# Fundamental group of the total space of an oriented $S^1$ fiber bundle over $T^2$

Let $$M$$ be the total space of an oriented $$S^1$$ fiber bundle over $$T^2$$.

Can we show the fundamental group of $$M$$ is nilpotent? More generally, how can we calculate the fundamental group of $$M$$ explicitly?

• Do you know any obstruction theory? Commented Aug 14, 2020 at 17:44
• @MichaelAlbanese : Thanks ! in that case, how do you prove that $M$ is a nilmanifold ? Commented Aug 14, 2020 at 18:58
• @Michael: According to math.stackexchange.com/questions/6466/subgroups-klein-bottle/…, $\pi_1(K)$ is not nilpotent, where $K$ is the Klein bottle. Since $K$ is an $S^1$ bundle over $S^1$, it follows easily that $S^1\times K$ is an $S^1$-bundle over $T^2$ for which $\pi_1$ is non-nilpotent. Hence, just saying $\pi_1(M)$ is an extension of $\mathbb{Z}^2$ by $\mathbb{Z}$ is not sufficient to conclude $\pi_1(M)$ is nilpotent. Commented Aug 14, 2020 at 19:54
• @MaximeRamzi: I might write an answer explaining this. Commented Aug 14, 2020 at 21:25
• @JasonDeVito: Nice example. It does follow the group is solvable though (in fact, $K\times S^1$ is a solvmanifold). Commented Aug 14, 2020 at 21:25

Here's a geometric view point which computes (a presentation of) $$\pi_1(M)$$. As a byproduct, we establish

The group $$\pi_1(M)$$ is nilpotent of at most $$2$$-steps. That is, $$[\pi_1(M),\pi_1(M)]$$ need not be trivial, but $$[\pi_1(M), [\pi_1(M),\pi_1(M)]]$$ is trivial.

(If $$M$$ is the trivial bundle, then $$\pi_1(M)\cong \mathbb{Z}^3$$ is abelian, i.e., one-step nilpotent. We will see below that if $$M$$ is non-trivial, then $$\pi_1(M)$$ is non-abelian.)

Because $$Diff(S^1)$$ deformation retracts to $$O(2)$$, a circle bundle is orientable iff it's principal. For a reasonable space $$X$$, principal circle bundles are classified by $$H^2(X)$$.

Now, decompose $$T^2$$ as a union of a small ball $$B$$ and the complement $$C$$ (enlarged slightly so there is overlap between $$B$$ and $$C$$) of the small ball. Note that $$B\cap C$$ deformation retracts to a circle.

Proposition 1: Every principal circle bundle $$M$$ over $$T^2$$ is obtained by gluing a trivial bundle $$B\times S^1$$ to a trivial bundle $$C\times S^1$$ via a map of their boundaries $$f:S^1\times S^1\rightarrow S^1\times S^1$$. Such an $$f$$ must have the form $$f(\theta, \phi) = (\theta, g(\theta) + \phi)$$ for some smooth map $$g:S^1\rightarrow S^1$$.

Proof:

Now, it's well known that $$C$$ deformation retracts to $$S^1\vee S^1$$, so $$H^2(C) = 0$$. Since $$B$$ is contractible, $$H^2(B) = 0$$ as well.

This means that any principal $$S^1$$ bundle over $$T^2$$ restricts to the trivial bundle on both $$B$$ and $$C$$. It follows that any principal $$S^1$$ bundle on $$T^2$$ is obtained as follows:

Glue $$C\times S^1$$ to $$B\times S^1$$ along their common boundary $$S^1\times S^1$$. The gluing map $$f:S^1\times S^1\rightarrow S^1\times S^1$$ must have the form $$f(\theta,\phi) = (\theta, g(\theta)+\phi)$$, where $$g:S^1\rightarrow S^1$$ is some smooth function. The form on the first factor is because the projection maps on $$C\times S^1$$ and $$B\times S^1$$ just project to the first factor, and these must match. The form on the second factor is because the bundle is $$S^1$$-principal. $$\square$$

Different choices of $$g$$ can lead to different bundles, but as is usual for bundles, homotopic $$g$$s lead to isomorphic bundles. So, we may as well focus on homotopy classes of $$g$$s, and there are precisely $$\pi_1(S^1)\cong \mathbb{Z}$$ of them. Representatives are given by $$g_k(\theta) = k\theta$$ for $$k\in \mathbb{Z}$$. For each such $$k$$, call the resulting total space $$M = E_k$$.

