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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 13th 2011
• (edited Apr 13th 2011)

I have touched formal group a bit, but don’t have time to do anything substantial.

I need to adjust some of the terminology that I had been setting up at cohesive (infinity,1)-topos related to infinitesimal cohesion : the abstract notion currently called “$\infty$-Lie algebroid” there should be called “formal cohesive $\infty$-groupoid”. The actual L-infinity algebroids are (just) the first order formal smooth $\infty$-groupoids.

While on the train I started expanding some other entries on this point, but I need to quit now and continue after a little interruption.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeApr 13th 2011
• (edited Apr 13th 2011)

In the usual context there is both an algebroid and a formal groupoid, te latter can be first order, second order, whatever. In any case they are different beasts there. I learned a version of the latter from Tomasz Maszczyk.

For example, for infinite neighborhood of a diagonal $\Delta$, there is a sequence of embeddings $\Delta\hookrightarrow\Delta^{(\infty)}\hookrightarrow X\times X$. Now consider a groupoid $G$ with $X = G_0$, with target and source put together into one map $G\to X\times X$. If one pulls it back one gets $G^{\infty}$ over $\Delta^{(\infty)}$ equipped with a composition induced from the composition on $G$, satisfying certain associativity etc. relations. This is sort of infinitesimal groupoid. And it is not an algebroid, just an infinitesimal version of the groupoid whose structure is as close as possible, even in structure, to the original.

• CommentRowNumber3.
• CommentAuthorzskoda
• CommentTimeApr 13th 2011

I mean $G^\infty$ which I mentioned above has the appeal that it is supported at $\Delta^\infty$, just as one wants: one has composition of things which are close one to another, that is belong to a neighborhood of the diagonal.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 13th 2011
• (edited Apr 13th 2011)

Yes, that’s what I said. A Lie algebroid is a first order instance of a formal groupoid.

Do we have a special term for “finite order formal”?

• CommentRowNumber5.
• CommentAuthorzskoda
• CommentTimeApr 13th 2011
• (edited Apr 13th 2011)

Urs, maybe you said, but just for the clearing-up record, I said that there exists a notion of infinitesimal groupoid of any order. Even the first order formal tangent groupoid is a different notion from the first order tangent algebroid. The matter is not in the order but in the packing of data. For a formal groupoid one still has analogues of the source and target map, but they get factored through the infinitesimal neighborhood and do not live on the whole of $X\times X$ like the original groupoid, nor live on the strict diagonal as the tangent algebroid.

Finite order formal, well, for me it is just $n$-th order truncation or the formal groupoid supported at $n$-th neighborhood.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeApr 13th 2011

Okay, so probably this is just a matter of repackaging data. At infinity-Lie algebroid is a fairly detailed discussion of how Lie algebroids are first order formal groupoids.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 10th 2018

I added to formal group the fundamental class of examples of a 1-dimensional formal group law, given by $\mu(x, y) = f^{-1}(f(x) + f(y))$ where $f(x)$ is any power series of the form $x + a_2 x^2 + a_3 x^3 +\ldots$ in $R[ [x] ]$. (That the universal formal group law arises in this way should be in some sense the content of Lazard’s theorem, right? So the same observation should probably go there.)

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeFeb 10th 2018
• (edited Feb 10th 2018)

Thanks, Todd!!

You seem to be looking into formal power series things now. I wonder what motivates you? But it’s great.

should be in some sense the content of Lazard’s theorem, right?

Yes, looks like it should. I haven’t thought about it this way. Or if I have, I forgot.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 10th 2018

The immediate source of this is that Jim Dolan is back in the States and he and I are talking about a bunch of things. So I’m digging underneath and establishing some infrastructure to help boost my understanding in these discussions, and there’s a good chance I’ll continue touching quite a few entries in the process.

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeFeb 10th 2018
• (edited Feb 10th 2018)

back in the States

A visit? Or moved back?

• CommentRowNumber11.
• CommentAuthorTodd_Trimble
• CommentTimeFeb 10th 2018

Well, I think moved back, but I don’t know the details.

• CommentRowNumber12.
• CommentAuthorDavidRoberts
• CommentTimeFeb 10th 2018

Thanks, Todd. Pass on my regards.