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

I will phrase this question in terms of attaching smooth manifolds along a submanifold, though it is certainly more general.

Let $M_1$ and $M_2$ be smooth $n$-manifolds (maybe closed, for simplicity), $N$ a closed $k$-manifold, $D$ a closed $(n-k)$-disk bundle over $N$ (so that $D$ is an $n$-dimensional manifold whose boundary is the associated $(n-k-1)$-sphere bundle), and suppose we have embeddings $f_i:D\rightarrow M_i$.

Now say $\tilde{M_i}=M_i\setminus f(N)$, let $E_0$ be the punctured-open-disk bundle associated with $D$, and let $\alpha:E_0\rightarrow E_0$ be defined by sending $v_x$ in the fiber of $x\in N$ to the point $(1-|v_x|)\frac{v_x}{|v_x|}$ in the same fiber (intuitively, $\alpha$ turns $E_0$ inside-out so that we can attach the manifolds with a "collar").

Then if we form the Topological pushout of $\tilde{M_1}$ and $\tilde{M_2}$ using the smooth embeddings $f_1|_{E_0}$ and $f_2|_{E_0}\circ \alpha$, this in fact produces a pushout in the smooth category. The resulting manifold $M$ then has a tangent bundle, and in fact this tangent bundle can be formed by attaching the tangent bundles of $M_1$ and $M_2$ using the same recipe.

So, finally, here is the question: is there a formula for characteristic classes of $M$ (maybe just restrict attention to Stiefel-Whitney, Chern, Pontryagin classes) in terms of the characteristic classes of $M_1$, $M_2$, and $N$ (and the embeddings $f_1$, $f_2$)? More generally, is there a similar formula for attaching arbitrary bundles over arbitrary topological spaces (i.e. we would form a topological pushout on the base spaces and indicate how the fibers would be identified over points that are glued together)?

share|improve this question
1  
If you want a formula you're going to have to go up to twisted coefficients, I think. For example, consider constructing a Moebius band as the union of two discs, glued together along two arcs in their boundaries. But with Stiefel-Whitney classes, there's no way of comparing the extensions so I don't think there's a formula. –  Ryan Budney Feb 8 '12 at 0:09
1  
Said another way, if you only look at the Stiefel-Whitney classes of $M_1, M_2$ and $N$ and the maps between them, there's no way of telling the Moebius band apart from a cylinder, when you write it as a union of two discs. –  Ryan Budney Feb 8 '12 at 0:26
    
Hey you. Have you managed to find a formula for the SW-classes of a connected sum $M_1\sharp M_2$ in terms of those of $M_1$, $M_2$ (the case when $N$ is a point)? This might be a good place to start. –  Mark Grant Feb 8 '12 at 13:05

1 Answer 1

up vote 2 down vote accepted

It depends what you mean by "formula", since you are talking about cohomology classes in different manifolds, so at the very least you need a way to relate them, which depends on context. So the "answer" is the Mayer-Vietoris sequence. Naturality of characteristic classes, together with the fact that the characteristic classes of (the tangent bundle of) a disk bundle $D(E)->M$ can be computed using $TD=TM\oplus E$ gives you some information which you then have to put together.

For characteristic numbers, there are formulas, eg Novikov additivity gives a formula for the signature of $M$ in terms of those of $M_1, M_2$ and $D$. Similarly for the Euler characteristic. If you glue $D\times I$ to $(M_1\cup M_2)\times I$ you get a bordism from $M_1\cup M_2$ to $M$ and so bordism invariance of characteristic numbers can be helpful.

share|improve this answer

Your Answer

 
discard

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.