Now that we're armed with a geometric picture of $$M$$, we can use Seifert-van Kampen to compute the fundamental group of $$E_k$$. Let's set up some notation.

First, $$\pi_1(C\times S^1)\cong F^2\times \mathbb{Z}$$ with $$F^2$$ a free group on two generators (say, $$a$$ and $$b$$). Let $$c$$ denote a generator of the $$\mathbb{Z}$$ factor.

Second, $$\pi_1(B\times S^1)\cong \mathbb{Z}$$, say generated by $$d$$.

Lastly, $$\pi_1((C\cap B)\times S^1)\cong \mathbb{Z}\times \mathbb{Z}$$, say generated by $$x$$ and $$y$$.

Because $$C$$ is more complicated than $$B$$, we'll view the $$(B\cap C)\times S^1$$ as living in $$C$$, which is then attached to $$B\times S^1$$ via $$f$$.

Proposition 2: The group $$\pi_1(E_k) \cong \langle a,b,c\, |[a,c], [b,c], [a,b]c^{-k}\rangle.$$

From the usual van Kampen argument for computing $$\pi_1$$ of a genus 2 surface, the inclusion $$(B\cap C)\times S^1\rightarrow C\times S^1$$ maps $$x$$ to the commutator $$[a,b]$$, and it maps $$y$$ to c.

But what is $$f_\ast:\pi_1((B\cap C)\times S^1)\rightarrow \pi_1(B\times S^1)$$? Well, it's enough to figure out the induces map to $$\partial (B\times S^1)\cong S^1\times S^1$$, because the inclusion $$\partial(B\times S^1)\rightarrow B\times S^1$$ obviously kills the first factor and is the identity on the second.

Well, factor the map $$(\theta,\phi)\mapsto (\theta, k\theta + \phi)$$ via the composition $$T^2\rightarrow T^3\rightarrow T^3\rightarrow T^2$$ where $$(\theta,\phi)\mapsto (\theta,\theta,\phi)\mapsto (\theta,k\theta,\phi)\mapsto (\theta,k\theta + \phi).$$ Then it's not too hard to see that $$x$$ maps to $$x + ky$$ in $$\partial(B\times S^1)$$, where the $$x$$ then maps to $$0$$, but $$ky$$ maps to $$kd$$. That is, $$x\in \pi_1((B\cap C)\times S^1)$$ maps to $$kd\in \pi_1(B\times S^1)$$. Similarly, $$y$$ maps to $$y$$.

Applying Seifert-van Kampen, we find a presentation for $$\pi_1(E_k)$$ is $$\langle a,b,c,d| [a,c],[b,c], [a,b]d^{-k}, cd^{-1}\rangle$$ which simplifies to $$\langle a,b,c| [a,c],[b,c], [a,b]c^{-k}\rangle,$$ (where the relations $$[a,c], [b,c]$$ come from $$\pi_1(C)$$ and the other relations come from Seifert-van Kampen). $$\square$$

As a quick check, when $$k=0$$ (so we get the trivial bundle), this is $$\pi_1(T^3)\cong \mathbb{Z}^3$$, as it should be.

Lastly, we claim this is nilpotent for any $$k$$.

Proposition 3: For $$k\neq 0$$, the presentation $$\langle a,b,c\,| [a,c], [b,c], [a,b]c^{-k}\rangle$$ defines a 2-step nilpotent group for every $$k$$.

Proof: First, $$[a,b] = c^k\neq 0$$, so this presentation is not 1-step nilpotent.

Since $$c$$ is central, the relation $$[a,b]c^{-k}$$ can be rewritten as $$abc^{-k} = ba$$. Multiplying on the right and left by $$a^{-1}$$ gives $$ba^{-1}c^{-k} = a^{-1}b$$, so $$ba^{-1} = c^k a^{-1} b$$. Similar relations exist with $$b^{-1} a$$ and $$b^{-1} a^{-1}$$. The point is that any word in $$a$$, $$b$$, and $$c$$ is equivalent to a word which is alphabetical: it is of the form $$a^s b^t c^u$$ for integers $$s,t,u$$.

Consider the obvious map $$\rho:\langle a,b,c\,|[a,c],[b,c], [a,b]c^{-k}\rangle \rightarrow \langle a,b,\,|[a,b]\rangle \cong \mathbb{Z}^2$$, obtained by setting $$c= e$$. We claim that $$\ker \rho = \langle c\rangle$$. It is obvious that $$c\in \ker \rho$$, so let's establish the other direction. Suppose $$w = a^s b^t c^u$$ is an alphabetical word in $$a,b,c$$ and that $$w\in \ker \rho$$. Then $$\rho(a^s b^t) = 0$$ which implies that $$s=t=0$$.

