May 23, 2006

Bunke on H, Part II

Posted by urs

Here is a transcript of the talk by U. Bunke which I mentioned in the last entry ($\to$).

The title of the talk was

Topological Stacks, T-Duality and 2-Periodic Sheaves

though, as it happens, it was really just about topological stacks and a way to define sheaves and cohomology on them.

(Similar constructions have been considered by algebraic geometers before, but not quite in the setup needed to understand twisted deRham cohomology, which is what motivates the present constructions.)

The talk was much more introductory than the version I heard before in Vienna. That doesn’t hurt.

I’ll try to reproduce a transcript of the notes that I have taken. As usual, personal comments by myself are set in italics.

1) Part I: the ordinary situation that we want to generalize

Let $V\to X$ be some real vector bundle over an oriented space $X$, which is classified by a map

(1)${h}_{V}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}X\to B\mathrm{SO}\left(n\right)\phantom{\rule{thinmathspace}{0ex}}.$

There is an element

(2)$\chi \in {H}^{n}\left(\mathrm{BSO}\left(n\right),ℤ\right)$

in the $n$th integral cohomology of $\mathrm{BSO}\left(n\right)$, such that its pullback under ${h}_{V}$ gives the Euler class $\chi \left(V\right)$ of $V$

(3)$\chi \left(V\right):=\left({h}_{V}{\right)}^{*}\chi \in {H}^{n}\left(X,ℤ\right)\phantom{\rule{thinmathspace}{0ex}}.$

But now consider what happens when $X$ is not orientable, such that the vector bundle $V$ is not classified by a map to $\mathrm{BSO}\left(n\right)$ but by a map to $\mathrm{BO}\left(n\right)$.

Applying the classifying functor $B\left(\cdot \right)$ to the sequence

(4)$1\to \mathrm{SO}\left(n\right)\to O\left(n\right)\to {ℤ}_{2}\to 1$

we obtain the 2-fold cover

(5)$\begin{array}{ccc}\mathrm{BSO}\left(n\right)& =& \mathrm{EO}\left(n\right)/\mathrm{SO}\left(n\right)\\ ↓\\ \mathrm{BO}\left(n\right)& =& \mathrm{EO}\left(n\right)/O\left(n\right)\end{array}$

(where of course $\mathrm{EG}$ denotes the universal $G$-bundle ober $\mathrm{BG}$). if we pull back our classifying map

(6)${h}_{V}:X\to \mathrm{BO}\left(n\right)$

along this double cover, we obtain a space $\stackrel{˜}{X}$

(7)$\begin{array}{ccccc}\stackrel{˜}{X}& \stackrel{\stackrel{˜}{{h}_{V}}}{\to }& \mathrm{BSO}\left(n\right)& =& \mathrm{EO}\left(n\right)/\mathrm{SO}\left(n\right)\\ f↓\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}& & ↓\\ X& \stackrel{{h}_{V}}{\to }& \mathrm{BO}\left(n\right)& =& \mathrm{EO}\left(n\right)/O\left(n\right)\end{array}\phantom{\rule{thinmathspace}{0ex}},$

which is nothing but the double cover of $X$. Notice, for later comparison, that hence the fiber of the projection

(8)$f:\stackrel{˜}{X}\to X$

(which is just a set of two points) is hence a $K\left({ℤ}_{2},0\right)$.

In any case, using the pullback we can again define the Euler class

(9)$\chi \left(V\right):=\left(\stackrel{˜}{{h}_{V}}{\right)}^{*}\chi \in {H}^{n}\left(\stackrel{˜}{X},ℤ\right)\phantom{\rule{thinmathspace}{0ex}}.$

We can regard integral cohomology ${H}^{n}\left(\stackrel{˜}{X},ℤ\right)$ as sheaf cohomology for the constant sheaf $ℤ$ pushed down to a point. In this sense we then have

(10)$\chi \left(V\right)\in {H}^{n}\left(\stackrel{˜}{X},ℤ\right)={H}^{n}\left(X,{f}_{*}{ℤ}_{\stackrel{˜}{X}}\right)\phantom{\rule{thinmathspace}{0ex}}.$

Next, the same observation in a more interesting example. Now we want to consider lifts to ${\mathrm{Spin}}^{ℂ}$-structures.

