Question from J. Milnor paper from 1968 about diffeomorphic manifolds In the article "A note on curvature and fundamental group"(1968)  by J. Milnor  the following side question arises:

where $G$ and $H$ are continuous (over $\mathbb{R}$) and discrete (over $\mathbb{Z}$) Heisenberg $3\times 3$ matrix group.
The fundamental group of orbit space G/H is isomorphic to the nilpotent group $H$ (and it follows from introductory facts about algebraic topology, e.g "Algebraic Topology - A First Course" W. Fulton, Corollary 13.16).
But I guess the author's doubt may not easy to answer, but the paper is known and maybe somebody knows the answer and could lighten up the problem.
*I supply an above entry with another statement (from "Treatise on Analysis" Volume III, Chapter XVI by J. Dieudonne, e.g statement 16.10.3) related to a unique differentiable structure on $G/H$:
Let $G$ be a Lie transformation group of a manifold $M$, $M/G$ the orbit space topologized by the finest topology for which the natural mapping $\pi: M\to M/G$ is continuous. Let
\begin{equation*}
D=\{(p,q)\in M\times M: \exists_{g\in G}\,\, p=g\cdot q\}
\end{equation*}
Then the following statements are true:
(i) $M/G$ is a closed Hausdorff space if and only if the subset $D\subset M\times M$ is closed.
(ii) There exists a differentiable structure on the topological space $M/G$ such that $\pi:M\to M/G$ is a submersion if and only if the topological subspace $D\subset M\times M$ is a closed submanifold.
In this case, the differentiable structure is unique and all $G$-orbits in $M$ have the same dimension.
 A: 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.
