MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a compact non-orientable surface, maybe with boundary, and let $\tilde X$ be the orienting cover of $X$. If I understand correctly, any smooth automorphism of $X$ lifts naturally to an automorphism of $\tilde X$ (since $\tilde X$ can be viewed as the space of couples (a point $x$ of $X$, an orientation of $T_xX$)). Moreover, composition lifts to composition and isotopic automorphisms lift to isotopic automorphisms. So we get a map $MCG(X)\to MCG(\tilde X)$ where $MCG$ stands for the mapping class group. (In the definition of the mapping class groups we do not require that maps or isotopies should be the identity on the boundary.)

  1. Is there a simple way to describe the kernel of the above map?

  2. Is it true that the mapping class group of a of non-orientable surface injects into the mapping class group of some orientable surface (not necessarily the orienting cover)?

  3. (a slightly unrelated question which nonetheless involves mapping class groups) Is it true that the mapping class group of an orientable surface without boundary injects into the mapping class group of some orientable surface with boundary?

share|cite|improve this question

3) is an open problem. With notation as in comments to Ryan's post, if $n=1$ and $g$ is at least two, then there's no embedding, as every finite subgroup of the target is cyclic. If $n=0$, then there is a $g'$ and an injection, see Aramayona, Leininger, Souto, Injections of mapping class groups, Geometry & Topology 13 (2009) 2523--2541. They also have references to what's known.

share|cite|improve this answer
Thanks, Richard! – algori Dec 3 '09 at 13:08

Your question (1) is pretty classical -- the Birman Hilden Annals paper from 1973 essentially answers it, no? I mean, they don't deal with the non-orientable case in their paper as far as I recall but the techniques still work.

Re (3), your notion of mapping class group for a surface with boundary would typically be called the mapping class group of a closed surface with marked points, in the literature I'm familiar with, denoted something like $MCG(X,n)$ for a closed surface $X$ and $n$ marked points -- in your case they would be circle boundary components.

In that regard there are extensions ($X$ a boundaryless surface)

$\pi_1 Diff(X) \to \pi_1 C_n X \to MCG(X,n) \to MCG(X) \to 0$

$C_n(X)$ is the configuration space of $n$ points in the surface $X$.

$\pi_1 Diff(X)$ is typically trivial but there are some non-trivial cases like the torus, sphere $\mathbb RP^2$, the Klein bottle. Some of these special cases give you injectivity $MCG(X,n) \to MCG(X)$ for $n$ small, but not many.

Anyhow, it's a natural map and is frequently useful. Are you really more interested in knowing exactly when there is such an embedding or not, or are you more interested in general relations? ie: do you have a reason for wanting to know the answer to these questions?

share|cite|improve this answer
I read part 3 the other way round: it asks whether there are any embeddings of MCG(X) into MCG(Y,n) for any X, Y, n. – HJRW Dec 3 '09 at 2:27
Ryan -- thanks! Will take a look at Birman-Hilden's paper tomorrow. Re the last question: the above sequence does not seem to help much, since I'm looking for a map in the other direction, i.e. from the MCG of a surface with no boundary to the MCG of a surface with boundary. And yes, there is a reason for that: I am trying to prove something about certain representations of mapping class groups. The case when there is no boundary is more difficult, as usual, I imagine. But it would follow from the easier case when there is a boundary if there is an injection $MCG(g,0)\to MCG (g',n)$. – algori Dec 3 '09 at 2:40
In 3, he's asking whether the extension by $\pi_1$ splits. – Greg Kuperberg Dec 3 '09 at 3:11
Ah, sorry about misreading your question. The map $MCG(M,n) \to MCG(M)$ has a splitting in several natural cases -- for the torus if $n \leq 4$, and the genus 2 surface if $n \leq 6$. This is because of the Dehn twist generators in these cases commute with the hyper-elliptic involution of the surface. In general I think that map is non-split -- sometimes $MCG(M)$ has bigger torsion subgroups than $MCG(M,n)$. But in general I don't know an obstruction for (or a construction of) an embedding $MCG(g,0) \to MCG(g',n)$ for $n>0$. – Ryan Budney Dec 3 '09 at 3:27
Rereading Richard's post I see my comment really should have $n=4$ for the torus and $n=6$ for the genus 2 surface. if $n<4$ for the torus or $n<6$ for the genus two surface there may not be a splitting since the mapping class group permutes the fixed points of the hyper-elliptic involution. – Ryan Budney Dec 3 '09 at 4:00

The second of your questions is answered in

Graham Hope and Ulrike Tillmann "On the Farrell cohomology of the mapping class group of non-orientable surfaces" Proc. Amer. Math. Soc. 137 (2009), no. 1, 393--400.

as Lemma 2.1, using results of Birman and Chillingworth. It can be realised as a subgroup of the mapping class group of its orientation cover. I think the injection is induced by the "lift diffeomorphisms to the orientation cover" map, which answers the first question also.

share|cite|improve this answer
Thanks, Oscar! For surfaces without boundary it works. I don't see how to apply Birman-Chillingworth's result to surfaces with boundary, since, as far as I understand, one of the conditions in the Birman-Chillingworth paper is that the mapping class group should be the outer automorphism group of $\pi_1$. – algori Dec 3 '09 at 13:07

You should have a look at the paper which deals with nonorientable surfaces with boundary.

share|cite|improve this answer

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.