Poincare Homology Sphere – Plumbing of Disk Bundles

In this post, we will provide a description of the Poincare homology sphere as a plumbing of disk bundles.

Plumbing Disk Bundles and Handle Decompositions

Definition. Let E_1 \to B_1 , E_2 \to B_2 be two disk bundles over surfaces and let D_1, D_2 be disks in B_1, B_2 respectively. A plumbing of two disk bundles over surfaces along D_1, D_2 is the space X = E_1 \cup_{\varphi} E_2 where \varphi: D_1 \times D^2 \to D_2 \times D^2 is given by

\varphi: (x,y) \mapsto (y,x)

Definition. A space X is a plumbing if it is obtained from finitely many such identification along disjoints disks that are in the bases.

Definition. A plumbing graph for a plumbing X is a graph \Gamma(V,E) where vertices are labeled by the disk bundles and edges correspond to plumbing between corresponding disk bundles.

Example. Let X = S^2 \times S^2 -D^4. Following the convention that S^2 = D_- \cup D_+, we see that

X =(D_-\times D_-) \cup ( D_-\times D_+) \cup (D_+\times D_-).

We have two disk bundles over spheres (D_-\times S^2) \cong S^2 \times D^2 and (S^2\times D_-)  \cong S^2 \times D^2 glued along D_-\times D_-.  It is clear that the gluing is base-to-fiber. Therefore, X is a plumbing with the plumbing graph consists of two vertices labeled S^2 \times D^2 and one edge connecting the two vertices. The space X also has a handle diagram

By reversing the roles of the handles and considering D_-\times D_- as a 2-handle of X, we can see that X is obtained by attaching a 2-handle D_+\times D_- to another two handle D_-\times D_-. The attaching circle \{0\} \times S^1 is isotopic to the meridian of the bundle surgery curve.

Lemma. The handle diagram for plumbing of disk bundles over the spheres with plumbing graph a tree is a link of unknots where each unknot corresponds to a vertex (and has framing number equals to the euler number of that vertex), and two unknots are linked when the corresponding vertices are joined by an edge.

Proof. We proceed by induction. When the plumbing graph has exactly one vertex, our plumbing is a disk bundle over one sphere. We have seen that the corresponding handle diagram for this plumbing consists of one circle labelled  by the Euler number for the plumbing.

In general, suppose that we add a vertex and edge (u,v) to our plumbing graph (in such a way that the graph remains a tree) where v corresponds to a new disk bundle E over a sphere with Euler number e. In a similar fashion to the previous example, the plumbing can be achieved by attaching a 2-handle to the 2-handle for u with framing e.  The attaching circle is again isotopic to the meridian. This completes the proof. ∎

We recall that a handle diagram for a 4-manifold is also a surgery diagram for its boundary. Therefore, the lemma leads us to a new description of the Poincare homology sphere.

Poincare Homology Sphere as a Plumbing

Recall that the Poincare homology sphere can be described as the boundary of the 4-manifold P with a handle diagram

It follows immediately from the previous lemma that the Poincare homology sphere is the boundary of the plumbing with the following plumbing graph.

Finally, we will give a description of the classification of disk bundles over surfaces.

Classification of Oriented Disk Bundles over Surfaces

Throughout, \Sigma will be a compact, closed, oriented surfaces. By restricting each fiber to the unit disk, we have a bijection between bundles over \Sigma:

{Oriented D^2 -bundles} \longleftrightarrow {\mathbb{R}^2 -bundles}.

For each oriented \mathbb{R}^2 -bundle E \to B, we can associated to it a \mathbb{C} -bundle E^{\mathbb{C}}\to B in a canonical way. We take an oriented vector bundle atlas \{U_i\} for E and identify U_i \times \mathbb{R}^2 with U_i \times \mathbb{C} via the formula

(p,(x,y)) \mapsto (p,x+iy) .

Therefore, we obtained a bijection between bundles over \Sigma:

{Oriented D^2 -bundles} \longleftrightarrow {\mathbb{C}^2 -bundles}.

Therefore, we will give a classification of \mathbb{C} -bundles up to homotopy.

Projective Space and the Universal Line Bundle

The inclusion \mathbb{C}^{n-1} \subseteq \mathbb{C}^n induces the inclusion \mathbb{CP}^{n-1} \subseteq \mathbb{CP}^n. Taking the direct limit, we get

\mathbb{CP}^{\infty}:= \lim_{}\mathbb{CP}^n.

The canonical line bundle over \mathbb{CP}^n is the bundle

E = \{(\ell,x) \mid \ell \text{ is a line in }\mathbb{C}^{n+1}  \text{ and } x \in \ell \}

Definition. The universal line bundle \gamma^1 is the canonical line bundle over \mathbb{CP}^{\infty}.

It is an important fact that every \mathbb{C} -bundle over a paracompact topological space is a pull-back of  \gamma^1 (unique up to homotopy). Furthermore,

Theorem. Two \mathbb{C}-bundles E, E' over a paracompact space B are isomorphic if and only if the induced maps \bar{f}: B \to \gamma^1 and \bar{f}': B' \to \gamma^1 are homotopic.

Consequently, the set of isomorphism classes of \mathbb{C} -bundle over \Sigma is in bijection with the set of homotopy classes of maps f: \Sigma \to \mathbb{CP}^\infty. Since \mathbb{CP}^\infty is a K(\mathbb{Z},2), the cohomology representation theorem says that homotopy classes of map [\Sigma \to \mathbb{CP}^\infty] is in bijection with H^2(\Sigma,\mathbb{Z}). Therefore, \mathbb{C} -bundle over \Sigma are classified by these cohomology classes.

