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

My question is:

For given k-dimension manifold M, Is there any way to produce a manifold N so that there exist a map $f:M\rightarrow N$, $degf=1$

The trivial method I know: The only trivial method I can image is when $M=M_1\#M_2$, then we can produce a deg=1 map from $M$ to $M_1$. Or for the torus, we can produce deg=1 map.

Topological Restrictiom I know: if $f:M\rightarrow N$, $degf=1$

(1)then every betti number of M must bigger than N.

(2)$f_*\pi_1(M)$ is onto.

edit: Another restriction is that the Gromov norm of $M$ must be at least as large as the Gromov norm of $N$. add by Zare

share|improve this question
2  
Set $N = M$ and use $f = $ identity? –  Vidit Nanda Aug 1 '12 at 8:21
3  
You can always take the sphere $N=S^n$ as a target. Take a spine $P$ of $M$ (i.e. a polyhedron such that $M\setminus P$ is an open ball $B$), send $P$ to the north pole of $S^n$ and send $B$ homeomorphically to the rest of $S^n$. –  Bruno Martelli Aug 1 '12 at 8:33
4  
Those are both examples of Siqi He's trivial construction with $M = M \# S^k = S^k \# M$. –  Douglas Zare Aug 1 '12 at 8:39
3  
Another restriction is that the Gromov norm of $M$ must be at least as large as the Gromov norm of $N$. –  Douglas Zare Aug 1 '12 at 12:28
1  
I think the intended question is "can one describe all the degree one maps from $M$ to another (closed oriented) manifold, up to homotopy equivalence ?". –  BS. Aug 1 '12 at 12:40

1 Answer 1

up vote 8 down vote accepted

In dimension 2, Edmonds proved that every degree one map between closed orientable surfaces is homotopic to a "pinch map", which takes a subsurface which is bounded by one component, and sends it to a disk. This is a special case of the connect sum method you describe, so that is really the only method in 2D.

In dimension 3, Yi Liu has proven that every closed orientable 3-manifold has a degree one map to finitely many other closed orientable 3-manifolds. I think it ought to be possible to make the proof into an algorithm. So given a closed manifold $M$, there should be an algorithm which produces finitely many manifolds $N$ such that there is a degree one map $M\to N$. It might also be possible to say something about the homotopy classes of such maps, but this hasn't been worked out. I should say also that there are many known 3-manifolds which have non-zero degree maps only to themselves or $S^3$.

Unfortunately, the proof doesn't give a prescription of how to produce such maps. There is a version of the pinching map for Seifert manifolds, by extending a pinch over the base orbifolds to the fibrations.

A related method for producing degree one maps is to observe that any knot complement has a degree one map to a solid torus which is the identity on the boundary torus. Thus, one can glue the knot complement and the solid torus to another manifold with torus boundary, and extend the degree one map as the identity over the manifold. This only works if you have a manifold with essential tori in it.

A third method allows you to specify the range, but not the domain. Take a 3-manifold $N$, and take a knot $K\subset N$ which is null-homotopic. Then one may perform a surgery on this knot with slope $\alpha$ to get a manifold $N_K(\alpha)$ which has a degree one map $N_K(\alpha)\to N$, by extending the meridian of the Dehn filling over a disk in $N$ which realizes the null-homotopy of $K$. This method is described in Section 3 of this paper.

share|improve this answer
    
Ian: All "nontrivial" constructions of degree 1 maps at the moment seem to have low-dimensional origin (I regard the "connected sum" or "map to sphere" or "product of degree 1 maps between product manifolds", constructions as "trivial"). One can ask then, for instance, if for every $n$ there are closed non-homeomorphic hyperbolic $n$-manifolds $M, N$ which admit a degree 1 map $M\to N$. (Same question if $M$ is hyperbolic and $N$ is just aspherical.) –  Misha Aug 2 '12 at 7:43
    
@Agol:Thank you for giving me so fantasitic answer! –  Siqi He Aug 2 '12 at 15:10

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.