Question from J. Milnor paper from 1968 about diffeomorphic manifolds

Solution 1:

In

J. Milnor, On the 3-dimensional Brieskorn manifolds, in “Knots, groups and 3- manifolds”, Ann. Math. Studies, vol. 84, Princeton Univ. Press 1975, pp. 175-224.

(free version here), Milnor kind of answers his own question. That is, he shows that if you change the powers $2,3,6$ to any powers $p,q,r$ with $p,q,r\geq 2$, then the resulting $3$-manifold $M_{p,q,r}$ is diffeomorphic to homogeneous space $G/H$. He further shows that if $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1$, then $G$ is nilpotent and $H$ is discrete. I'm willing to bet that he knew the answer to the specific case of $(p,q,r) = (2,3,6)$ by then, but this specific case does not appear to be addressed in that paper.

That said, I'll sketch a proof that the answer to Milnor's question is yes, $M^3:=M_{2,3,6}$ is diffeomorphic to $G/H:=H_3(\mathbb{R})/H_3(\mathbb{Z})$.

A manifold $N$ is called prime if its only expression as a connect sum is trivial. That is, if $N = A\sharp B$ for manifolds $A$ and $B$, then $A = S^n$ or $B = S^n$. Here, when I write "=", I'm thinking "homeomorphic". Note that Moise showed that for $3$-manifolds, "homeomorphic" and "diffeomorphic" coincide.

Proposition: Suppose $N$ is a closed $3$-manifold and $\pi_1(N)$ is nilpotent. Then $N$ is prime.

Proof: Suppose $N = A\sharp B$. A simple application of Seifert-van Kampen shows $\pi_1(N) \cong \pi_1(A)\ast \pi_1(B)$. We claim that this forces at least one of $\pi_1(A)$ or $\pi_1(B)$ to be trivial. If $\pi_1(N)$ is trivial, then this is obvious, so assume $\pi_1(N)$ is non-trivial.

Because $\pi_1(N)$ is non-trivial and nilpotent, it has a non-trivial center. On the other hand, in a free product of non-trivial groups, the center is trivial. Thus, $\pi_1(N)\cong \pi_1(A)\ast \pi_1(B)$ implies one of $\pi_1(A)$ or $\pi_1(B)$ is trivial. Supposing without loss of generality it's $\pi_1(A)$ which is trivial, by the solution to the Poincare conjecture (due to Perelman), we know $A = S^3$. Thus $N$ is prime. $\square$

In addition, both $M^3$ and $G/H$ are orientable. For $M^3$, this follows from Lemma 7.1 of Milnor's paper I cited above. This lemma asserts that if $lcm(p,q) = lcm(p,r) =lcm(q,r)$, then $M_{p,q,r}$ is the total space of a principal $S^1$-bundle over an orientable surface. From the result and, e.g., the Gysin sequence, it follows easily that $H^3(M^3)\cong \mathbb{Z}$ so $M^3$ is orientable. For the Heisenburg homogeneous space, this answer shows that $H^3(G/H)\cong \mathbb{Z}$ so it's orientable as well.

Finally, note that since $\pi_1(M^3)\cong \pi_1(G/H)\cong \pi_0(H)$, the fundamental group is infinite. Thus, neither $M^3$ nor $G/H$ can be covered by $S^3$. That is, they are not lens spaces.

Now, we use Theorem 2.2 of

M. Aschenbrenner, S. Friedl, and H. Wilton. 3-Manifold Groups EMS Series of Lectures in Mathematics, vol. 20, European Mathematical Society (EMS), Zürich, 2015. (arXiv version)

Theorem: (My paraphrase of it) Let $N$ and $N′$ be two orientable, closed, prime 3-manifolds with isomorphic fundamental groups. If $N$ and $N'$ are not lens spaces, then $N$ and $N'$ are homeomorphic.

For both $M^3$ and $G/H$, we have verified each of these hypothesis. As a result, $M^3$ and $G/H$ are homeomorphic, hence diffeomorphic.