From Morse Theory to Handlebody Decomposition

Throughout, we will suppose that M is a smooth, compact manifold and f:M \to \mathbb{R} is Morse. For any x \in \mathbb{R}, we denote M^x := f^{-1}((-\infty,x]).

Theorem. Suppose a < b and M^b - M^a contains no critical points then M^a is a smooth deformation retract of M^b.

Proof. Fix a Riemannian metric \langle \ , \ \rangle on M, say from an embedding M \to \mathbb{R}^k.  The Riemannian metric defines an isomorphism T^*M \to TM. Therefore, we can define \nabla f \in TM satisfying \langle X, \nabla f \rangle = df (X).

Since the critical points of f are isolated, there exists an open neighborhood U that separates M^b - M^a from the critical points.  We define a bump function \rho: M \to \mathbb{R} supported in U. In particular, \rho(p) is given by \frac{1}{\langle \triangledown f, \triangledown f\rangle} for all p \in M^b - M^a and 0 for all p \in M - U.

Now we define a vector field X \in TM by X_q = \rho(q) (\triangledown f)_q. Let \phi_t be the flow associated to X_q. In particular, \phi_t(q) = \gamma(t) where \gamma satisfies the following ODE

\frac{d}{dt} \gamma(t) = X_{\gamma(t)} and \gamma(0) = q.

For a fixed q,  we have

\frac{d}{dt}\phi_t(q) = X_{\gamma(t)}.

Therefore, for |f|\leq |b-a| and q\in M^b-M^a , we have

\frac{d}{dt}(f\circ \phi_t(q)) = df (X_{\gamma(t)}) = \langle \triangledown f, X \rangle =1.

We claim that \phi_{b-a}: M^a \to M^b is a diffeomorphism. Observe that for all q \in M^a, we have

\int_0^{b-a} \frac{d}{dt}(f\circ \phi_t(q)) dt = b-a.

On the other hand, \int_0^{b-a} \frac{d}{dt}(f\circ \phi_t(q)) dt  = f(\phi_{b-a}(q)) - f(q). Since f(q) \leq a, we must have f(\phi_{b-a}(q))<b. Thus, \phi_{b-a}(q) \subseteq M^b for all q \in M^a. To see surjectivity, we can calculate similarly with \phi_{a-b}.

To obtain the deformation retract, we define r_t: I \times M^b \to M^b to be q if f(q) \leq a  and to be \phi_{t(a-f(q))}(q) if a \leq f(q) \leq b. This completes the proof.∎

 Theorem. Suppose that p \in M is the only critical point with the critical value c with index \lambda. Then there exists \epsilon > 0 such that M^{c +\epsilon} is diffeomorphism to M^{c-\epsilon} \cup H for H a single \lambda-handle.

Proof. Let U be neighborhood of p such that in coordinate (u_1,\dots,u_n) we have p = (0,\dots,0) and

f(\vec{u}) = c- u_1^2 - \dots - u_\lambda^2 + u_{\lambda + 1}^2 + \dots + u_n^2 = c - \eta(\vec{u}) + \rho(\vec{u}).

Choose \epsilon > 0 such that f^{-1}([c-\epsilon,c+\epsilon]) only contains p as a critical point and the closed ball B(0,2\epsilon) is in U. Let \mu be a smooth bump function satisfying \mu(0) > \epsilon, \mu(r) = 0 for all r \geq 2 \epsilon and -1 < \mu(r) \leq 0 for all r.

Define F: M \to \mathbb{R} such that F = f on M - U and F(q) = f(q)- \mu(q)(\eta(q)+2\rho(q)) otherwise.

Observe that F^{-1}((-\infty,c+\epsilon]) = M^{c+\epsilon} since F and f only differ when \eta + 2\rho \leq 2 \epsilon.  Particularly, in this region, we have F \leq f \leq c + \eta/2 + \rho \leq c + \epsilon.

Exercise. F has the same critical points and is Morse.

Observe that F^{-1}([c-\epsilon,c+\epsilon]) contains no critical points. Since F^{-1}([c-\epsilon,c+\epsilon]) \subseteq f^{-1}([c-\epsilon,c+\epsilon]), the only possibility is p.  But F(p) = c - \mu(0) < c - \epsilon .

By the previous theorem, M^{c+\epsilon} = F^{-1}((-\infty,c+\epsilon]) deformation retracts to F^{-1}((-\infty,c-\epsilon]). Now observe that we can write

F^{-1}((-\infty,c-\epsilon])= M^{c-\epsilon} \cup H

where H = F^{-1}((-\infty,c-\epsilon]) \cap f^{-1}([c-\epsilon,\infty)) is a closed subset of U.

The \lambda-cell D^{\lambda}, given by \eta < \epsilon and \rho =0, is contained in H. Indeed, for q \in D^{\lambda}, f(q) = c-\eta \geq c - \epsilon. Next since \frac{dF}{d\eta} < 0, we have F(q) \leq F(p) = c - \epsilon.  Thus, D^{\lambda} \subset H.

In fact H is topologically D^{\lambda} \times D^{n-\lambda}. Furthermore, H \cap M^{c-\epsilon} is homeomorphic to S^{\lambda} \times D^{n-\lambda}. (See Milnor’s Morse Theory for details). ∎

Corollary. M is smooth, compact and has a handle decomposition.

