Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If you have two manifolds $M^m$ and $N^n$, how does one / can one decompose the diffeomorphisms $\text{Diff}(M\times N)$ in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$? Is there anything we can say about the structure of this group? I have looked in some of my textbooks, but I haven't found any actual discussion of the manifolds $\text{Diff}(M)$ of a manifold $M$ other than to say they are "poorly understood."

Can anyone point me to a source that discusses the manifold $\text{Diff}(M)$? My background is in physics, and understanding the structure of these kinds of groups is important for some of the the things we do, but I haven't seen any discussion of this in my differential geometry textbooks.

share|improve this question
As a general comment, there's no good reason to expect that the structure of Aut(A x B) is determined by the structure of Aut(A) and Aut(B) in an arbitrary category, where x denotes the product. The product, by definition, only lets you decompose functions into it; it doesn't say anything about how to decompose functions out of it. (This is essentially the same reason that while the maps A -> 1 are trivial by definition, where 1 is the terminal object, the maps 1 -> A can be interesting.) –  Qiaochu Yuan Mar 13 '10 at 6:57
(Even in the special case where x is a biproduct, one still has to consider Hom(A, B) and Hom(B, A).) –  Qiaochu Yuan Mar 13 '10 at 7:18
From Ryan's comment you can see that just understanding the homotopy type of such a space is not well understood. –  Sean Tilson Mar 29 '10 at 4:46
To expand on Qiaochu's comment: in the category of groups, let $A1=B1$ be trivial groups , and let $A2=B2$ be groups of order $2$ . All four of these groups have trivial automorphism groups, but $\operatorname{Aut}(A_1 \times B_1)$ is trivial whereas $\operatorname{Aut}(A_2 \times B_2) \cong \operatorname{GL}_2(\mathbb{F}_2) \cong S_3$. –  Pete L. Clark Aug 3 '10 at 1:50

3 Answers 3

up vote 27 down vote accepted

The homotopy-type of the group of diffeomorphisms of a manifold are fairly well understood in dimensions $1$, $2$ and $3$. For a sketch of what's known see Hatcher's "Linearization in three-dimensional topology," in: Proc. Int. Congress of. Math., Helsinki, Vol. I (1978), pp. 463-468.

Similarly, the finite subgroups of $Diff(M)$ are well understood in dimensions $3$ and lower. Hatcher's paper is a good reference for that as well, when combined with a few semi-recent theorems.

If you're interested in general subgroups of $Diff(M)$, there's still a fair bit of discussion going on just for subgroups of $Diff(S^1)$, as it contains a pretty rich collection of subgroups.

In high dimensions there's not much known. For example, nobody knows if $Diff(S^4)$ has any more than two path-components. See for example this little blurb. Some of the rational homotopy groups of $Diff(S^n)$ are known for $n$ large enough.

I wrote a survey on what's known about the spaces $Diff(S^n)$, and spaces of smooth embeddings of one sphere in another $Emb(S^j,S^n)$ a few years ago, here.

Getting back to your earlier question, groups of diffeomorphisms of connect-sums can be pretty compicated objects. In dimension $2$ it's already interesting. For example, $Diff(S^1 \times S^1)$ has the homotopy-type of $S^1 \times S^1 \times GL_2(\mathbb Z)$. Diff of a connect-sum of $g$ copies of $S^1 \times S^1$ has the homotopy-type of a discrete group provided $g>1$, this is called the mapping class group of a surface of genus $g$. It's a pretty complicated and heavily-studied object. In the genus $g=2$ case this group is fairly similar to the braid group on $6$ strands.

In dimension $3$, it's an old theorem of Hatcher's that $Diff(S^1 \times S^2)$ doesn't have the homotopy-type of a finite-dimensional CW-complex, as it has the homotopy-type of $O_2 \times O_3 \times \Omega SO_3$. I've been spending a lot of time recently, studying the homotopy-type of $Diff(M)$ when $M$ is the complement of a knot in $S^3$, and knot complements in general. The paper of mine I linked to goes into some detail on this.

