## August 15, 2005

### CFT, Gerbes and K-Theory in Oberwolfach, II

#### Posted by Urs Schreiber

Oberwolfach is the mixture of a Zen monastery with a mathematical bioreactor. Nice experience.

Monday’s morning sessions are just over. Hisham Sati and Jarah Evslin lectured on ‘M-theory for Mathematicians’. That’s an ambitious title, especially when the talk starts with the statement that its topic is not defined. They gave an overview of stuff that every physicists ‘knows’, in a way, but the mathematician’s questions soon revealed that concepts like ‘degrees of freedom’ or even ‘theory’ are not entirely easy to make precise.

Maybe if the talk would have been titled ‘11D SUGRA fro Mathematicians’ it might have been easier for everybody involved. This way we kind of enevitably ended with rather vague statements about ‘M-theory’ that made one mathematician from the audience conclude for himself that:

This is all very sociologic. It has nothing to do with science.

(Oh dear, now I have written something after all that might attract some traffic ;-)

Jarah prepared a discussion about the supergravity 3-form by reviewing how electric and magnetic charges in Maxwell theory behave and how the Dirac quantization condition arises from quantizing.

This standard discussion, where you demand the integral of the field-strength $F$ over any 2-cycle to be integral is really the analog for 0-gerbes = bundles of Murray’s construction of the tautological bundle gerbe from any integral 3-form $H$.

I remarked that if a similar argument is applied to the fieldstrength ${G}_{4}$ of the supergravity ${C}_{3}$, it would construct the bundle 2-gerbe that ${C}_{3}$ is ‘expected’ to be the connection of. I said that because I am eager to understand what string physics really tells us about the global nature of ${C}_{3}$. Brano Jourčo and Paolo Aschieri have particularly made the point, last year, that it should be the connection of a 2-gerbe (and that the boundary of the M2-brane hence has to couple to a nonabelian 1-gerbe), but it seems that this is argument is either not widely known or perhaps not appriciated. Clarification is needed, it seems.

After the talk I had more discussion about this point with Hisham Sati and Danny Stevenson. Seems like the right kind of people are assembled here to think about this particular question at the intersection of ‘M-theory’ and higher gauge theory.

Another meme with is beginning to spread is the search for (different notions of ) categorified vector bundles. I think I now know at least four different and probably inequivalent definitions that all seem to have their special use. (The issue here is, similar to quantization, that categorifying structure $S$ depends on which of several equivalent incarnations of $S$ you start with.) For instance the definition which Danny Stevenson and John Baez have come up with, where you look at category objects in the category of vector bundles, seems to be the right thing to finally get a crisp category-language definition of ($p$)-algebroids. I have the very strong feeling that it must be true that an algebroid, (a vector bundle $E$ over $M$ with an anchor bundle morphism to $\mathrm{TM}$ and a Lie bracket on the sections of $E$) is the same as the tangent 2-bundle of a given groupoid with the space of morphisms ‘divided out’ by the group action. All I need to figure out is how this ‘dividing out’ works technically. I bet once one has it this will be obvious in retrospect. Should be like concentrating on the ‘left invariant vector fields’ on the groupoid. If anyone knows how to to this, PLEASE let me know!

More later…

Posted at August 15, 2005 12:40 PM UTC

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## 1 Comment & 1 Trackback

### Re: CFT, Gerbes and K-Theory in Oberwolfach, II

If anyone knows how to to this,

It’s really sort of obvious, once you think about it. Even better, it is explained in books like in section 3.5 of

K. Mackenzie
General Theory of Lie Groupoids and Lie Algebroids
Cambridge Univ. press (2005).

It works like this.

Let $G$ be the morphism space of a Lie groupoid over the object space $M$. Let $G\stackrel{s,t}{\to }M$ be the source and target maps.

Let ${G}_{x}\subset G$ be the s-fibre of $G$ at $x$, which is the subspace of $G$ of all morphisms starting at $x$.

Consider any morphism

(1)$x\stackrel{g}{\to }y$

in $G$. Composing it with any morphism in ${G}_{y}$ gives morphism in ${G}_{x}$ and since everything is smooth and invertible this is a diffeomorphism

(2)${L}_{g}:{G}_{t\left(g\right)}\to {G}_{s\left(g\right)}\phantom{\rule{thinmathspace}{0ex}}.$

Let a vertical vector $V\in {T}_{x}G$ be a vector which is tangent to ${T}_{x}{G}_{x}$, i.e. such that

(3)$V\in {T}_{x}{G}_{x}\subset {T}_{x}G\phantom{\rule{thinmathspace}{0ex}}.$

Flows of fields of such vectors don’t move the source object of morphisms.

Now, a left-invariant vector field $X$ on $G$ is one which is invariant under the above diffeomorphism ${L}_{g}$ such that for any two composable morphisms

(4)$x\stackrel{g}{\to }y\stackrel{h}{\mathrm{to}}z$

we have

(5)$X\left(\mathrm{gh}\right)={\mathrm{dL}}_{g}\left[X\left(h\right)\right]\phantom{\rule{thinmathspace}{0ex}}.$

One can see that all such left-invariant vector fields on $G$ are completely specified by their values on the identity morphisms $\left\{{\mathrm{Id}}_{x}\right\}\simeq M$.

Hence one forms the bundle

(6)$\mathrm{AG}={\cup }_{x\in M}{T}_{{\mathrm{Id}}_{x}}{G}_{x}$

of all vertical tangent vectors sitting on identity morphisms over $G$. This is the bundle of our algebroid that we are looking for.

The point is that any section $X\in \Gamma \left(\mathrm{AG}\right)$ of $\mathrm{AG}$ uniquely coresponds to a left-invariant section $\stackrel{⇀}{X}$ of $\mathrm{TG}$ defined by

(7)$\stackrel{⇀}{X}\left(g\right):={\mathrm{dL}}_{g}\left[X\left(t\left(g\right)\right)\right]\phantom{\rule{thinmathspace}{0ex}}.$

So this means how to put a Lie bracket on sections of $\mathrm{AG}$: we just send any section of $\mathrm{AG}$ to its associated left-inavariant section of $\mathrm{TG}$, take the ordinary verctor field bracket there and from the resulting left-invariant vector field we keep just the elements over the identity morphism:

(8)$\left[X,Y\right]:=\left[\stackrel{⇀}{X},\stackrel{⇀}{Y}\right]{\mid }_{\mathrm{Id}}\phantom{\rule{thinmathspace}{0ex}}.$

Finally, one can check that the differential of the source map

(9)$\mathrm{AG}\to \mathrm{TG}\stackrel{\mathrm{ds}}{\to }\mathrm{TM}$

is a bundle morphism and turns $\mathrm{AG}\to M$ with the above bracket on its sections indeed into a Lie algebroid.

I also wanted to discuss an example, but now it’s already dinner time. What I next need to figure out is what the algebroid of the groupoid of thin homotopy equivalence classes of paths in some manifold looks like. It should be just isomorphic to $\mathrm{TM}$ itself, I believe.

Posted by: Urs on August 18, 2005 5:42 PM | Permalink | Reply to this
Read the post Differentiating Lie Groupoids to Lie Algebroids
Weblog: The n-Category Café
Excerpt: On an attempt to find a nice way to formulate the definition of the Lie algebroid obtained from a Lie groupoid.
Tracked: March 14, 2007 9:50 PM

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