Framing Revisited

When we build M \cup_{\varphi} H, we have an attaching map \varphi: S^\lambda \times D^{n-\lambda} \to \partial M . How may we specify \varphi?

Exercise. If \varphi and \varphi' are isotopic attaching maps then M \cup_{\varphi} H and M \cup_{\varphi'} H are diffeomorphic.

Therefore, we need to specify isotopy classes of \varphi:S^{\lambda} \times D^{n-\lambda} \to \partial M. First we observe that \varphi picks out a trivialization of the normal bundle to S^\lambda. Conversely, using a tubular neighborhood \nu(S^{\lambda}), a trivialization of \nu(S^{\lambda}) for a \lambda-knot gives an embedding S^\lambda \times D^{n-\lambda}. Since isotopic trivialization gives isotopic embedding, so the embedding \varphi is classified by \varphi_0: S^{\lambda} \to \partial M a knot and f: \nu S^{\lambda} \to S^{\lambda} \times \mathbb{R}^{n-\lambda} a trivialization which also picks out for each point q an n-\lambda frame.

Remark: Two trivializations f and g are different by an element in \pi_{\lambda}(GL(n-\lambda,\mathbb{R})). Therefore, we have a bijection between framing of S^{\lambda} up to isotopy and \pi_{\lambda}(GL(n-\lambda,\mathbb{R}) given a choice of base framing.


Branched Covers

Branched Covers

Definition 11. Let M,N be compact k-dimensional manifolds. Let A\subseteq M and B\subseteq N be codimension 2 submanifolds (closed and embedded). A continuous map f:M\to N is a branched cover over B if f(A)=B and f|_{M\setminus A} is a covering map. The degree of this cover is called the branch order.

Example. Let M=N=\mathbb{C}\cup\{\infty\} (Riemann sphere) and A=B=\{0,\infty\}.

Sphere displaying the map z to z^q
f(z)=z^q is a branched cover of degree q

Every Riemann surface is a branch cover over \hat{\mathbb{C}} with finitely many branched points.

Definition 12. The monodromy of a branched cover f:M\to N branched over B is the monodromy of the cover in f:M\setminus A\to N\setminus B.

Remark. Properties are often assigned to f based on monodromy. For example, a cyclic branched cover is branched cover with a cyclic monodromy group.

Motivation. Do cyclic branched covers of S^3 over a knot exist? How do we build X_q\to S^3 over K? We approach this by building covers of S^3\setminus K.

The universal abelian cover is the covering corresponding to the commutator subgroup. This is seen to be an infinite cyclic cover by recalling H_1(S^3\setminus K)  = \mathbb{Z}  \quad\Rightarrow\quad  \pi_1(S^2\setminus K)^{\mathrm{ab}}  = \mathbb{Z}

Build the universal abelian cover \widetilde{X}. Fix a Seifert surface \Sigma for K. Let \mathring{N} \hookrightarrow S^3 be \Sigma\times (-1,1), the interior of an open bicollar. Let Y=S^3\setminus \mathring{N}\setminus K. Observe that \partial Y has two connected components (one for each side of the bicollar), and each is homeomorphic to \mathring{\Sigma}. Call them \partial Y^{+} and \partial Y^{-}.

Take Y_i to be countably many copies of Y. Define \widetilde{X}:=\bigcup Y_i where \partial Y_i^{+} is glued to \partial Y_{i+1}^{-}. Each Y_i maps to S^3\setminus K by gluing \partial Y_i^{+} and \partial Y_i^{-} together. This map is a (one-to-one) covering \mathring{Y_i}\to (S^3\setminus K)\setminus \Sigma. Use these to map \widetilde{X}\to S^3\setminus K.

By construction, \widetilde{X} is a cover, and the shift Y_i\to Y_{i+1} generates the monodromy of the cover which is isomorphic to \mathbb{Z}.

Remark. Regular covers are classified by their monodromy.

Corollary 10. This $\widetilde{X}$ is the universal abelian cover.

Corollary 11. The space X_q = \bigcup_{i=1}^{q-1} Y_i with \partial Y_i^{+} glued to \partial Y_{i+1(\mathrm{mod} q)}^{-} is the order q cyclic cover of S^3\setminus K.

Observe that \rho^{-1}(\partial N(K)) \subseteq X_q for the covering \rho:X_q\to S^3\setminus K is connected, so \partial X_q is a torus and the cover \rho(\partial X_q)\to \partial (S^3\setminus N(K)) is the map S^1\times S^1\xrightarrow{\mathrm{id}\times(q\text{-cover})} S^1\times S^1. (The q copies of the longitude break the meridian into q-fold pieces. All longitudes collapse to a single longitude, and each piece of the meridian covers a single meridian.)

To build \Sigma_q\to S^3 (the q-fold branched cover over K), take X_q \sqcup_{\text{glue}} S^1\times D^2, the covering map on X_q as before, and S^1\times D^2 \xrightarrow{\mathrm{id}\times(z\mapsto z^q)} N(K) covers with the core curve going to K.

Exercise 12. Classify branch covers over the unknot. (Hint: they’re lens spaces.)

Description 2: Poincaré Homology Sphere via Branched Covers

The Poincaré sphere Q is the 5-fold cyclic branched cover over the trefoil.

We will use the surgery presentations of knots. We begin by defining h:S^3\setminus T\to S^1\setminus T that takes K to the unknot.

