Mapping Class Group of Annulus

584 Views Asked by At

I studied the construction of mapping class group for an annulus, primarily from "A primer of Mapping class group" (Page 51-52 https://www.maths.ed.ac.uk/~v1ranick/papers/farbmarg.pdf). However, I need more details to make sense what explained there, and I found another nice similar construction on the following document https://www3.nd.edu/~andyp/teaching/2014SpringMath541/TorelliBook.pdf on page 6.

The idea is to construct a group isomorphism form $$\mbox{Mod}(\mathbb{A}) \rightarrow \mathbb{Z}.$$ The constrution using universal covering spaces and lifting a map. The construction and surjectivity of the group homomorphism can be understandable. However, the injectivity is challenging. Both sources mentioned deck transformation, and comutativity of deck transformation and homotopy which allow projecting homotopy on covering space to the annulus $\mathbb{A}.$

To be specific, I cannot understand the discussion of injectivity mentioned on both sources (For the primer book, last paragraph of page 51, for the 2nd document, the injectivity paragraph on the bottom page 6). Moreover, it is hard to find an algrbraic topology book doing the relation between deck transformation and homotopy.

The construction is quite long and detialed, so I apologize for not typing it all here, but refers to the reference instead (I prefer notation and construction explain in the 2nd one https://www3.nd.edu/~andyp/teaching/2014SpringMath541/TorelliBook.pdf).

With notation used in the 2nd document (page 6), can anyone help pointing out why "the straight line homotopy from $$\tilde{H} \ \mbox{ to id}$$ commutes with deck group" and why "this fact allows projecting of the isotopy on the above covering space, the infinite strip $\tilde{\mathbb{A}} = \mathbb{R} \times [0,1],$ down to the lower space annulus $\mathbb{A}$" ?

Thank you very much.

Also, if there is a good reference of algebraic topology books dealing with deck transformation and lifting/projecting homotopy/isotopy, please leave the list in the comment.

1

There are 1 best solutions below

0
On

This answer will use notation from Andy Putman's book, your second source.

I'm not sure what you mean by "the relation between deck transformations and homotopy" if you cannot find it in Algebraic Topology textbooks. In particular, Hatcher has all the facts I use. The idea is that deck transformations are automorphisms of the covering map, hence should be compatible (in a suitable sense) with any definitions involving the cover.

More explicitly, $\tilde H$ commutes with the deck transformation $z \mapsto z+n$, since $\tilde H(z) + n$ and $\tilde H(z+n)$ are both lifts of $H$ that map $0$ to $n$. Therefore $\tilde H_t$ commutes with $z \mapsto z+n$ since $$\tilde H_t(z+n) = (1-t) \tilde H(z+n) + t (z+n) = (1-t)(\tilde H(z) + n) + t(z+n) =\tilde H_t(z) + n.$$

But this means that $p (\tilde H_t(z))$ depends only on $p(z)$ and not its lift $z$, so $H_t(p(z)) = p (\tilde H_t(z))$ is well-defined. It should be straightforward to check that this is a homotopy from $H$ to $\mathrm{id}$.