Given any commutator $$[w_1,w_2]$$ in words $$w_1$$ and $$w_2$$, we see that $$\rho([w_1,w_2]) = [\rho(w_1), \rho(w_2)] = 0$$ because $$\mathbb{Z}^2$$ is abelian. Thus, the first derived subgroup of this presentation is a subgroup of $$\ker \rho = \langle c\rangle$$. Since $$c$$ is central, the second derived subgroup is trivial. $$\square$$

Let me use $$T$$ to denote the torus. Oriented circle bundles over $$T$$ (indeed over any closed, connected surface) are classified up to orientation preserving bundle isomorphism by an integer known as the Euler number. To define this number, choose a standard cell decomposition of $$T$$ with a single vertex, two edges, and a single 2-cell. Let $$T^{(i)}$$ denote the $$i$$-skeleton and let $$E^{(i)}$$ denote the bundle restricted to $$T^{(i)}$$. Labelling the 1-cells of $$T^{(1)}$$ as $$a,b$$ so that the 2-cell is attached along the curve $$\gamma : S^1 \to T$$ which is expressed as a concatenation of edges of the form $$aba^{-1}b^{-1}$$.

Here's an obstruction theory description of the Euler number. Roughly speaking the pullback bundle $$\gamma^* E$$ is a trivial circle bundle over $$S^1$$, but the structure of the bundle $$E \mapsto T$$ gives two trivializations of $$\gamma^* E$$. Comparing those two trivializations produces the Euler number: any two trivializations of the same circle bundle over $$S^1$$ differ by a Dehn twist around any one of the circle fibers, and the Dehn twist exponent is the Euler number.

First, $$E^{(1)}$$ is a trivial circle bundle over $$T^{(1)}$$, and so we may pick a bundle isomorphism $$E^{(1)} \approx T^{(1)} \times S^1$$ Pulling this back under the map $$\gamma : S^1 \to T^{(1)}$$ gives the first trivialization of $$\gamma^* E$$.

Second, consider an orientation preserving characteristic map $$\chi : D^2 \to T$$ for the 2-cell, whose restriction to the boundary circle $$S^1$$ is the curve $$aba^{1}b^{-1}$$. The pullback bundle $$\chi^* E$$ is a trivial bundle over $$D^2$$ (because $$D^2$$ is contractible), and so we may choose a bundle isomorphism $$\chi^* E \approx D^2 \times S^1$$ Since $$\gamma$$ is defined on $$\partial D^2 = S^1$$, one may pullback this trivialization of $$\chi^* E$$ to give the second trivialization of $$\gamma^* E$$.

From this description, if the Euler number is equal to $$X$$ then one can use Van Kampen's theorem to obtain the presentation $$\pi_1(E) = \langle a, b, f \mid a b a^{-1} b^{-1} = f^X, \,\, a f a^{-1} f^{-1} = b f b^{-1} f^{-1} = \text{Id} \rangle$$ and from this the nilpotency of $$\pi_1(E)$$ clearly follows.

There's a nontrivial theorem going on here: one must prove that the Euler number is well-defined independent of choice (the main choice being the trivialization of $$E^{(1)}$$), and that two oriented circle bundles over $$T$$ are isomorphic as oriented bundles if and only if they have the same Euler number. Those proofs are where the true "obstruction theory" arguments take place.

• Funny that we posted essentially isomorphic answers. Since yours was first, I'm happy to delete mine, but I'm also happy to leave it. Thoughts? Also, does your presentation for $\pi_1(E)$ need extra relations asserting that $f$ commutes with $a$ and $b$? It's not clear to me that these relations follow from the relation you have. If $f$ does not commute with $a$ or $b$, then it's not clear to me that $\pi_1(E)$ is nilpotent (and it's not clear to me that we reached the same answer for $\pi_1$.) Commented Aug 14, 2020 at 19:09
• Oops, yes, I left out the commutation relations. I'll fix that. Commented Aug 14, 2020 at 20:39
• I think our answers are nicely complementary, to be honest. Commented Aug 14, 2020 at 20:50
• Then I will leave mine. Commented Aug 14, 2020 at 21:01

Orientable circle bundles over $$X$$ are classified by $$H^2(X; \mathbb{Z})$$ via the first Chern class (or Euler class). In particular, orientable circle bundles over $$T^2$$ are classified by $$H^2(T^2; \mathbb{Z}) \cong \mathbb{Z}$$. Let $$M_r$$ denote the total space of the orientable circle bundle over $$T^2$$ with first Chern class $$r$$; this is what is referred to as the Euler number in Lee Mosher's answer. For $$r \neq 0$$, the manifolds $$M_r$$ and $$M_{-r}$$ are diffeomorphic, but have opposite orientations. Below we give a construction of the manifolds $$M_r$$. This is more than you ask for, but it provides a different point of view than the other answers.