Surgery -1 on this knot in S^3\setminus S^1 is S^3\setminus\{\text{figure-}8\}

Note that outside of T, we can twist the knot to reverse the crossing.  The tube T goes through the Seifert surface \Sigma.  This surgery may introduce genus onto \Sigma.

Definition 13. A surgery presentation for a knot K is a knot (or link) with surgery coefficients in S^3\setminus S^1 so that the result of the surgery is S^3\setminus K.

Theorem 12. Every knot has a surgery presentation.

Exercise 13. Prove the above theorem. (This is guided by a series of exercises in Rolfsen 6.D.)

To construct X_q of the surgery presentation,

3-cover to form link of unknots

To get \Sigma_q, glue in a solid torus to \partial X_q so that the Seifert curves are longitudes and the other direction are meridians.

Example The Whitehead link with a -1 twist is a surgery presentation of the trefoil. Using this, we get a surgery for \Sigma_5 over the trefoil:

Sigma-5 over the trefoil
\Sigma_5 over the trefoil


Framing and Dehn Surgery

Dehn Surgery

Definition 6. Let K be a knot in S^3, a Dehn surgery on K is the manifold

M=(S^3 - N(K)) \sqcup_h S^1 \times D^2

for some homeomorphism h: \partial(S^1 \times D^2)\xrightarrow{\sim} \partial (S^3 - N(K)) .

The process of gluing in the solid torus can be broken down into two steps: gluing in the meridian disk along its boundary and then gluing in a 3-cell. Since gluing a 3-cell does not change the fundamental group of the space, the fundamental group of the resulting manifold is determined by the image of \partial(\{*\}\times D^2) in \partial (S^3 - N(K)).

In order to compute the fundamental group explicitly, we need a good way to describe the gluing homeomorphism h. To this end, we will introduce a framing of a solid torus which provides us with “coordinates” on \partial (S^3 - N(K)).

Definition 7. A meridian \mu in \partial (S^1 \times D^2) is a simple closed curve that bounds a disk in the solid torus but . A longitude in \partial (S^1 \times D^2) is a simple closed curve isotopic in S^1 \times D^2 to the core curve.

Exercise 7. Show that a simple closed curve \mu is a meridian in S^1 \times D^2 if and only if it is null-homotopic. Show that there are infinitely many isotopy classes of longitude in \partial(S^1 \times D^2).

It follows from the above exercise that there is no canonical choice for the longitude of the solid torus. However, in our application to Dehn-surgery on the knot complement, we will use Seifert surface for a knot K to define the preferred longitude.

Seifert Surface

Definition 8. A Seifert surface for a knot K \subset S^3 is a connected, bicollared, compact surface \Sigma \subseteq S^3 such that \partial \Sigma = K.

Theorem 7. Every knot K\subset S^3 has a Seifert surface.

Proof. We first orient K. We modify all crossings as in the following picture.

We have a finite collection of disjoint circles. In particular, we get the following diagram for the trefoil knot.

For each circle, we attach a disk. Then we glue these disks over crossings with twisted strips as in the following diagram.

Thus, we obtain a surface \Sigma whose boundary is the knot. To see that it is also bicollared, we color each disk (prior to gluing) with red if the boundary is oriented CCW and blue otherwise. By the Jordan’s curve theorem, any pair of circles at each crossing must either be nested or unnested. In either case, the gluing of the disk over the crossing is still consistent with the coloring scheme (see below).

 Thus, the resulting surface \Sigma is bicollared. Hence, it is a Seifert surface for the knot K. ∎

Definition 9. Let \overline{N(K)} be an embedded solid torus in S^3. The preferred longitude \lambda of \overline{N(K)} is \Sigma \cap \partial \overline{N(K)} where \Sigma is a Seifert surface for K.

We remark that Seifert surfaces are not unique. However, the preferred longitude \lambda is unique in the following sense.

Lemma 8. The preferred longitude \lambda for K is the unique simple closed curve up to isotopy in \partial(S^3 -N(K)) such that [\lambda] = 0 in H_1(M(K)).

Proof. By construction, \lambda is the boundary of the Seifert surface, and therefore is homologically trivial. For uniqueness, we consider the Mayer-Vietoris sequence for S^3 = M(K) \cup N(K). In particular, we have

H_2(S^3) \to H_1(\partial (N(K))) \to H_1(N(K))\oplus H_1(M(K)) \to H_1(S^3)

0 \to \langle[\lambda],[\mu]\rangle \to \langle [\lambda]\rangle\oplus \langle [\mu]\rangle \to 0.

Let \gamma be a simple closed curve in \partial (N(K)) that is homologically trivial in H_1(M(K)). In H_1(N(K)) \oplus H_1(M(K)), we have [\gamma] = a [\lambda] for some a \in \mathbb{Z}. Since \gamma is a simple closed curve, we must have [\gamma] = \pm[\lambda]. ∎

A different way to define the same preferred longitude for a knot is via the linking number between \lambda and K. In particular, \text{lk}(\lambda, K)=0. For more details, see Rolfsen – Knots and Links. The notion of the preferred longitude can also be extended to links by choosing one preferred longitude for every link component. Finally,  we should note that the “natural” choice of the simple closed curve running parallel to the knot K is not always the preferred longitude.


The red curve is not the preferred longitude since it has non-zero linking number with knot. The orange curve indicates the two disk components used to build the Seifert surface. The blue curve is the preferred longitude.

