## February 27, 2009

### Question on Geometric Function Theory

#### Posted by Urs Schreiber

I am further thinking about some issues which we discussed at the recent entry Ben-Zvi on geometric function theory (see $n$Lab: geometric function theory for some context).

From where I am coming the site over which we are looking at our generalized spaces in form of $\infty$-stacks is not an algebraic site and the $(\infty,1)$-stack QC of quasicoherent sheaves which Ben-Zvi-Francis-Nadler use so fruitfully as a model for nice geometric functions on generalized spaces is not manifestly available.

There is another construction naturally desiring to take its place, though, and I am wondering how the two perspectives would connect.

Suppose we have fixed some $(\infty,1)$-category $\hat C$ whose objects we regard as generalized spaces ($\infty$-stacks), whose morphisms we regard as cocycles and whose 2-morphisms as coboundaries, etc, so that for $X$, $A$ in $\hat C$ we express the cohomology on $X$ with coefficients in $A$ as $H(X,A) := \pi_0(\hat C(X;A)) = Ho_C(X,A)$.

A cocycle $g \in \hat C(X,A)$ is interpreted as classifying an $A$-principal bundle on $X$, whose total “space” $P \to X$ is the homotopy pullback of $pt \to A$ along $g$.

Suppose that $\hat C$ sits inside an $(\infty,2)$-category $\hat C'$ which hosts also the corresponding associated bundles (higher vector bundles) which may have non-invertible morphisms between them (which are 2-morphisms between cocycles=1-morphisms in $\hat C'$). A representation of $A$ is supposed to be some 1-morphism $\rho : A \to \infty Vect$ in $\hat C'$ with $\infty Vect$ not in $\hat C$ and the pullback along the composition of that with the cocycle $g$ gives the total space $E$ of the associated bundle.

Ordinary sections of $E$ are in $Hom_{\hat C'(X,Vec)}(pt, \rho \circ g)$ and one can see that under homotopy pullback of the point every such section canonically gives rise to a span of total spaces of the form

$\array{ && Q \\ & \swarrow && \searrow \\ \Omega_{pt} \infty Vect &&&& E }$

where on the left we have the based loop object of $\infty Vect$ at the point at which we work.

Not all spans of this form arise from ordinary sections of $E$ this way: if we allow $Q$ here to be arbitrary such a span encodes general spaces $U \to E$ over $E$ equipped with an $\Omega \infty Vect$-valued cocycle on them.

To better see where we are, suppose we look at $n=2$ where $E$ has as fibers 2-vector spaces and set $\infty-Vect := 2Vect := Ch(Vect)-Mod$ in the above, equipped with the canonical point. Then $\Omega \infty Vect = Ch(Vect)$ and an $\Omega \infty-Vect$-cocycle is a complex of vector bundles.

So in this case generalized section spans as above arrange themselves naturally into a structure $C(E)$ whose

- objects are pairs consisting of a space $U \to E$ and a complex of vector bundles $V \to U$ on $U$;

- morphisms are pairs consisting of a commuting triangle $\array{ U &&\stackrel{f}{\to}&& U' \\ & \searrow && \swarrow \\ && E }$

together with a morphism $V \to f^* v'$ of the corresponding complexes of vector bundles. For other choices of $\infty Vect$ we get accordingly other structure than complexes of vector bundles on the spaces $U$, $U'$.

We may also forget for the time being the way we obtained the space $E$ here as the total space of some associated bundle and consider this construction $C(X)$ for all objects (spaces) $X$ in $\hat C$.

Now I am coming to my question: it seems that this gadget $C(X)$ wants to play the role of of the right notion of nice geometric functions on $X$:

- whatever it is ($(\infty,1)$-category or the like), it is monoidal, thanks to the fact that it consists of maps to the (in general directed) loop space object $\Omega \infty Vect$.

- it is naturally the thing acted on via pull(-tensor-)push by bi-branes, i.e. by those spans in $\hat C'$

$\array{ && Q \\ & \swarrow && \searrow \\ E_1 &&&& E_2 }$

which in the pull-push realization of QFT are supposed to act on them.

In fact, since $C(X)$ should really be just the thing of span-morphism from $\Omega \infty Vect$ into $X$, the action of further spans on this is much like the action of the category of spans on its under-category under $\Omega \infty Vect$, if you see what I mean.

In any case, be that as it may: I am wondering how the “functions” $C(X)$ for the special case spelled out above, i.e. for $\infty Vect := Ch(Vect)-Mod$ would relate to the concept of “function” given by complexes of coherent sheaves.

From one point of view, $C(X)$ is really the fibred category associated with the stack of complexes of vector bundles on the over-category $\hat C/X$. There is a canonical morphism from that to sheaves on $X$. What is the image of that map when we are working over the algebraic site?

Posted at February 27, 2009 11:06 AM UTC

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### Homotopy Limits in SSet Cat and QuasiCat

Here is another question, a straightforward one. The answer is certainly in one of the books on my desk, but maybe you can give me the answer before I find it myself:

Suppose $C$ is a category enriched in Kan complexes, $K$ is a small category regarded as a category enriched in discrete Kan complexes and $j : K \to C$ is a $K$-diagram in $C$.

Then there is the homotopy limit over $j$ in its incarnation as a weighted $SimpSet$-limit with weight $W : K \to SSet$, $k \mapsto N(K/k)$ (as described here).

On the other hand, we can regard $C$ as a quasi-category after passing to its simplicial nerve. Then there is Joyal’s definition of limit over $j$ as the terminal object of the over-quasi-category $C_{/j}$.

I’d expect that these two notions of (homotopy) limit of $j$ coincide generally (if either exists). Is that right?

Posted by: Urs Schreiber on March 2, 2009 10:46 AM | Permalink | Reply to this

### Re: Homotopy Limits in SSet Cat and QuasiCat

Yes; see section 4.2.4 in Lurie’s book.

Posted by: Mike Shulman on March 2, 2009 7:56 PM | Permalink | Reply to this

### Re: Homotopy Limits in SSet Cat and QuasiCat

Yes; see section 4.2.4 in Lurie’s book.

Ah, right. Thanks.

Have now added this to $n$Lab: limit in quasi-categories.

Posted by: Urs Schreiber on March 2, 2009 9:05 PM | Permalink | Reply to this

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