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Though I am a beginner to differential topology, pardon me for something very basic. Here is my attempt!

H(The set of $3 \times 3$ unipotent matrices over $\mathbb{R}$, Heisenberg group) is homeomorphic to $\mathbb{R}^3$. It suffice to show that $\mathbb{R}^3/\mathbb{Z}^3$ is homeomophic to $T^3=\mathbb{S}^1\times\mathbb{S}^1\times\mathbb{S}^1$ Now define a map $f:\mathbb{R}^3/\mathbb{Z}^3\to S^1\times S^1\times S^1$ as follows:

$f([(x,y,z)])=q\circ p^{-1}([(x,y,z)])=q(x+n,y+n,z+n)=(e^{2\pi i(x+n)},e^{2\pi i(y+n)},e^{2\pi i(z+n)})=(e^{2\pi iz},e^{2\pi iy},e^{2\pi iz})$(As cosine and sine are $2\pi$ periodic functions), where p is the canonical quotient map from $\mathbb{R}^3$ to $\mathbb{R}^3/\mathbb{Z}^3$ and q is defined as $q(x,y,z)=(e^{2\pi ix},e^{2\pi iy}, e^{2\pi iz})$. Clearly $f$ is well defined. Now the continuity of $f$ follows by continuity of $p,q$ and openness of $p$(as $p$ is projection under a group action $\mathbb{Z}$). Surjectivity follows as given a point $a\in T^3$, $\exists b \in \mathbb{R}^3$ such that $q(b)=a$(q is surjective) and then $p(b)\in\mathbb{R}^3/\mathbb{Z}^3$. Clearly $f$ is one-one by definition. Hence, by universal mapping property of quotient map, $q$ induces a homeomorphism $f$. Therefore, $H/\Gamma\simeq \mathbb{R}^3/\mathbb{Z}^3\simeq T^3$($\Gamma$ is the discrete Heisenberg group).

But even I am really not sure whether this really works or not! As the action is defined as: \begin{bmatrix} a\\ b\\ c \end{bmatrix} \begin{bmatrix} m\\ n\\ o \end{bmatrix} =\begin{bmatrix} x+m\\ n+ao+b\\ c+o \end{bmatrix}

The second slot is not just the usual translation so the action may be different then the usual action of $\mathbb{Z}^3$ on $\mathbb{R}^3$. Any help is appreciated!

Infinity
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  • It might be cleaner to show that $\mathbb{R}^3/\mathbb{Z}^3 \cong (\mathbb{R}/\mathbb{Z})^3$ and then separately show that $\mathbb{R}/\mathbb{Z}\cong S^1$. – William Feb 24 '19 at 03:17
  • Even then it's $\mathbb{S}^1\times \mathbb{S}^1\times \mathbb{S}^1$. I don't really see what this gets me out of the weird action – Infinity Feb 24 '19 at 03:20
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    What are you trying to prove? It's not true that this manifold is homeomorphic to $T^3$, because their fundamental groups (the discrete Heisenberg group and $\mathbb{Z}^3$) are not isomorphic. – Qiaochu Yuan Feb 24 '19 at 03:46
  • Okay, they what is the quotient space actually? I am interested to know what exactly is the space $G/\Gamma$ – Infinity Feb 24 '19 at 03:52
  • Have you seen https://math.stackexchange.com/questions/434384/the-heisenberg-manifold ? – Tyrone Feb 25 '19 at 10:46
  • Yeah but even that was not that helpful @Tyrone – Infinity Feb 26 '19 at 21:29

1 Answers1

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Even though this is very late, I happened to stumble across a problem regarding this group, and I could really relate to the struggle of the OP trying to find something "elementary" regarding this particular 3-manifold. Since I've managed to figure it out, I thought it could be useful to leave this as a future reference. In particular, I'll prove that the given quotient space is a compact differentiable manifold of dimension 3.