Surgery on a Knot

Let \lambda and \mu be the (preferred) longitude and the meridian of N(K) as discussed in previous sections. We orient the meridian \mu using the right-hand rule with respect to the orientation of the knot. The longitude \lambda is oriented so that the algebraic intersection number between \lambda and \mu is +1. The homeomorphism h: \partial(S^1 \times D^2)\xrightarrow{\sim} \partial (S^3 - N(K)) is now specifed by h(\partial(\{*\}\times D^2)) = J. For two coprime integers a,b, we can specify a curve J = a \lambda + b \mu with some sign ambiguity for a,b. However, the ratio b/a is well-defined.

Definition 10. The ratio r = b/a described above is called the surgery coefficient. We denote the manifold obtained by b/a surgery on M(K) by M(K,b/a).

We remark that if \tilde{K} is the mirror image of K, then M(K,b/a) is homeomorphic to M(\tilde{K},-b/a) via an orientation-preserving homeomorphism.

Lemma 9. H_k(M(K,b/a)) is \mathbb{Z} for k = 0,3, \mathbb{Z}/b for k = 1 and 0 otherwise.

Proof. For the first homology, it suffices to compute H_1(S^3 - N(K) \cup_h h(\{*\}\times D^2)). We have the Mayer-Vietoris sequence for reduced homology

H_1(\partial S^2) \to H_1(S^3 - N(K)) \oplus H_1(D^2) \to H_1(M(K,b/a)) \to 0

\langle [c]\rangle \to \langle[\mu] \rangle \oplus 0 \to H_1(M(K,b/a)) \to 0

where the first map is given by [c] \mapsto b [\mu]. By exactness, H_1(M(K,b/a)) \cong \mathbb{Z}/b\mathbb{Z}.

Exercise 8. Complete the proof by computing H_k(M(K,b/a)) for k=2,3.

Description 1: Poincare Homology Sphere as Dehn’s surgery on knot

The Poincare homology sphere first described by Dehn was Q = M(T_{2,3},1) which is the +1 surgery on the right trefoil knot. To distinguish Q and S^3, we compute the fundamental group of Q.

M(T_{2,3}) with the surgery curve \lambda in blue and [\lambda] = d^{-1}.
With respect to the generators x,y,z in the above diagram, we have

\pi_1(M(T_{2,3})) = \langle x,y,z \mid zxz^{-1}y^{-1},xyx^{-1}z^{-1},yzy^{-1}x^{-1}   \rangle.

By substituting z = xyx^{-1}, we get

\pi_1(M(T_{2,3})) =\langle x,y \mid xyx = yxy\rangle.

Since the surgery coefficient is +1, the relation coming from gluing the meridian disk is precisely yx^{2}yx^{-3}=1. Thus, we have

\pi_1(Q) =\langle x,y \mid xyx=yxy, yx^{2}yx^{-3} = 1 \rangle.

Now substitute w = xy, we get

\pi_1(Q) = \langle x,w \mid (xw)^2= w^3=x^5 \rangle.

Exercise 9. Use the presentation of \pi_1(Q) to show that H_1(Q) = 0.

Exercise 10. Recall that I=\langle r,s,t \mid r^2,s^3,t^5,rst\rangle. Show that the formula x\mapsto s and w \to t defines a 2-to-1 onto homomorphism \pi_1(Q) \to I, and so |\pi_1(Q)|=120.

Exercise 11.  Show that \pi_1(Q) is neither S_5 nor I_h.

I. Knots and Links

Basic Definitions

Definition 4. A knot is a smooth (or equivalently PL) embedding S^1 \to S^3 whose image we obtain denote as K. Two knots are equivalent if they are isotopic to each other. The knot corresponding to the embedding of S^1 into S^3 \cong \mathbb{R}^3 \cup \{\infty\} as the unit circle is called the unknot.

Example. The trefoil knot.

The image is taken from the Wikipedia page about the trefoil knot.

One of the first motivating questions of knot theory is to determine if a knot is actually knotted. Furthermore, we ask if we can distinguish two knots from each other. Since knots are considered up to isotopy, the knot complement M(K):=S^3 - K gives us an invariant of knots. In fact, the following theorem by Gordon and Luecke tells us that M(K) is the complete knot invariant.

Theorem (Gordon – Luecke). Two knots J and K are equivalent if and only if M(J) and M(K) are homeomorphic via an orientation preserving homeomorphism.

In order to understand knots, we should study the topological space surrounding it. This allows us to use algebraic invariant such as the homology groups and the fundamental group to study knots. Unfortunately, the homology groups H_*(M(K)) does not help much.

Proposition 2. For any knot K \subset S^3, the homology groups H_k(M(K)) is \mathbb{Z} for k = 0,1 and 0 otherwise.

Proof. We have H_0(M(K))\cong \mathbb{Z} since the M(K) is connected. We shall use the Mayer-Vietoris sequence to compute the higher homology groups.

Let X be a regular neighborhood of K in S^3 and consider the Mayer-Vietoris sequence for S^3 = M(K) \cup X. Note that the intersection M(K)\cap  X  = X - K is homotopic to the solid torus without the core circle. Therefore, X-K deformation retracts onto the torus T. Also observe that X deformation retracts onto its core circle. So for dimension 1, we have the following Mayer-Vietoris sequence

H_2(S^3) \to H_1(T) \to H_1(M(K)) \oplus H_1(X) \to H_1(S^3).