For $$r = 0$$, we have $$M_0 = T^3$$ and $$\pi_1(T^3) \cong \mathbb{Z}^3$$ which is abelian and hence nilpotent.

For $$r \neq 0$$, consider the quotient $$H(3, \mathbb{R})/\Gamma_r$$ where

$$H(3, \mathbb{R}) = \left\{\begin{bmatrix} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1\end{bmatrix} : x, y, z \in \mathbb{R}\right\}$$

is the three dimensional Heisenberg group, and

$$\Gamma_r = \left\{\begin{bmatrix} 1 & a & \frac{c}{r}\\ 0 & 1 & b\\ 0 & 0 & 1\end{bmatrix} : a, b, c \in \mathbb{Z}\right\}$$

is a discrete subgroup. Note that $$H(3, \mathbb{R})$$ is diffeomorphic to $$\mathbb{R}^3$$, but they are not isomorphic as Lie groups because $$H(3, \mathbb{R})$$ is not abelian while $$\mathbb{R}^3$$ is. For $$r = 1$$, the subgroup $$\Gamma_1$$ is precisely the three-dimensional integral Heisenberg group $$H(3, \mathbb{Z})$$ and the quotient $$H(3, \mathbb{R})/H(3, \mathbb{Z})$$ is known as the three-dimensional Heisenberg manifold.

For each $$r \neq 0$$, the quotient $$H(3, \mathbb{R})/\Gamma_r$$ is orientable as the nowhere-zero three-form $$dx\wedge dy\wedge dz$$ on $$H(3, \mathbb{R})$$ is invariant under the action of $$\Gamma_r$$. Moreover, there is compact fundamental domain for the action of $$\Gamma_r$$ (see here for the $$r = 1$$ case), and hence $$H(3, \mathbb{R})/\Gamma_r$$ is compact. One can show that the map $$H(3, \mathbb{R})/\Gamma_r \to \mathbb{R}^2/\mathbb{Z}^2$$ given by $$A + \Gamma_r \mapsto (x, y) + \mathbb{Z}^2$$ is a submersion, so it follows from Ehresmann's theorem that $$H(3, \mathbb{R})/\Gamma_r$$ is the total space of an orientable circle-bundle over $$T^2$$.

Note that $$\pi_1(H(3, \mathbb{R})/\Gamma_r) \cong \Gamma_r$$. As

$$\begin{bmatrix} 1 & a_1 & \frac{c_1}{r}\\ 0 & 1 & b_1\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & a_2 & \frac{c_2}{r}\\ 0 & 1 & b_2\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & a_1 + a_2 & \frac{c_1}{r} + a_1b_2 + \frac{c_2}{r}\\ 0 & 1 & b_1 + b_2\\ 0 & 0 & 1\end{bmatrix},$$

we see that

$$[\Gamma_r, \Gamma_r] = \left\{\begin{bmatrix} 1 & 0 & c\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix} : c \in \mathbb{Z}\right\}$$

and hence $$[[\Gamma_r, \Gamma_r], \Gamma_r]$$ is trivial, i.e. $$\Gamma_r$$ is two-step nilpotent.

Is $$H(3, \mathbb{R})/\Gamma_r \to T^2$$ the Chern class $$r$$ orientable circle bundle (as notation might suggest)? In order to answer this, note that it follows from the Gysin sequence applied to the circle bundle $$M_r \to T^2$$ that $$H_1(M_r; \mathbb{Z}) \cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/r\mathbb{Z}$$. On the other hand

$$H_1(H(3, \mathbb{R})/\Gamma_r; \mathbb{Z}) \cong \pi_1(H(3, \mathbb{R})/\Gamma_r)^{\text{ab}} \cong \Gamma_r^{\text{ab}} \cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/r\mathbb{Z}$$

where the last isomorphism follows from the description of $$[\Gamma_r, \Gamma_r]$$ above. Therefore, up to a choice of orientation, we see that $$H(3, \mathbb{R})/\Gamma_r \to T^2$$ is the Chern class $$r$$ orientable circle-bundle and hence $$M_r$$ is diffeomorphic to $$H(3, \mathbb{R})/\Gamma_r$$. Hence $$\pi_1(M_r) \cong \pi_1(H(3,\mathbb{R}/\Gamma_r) \cong \Gamma_r$$ is two-step nilpotent.

