\documentclass{amsart} \input{diagrams} \begin{document} \title{Typical Problems for Qualifying Exam (with Solutions)} \author{Danny Calegari} \maketitle \vskip 24pt \section{Major Area: Algebraic Topology} \begin{itemize} \item{What is $\pi_2$ of $S^2 \vee S^1$?} It's the free ${\Bbb Z}[t,t^{-1}]$ module on one generator, generated by the characteristic map of $S^2$ into the $S^2$. This can be seen from the fact that $\pi_2(X) = \pi_2(\tilde X)$ for $\tilde X$ the universal cover of $X$, by the homotopy lifting criterion. \item{Calculate $\pi_1(X)$ where $X$ is the three manifold obtained from $T^2 \times I$ by identifying the opposite faces by the glueing map $(1,0) \to (2,1), (0,1) \to (1,1)$} This is an HNN-extension of $$\pi_1(T^2 \times I) = {\Bbb Z} \oplus {\Bbb Z}$$ A presentation for it is given by $$$$ \item{Show that the free group on two generators contains the free group on $n$ generators with finite index} $F_2$ is the fundamental group of the figure 8. This is finitely covered by a ${\Bbb Z}/(n-1){\Bbb Z}$ by the wedge on $n$ circles, whose fundamental group is $F_n$. Hence there is an injection $i:F_n \to F_2$ such that the index is $n-1$. \item{Show that every subgroup of a free group is free} Let $F_n$, the free group in question, be given as $\pi_1(X)$ where $X$ is a wedge of $S^1$'s. Let $G$ be a subgroup of $F_n$, and let $\tilde X$ be the covering space of $X$ induced by $G$. Then $\pi_1(\tilde X)$ is free, since it is a $1$-complex. But $\pi_1(\tilde X) \cong G$, and we are done. \item{Give an example of a pair $(X,A)$ such that $\pi_i(X,A) \ne \pi_i(X/A)$ for some $i$} Let $X=D^2$, $A=S^1 = \partial D^2$. Then $X$ is contractible, and $\pi_1(A)= {\Bbb Z}$, $\pi_i(A) = 0$ for $i \ne 1$. So by the homotopy exact sequence, $\pi_2(X,A) = {\Bbb Z}$, $\pi_i(X,A) = 0$ for $i \ne 2$. But $X/A = S^2$, which has $\pi_3(S^2) = {\Bbb Z} \ne 0$, as required. \item{Give an example of two spaces with equal homotopy groups but unequal homology groups} Let $X = P^2 \times S^3$ and $Y = S^3 \times P^2$. Then $\pi_1(X)=\pi_1(Y) = {\Bbb Z}/2{\Bbb Z}$. Moreover, $\tilde X = \tilde Y = S^2 \times S^3$ so all their higher homotopy groups are equal. However, $X$ is unorientable and $Y$ is orientable, so since they are closed compact $5$-manifolds, $$H_5(Y;{\Bbb Z}) = {\Bbb Z}, H_5(X;{\Bbb Z}) = 0$$ \item{Give an example of two spaces with the same cohomology groups but with a different ring structure} Let $X$ be ${\Bbb CP}^2$ and $Y$ be $S^4 \vee S^2$. Then the ring structure of $H^*(Y)$ is trivial, whereas if $\alpha$ generates $H^2({\Bbb CP}^2)$ then $$\alpha \cup \alpha [{\Bbb CP}^2] = 1$$ \item{Show that a compact surface with sectional curvature positive everywhere is homeomorphic to $S^2$} Embed the surface isometrically in some large dimensional euclidean space. Then take a morse function on the surface given by projection onto some line through the origin. Since the sectional curvature is everywhere positive, there can be no points of index $1$. Consequently, $H_1(S)=0$ and $S$ is a sphere. \item{Calculate the homology (with coefficients in ${\Bbb Z}$) of the Lens space $L(a,b)$. This is a compact orientable closed $3$-manifold, so $H_3 = H_0 = {\Bbb Z}$. It is obtained from $S^3$ by a ${\Bbb Z}/a{\Bbb Z}$ action, so $H_1 = H^2 = {\Bbb Z}/a{\Bbb Z}$. By the universal coefficient theorem, and the fact that $H^3 = H_0 = {\Bbb Z}$, we have that $H_2 = 0$. \item{Prove that for any orientable compact $3$-manifold $M$ with boundary $\partial M$ that half the first rational homology of $\partial M$ is killed by the inclusion into $M$.