2012-11-12T18:21:50Z

Yes you're right. I meant compact hausdorf everywhere. I'm just used to the French/Bourbakist convention to call "compact" a compact Hausdorff space, and quasi-compact a space which is compact but possibly non Hausdorff.

Comment by Simon Henry

2012-09-27T12:41:08Z

exactly, I'm thinking of U, q<0 and q<1 as truth value. and $1_U$ is the indicator function. Maybe I should edit to replace $\cup$ by $\vee$ and $\cap$ bu $\wedge$.

Comment by Simon Henry

2012-09-25T22:31:56Z

I confirm : this is equivalent.

Comment by Simon Henry

2012-09-25T14:52:29Z

Thank you for this reference. I edited my question to include the definition of a cut.

Comment by Simon Henry

2012-09-25T07:33:48Z

That might be a silly question but : what is the semi group structure on $\beta \mathbb{Z}$ ? as far as i know cartesian product of ultrafilter are no longer ultrafilter, so the sum of two ultrafilter has no reason to be an ultrafilter in general ?

Comment by Simon Henry

2012-09-23T13:02:49Z

actualy my idea come more from the fact that compact, hausdorf, extremally disconected space (stonean space) are exactly the stone-Cech compactification of boolean local (I think it's in Johnstone's Stone Space too, isn't it ? ). So i just had to add an hypothesis of existence of "points" for the local of clopen to be able to conclude that the starting boolean local was a discret space. But i guess those two result are closely related.

Comment by Simon Henry

2012-09-23T07:21:23Z

I think this might be correct (i would have to check it...) : X is the stone-Cech compactification of a discret space if and only if : X is compact, Hausdorf, extremally disconected (en.wikipedia.org/wiki/…) and has a dense subset open "open point" (point x such that {x} is open ). is this interesting for you ?