Or equivalently,

0 \to \mathbb{Z}^2 \to H_1(M(K)) \oplus \mathbb{Z}  \to 0,

which implies that H_1(M(K))\cong \mathbb{Z}. The rest of the proof is left as an exercise.

Exercise 3. Complete the proof of the proposition above.

The Knot Group

On the other hand, the fundamental group \pi_1(M(K)), which is also referred to as the knot group, gives us a better invariant for knots. It gives us a way to distinguish a large number of knots from the unknot. In fact, we will use the fundamental group to give an example of an infinite family of distinct knots, namely the torus knot.

Before defining the torus knot, we recall that a meridian of a solid torus is any simple closed curve that bounds a disk. A longitude of a solid torus is any simple closed curve that is isotopic to the core curve.

Definition 5. Let p,q be relatively prime integers. A knot T_{p,q} is a knot that winds p times around the longitude and q times around the meridian of the standard solid torus in \mathbb{R}^3. We denote the knot group of T_{p,q} as G_{p,q}.

Lemma 3. G_{p,q} \cong \langle x,y \mid x^p = y^q \rangle.

Proof.  For brevity, put K := T_{p,q}. We apply van-Kampen’s theorem to the neighborhoods of X - K and Y - K where X is the solid torus in S^3 whose boundary contains K and Y:= \overline{S^3 - X}. Observe that Y is also a solid torus. To see this, observe that the meridian (resp. longtitude) of Y is the longitude (resp. meridian) of X (drawn as the orange solid torus in the picture) (see picture).

The image is generated by Mathematica.

Note that X - K (resp. Y- K) still deformation retracts onto its core circle. Therefore, \pi_1(X-K) = \langle x \rangle and  \pi_1(Y-K) = \langle y \rangle where x (resp. y) is the homotopy class of the longitude in X (resp. Y). The intersection of the two neighborhoods of X-K and Y-K deformation retracts onto T-K which is homotopic to the annulus. Thus, \pi_1(T-K) = \langle a\rangle where a is the homotopy class of the core curve of the annulus.

Since the core curve of the annulus is isotopic to K, we get an embedding \alpha (resp. \beta) of \pi_1(T-K) into \pi_1(X-K) (resp. \pi_1(Y-K)).  By the definition of the torus knot K and the fact that the roles of longitude and meridian of X and Y are switched, we get explicit formulas:

\alpha:  a \mapsto x^p,\quad\text{and}\quad \beta: a \mapsto y^q.

By van-Kampen’s theorem, we get

G_{p,q} \cong \langle x,y \mid x^p = y^q \rangle

as desired. ∎

We remark that T_{\pm 1,q} and T_{p,\pm 1} are the unknots and that T_{p,q} and T_{q,p} are equivalent.

The following theorem by Schreier shows that the unordered pair p,q distinguishes T_{p,q}.

Theorem 4. Let 1 < p < q then G_{p,q} is determined by p,q.

Proof. For brevity, put G:= G_{p,q}Z:= Z(G) the center of G, and C the subgroup generated by x^p. Since x^p = y^q by viewing x^p as either a power of x or y, we see that x^p commutes with both x and y. Therefore, C \subseteq Z which implies that C is a normal subgroup of G. Observe that G/C \cong \mathbb{Z}/p\mathbb{Z} * \mathbb{Z}/q\mathbb{Z}. Using the normal form of elements of  \mathbb{Z}/p\mathbb{Z} * \mathbb{Z}/q\mathbb{Z}, we can show that the center of \mathbb{Z}/p\mathbb{Z} * \mathbb{Z}/q\mathbb{Z} is trivial. Hence, C \supseteq Z. Thus, C = Z. Hence, G/Z \cong \mathbb{Z}/p\mathbb{Z} * \mathbb{Z}/q\mathbb{Z}. The universal property of the free product * tells us that the group \mathbb{Z}/p\mathbb{Z} * \mathbb{Z}/q\mathbb{Z} is uniquely determined by the pair 1<p<q. Hence, G_{p,q} is determined by the pair 1<p<q.∎

Corollary 5. There are infinitely many knots.

The knot group is not a complete knot invariant. The knots T_{p,q} and T_{p,-q} are not equivalent. Dehn showed that T_{2,3} and T{2,-3} are not equivalent.

Exercise 4. Find (or write) a contemporary account of the above fact (using group theory).

We should also mention that there are other non-trivial knots that are not torus knots, for example the figure-8 knot.

The Wirtinger Presentation

We will conclude this post by introducing the Wirtinger presentation of the knot group which gives us a systematic way to obtain a presentation of knot groups. The idea for the proof is to reconstruct M(K) and compute \pi_1 at each stage. We shall describe this construction and use the trefoil knot as an example.

At each crossing of any knot K, we may assume that the knot strands start and terminate at some points on the xy-plane where these points are in a small neighborhood of the crossing. Furthermore, we assume that strands lie entirely on the upper half space z \geq 0. The following picture illustrate the knot positioning for the trefoil knot

 The first piece, A of M(K) is taken to be the upper half-ball of S^3 intersecting M(K). This is homeomorphic to B^3 - \{k \text{ over passing strands}\}. The fundamental group of this space is the free group on k letters x_1, \dots, x_k generated by loops going around these strands.

For the first crossing, we glue into A a 3-ball minus a strand B_1 = B^3 -\{1 \text{ strand} \} as in the following picture.