From the perspective of differential geometry, the homotopy-type of $Diff(S^n)$ is rather interesting as it's closely related to the homotopy-type of the space of "round Riemann metrics" on $S^n$. This is a classic construction, is outlined in my paper but it goes like this: $Diff(S^n)$ has the homotopy type of a product $O_{n+1} \times Diff(D^n)$ where the diffeomorphisms of $D^n$ are required to be the identity on the boundary -- this is a local linearization argument. $Diff(D^n)$ has the homotopy-type of the space of round metrics on $S^n$. The idea is that any two round metrics are related by a diffeomorphism of $S^n$. So $Diff(S^n)$ acts transitively on the space of round metrics (with a fixed volume, say), and the stabilizer of a round metric is $O_{n+1}$ basically by the definition of a round metrics. Kind of silly but fundamental.

share|improve this answer

When $N$ is the circle there's a sort of answer. In fact there's a whole chapter in the book

Burghelea, Dan; Lashof, Richard; Rothenberg, Melvin: Groups of automorphisms of manifolds. With an appendix ("The topological category'') by E. Pedersen. Lecture Notes in Mathematics, Vol. 473. Springer-Verlag, Berlin-New York, 1975.

dedicated to showing that after looping once, the space $\text{Diff}(M\times S^1)$ splits up to homotopy as (the loops of) $$ \text{Diff}(M\times I) \times B\text{Diff}(M\times I) \times \eta(M) , $$ where the middle term is a non-connective one-fold delooping of $\text{Diff}(M\times I)$ and $\eta(M)$ is the mysterious "nil-term" (when writing $\text{Diff}(W)$ for a manifold $W$ with boundary, the convention is that the diffeomorphisms are to preserve the boundary pointwise). In particular, once gets a decomposition on the level of homotopy groups. (This theorem is an analog of the Bass-Heller-Swan type result which says $K(R[t]) \simeq K(R) \times BK(R) \times \eta(R)$.)

One can say something about the homotopy type the nil-term in the concordance stable range, roughly, $\dim M/3$.

Furthermore, $\text{Diff}(M\times I)$ sits in a fibration sequence $$ \text{Diff}(M\times I) \to C(M) \to \text{Diff}(M) $$ where $C(M)$ is the topological group of concordances of $M$. After inverting 2, this sequence is homotopically trivial and $\pi_k(\text{Diff}(M\times I))$ can be identified with the invariant part of the $\Bbb Z_2$-action on $\pi_k(C(M))$ induced by conjugating a concordance with the diffeomorphism which turns $M\times I$ upside down ($(x,t) \mapsto (x,1-t)$). Lastly, $\pi_k(C(M))$ can be studied via algebraic $K$-theory methods when $k$ is within the concordance stable range.

share|improve this answer
Hi John. This result about the fibration being trivial in the stable range after inverting 2, what's the reference for that? Is that Igusa? –  Ryan Budney Jan 24 '11 at 2:10
No, It's not Igusa. I do remember Kiyoshi having attributed it to Hatcher. It might be in Hatcher's paper from the 1976 Stanford conference. I realize now that might also have to loop the fibration once to get the correct statement. –  John Klein Jan 24 '11 at 4:44

I'm not an expert, but my impression was that you can't reasonably expect anything like a decomposition in general. Here is a big list of references on automorphisms of manifolds, compiled by Andre Henriques. The wikipedia page has a brief discussion of diffeomorphism groups, and it requires somewhat less background.

share|improve this answer
Your 1st link is broken. –  Ryan Budney Mar 13 '10 at 6:52
How odd. I just tried it on 3 different browsers and 2 different computers and it worked each time. I'll try to put it in a comment. web.archive.org/web/20070208084859/http%3A//… –  S. Carnahan Mar 13 '10 at 7:21
I get a "403 Forbidden" error message. It appears I'm banned from that webpage. Strange. –  Ryan Budney Mar 13 '10 at 7:23
Perhaps the administrators of archive.org are hockey fans... –  S. Carnahan Mar 13 '10 at 7:53
André has it at his new home: staff.science.uu.nl/~henri105/talbotrefs.html –  Ben Wieland May 17 '10 at 21:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.