# 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? Aug 14 '20 at 17:44
• @MichaelAlbanese : Thanks ! in that case, how do you prove that $M$ is a nilmanifold ? Aug 14 '20 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. Aug 14 '20 at 19:54
• @MaximeRamzi: I might write an answer explaining this. Aug 14 '20 at 21:25
• @JasonDeVito: Nice example. It does follow the group is solvable though (in fact, $K\times S^1$ is a solvmanifold). Aug 14 '20 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. $$\square$$

• In place of $Diff(\mathbb{R})$, did you mean $Diff(S^1)$? Aug 15 '20 at 3:30
• @MichaelAlbanese: Yes, I saw that earlier, but didn't think it was worth the edit bump to change it. Aug 15 '20 at 3:57
• I edited my answer, so I took the opportunity to make the edit. I hope you don't mind. Aug 15 '20 at 15:01
• @MichaelAlbanese: I don't mind at all. Thanks! Aug 15 '20 at 15:16

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 $\pi_1$.) Aug 14 '20 at 19:09
• Oops, yes, I left out the commutation relations. I'll fix that. Aug 14 '20 at 20:39
• I think our answers are nicely complementary, to be honest. Aug 14 '20 at 20:50
• Then I will leave mine. Aug 14 '20 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.

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. Aug 14 '20 at 16:57