Observe that the intersection between A and B_1 is a twice punctured disk with the fundamental group F_2 generated by \beta_1 and \beta_2, winding around each puncture. In B_1, the product of these two loops (oriented appropriately) is trivial. By van-Kampen, we get only one additional relation coming in two flavors

Each strand starts at a crossing and end at a crossing, so we have k crossings in total. By applying the same argument above, we get

\pi_1(A \cup B_1 \cup \dots \cup B_k) = \langle x_1,\dots, x_k \mid r_1,\dots,r_k\rangle.

Now observe that the boundary of A is S^2  - \{2k \text{ points}\}. By gluing B_i, we replace a neighbor hood of two punctured points on the  xy\text{-plane} by the hemisphere of B_i as the boundary. Thus, the boundary of A \cup B_1 \cup \dots \cup B_k is just S^2. Observe that by gluing a 3-cell as the lower half-space, we keep the fundamental group unchanged and obtain M(K). Hence, we obtained the Wirtinger presentation of \pi_1(M(K)).

Theorem 6. The knot group of K has the Wirtinger presentation \pi_1(M(K)) = \langle x_1,\dots, x_k \mid r_1,\dots,r_k\rangle.

Exercise 5. Compute the Wirtinger presentation of the trefoil knot.

Exercise 6. Find an explicit isomorphism between G_{2,3} and the Wirtinger presentation of the trefoil knot.

An Introduction and The Icosahedral Group

Introduction and Overview

We begin with Poincaré’s point of view on topology, which is essentially polyhedral. For him, a topological space is constructed by gluing polyhedra along faces. What matters are the matrices obtained from the boundary maps and their numerical invariants, the Betti and torsion numbers. Thanks to the work of Noether we now understand that these are better thought of as homology groups, and the Betti and torsion numbers are the invariants coming from the classification of finitely generated abelian groups. At the time, the emphasis was on determining complete set of numbers to classify topological spaces, and Poincaré conjectured that the Betti and torsion numbers were that set, at least for the three-sphere. The homology sphere provided a counterexample to this initial conjecture and led Poincaré to formulate the more famous (and true!) version of his conjecture. (The Lens spaces were known at the same time and also furnish a counterexample, but their classification up to homeomorphism was not yet known.)

the poincaré conjecture

Poincaré conjectured that a closed orientable 3-manifold M was homeomorphic to the 3-sphere \mathbb{S}^3 if and only if \pi_1(M) = 1. Perelman resolved this conjecture in 2003, and in the process proved a much stronger statement, that if M, N are irreducible closed orientable 3-manifolds that are not Lens spaces, then \pi_1(M) \cong \pi_1(N) if and only if M and N are homeomorphic.

We are not going to use fundamental group isomorphisms to show that our different constructions of the Poincaré sphere are homeomorphic. Geometrization is a very powerful tool, but appealing to it obscures the interrelations among the different constructions we will consider. As some of these constructions are being generalized to settings without such a powerful tool it is valuable to understand them directly. It is also fun!

The one place we will use the Poincaré conjecture is in knowing that there are no “fake \mathbb{S}^3s”. Older results in 3-manifold theory are often stated up to connect-sum with a (now known to not exist) counterexample to the Poincaré conjecture. We’ll save our breath and say no more about the issue.

a word on category

Poincaré worked with finite simplicial complexes, and gave an inadequate proof that this was not a limitation (Stillwell 2012). Some of the constructions we will consider are simplicial in nature; others will construct the homology sphere as a smooth manifold. Further, the idea of a piecewise linear (PL) structure, where the link of every vertex is a topological sphere, is required for our simplicial complexes to be manifolds. To pass between our different constructions we must know how these categories relate, that is,

  • Is every triangulation a PL structure?
  • Is every manifold PL?
  • Is every PL manifold smooth?

In general the answer is no to all of the above, and even when the answer is yes it might not be in a unique way (the Poincaré sphere can be used to construct some of the counterexamples). However in this course we will primarily be working in dimension 3, where Moise proved that every topological 3-manifold has a PL structure unique up to PL homeomorphism (Moise 1952), and Whitehead proved that every PL 3-manifold has a unique smooth structure (Whitehead 1961). As a result we will for the most part glibly ignore issues of category. When dealing with 4-manifolds, however, we will exercise due care.

Moise, Edwin E. Affine structures in 3-manifolds. V. The triangulation theorem and Hauptvermutung. Ann. of Math. (2) 56, (1952). 96–14. MR0048805

Whitehead, J. H. C. The immersion of an open 3-manifold in euclidean 3-space. Proc. London Math. Soc. (3) 11 1961 81–90. MR0124916

The Big Picture

See the syllabus post for an overview of the big picture of the course.

The Icosahedral (Dodecahedral) Group

In constructing the Poincaré sphere we will want to know that the result, while homologically a sphere, is not homeomorphic to the sphere. We will do that by showing that the fundamental group is non-trivial. A good way to show a group is non-trivial is to find a surjection from it to a group known to be non-trivial. We will use the rotational icosahedral group as the target group, and the combinatorics of the icosahedron will appear again and again in the course.

Definition 1. The icosahedral group I_h is the isometry group of the regular icosahedron in euclidean 3-space, and the rotational icosahedral group I is the orientation preserving isometry group.

This is a natural geometric definition, but doesn’t lend itself immediately to a presentation. One could produce a presentation in an ad-hoc fashion, starting with vertex stabilizers and so on, but we will take a different approach.

