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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeNov 3rd 2010
• (edited Nov 3rd 2010)

Domenico, Jim and myself have been trying to write out in LaTeX some of the things that we have been discussing at infinity-Chern-Weil theory.

There is now a pdf document:

Domenico, Urs, Jim, Cocycles for differential characteristic classes (schreiber)

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeNov 3rd 2010
• (edited Nov 3rd 2010)

Typo: Chern-Smimons, page 53 as of the moment. And mathrm without backslash in 4.2.12. Kan compex in 4.2.19.

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeNov 3rd 2010

There is an interesting thing with Poisson-Lie structure relevant for physics. A Lie group is Poisson-Lie if it is a Poisson manifold and the multiplication is a Poisson map. At infinitesimal level there is a notion of Lie bialgebra, the tangent bialgebra to a Poisson-Lie group is a Lie bialgebra. Now the Lie bialgebra here can be considered as a Lie algebra with a additional cocycle with values in the tensor square. Now, conversely, if we want to integrate a Lie bialgebra, then we can choose a connected, simply connected Lie group integrating a Lie algebra and the cocycle with integrate as well. But suppose that we start with a not-simply connected Lie group with the cocycle on the Lie algebra. Then one can not do this in general. I am a bit uneasy with this, and think that some sort of higher categorical repair could do something about the questionable integration of the cocycle, if I understand it in a bit generalized way.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeNov 3rd 2010

Typo: Chern-Smimons, page 53 as of the moment. And mathrm without backslash in 4.2.12.

Thanks, Zoran! I fixed these in our local version for the moment.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeNov 3rd 2010

Now the Lie bialgebra here can be considered as a Lie algebra with a additional cocycle with values in the tensor square.

That’s a nice fact. Could you record that in some nLab entry?

But suppose that we start with a not-simply connected Lie group with the cocycle on the Lie algebra. Then one can not do this in general. I am a bit uneasy with this, and think that some sort of higher categorical repair could do something about the questionable integration of the cocycle, if I understand it in a bit generalized way.

good question. That’s a special case of a general question that I don’t have a general good answer for. The Lie integration of $L_\infty$-algebra cocycles lives naturally on some higher connected covers. The general question is: to which extent does that push down again to one of the quotients? I need to think more about this.

• CommentRowNumber6.
• CommentAuthorzskoda
• CommentTimeNov 3rd 2010
• (edited Nov 3rd 2010)

Probably I would not be of much help, but I would like to know how to possibly start thinking about this. I see a question and have some intuition about the answer, but not a cue of the strategy…

Yes, I will put the quoted fact in some entry. Today or soon.

I added above: typo Kan compex in 4.2.19.

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 3rd 2010
• (edited Nov 3rd 2010)

typo Kan compex in 4.2.19.

Thanks, have fixed that in the local copy.

I might need a more quiet moment to recollect my thoughts on this, but a first observaiton is this:

we are asking for the reverse process of forming fractional characteristic maps: traditionally we start with a cocycle $\mathbf{B}G \to \mathbf{B}^n (\mathbb{Z} \to \mathbb{R})$ and find that after refining to a higher cover $\mathbf{B}\hat G$ it can be quotiented

$\array{ \mathbf{B}\hat G &\stackrel{\frac{1}{k} c}{\to}& \mathbf{B}^n (\mathbb{Z} \to \mathbb{R}) \\ \downarrow && \downarrow^{\cdot k} \\ \mathbf{B}G &\stackrel{c}{\to}& \mathbf{B}^n (\mathbb{Z} \to \mathbb{R}) } \,.$

$\array{ \mathbf{B}\hat G &\stackrel{c}{\to}& \mathbf{B}^n (\mathbb{Z} \to \mathbb{R}) \\ \downarrow \\ \mathbf{B}G }$

(or similar) and ask if we can complete it.

If we put ourselves in the context of an $\infty$-topos, then of course we have the universal completion by the pushout. So the question is: can we usefully project out from the pushout an abelian class

$\mathbf{B}G \coprod_{\mathbf{B}\hat G} \mathbf{B}^n (\mathbb{Z} \to \mathbb{R}) \to A \,.$
• CommentRowNumber8.
• CommentAuthorzskoda
• CommentTimeNov 3rd 2010

Thanks, this is a useful point of view. Thanks for writing it down clearly.

I added few references into Pontrjagin class.