# Converse to Hopf degree theorem

Below, I mean smooth oriented closed connected manifolds and smooth maps (but am happy to hear about the topological category, or unoriented manifolds, etc instead).

Say that $$X^n$$ has the Hopf property if two maps $$f_0,f_1 : M^n\to X^n$$ are homotopic if and only if they have the same degree.

Say that $$X$$ has the self-Hopf property if the Hopf property holds for $$M=X$$.

The Hopf degree theorem says that $$S^n$$ has the Hopf property. It's easy to see that $$T^n$$ doesn't have the (self-)Hopf property. My question is:

If $$X$$ has the (self-)Hopf property is it homeomorphic to $$S^n$$?

• I suppose you will want both $M$ and $X$ to be connected. May 5, 2022 at 16:18

See the second half of the answer for a complete characterisation of closed orientable manifolds with the Hopf property.

Note that $$X$$ having the Hopf property is equivalent to the injectivity of $$\deg : [M, X] \to \mathbb{Z}$$ for every $$M$$; here $$[M, X]$$ denotes the free homotopy classes of maps $$M \to X$$.

If $$X$$ is simply connected, then for $$M = S^n$$ we have $$[S^n, X] = \pi_n(X)$$ - in general we would obtain a quotient of $$\pi_n(X)$$ by an action of $$\pi_1(X)$$. With respect to the group operation $$\ast$$ in $$\pi_n(X)$$, we have $$\deg([f]\ast [g]) = \deg([f]) + \deg([g])$$, so $$\deg : \pi_n(X) \to \mathbb{Z}$$ is a group homomorphism which must be injective if $$X$$ has the Hopf property.

If $$X$$ is a simply connected manifold with the Hopf property, then either $$\pi_n(X) = 0$$ or $$\pi_n(X) \cong \mathbb{Z}$$. Moreover, if $$\pi_n(X) \cong \mathbb{Z}$$, then $$X$$ is a rational homology sphere.

For the final claim, note that $$\pi_n(X) \cong \mathbb{Z}$$ means that there exists a map $$S^n \to X$$ of non-zero degree. It then follows from Poincaré duality that $$X$$ must be a rational homology sphere, see here.

Example: This observation can be used to show that many non-trivial products involving simply connected spheres do not have the Hopf property. For instance, for any simply connected closed manifold $$Y$$, the product $$S^2\times Y$$ does not have the Hopf property. To see this, let $$n = \dim Y > 1$$ and note that $$\pi_{n+2}(S^2\times Y) \cong \pi_{n+2}(S^2)\oplus\pi_{n+2}(Y) \cong F\oplus\pi_{n+2}(Y)$$ for some non-trivial finite group $$F$$.

Thanks to Nick L's observation in the comments below, we have a complete characterisation of closed orientable manifolds with the Hopf property.

$$X$$ has the Hopf property if and only if $$X$$ is homeomorphic to $$S^n$$.

Proof: If $$X$$ is homeomorphic to $$S^n$$, then it has the Hopf property by the Hopf degree theorem.

Suppose that $$X$$ is not homeomorphic to $$S^n$$. Then there is $$0 < k < n$$ such that $$\pi_k(X) \neq 0$$. Choose an essential map $$f : S^k \to X$$ and define $$F : S^k\times S^{n-k} \to X$$ by $$F(x, y) = f(x)$$. If $$i : S^k \to S^k\times S^{n-k}$$ denotes an inclusion into the first factor, then $$F\circ i = f$$, so the composition

$$\pi_k(S^k) \xrightarrow{i_*} \pi_k(S^k\times S^{n-k}) \xrightarrow{F_*} \pi_k(X)$$

is precisely $$f_* : \pi_k(S^k) \to \pi_k(X)$$ which is determined by $$[\operatorname{id}] \mapsto [f]$$. As $$f_* \neq 0$$, we see that $$F_* \neq 0$$ and hence $$F : S^k\times S^{n-k} \to X$$ is essential. Now note that $$F = f\circ\operatorname{pr}_1$$ where $$\operatorname{pr}_1 : S^k\times S^{n-k} \to S^k$$ denotes projection onto the first factor. Since $$F$$ factors through $$S^k$$ and $$k < n$$, the map $$F$$ has degree zero. As $$F$$ has degree zero but is not nullhomotopic, $$X$$ does not have the Hopf property. $$\square$$

