# Converse of Poincaré-Hopf theorem

Let $M$ be a connected, compact, oriented manifold of dimension $n<7$. If any two maps $M \to M$ having equal degrees are homotopic, must $M$ be diffeomorphic to the $n$-sphere?

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Of course not. Consider a homotopy-sphere for example. –  Ryan Budney Jul 21 '13 at 7:01
There's infinitely many examples that aren't even homotopy spheres. Just take any manifold so that the group of homotopy-classes of homotopy self-equivalences is trivial. For example, a hyperbolic manifold with trivial isometry group. –  Ryan Budney Jul 21 '13 at 7:25
@Ryan: As I am sure you know, in dimensions less than seven and different from four, any homotopy sphere is diffeomorphic to the standard sphere; in dimension four, the smooth Poincaré conjecture is unresolved. –  Ricardo Andrade Jul 21 '13 at 7:47
It follows from Mostow rigidity. There's plenty of symmetry-less hyperbolic 3-manifolds. Every finite group is the isometry group of a hyperbolic 3-manifold, and the trivial group is the isometry group of infinitely many. –  Ryan Budney Jul 21 '13 at 10:22
@Ryan: in the hyperbolic manifold example, how about non-homotopic zero-degree maps? –  Sergei Ivanov Jul 21 '13 at 18:21

$\newcommand{\ZZ}{\mathbb{Z}}$$\newcommand{\CC}{\mathbb{C}}A simple counter-example is given by M = \CC P^3. Recall first that the cohomology ring of \CC P^3 is a truncated polynomial algebra:$$ H^\ast(\CC P^3;\ZZ) = \ZZ[x]/(x^4) $$with x \in H^2(\CC P^3;\ZZ). The set of homotopy classes of self-maps of \CC P^3 is$$ [\CC P^3,\CC P^3] = [\CC P^3,\CC P^\infty] = H^2(\CC P^3;\ZZ) = \ZZ $$where a self-map f of \CC P^3 is taken to the unique integer k\in\ZZ such that f^\ast x = kx. The first isomorphism above, [\CC P^3,\CC P^3] = [\CC P^3,\CC P^\infty], follows by cellular approximation from: • the source \CC P^3 is a CW-complex of dimension six; • the target \CC P^3 is the skeleton of dimension seven of the usual CW-structure on \CC P^\infty. Consequently, a self-map f:\CC P^3 \to \CC P^3 corresponding to the integer k \in \ZZ has degree k^3: we have f^\ast x = kx, therefore$$ f^\ast(x^3) = (f^\ast x)^3 = (kx)^3 = k^3 x^3 $$Since k\mapsto k^3 is an injective map \ZZ\to\ZZ, any two self-maps of \CC P^3 which have the same degree are homotopic. On the other hand, \CC P^3 is not diffeomorphic, or even homotopy equivalent, to S^6. Further example added later: \newcommand{\To}{\longrightarrow}$$\newcommand{\Hom}{\operatorname{Hom}}$$\newcommand{\QQ}{\mathbb{Q}}$$\newcommand{\RR}{\mathbb{R}}$Another counter-example is given by any real projective space $\RR P^n$ of odd dimension $n>1$. So we get the counter-examples $\RR P^3$ and $\RR P^5$ in dimensions less than seven.

I have failed to find any reference with a proof that two self-maps of $\RR P^n$ which have the same degree are homotopic, although this result is stated without proof as theorem 2.1 in this article (McGibbon, Self-maps of projective spaces, Transactions of the A.M.S. 271 (1), 1982, pages 325-346). For completeness, I will give a proof below. The proof is more homotopy theoretical than for the case of $\CC P^3$ above, yet also uses fairly standard techniques.

[The rest of this answer is rather long, and you may ignore it if you are not interested in the proof for $\RR P^n$.]

Proof that two self-maps of $\RR P^n$ (with $n>1$ odd) are homotopic if they have the same degree

Assume that $n$ is an odd integer with $n>1$. I will actually prove the stronger result that two basepoint-preserving self-maps of $\RR P^n$ which have the same degree are homotopic via a basepoint-preserving homotopy. More precisely, I will prove that the map which takes (the pointed homotopy class of) a self-map to its degree $$\deg : [\RR P^n,\RR P^n]_\ast \To \ZZ$$ is a bijection. Let us start with a lemma which will be useful later on.