Recall that ${\mathrm{Spin}}^{ℂ}\left(n\right)$ is a group that fits into the exact sequence

(11)$1\to U\left(1\right)\to {\mathrm{Spin}}^{ℂ}\left(n\right)\stackrel{\pi }{\to }\mathrm{SO}\left(n\right)\to 1\phantom{\rule{thinmathspace}{0ex}}.$

More precisely, let $x$ be the image of a vector $\stackrel{⇀}{x}\in {ℝ}^{n}$ under the map that sends vectors to the respective Clifford algebra element associated to the standard scalar product on ${ℝ}^{n}$. Then

(12)${\mathrm{Spin}}^{ℂ}\left(n\right)=\left\{\lambda \phantom{\rule{thinmathspace}{0ex}}{x}_{1}\cdots {x}_{2k}\phantom{\rule{thinmathspace}{0ex}}\mid \phantom{\rule{thinmathspace}{0ex}}\lambda \in U\left(1\right)\phantom{\rule{thinmathspace}{0ex}},\mid x\mid =1\right\}$

is the group of even orthogonal Clifford elements together with a complex phase.

Of course we are now interested in lifting $\mathrm{SO}\left(n\right)$ structures to ${\mathrm{Spin}}^{ℂ}$-structures. Some $\mathrm{SO}\left(n\right)$-bundle $V\to X$ with classifying map ${h}_{V}$ can be lifted to a ${\mathrm{Spin}}^{ℂ}\left(n\right)$-bundle if a map $\stackrel{˜}{{h}_{V}}$ in

(13)$\begin{array}{ccc}X& \stackrel{\stackrel{˜}{{h}_{V}}}{\to }& B{\mathrm{Spin}}^{ℂ}\left(n\right)\\ =& & \phantom{\rule{thickmathspace}{0ex}}↓B\pi \\ X& \stackrel{{h}_{V}}{\to }& \mathrm{BSO}\left(n\right)\end{array}$

exists.

In general it won’t. So, as before, we may ask for the pullback space $\stackrel{˜}{X}$ such that we do have

(14)$\begin{array}{ccc}\stackrel{˜}{X}& \stackrel{\stackrel{˜}{{h}_{V}}}{\to }& B{\mathrm{Spin}}^{ℂ}\left(n\right)\\ f↓\phantom{\rule{thickmathspace}{0ex}}& & \phantom{\rule{thickmathspace}{0ex}}↓B\pi \\ X& \stackrel{{h}_{V}}{\to }& \mathrm{BSO}\left(n\right)\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

The question is: what is $\stackrel{˜}{X}$ like?

From the exact sequence which ${\mathrm{Spin}}^{ℂ}\left(n\right)$ sits in we know that the fiber of $B\pi$ is a $BU\left(1\right)\simeq K\left(ℤ,2\right)$. Hence so is the fiber of $\stackrel{˜}{X}\stackrel{f}{\to }X$. So $\stackrel{˜}{X}$ is a $K\left(ℤ,2\right)$-bundle over $X$. These are classified by maps $X\to BK\left(ℤ,2\right)\simeq K\left(ℤ,3\right)$, hence by an element in ${H}^{3}\left(X,ℤ\right)$.

For the lift under consideration, this class is the third Stiefel-Whitney class ${W}_{3}\left(V\right)$ of our vector bundle $V$.

(Here is a personal comment on this: of course, in other words, the $\mathrm{BU}\left(1\right)$-bundle $\stackrel{˜}{X}\stackrel{f}{\to }X$ is “the same” ($\to$) as the $U\left(1\right)$-gerbe that is known as the lifting gerbe obstructing the lift of our $\mathrm{SO}\left(n\right)$-bundle to a ${\mathrm{Spin}}^{ℂ}\left(n\right)$-bundle.)

Anyway, what is most important for our goal is the following observation on the cohomology of the bundles appearing here.

Similarly as in the previous example, given any (integral, say) cohomology class on $B{\mathrm{Spin}}^{ℂ}\left(n\right)$

(15)$\kappa \in {H}^{*}\left(B{\mathrm{Spin}}^{ℂ}\left(n\right)\right),ℤ\right)$

we may, if it exists, use the lift $\stackrel{˜}{{h}_{V}}$ to pull it back to $X$, obtaining a class

(16)$\kappa \left(V\right):=\left(\stackrel{˜}{{h}_{V}}{\right)}^{*}\kappa$

characterizing the vector bundle $V$.