Let $H=\left\{ \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} \in \operatorname{Mat}_{3\times 3}(\mathbb{R}) : a,b,c \in \mathbb{R} \right\}$ be the real Heisenberg group with the Euclidean topology induced by $\mathbb{R}^9$ and let $\Gamma$ be the discrete Heisenberg group, i.e. the set of all matrices in $H$ with integer entries. Consider on $H$ the equivalence relation $\sim$ defined by $$A'\sim A \iff \text{there exists } \gamma\in \Gamma \text{ satisfying } A'=\gamma A.$$ We want to prove that the quotient space $M=H/\Gamma$ is a topological 3-manifold (i.e. that it is second-countable, Hausdorff and locally 3-Euclidean) which admits a differential structure.

$\textbf{M is second-countable:}$ Let's first prove that $\sim$ is an open relation, i.e. that for any open set $U\subset H$, also $[U]=\bigcup_{A\in U}[A]$ is open in $H$. Let $\mu_{\gamma}\colon H\to H$ be the left multiplication by $\gamma\in \Gamma$, i.e. $\mu_{\gamma}(A)=\gamma A$ for any $A\in H$. Since $\Gamma$ is a subgroup of $H$ and both $\mu_{\gamma}$ and its inverse $\mu_{\gamma}^{-1}=\mu_{\gamma^{-1}}$ are continuous by definition of matrix multiplication, we have that $\mu_{\gamma}$ is a homeomorphism. Thus, for any open set $U\subset H$, $$[U]=\bigcup_{A\in U}[A] = \bigcup_{A\in U}\{\gamma A: \gamma\in \Gamma\}=\bigcup_{A\in U} \bigcup_{\gamma\in \Gamma}\{\gamma A\}=\bigcup_{\gamma \in \Gamma} \mu_{\gamma}(U)$$ is open as it is the union of open sets (since $\mu_{\gamma}$ is a homeomorphism and homeomorphism are open maps). This proves that $\sim$ is an open relation, thus the canonical map $\pi\colon H\to M$, $A\mapsto [A]$ is also open. Since $H$ has a numerable basis $\mathcal{B}=\{B_i : i\in I\}$ inherited as a subspace of the Euclidean space $\mathbb{R}^9$, we have that $\mathcal{B}'=\{\pi(B_i): i\in I\}$ is a numerable basis for $M$, i.e. $M$ is second-countable.

$\textbf{M is Hausdorff:}$ Since $\sim$ is an open relation, in order to prove that $M$ is Hausdorff it suffices to prove that $R=\{(A,A'): A\sim A'\}$ is closed in $H\times H$. Let $\iota\colon \mathbb{R}^3\to H$ be the "inclusion map" $$(a,b,c)\mapsto \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$ (which can be easily checked to be a homeomorphism) and let $f\colon H\times H \to H$, $(A,A')\mapsto A'A^{-1}$. Since $$A'\sim A \iff A'=\gamma A \iff A'A^{-1}=\gamma\in \Gamma \iff f(A,A')\in \Gamma$$ we have that $R=f^{-1}(\Gamma)$. Since $f$ is continuous and $\Gamma=\iota(\mathbb{Z}^3)\subset H$ is closed as $\mathbb{Z}^3\subset \mathbb{R}^3$ is closed, we conclude that $R\subset H\times H$ is closed, thus the quotient space $M$ is Hausdorff.

