# Degrees of self-maps of aspherical manifolds

In "Infinitesimal computations in Topology", Publ IHES, page 318, Dennis Sullivan writes "Recall any self-mapping of a Riemann surface of genus $g>1$ either has degree $0$ or degree $\pm 1$." There is a footnote to that sentence, saying this statement is more generally "[t]rue for a closed $K(\pi,1)$-manifold of non-zero Euler characteristic." Why is that true (even the surface case is a mystery to me)?

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en.wikipedia.org/wiki/Riemann–Hurwitz_formula should be useful, however, as in Richard Kent's answer, it must be showed that a map of degree bigger than $1$ is homotopic to a covering map. – rpotrie Oct 14 '10 at 16:44
@Rpotrie: Due to A. Edmonds, every surface map can be homotopically decomposited to pinchings and branch coverings. And pinchings do not change the mapping degree. So the Hurwitz formula of branch coverings is sufficient to show the degree cannot be larger than 1. – X.M. Du Oct 16 '10 at 10:28

That is because surfaces of higher genus have a non-zero simplicial volume. It is a general fact that all selfmaps of closed manifolds with non-zero simplicial volume have degree either $-1,0$ or $+1$. (see for example here)

I do not know about the other cases which are mentioned in the footnote.

EDIT: As Bruno Martelli pointed out in a comment, Gromov conjectured that closed aspherical manifolds of non-zero Euler characteristic have non-zero simplicial volume. Hence, the same argument is likely to apply in many cases which are interesting.

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Thanks, that answers the question for surfaces and for hyperbolic manifolds as well. – Johannes Ebert Oct 14 '10 at 16:44
I think there is an open conjecture of Gromov which says that every aspherical manifold with non-zero Euler characteristic should have a strictly positive simplicial volume. If the conjecture is true, then Andreas' argument extends to every aspherical manifold. – Bruno Martelli Oct 14 '10 at 19:25

Here's an argument that the map on fundamental groups $\phi:M\to M$ is surjective if $deg(\phi)\neq 0$, where $M$ is an $n$-dimensional closed orientable manifold and $\chi(M)\neq 0$. Suppose $deg(\phi)\neq 0$, then there exists a finite-sheeted cover $\tilde{M}\to M$ such that $\phi_{\#}(\pi_1(M))=\pi_1(\tilde{M})$ (as noted by Richard Kent, if $\tilde{M}\to M$ were an infinite cover, then $deg(\phi)=0$). Consider the lift $\tilde{\phi}:M\to\tilde{M}$. Then $deg(\tilde{\phi})\neq 0$ as well. Then $\tilde{\phi}^\ast:H^n(\tilde{M},\mathbb{Q})\to H^n(M,\mathbb{Q})$ is an isomorphism of vector spaces. By Poincare duality, for any $\alpha\in H^k(\tilde{M},\mathbb{Q})$ there exists $\beta\in H^{n-k}(\tilde{M},\mathbb{Q})$ such that $\alpha\cup\beta = [\tilde{M}]$. Then $\tilde{\phi}^\ast(\alpha\cup\beta)=\tilde{\phi}^\ast(\alpha)\cup\tilde{\phi}^{\ast}(\beta) = \tilde{\phi}^\ast[\tilde{M}]\neq 0$, so $\tilde{\phi}^\ast(\alpha)\neq 0$. Thus $\tilde{\phi}^\ast$ is an injection from $H^\ast(\tilde{M},\mathbb{Q})\hookrightarrow H^\ast(M,\mathbb{Q})$. But the covering projection $\tilde{M}\to M$ induces an injection $H^\ast(M,\mathbb{Q})\hookrightarrow H^\ast(\tilde{M},\mathbb{Q})$, so we see that $H^\ast(\tilde{M},\mathbb{Q})\cong H^\ast(M,\mathbb{Q})$ (as graded vector spaces), and therefore $\chi(\tilde{M})=\chi(M)$. Thus the cover $\tilde{M}\to M$ is degree one, and we see that $\phi_{\#}:\pi_1(M)\to \pi_1(M)$ is a surjection.