Lemma: A map $f:\RR P^n \to \RR P^n$ (with $n>1$) has even degree if and only if $f$ induces the trivial map on fundamental groups.

Proof: This proof involves some cohomological computations similar to the ones done earlier for $\CC P^3$. Note that $H^\ast(\RR P^n;\ZZ/2) = \ZZ/2[x]/(x^{n+1})$ where $x$ is in degree $1$. Since $\pi_1(\RR P^n)=\ZZ/2$, a self-map $f$ of $\RR P^n$ induces the trivial homomorphism on $\pi_1$ if and only if it induces the trivial homomorphism on $H^1(-;\ZZ/2)$, i.e. if and only if $f^\ast x = 0$. Letting $f^\ast x = k x$ with $k\in\ZZ/2$, then $f^\ast(x^n) = k^n x^n = k x^n$, so $k = \deg f \mod 2$. The result follows. ■

A cofibre sequence involving $\RR P^n$. Our starting point is the following description of $\RR P^n$. The space $\RR P^n$ is obtained by attaching a $n$-cell to $\RR P^{n-1}$ whose attaching map is the standard covering map $\pi : S^{n-1} \to \RR P^{n-1}$. In short, $\RR P^n$ is the homotopy cofibre of the map $\pi$. The associated homotopy cofibre sequence — extended to the right by one term — is: $$S^{n-1} \overset{\pi}{\To} \RR P^{n-1} \overset{i}{\To} \RR P^n \overset{q}{\To} S^n$$ Here, the first map $\pi$ is the standard quotient map, the second map $i$ is the inclusion, and the third map $q$ collapses $\RR P^{n-1}$ to a point.

The associated exact sequence in homotopy. Applying the functor $[-,\RR P^n]_\ast$ to the above cofibre sequence produces an exact sequence: $$[S^n,\RR P^n]_\ast \overset{q^\ast}{\To} [\RR P^n,\RR P^n]_\ast \overset{i^\ast}{\To} [\RR P^{n-1},\RR P^n]_\ast \overset{\pi^\ast}{\To} [S^{n-1},\RR P^n]_\ast$$ This is an exact sequence of pointed sets. Moreover, the leftmost term is a group, and the sequence verifies a stronger equivariant exactness property at the term $[\RR P^n,\RR P^n]_\ast$ which will be explained and used later.

Calculation of the terms of the exact sequence. First, let us calculate all terms of the sequence other than $[\RR P^n,\RR P^n]_\ast$:

• we have isomorphisms $[S^n,\RR P^n]_\ast = \pi_n(\RR P^n) = \pi_n(S^n) = \ZZ$ as groups (by using the covering map $S^n \to \RR P^n$ which induces an isomorphism on higher homotopy groups);

• similarly, $[S^{n-1},\RR P^n]_\ast = \pi_{n-1}(\RR P^n) = \pi_{n-1}(S^n) = 0$;

• $[\RR P^{n-1},\RR P^n]_\ast = [\RR P^{n-1},\RR P^\infty]_\ast = \widetilde{H}^1(\RR P^{n-1};\ZZ/2) = \ZZ/2$ by using cellular approximation again.

Summarizing, the previous exact sequence of pointed sets is $$\ZZ \overset{q^\ast}{\To} [\RR P^n,\RR P^n]_\ast \overset{i^\ast}{\To} \ZZ/2 \overset{\pi^\ast}{\To} 0$$ In particular, $i^\ast$ is a surjective function.

Description of the arrow $i^\ast$ of the exact sequence. The next step is to identify the function $i^\ast$ appearing in the exact sequence. Consider the following isomorphisms characterizing the target of $i^\ast$: $$[\RR P^{n-1},\RR P^n]_\ast = \widetilde{H}^1(\RR P^{n-1};\ZZ/2) = \Hom\bigl(\pi_1(\RR P^{n-1}),\ZZ/2\bigr) = \Hom\bigl(\pi_1(\RR P^{n-1}),\pi_1(\RR P^n)\bigr)$$ Recall that the inclusion $i:\RR P^{n-1}\to \RR P^n$ induces an isomorphism on $\pi_1$. Therefore, the function $$i^\ast : [\RR P^n,\RR P^n]_\ast \To \Hom\bigl(\pi_1(\RR P^{n-1}),\pi_1(\RR P^n)\bigr) = \ZZ/2$$ takes $f:\RR P^n \to \RR P^n$ to $0\in\ZZ/2$ if and only if $f$ induces the trivial homomorphism on fundamental groups. The lemma at the beginning of this proof now implies that $i^\ast$ takes a map $f$ to its degree mod $2$, $\deg_2 f$.