The class $\kappa$ of interest is the one obtained as follows. Consider the map

(17)$\begin{array}{ccc}{\mathrm{Spin}}^{ℂ}& \to & U\left(1\right)\\ \lambda \phantom{\rule{thinmathspace}{0ex}}{x}_{1}\cdots {x}_{2k}& ↦& {\lambda }^{2}\end{array}\phantom{\rule{thinmathspace}{0ex}}.$

Applying the classifying functor $B$ to this yields a map

(18)$B{\mathrm{Spin}}^{ℂ}\left(n\right)\stackrel{B\mu }{\to }BU\left(1\right)\simeq K\left(ℤ,2\right)\phantom{\rule{thinmathspace}{0ex}}.$

Now, in ${H}^{*}\left(K\left(ℤ,2\right),ℤ\right)\simeq ℤ\left[{c}_{1}\right]$ lives the Chern class ${c}_{1}$ of the universal $U\left(1\right)$-bundle $EU\left(1\right)\to BU\left(1\right)$. Pulling this back along $B\mu$ yields

(19)$\kappa :=\left(B\mu {\right)}^{*}{c}_{1}\phantom{\rule{thinmathspace}{0ex}}.$

(That’s of course important for various reasons, but actually for the present purpose it does not really matter. I keep this example for the sake of completeness.)

More important for us is the case where ${h}_{V}$ does not lift and the nontrivial $\stackrel{˜}{X}\to X$ pullback bundle enters the game. In this case

(20)$\kappa \left(V\right)=\left(\stackrel{˜}{{h}_{V}}{\right)}^{*}\kappa \in {H}^{*}\left(\stackrel{˜}{X},ℤ\right)$

lives in the integral cohomology of $\stackrel{˜}{X}$.

How is that related to cohomology on $X$? In formulas, the answer is

(21)${H}^{*}\left(\stackrel{˜}{X},ℤ\right)\simeq {H}^{*}\left(X,{\mathrm{Rf}}_{*}\phantom{\rule{thinmathspace}{0ex}}{f}^{*}{ℤ}_{X}\right)\phantom{\rule{thinmathspace}{0ex}}.$

Here ${\mathrm{Rf}}_{*}$ is the right derived functor obtained from $f$, ${ℤ}_{X}$ is the constant $ℤ$-sheaf on $X$ and this is to be understood in terms of sheaf cohomology.

(Apparently this last step is some obvious, elementary fact, but, unfortunately, I don’t really know how this works in detail. Sorry. )

The point is that using this ${\mathrm{Rf}}_{*}\circ {f}^{*}$ construction we obtain a notion of twisted cohomology

(22)${\mathrm{Tw}}_{\stackrel{˜}{X}\to X}\left({ℤ}_{X}\right)\phantom{\rule{thinmathspace}{0ex}},$

where the twist is given by the bundle $\stackrel{˜}{X}\to X$, which controls the lift.

So this is a way to twist cohomology by means of a bundle. What we are after is twisting cohomology by a gerbe.

So we have diagrams of the form

(23)$\begin{array}{ccc}\stackrel{˜}{X}& \stackrel{\stackrel{˜}{{h}_{V}}}{\to }& B{\mathrm{Spin}}^{ℂ}\left(n\right)\\ f↓\phantom{\rule{thickmathspace}{0ex}}& & \phantom{\rule{thickmathspace}{0ex}}↓B\pi \\ X& \stackrel{{h}_{V}}{\to }& \mathrm{BSO}\left(n\right)\end{array}$

in ordinary (smooth/topological) spaces, which we would like to understand in the situation where all spaces are replaced by stacks and all morphisms by maps of stacks.

So in particular, in that case the diagram $\begin{array}{c}\stackrel{˜}{X}\\ ↓\\ X\end{array}$ becomes a morphism of stacks. If $X$ is the space $X$ itself regarded as a stack and $\stackrel{˜}{X}=G$ is a $U\left(1\right)$-gerbe over it - and if we can make sense of the notion of cohomology on (topological) stacks - then we should obtain a notion of cohomology twisted (not by a bundle $\stackrel{˜}{X}\to X$ as above, but) by a gerbe $G$.

This is the twisted cohomology we are after. For the correct setup it will be (non-canonically) isomorphic to the $H$-twisted deRham cohomology on $X$.

I’ll reproduce my notes on this categorification of the above setup in the next entry.

Posted at May 23, 2006 4:35 PM UTC

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