Timeline for What is the 2-category whose 0-objects are Lie algebroids?
Current License: CC BY-SA 2.5
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 6, 2010 at 2:07 | vote | accept | Theo Johnson-Freyd | ||
| Aug 3, 2010 at 20:52 | comment | added | Theo Johnson-Freyd | (Above, "pr" should be "pt".) | |
| Aug 3, 2010 at 20:51 | comment | added | Theo Johnson-Freyd |
(continuation) if $X$ is any space, then the pair groupoid $X\times X \rightrightarrows X$ is equivalent in the bicategory of groupoids to $\{\rm pt\}$, whereas the tangent algebroid to $X\times X\rightrightarrows X$ is the tangent bundle ${\rm T}X \to X$. The (co)homology of this algebroid if the de Rham (co)homology of $X$, which is not equivalent to the de Rham (co)homology of $\{\rm pr\}$ unless $X$ is contractible. Since n-functors should take n-equivalences to n-equivalences, there is no chance that all the things I thought were functors are.
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| Aug 3, 2010 at 20:49 | comment | added | Theo Johnson-Freyd | This is great, and ultimately I think the correct answer. I realized that I had in my mind two or three functors, and not all of them could possibly be functors. Namely, I had imagined that there was (1) a functor from the bicategory of Lie groupoids to some bicategory of Lie algebroids that on objects took a groupoid to its tangent algebroid; (2) a functor from the bicategory of Lie algebroids to differential chain complexes that on objects took a Lie algebroid to its BRST complex; (3) had a (n?) functor from chain complexes to graded vector spaces taking homology. But (continued) | |
| Aug 2, 2010 at 19:43 | comment | added | Chris Schommer-Pries | Right. Let's see, the pair groupoid of M gets associated to the Lie algebroid over M given by TM with anchor map the identity. Since the pair groupoid is equivalent to the point, this Lie algebroid is equivalent to the zero Lie algebroid. This implies, for example, that if our cover is a trivial bundle $M \times X \to X$, and we pull-back the Lie algebroid A over X to $X \times M$, then we get $A \oplus TM$ with the obvious anchor map to $TX \oplus TM$. These will also be equivalent. In fact I think the formula $A \times_{TX} TU$ holds generally for any surjective submersion $U \to X$. | |
| Aug 2, 2010 at 15:45 | comment | added | Konrad Waldorf | Thanks for the verification! And the Cech groupoid of a general surjective submersion? For example the one of a principal $G$-bundle $pi:P\to X$. Isn't the Lie algebroid here the vertical tangent space? What I want to say is that you probably get trivial effects when you have a covering with discrete fibres, but can have non-trivial ones when the fibres have dimensions greater than zero. | |
| Aug 2, 2010 at 12:19 | comment | added | Chris Schommer-Pries | Konrad, you're right! That was a mistake. I corrected it to "trivial" Lie algebroid since in that example you get the zero Lie algebroid. Here is a link if people need a reminder about the construction: en.wikipedia.org/wiki/… . I don't think this effects the rest of the answer significantly. The "pull-back" Lie algebroid is $A \times_{TX} TU$, where we use the map $df:TU \to TX$. This is still a vector bundle since $U \to X$ is a surjective submersion. | |
| Aug 2, 2010 at 12:09 | history | edited | Chris Schommer-Pries | CC BY-SA 2.5 |
corrected mistake.
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| Aug 2, 2010 at 6:48 | comment | added | Konrad Waldorf | Chris, I am admittedly very slow in understanding your answer! Can you explain why the Lie algebroid of the Cech of the cover U is TU? I just see that TU receives an injective map from that Lie algebroid. | |
| Aug 1, 2010 at 13:39 | history | edited | Chris Schommer-Pries | CC BY-SA 2.5 |
Major changes. A full answer to the question added.
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| Jul 31, 2010 at 19:15 | history | answered | Chris Schommer-Pries | CC BY-SA 2.5 |