In conclusion, the exact sequence becomes: $$\ZZ = [S^n,\RR P^n]_\ast \overset{q^\ast}{\To} [\RR P^n,\RR P^n]_\ast \overset{\deg_2}{\To} \ZZ/2$$ where the second arrow $\deg_2$ returns the degree mod $2$ of a self-map of $\RR P^n$. We know from before that $\deg_2 = i^\ast$ is surjective.

Equivariant exactness property of the exact sequence. At this point, we require the special equivariant exactness property of the long exact sequence in homotopy associated to a cofibre sequence. In this case, it states:

• the group $\ZZ = \pi_n(\RR P^n) = [S^n,\RR P^n]_\ast$ (with its usual group structure) acts on $[\RR P^n,\RR P^n]_\ast$;

• $f$ and $g$ map to the same element via $\deg_2 : [\RR P^n,\RR P^n]_\ast \to \ZZ/2$ if and only if there exists $\alpha\in [S^n,\RR P^n]_\ast$ such that $\alpha\cdot f = g$ in $[\RR P^n,\RR P^n]_\ast$.

The action of $\alpha \in [S^n,\RR P^n]_\ast$ takes $f \in [\RR P^n,\RR P^n]_\ast$ to: $$\alpha\cdot f : \RR P^n \overset{\sigma}{\To} S^n \vee \RR P^n \overset{\alpha+f}{\To} \RR P^n$$ where the map $\sigma$ simply "pinches a small spherical bubble off of $\RR P^n$". Note that $\deg(\alpha\cdot f) = \deg\alpha + \deg f$.

Conclusion of the proof. We are now ready to finish the proof. First we will show the injectivity of the degree function. Assume that $f$ and $g$ are pointed self-maps of $\RR P^n$ which have the same degree. Then their degrees mod $2$ also coincide and, by the exactness property above, there exists $\alpha$ such that $\alpha\cdot f = g$. But $$\deg f = \deg g = \deg(\alpha\cdot f) = \deg\alpha + \deg f$$ so that $\deg\alpha = 0$. It follows that $\alpha = 0$ in $\pi_n(\RR P^n)$, since elements in $\pi_n(\RR P^n) = \pi_n(S^n)$ are determined by their degree (here we require that $n$ is odd). Thus $g = \alpha\cdot f = f$ in $[\RR P^n,\RR P^n]_\ast$.

Finally, we show that any integer is the degree of some map $\RR P^n \to \RR P^n$. Recall that the function $\deg_2 : [\RR P^n,\RR P^n]_\ast \to \ZZ/2$ taking a map to its degree mod $2$ is surjective. In other words, there are self-maps of $\RR P^n$ with even and odd degrees. Since $\deg(\alpha\cdot f) = \deg\alpha + \deg f$, the desired conclusion follows by observing that any even integer is the degree of some map $\alpha : S^n \to \RR P^n$ — which holds because the covering map $S^n \to \RR P^n$ has degree $2$ and induces an isomorphism on $\pi_n$.

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Beautiful. ${}$ –  Mariano Suárez-Alvarez Jul 21 '13 at 22:41
To complete @RyanBudney's suggestion (and @Misha's comment), there are 3-manifolds in the snappea census which have trivial isometry group, so in particular have no non-trivial degree 1 self-maps, and which have trivial first betti number. If the map is degree $>0$, then this follows from Gromov's proof of Mostow rigidity (using the Gromov norm, the degree must $=\pm 1$, and the map is homotopic to an isometry). If a degree zero map from a 3-manifold to itself is homotopically non-trivial, then it must surject the fundamental group of some covering space. If this covering space is finite-index, then since the group is co-finitely Hopfian, this map must induce an isomorphism on fundamental group, and thus must be degree $\pm 1$, a contradiction. Thus, the image group must be infinite index. These groups have positive betti number, a contradiction.