Definition 2. The triangle group \Delta(\ell, m, n) is the group generated by reflections in a triangle with angles \pi/\ell, \pi/m, \pi/n.

These groups can be realized as reflections in the edges of geodesic triangles in one of the three model plane geometries according to the total angle required. The triangle group is generated by reflections in the three sides, which we will call a, b, c, and presentation

\Delta(\ell, n, m) = \langle a, b, c | a^2, b^2, c^2, (ab)^\ell, (bc)^m, (ca)^n\rangle

The elements r= ab, s=bc, and t=ca correspond to rotational symmetries of order \ell, m, and n about the vertices of the triangle, and they satisfy the equation rs = t^{-1}.

Definition 3. The rotational triangle group \Delta^+ (\ell, m, n) is the subgroup of \Delta(\ell, m, n) generated by r, s, and t from above. It has a presentation

\Delta^+(\ell, m, n) = \langle r, s, t | r^\ell, s^m, t^n, rst \rangle

Lemma 1. The icosahedral group is isomorphic to \Delta(2, 3, 5) and the isomorphism preserves the rotational subgroups.

Proof. The restriction to rotations will be immediate from the geometry of the proof, which is by picture:

The barycentric subdivision of the spherical icosahedron, which provides a model of \Delta(2, 3, 5). The image was created using KaleidoTile by Jeff Weeks.

In the barycentric subdivision of the icosahedron (equivalently of the dodecahedron), fattened to be spherical, each triangle is a 2, 3, 5 triangle. The triangle group acts freely and transitively on the triangles, which includes it in the isometry group. Conversely, an isometry is determined by its action on a single triangle, so must be in the image of the inclusion of the triangle group. ∎

The icosahedral group, from its action on the space of triangles, has 120 elements. It is natural to ask if it is isomorphic to S_5.

Exercise 1. Prove S_5 and I_h are not isomorphic.

Exercise 2. Prove that I is isomorphic to A_5.

The icosahedral group is interesting on its own, for more you could read Klein’s book (1956).

Klein, Felix. Lectures on the icosahedron and the solution of equations of the fifth degree. Translated into English by George Gavin Morrice. Second and revised edition. Dover Publications, Inc., New York, N.Y., 1956. xvi+289 pp. MR0080930

Edited to add ‘irreducible’ in the classification of closed 3-manifolds by fundamental group. Thanks to Dave Futer.


This is our rough schedule of topics. The choice of order is based mostly on my feelings. The course Canvas page will have the calendar of recommended readings in advance of class meetings.


Before stating his famous conjecture Poincaré claimed, in the second supplement to Analysis situs, that a 3-manifold has the homology groups of the 3-sphere if and only if it is homeomorphic to the 3-sphere. Later, in the fifth supplement, and after almost decade of correspondence with Heegaard and other early topologists, Poincaré discovered a counterexample: his famous homology sphere, a 3-manifold that is not homeomorphic to the 3-sphere but has the same homology groups. This discovery led to the well-known the Poincar ́e conjecture: a 3-manifold with trivial fundamental group is homeomorphic to the 3-sphere. Since then, the Poincaré homology sphere has been a central example in low-dimensional toplogy. We will sample themes (constructions, classification theorems, and structure theorems) from the past century and a quarter of low-dimensional topology, using the Poincaré homology sphere as our guide.


Knowledge of the fundamental group, van Kampen’s theorem, homology, and covering space theory, as one would gain in an introductory algebraic topology course.

I. Knots, links and surgery

We will start by introducing the fundamental group of a knot or link complement, and the quotient groups obtained by Dehn surgery. We will also introduce the intimately related idea of a branched cover of a knot or link, and show that every branched cover can be obtained by a surgery. By studying the Lickorish calculus of link surgery diagrams we will present the Poincaré sphere as a surgery on a knot, a (particular) link, and as a branched cover; as well as study surgery as a general technique for producing homology spheres.

II. Four Manifolds

The certain operations and surgeries in the Lickorish calculus are 3-dimensional shadows of operations on 4-manifolds, known as the Kirby calculus. We begin this section by introducing Kirby calculus and using it as a bridge to view the Poincaré sphere as the boundary of a compact 4-manifold. The sojourn in four dimensions continues with the plumbing construction, followed by a realization of the Poincaré sphere as the link of a singularity of an algebraic surface. The relationship between the algebraic surface and the SU(2) representation of the fundamental group will lead us to study the next section.

III. Group actions and the icosahedral group

In this section we will give the classic construction of the Poincaré sphere as a gluing of a dodecahedron, by studying the action on S^3 coming from the previous section. We will spend some time investigating other manifolds constructed as identification spaces and as quotients of actions, and the relation to geometric structures.

IV. 3-manifolds

The considerations of the previous three sections all deal with three manifolds in some way or another. In this section we will study overtly 3-manfiold techniques. First, in 3 dimensions a polyhedral identification space can be decomposed into a Heegaard splitting; this was Poincaré’s original construction, though we will also see a different Heegaard splitting, and spend some time discussing a calculus of moves on splittings. To unify several previous constructions we will then study Seifert fibered 3-manifolds and describe the Seifert fibering of the Poincaré sphere.

V. Modern topics

Time permitting, it would be nice to discuss some of the following: the idea of an L-space (the Poincaré sphere is an example),the conjectural relationship between foliations, orderings on the fundamental group, and L-spaces; open book decompositions and contact topology; Ricci flow in positive curvature.