If $\pi_1(M)$ is Hopfian, then $\phi_{\#}$ is an isomorphism, and we conclude that $\phi$ is a homotopy equivalence when $M$ is a $K(\pi,1)$, and therefore $deg(\phi)=\pm 1$. However, $\pi_1(M)$ might not be Hopfian. Assume $n\geq 4$ (since $n=2$ is Hopfian, and is taken care of in Richard Kent's answer) and $M$ is aspherical. Then $Ker(\phi_{\#})$ is finitely normally generated (since $\pi_1(M)$ is finitely presented). Choose a link $L\subset M$ such that $Ker(\phi_{\#})$ is normally generated by $\pi_1$ of the components of $L$. We surger $M$ by adding 2-handles along the components of $L$ to get $M'$ such that $\pi_1(M')=\pi_1(M)$ ($M'$ might not be aspherical). But we may extend the map $\phi_{|M-\mathcal{N}(L)}:M-\mathcal{N}(L)\to M$ to a map $\phi': M' \to M$ by mapping the attached 2-handles into $M$, which is possible since each component of $L$ maps to a contractible loop in $M$. Since the cores of the 2-handles are codimension $\geq 2$, we see that $deg(\phi')=deg(\phi)$. But since $\phi'_{\#}:\pi_1(M')\to \pi_1(M)$ is an isomorphism, $\phi$ is homotopic to the classifying map, there's a gap here so we conclude that $deg(\phi')=\pm 1$, so $deg(\phi)=\pm 1$.

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Great answer! Could you please expend a bit the last two lines? Why "$\phi'$ is homotopic to the classifying map" implies $deg(\phi')=\pm 1$? Also, why $Ker(\phi_{\#})$ is finitely normally generated? – Dmitri Oct 15 '10 at 9:13
@ Dmitri: since $\pi_1(M)$ is finitely presented, we only have to kill finitely many elements of $ker(\phi)$ before we get a presentation for $\phi(\pi_1(M))\cong \pi_1(M)$ (just enough to get one of each relation for $\pi_1(M)$). As for your second question, this points out a gap in my argument. The classifying map of a manifold to an aspherical manifold with the same fundamental group need not be degree one. So I don't know how to complete the argument. – Ian Agol Oct 15 '10 at 17:30
I found a paper which for $n=4$ implies that the classifying map is degree one if and only if an invariant of the Postnikov tower $k_3(M)$ vanishes. ams.org/mathscinet-getitem?mr=1809909 The paper doesn't give any examples where $k_3(M)$ is non-vanishing. In any case, maybe surgery is not the right approach. – Ian Agol Oct 15 '10 at 18:00

For $K(\pi, 1)$'s, the answer is:

Because Euler characteristic is multiplicative under covering spaces.

Edit: As pointed out in the comments, I was assuming the map was $\pi_1$--injective.

Here's an elementary proof of what you want for surfaces:

The fundamental group of $M$ has rank $2g$, where $g$ is the genus of $M$. So the image $H$ of $\pi_1(M)$ has rank at most $2g$.

If the index of $H$ in $\pi_1(M)$ is infinite, then $f$ has degree zero, as it lifts to a map to a noncompact surface.

So we can assume that $H$ has finite index $m$ in $\pi_1(M)$.

By multiplicativity of Euler characteristic under covers, $H$ is the fundamental group of a surface of genus gm - m + 1, which has rank $2gm - 2m +2$. This is a contradiction unless $m = 1$, in which case the map is surjective on the fundamental group. Since surface groups are hopfian (since they are residually finite), $f$ is injective on the fundamental group, and so $f$ is a homotopy equivalence. So it has degree $\pm 1$.

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Why is every self-map of non-zero degree homotopic to a covering? One needs that it is injective on $\pi_1$. Is that automatic? – Andreas Thom Oct 14 '10 at 16:38
Richard, could you please explain how you know that such a map would homotopic to a covering, if it is of non-zero degree? – Dmitri Oct 14 '10 at 16:39
Oops. I actually wasn't thinking about showing that it's homotopic to a covering, but I think I was assuming it was $\pi_1$--injective, and probably secretly thinking I was working with a surface. In that case, if the degree is nonzero and not $\pm 1$, then you lift the map to a finite cover, and get a homotopy equivalence between spaces with different Euler characteristics. Sorry, there is quite a bit more to be said. – Richard Kent Oct 14 '10 at 16:47
If the induced map $\pi_1 (f)$ on the fundamental group is injective, then the map $B\pi_1 (f): B \pi_1 (M) \to B\pi_1 (M)$ is homotopy equivalent to $f$ and it is a covering (up to homotopy); this is enough to deduce the multiplicativity of the Euler characteristic. – Johannes Ebert Oct 14 '10 at 16:56
@Johannes Also, in your argument, ker(f) has infinite index, so you can't use multiplicity of Euler characteristic. – Richard Kent Oct 14 '10 at 21:28