• I think your answer can completely resolve the question. If $\pi_k \neq 0$ for $0 < k < \dim(X)$ then define a map from $S^{k} \times S^{\dim(X)-k}$ which maps $S^{k}$ non-trivially and contracts $S^{\dim(X)-k}$. Then it has degree 0 and is non-trivial, hence $\pi_k=0$ in this range. Hence by math.stackexchange.com/questions/4223840/… the manifold has to be homeomorphic to a sphere. May 5, 2022 at 19:31
• This is a very nice piece of mathematics, thank you to both of you :) May 5, 2022 at 21:19
• @NickL: Very nice. I have included an expanded version of the argument in my answer, I hope you don't mind. May 6, 2022 at 0:55
• amazing, thank you both! May 6, 2022 at 14:33

$$\mathbb{CP}^n$$ has the self-Hopf property for $$n$$ odd.

See Theorem 2.2 of Self Maps of Projective Spaces C. A. McGibbon Transactions of the American Mathematical Society Vol. 271, No. 1 (May, 1982), pp. 325-346 (22 pages) available here https://www.jstor.org/stable/1998769?seq=2

By the way it is stated in the topological category but I believe the smooth result follows see https://math.stackexchange.com/questions/1028457/smooth-homotopy. Note that each class has a smooth representative (I think it is the map given by $$z_i \mapsto z_i^{\lambda}$$ in homogeneous coordinates, $$\lambda$$ corresponding to McGibbbon's notation )

• Why $n$ odd? Surely this just follows from the identifications $$[\mathbb{CP}^n, \mathbb{CP}^n] = [\mathbb{CP}^n, \mathbb{CP}^{\infty}] = [\mathbb{CP}^n, K(\mathbb{Z}, 2)] = H^2(\mathbb{CP}^n; \mathbb{Z}) \cong \mathbb{Z}.$$ May 5, 2022 at 18:07
• Ah, I see. Via the identifications above, a map $f : \mathbb{CP}^n \to \mathbb{CP}^n$ corresponds to $k \in \mathbb{Z}$ with $f^*\alpha = k\alpha$ where $\alpha$ is a generator for $H^2(\mathbb{CP}^n; \mathbb{Z})$. The degree of $f$ is $k^n$. If $n$ is odd, the map $k \to k^n$ is injective, but this is not the case for $n$ even as $(-k)^n = k^n$. May 5, 2022 at 18:12
• Very interesting thank you! May 6, 2022 at 14:35

A manifold $$M$$ of dimension $$>1$$ with $$H^1(M) \neq 0$$ does not have the self-Hopf property. For if so, then there is a map $$f: M \to S^1$$ that induces a non-trivial homomorphism $$f_*: H_1(M) \to \mathbb{Z}$$. Compose this with a map $$g: S^1 \to M$$ carrying a non-torsion homology class to get a map $$M \to M$$ that induces a non-trivial map in homology and hence is not null-homotopic. But it factors through $$S^1$$ and so has degree $$0$$.

This was a counterexample to an attempted proof that a hyperbolic manifold with trivial isometry group has the self-Hopf property. (Hyperbolic manifolds with trivial isometry group exist, at least in dimension $$3$$.) The argument is that a map of non-zero degree must in fact have degree one (using Gromov's norm) and hence are homotopic to an isometry. (This uses Gromov's proof of Mostow rigidity; see Haagerup, Uffe; Munkholm, Hans J. Simplices of maximal volume in hyperbolic n-space. Acta Math. 147 (1981), no. 1-2, 1–11.) So by the hypothesis on the isometry group must be homotopic to the identity. But the previous paragraph says that this breaks down for degree 0.

• Very interesting thank you! May 6, 2022 at 14:34
• The same argument shows that $S^k\times Y$ does not have the self-Hopf property. If $\operatorname{pr}_1 : S^k\times Y \to S^k$ denotes projection onto the first factor and $i : S^k \to S^k\times Y$ denotes an inclusion as the first factor, then $i\circ\operatorname{pr}_1 : S^k\times Y \to S^k\times Y$ has degree zero since it factors through $S^k$, but it is not nullhomotopic as it induces a non-trivial map on $H_k$. More generally, if $X$ is $(k-1)$-connected and there is $f : X \to S^k$ with $f^*$ non-zero on $H^k$, then $X$ does not have the self-Hopf property. May 6, 2022 at 14:50