$\textbf{M is locally 3-Euclidean:}$ The key claim is that for any $A\in H$ there exists an open neighborhood $U_A \subset H$ of $A$ which intersects each orbit of the action of $\Gamma$ on $H$ at most once. To prove this, let's start by examining what each orbit looks like. Suppose that $B'\sim B\Rightarrow B'=\gamma B$ for some matrices $$ B=\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}, \ B'=\begin{pmatrix} 1 & a' & c' \\ 0 & 1 & b' \\ 0 & 0 & 1 \end{pmatrix}, \ \gamma=\begin{pmatrix} 1 & p & r \\ 0 & 1 & q \\ 0 & 0 & 1 \end{pmatrix}.$$ We can check that this is equivalent to the system of equations $$\begin{cases} a'= a+p \\ b'=b+q \\ c'= c+r+bp\end{cases} \ \ (\star)$$ where $p,q,r\in \mathbb{Z}$. Thus, letting $x=\iota^{-1}(A)\in \mathbb{R}^3$, we have that $U_A=\iota(D(x,1/2))$ has the required properties, where $D(x,1/2)=\{x'\in \mathbb{R}^3: |x'-x|< 1/2\}$. Indeed, it's clear that $U_A\subset H$ is an open neighbourhood of $\iota(x)=A$. For the sake of contradiction, suppose there existed $B,B'\in U_A$ such that $B'\sim B$, and let $y=\iota^{-1}(B),$ $z=\iota^{-1}(B')\in \mathbb{R}^3$. Since $$|y-z|\leq |y-x|+|x-z|<1/2 + 1/2 = 1,$$ looking at $(\star)$ we must also have that $|p|=|a'-a|<1$ and $|q|=|b'-b|<1$, which forces $p=q=0$ as $p$ and $q$ are integers. Plugging this back into the third equation of $(\star)$, for the same reason $c'=c+r\Rightarrow |r|=|c'-c|<1$ forces $r=0$. Since $(a',b',c')=(a,b,c)$, this proves that $B'=B$, i.e. $U_A$ contains at most one element from each orbit of the action of $\Gamma$ on $H$.

Why is this useful? Consider the canonical map $\pi\colon H\to M$. We already know from before that $\pi$ is a continuous open map; however, what we just proved shows that the restriction $\pi_{|_{U_A}}\colon U_A\to M$ is also injective, thus it is a homeomorphism onto its image. Pick an arbitrary point $v=\pi(A)\in M$, and let $V=\pi(U_A)$ where $U_A$ is defined as above. Then, $\varphi_V=\iota^{-1}\circ \pi_{|_{U_A}}^{-1}\colon V\to D(x,1/2)$ is a homeomorphism (as it is a composition of homeomorphisms), thus $V\subset M$ is an open neighbourhood of $v$ which is homeomorphic to an open set of $\mathbb{R}^3$, i.e. $M$ is locally $3$-Euclidean.

$\textbf{M has a differential structure:}$ Let $\mathcal{A}$ be the set of all charts $(V,\varphi_V)$ defined as above. We claim that $\mathcal{A}$ is an atlas for $M$. Clearly, the union of all $V$'s is $M$, thus it suffices to show that the charts are differentiably compatible. Let $(V,\varphi_V)$, $(V',\varphi_{V'})\in \mathcal{A}$ and suppose that $V\cap V'\neq \emptyset$ (otherwise it's trivial). Then, for any $x\in \varphi_V(V\cap V')$ we have that $$\varphi_{V'}\circ \varphi_V^{-1}(x)=(\iota^{-1}\circ \pi_{|_{V'}}^{-1})\circ (\iota^{-1}\circ \pi_{|_{V}}^{-1})^{-1}(x) =\iota^{-1}\circ \pi_{|_{V'}}^{-1} \circ \pi_{|_{V}} \circ \iota(x) = x $$ is obviously smooth and similarly $\varphi_{V}\circ \varphi_{V'}^{-1}(x)=x$ for any $x\in \varphi_{V'}(V\cap V')$, thus $\mathcal{A}$ is an atlas and $M$ is a differentiable $3$-manifold.

$\textbf{M is compact:}$ This follows from the fact that any $A=\begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\in H$ is equivalent to $$A'=\begin{pmatrix} 1 & \{a\} & \{c-b\lfloor a\rfloor\} \\ 0 & 1 & \{b\} \\ 0 & 0 & 1 \end{pmatrix}$$ where $\lfloor \cdot \rfloor$ and $\{\cdot \}$ denote integer and fractional part, respectively. The proof of this fact follows from plugging in the values in $(\star)$, which shows that $p,q$ and $r$ are actually integers, thus $A'\sim A$. Since all entries of $A'$ are real numbers in $[0,1]$, we see that we can restrict $\pi$ to the compact "unit cube" $\iota([0,1]^3)\subset H$ and its image is still all $M$, thus $M$ is compact (as $\pi$ is continuous and surjective). $\square$

Federico Clerici
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