To see that the description of $$\pi_1(M_r)$$ is consistent with the ones given by Jason DeVito and Lee Mosher, note that there is an isomorphism $$\Gamma_r \cong \langle a, b, c \mid [a, c] = [b, c] = 1, [a, b] = c^r\rangle$$ given by

$$\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix} \mapsto a,\quad \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix} \mapsto b,\quad \begin{bmatrix} 1 & 0 & \frac{1}{r}\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix} \mapsto c.$$

It follows from this presentation that

$$\Gamma_r^{\text{ab}} \cong \langle a, b, c \mid [a, b] = [a, c] = [b, c] = c^r = 1\rangle \cong \langle a\rangle\oplus\langle b\rangle\oplus\langle c \mid c^r = 1\rangle \cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}/r\mathbb{Z}$$

which agrees with the calculation above.

In general, manifolds which arise this way always have nilpotent fundamental group. More precisely, for every $$r$$, the manifold $$M_r$$ is of the form $$N/\Gamma$$ where $$N$$ is a simply connected nilpotent Lie group and $$\Gamma < N$$ is a discrete subgroup; that is, $$M_r$$ is a nilmanifold (every compact nilmanifold can be realised as such a quotient). As $$\pi_1(N/\Gamma) \cong \Gamma < N$$ and $$N$$ is nilpotent, it follows that the fundamental group of a compact nilmanifold is always nilpotent. The non-trivial part of this answer was to show that orientable circle bundles over $$T^2$$ are indeed nilmanifolds.

• Your $M_r$ is the mapping torus of $\begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}$. Is it easy to see that this is not diffeomorphic to the mapping torus of $\begin{bmatrix} -1 & r \\ 0 & -1 \end{bmatrix}$? I know for $r=0$ the first mapping torus is trivial $T^3$ while the second mapping torus has holonomy/monodromy cyclic order 2 and is the unit tangent bundle of the Klein bottle. But I don't know how to distinguish the mapping tori in general. Is it it easy to see they have different fundamental groups? Or some other topological invariant distinguishes them? Commented Jan 30, 2022 at 18:02
• It's not obvious to me, but it may be of interest to note that since $\begin{bmatrix} -1 & r \\ 0 & -1 \end{bmatrix}^2 = \begin{bmatrix} 1 & -2r \\ 0 & 1 \end{bmatrix}$, such a mapping torus is double covered by $M_{-2r}$. Commented Feb 1, 2022 at 22:54
• Another follow-up question: The $N_r$ are all principal circle bundles over the torus $T^2$. Are they also principal $T^2$ bundles over the circle? Or is it clear that every principal $T^2$ bundle over the circle is trivial? Commented Feb 11, 2022 at 16:00
• @IanGershonTeixeira: Every principal $T^2$-bundle over the circle is trivial. If you know about classifying spaces, this is clear. Commented Feb 11, 2022 at 16:16
• Oh oops you're right of course. It is exactly this question math.stackexchange.com/questions/36444/… just generalized to show that any principal $T^n$ bundle over the circle is trivial using the additional fact that the classifying space of a product is $B(G\times G)\cong BG \times BG$ so $BT^n\cong (\mathbb{C}P^\infty)^n$ is still simply connected and all homotopy class of maps from $S^1$ are still trivial. Commented Feb 11, 2022 at 16:37

You have the long exact sequence of the fibration $$\pi_2(T^2) \to \pi_1(S^1) \to \pi_1(M) \to \pi_1(T^2) \to 0$$ Happily, $$\pi_2(T^2)$$ vanishes (the universal cover is contractible) and so your group is some extension of $$\mathbb{Z}^2$$ by $$\mathbb{Z}$$. We can't know which one without more knowledge of the geometry of $$M$$ (it might be that all such groups are nilpotent -- I don't know!)

• One almost trivial comment: If the image of $\pi_1(S^1)$ in $\pi_1(M)$ happens to land in the center of $\pi_1(M)$, then $\pi_1(M)$ is nilpotent. Commented Aug 14, 2020 at 16:57