} Assume first that $\partial M$ is connected and proceed by induction. By the long exact sequence, if the ranks of $H_2(M), H_2(M,\partial M), H_1(\partial M), H_1(M), H_1(M,\partial M)$ are denoted by $a,b,c,b,a$ then we have $2a-2b+c=0$ so $c = 2(b-a)$. But if $k$ is the dimension of the kernel of the inclusion $H_1(\partial M) \to H_1(M)$ then $k-b+a=0$ so $k = b-a = c/2$, as required. As a consequence, $|H_1(M)|= \infty$ for orientable $3$-manifolds with non-sphere boundary. \item{Show that an element of $H_{n-1}$ for an orientable $n$-manifold is represented by a smoothly embedded $n-1$-manifold} By Poincar\' e duality, we have that $H_{n-1}(M;{\Bbb Z}) \cong H^1(M; {\Bbb Z})$. Now elements of $H^1(M)$ are precisely given by homotopy classes of maps of $M$ into $S^1$. For each such homotopy class, choose a smooth representative which is transverse to some fixed point in $S^1$. Then pulling back the inverse image of this point, we get a smooth ${n-1}$-submanifold of $M$ which represents the homology class in question. Note: this technique also works for elements of $H_{n-2}$, by considering smooth maps into ${\Bbb CP}^m$ for $m$ sufficiently large which are transverse to the ``equatorial" ${\Bbb CP}^{m-1}$. \item{Calculate the first Cech cohomology of $S^1$ with coefficients in the sheaf $\mathcal{S}$ where $\Gamma(\mathcal{S},U) = {\Bbb Z}$ for each $U$, and the restriction maps are $\rho_{U_1,U_1\cap U_2} = 1$, $\rho_{U_1,U_1\cap U_3} = 1$, $\rho_{U_2,U_2\cap U_3} = 1$, $\rho_{U_2,U_2\cap U_1} = 1$, $\rho_{U_3,U_3 \cap U_1} = -1$, $\rho_{U_3,U_3 \cap U_2} = 1$. The coboundary map takes a triple $(a,b,c)$ to $(a-b,a+c,b-c)$ which is zero iff $a=b=c=-a$, which is to say, iff $(a,b,c) = (0,0,0)$. Hence the kernel is trivial and $H^0(S^1;\mathcal{S}) = 0$. Moreover, it is clear that this map is surjective, so $H^1(S^1;\mathcal{S})=0$, as required. \item{If a CW complex $\Sigma$ satisfies $H_2(\Sigma) = {\Bbb Z} \oplus {\Bbb Z}$ and $H_i(\Sigma) = 0$ for all $i \ne 2$ then show that $\Sigma$ is homotopy equivalent to $S^2 \vee S^2$.} Firstly, we know by Hurewicz' theorem that $\pi_2(\Sigma) = {\Bbb Z} \oplus {\Bbb Z}$ so there is a map $S^2 \vee S^2 \to \Sigma$ inducing an isomorphism on $\pi_2$. By the long exact sequence of homology, using the mapping cylinder we see that $H_i(M_f,S^2 \vee S^2) = 0$ for all $i$. Hence by the relative Hurewicz theorem, $\pi_1(M_f,S^2 \vee S^2) = 0$ for all $i$. Hence $f$ induces an isomorphism on $\pi_i$ for all $i$, so by Whitehead's theorem, $f$ is a homotopy equivalence. \item{Show that if $G$ is a finitely generated, finitely presented group, then $G$ is the fundamental group of some compact $4$-manifold} We can certainly produce a free group on $n$-generators by adding a collection of $1$-handles to a $0$-handle. Then we need to know that a $2$-handle can be added giving any relation required. But this is easy, since we can certainly find a simple closed curve in the boundary of any four manifold representing any element of the fundamental group, since a map from $S^1$ into a $3$-manifold can be homotoped to be non-intersecting. Hence such a compact $4$-manifold exists, and even one without any $3$-handles or $4$-handles, so that $H_3(M)=H_4(M)=0$ and $H_2$ is torsion free, by the universal coefficient theorem. \item{Show that a simply connected differentiable manifold is orientable} If $M$ were not orientable, it would admit a connected double cover given by the principle ${\Bbb Z}/2{\Bbb Z}$ structure bundle associated to the $GL_n({\Bbb R})$ bundle giving the glueing instructions for some good cover. hence $\pi_1(M)$ would contain ${\Bbb Z}/2{\Bbb Z}$ as a subgroup, contradicting simply-connectedness. \item{Classify $S^3$ bundles over $S^5$} These are given by homotopy classes of maps $[S^5,BS^3]_0$, which is equal to $\pi_5(BS^3)$, which by the homotopy exact sequence of a fibration is equal to $\pi_4(S^3) = {\Bbb Z}/2{\Bbb Z}$. \item{Show that any two embeddings of a connected closed set $X$ in $S^2$ has homeomorphic complements $C_1$, $C_2$} The complement is a collection of open disks in $S^2$, which are classified by $H_0(C_1)$. But by Alexander duality, $H_0(C_1) \cong H^1(X) \oplus {\Bbb Z} \cong H_0(C_2)$ so we are done. \item{Show that ${\Bbb CP}^2$ does not cover any manifold other than itself} We show in fact that ${\Bbb CP}^2$ does not admit any homeomorphism without fixed points. For, by the Lefschetz theorem, for any homeomorphism $f$, $\Lambda(f) = 1 \pm 1 \pm 1 \ne 0$, so $f$ has some fixed points. \item{Calculate the rational homology of $P^2 \times S^3$} By the Kunneth theorem, and the universal coefficient theorem, this is equal to the rational cohomology of $P^2 \times S^3$ which is just equal to ${\Bbb Z}$ when $i = 0,3$ and $0$ elsewhere. \item{If $M$ is a non-orientable closed compact $n$-manifold, show that $H_{n-1}(M;{\Bbb Z})$ contains a ${\Bbb Z}/2{\Bbb Z}$ term.} $M$ is ${\Bbb Z}/2{\Bbb Z}$-orientable, so $H_n(M;{\Bbb Z}/2{\Bbb Z}) = {\Bbb Z}/2{\Bbb Z}$. By the universal coefficient theorem, this is equal to $Tor(H_{n-1},{\Bbb Z}/2{\Bbb Z})$ and the theorem follows. \end{itemize} \section{Special Topic: 3-Manifolds and Geometrization} \begin{itemize} \item{Suppose $M$ is an orientable compact irreducible manifold with $\pi_1(M)$ free. What is $M$?} Since $M$ is irreducible and orientable, we must have $\pi_2(M) = 0$. Moreover, $\pi_1(M)$ is infinite so the universal cover of $M$ is non-compact and has $H_3(M) = 0$. Hence by the Hurewicz theorem, $M$ is a $K(\pi,1)$ and is homotopy equivalent to a wedge of circles by some map $f$. In particular, $M$ has boundary. Arrange for $f$ to be transverse to selected points on $f$. Then the pullback of the intersection can be made incompressible, and therefore since $f$ is a homotopy equivalence, must consist of a collection of disks. By cutting along these disks inductively we reduce the rank of $\pi_1(M)$ until we end up with a homotopy $3$-cell. Alternatively, since $\partial M$ is not an $S^2$ unless $M$ is a $B^3$, it is compressible, so we can find a compressing disk (by the loop theorem) and cut along this. The resulting manifold still has a free fundamental group, so we continue inductively until we are left with a homotopy $3$-cell. Since $M$ was prime, this must in fact be $B^3$, and we have that $M$ is a handlebody, as required. \item{Consider a $+k$ surgery on a trefoil. How can this be related to a Heegard splitting?} A trefoil has tunnel number 1, given by the addition of an obvious arc $\alpha$. Adding a thickened neighborhood of this to the solid torus glued in under surgery gives a genus $2$ handlebody. By the definition of tunnel number, the complement is unknotted and therefore also a genus $2$ handlebody, so this defines a Heegard splitting of $+k$ surgery on the trefoil. \item{Why is it sufficient to specify a homotopy class of essential curves in a torus to define a Dehn surgery?} The first fact is that the mapping class group of the torus is isomorphic to $SL(2,{\Bbb Z})$. This follows by showing that any homeomorphism homotopic to the identity is isotopic to the identity, which can be established by a handle-straightening argument. Then a homotopy class of essential curves describes where the meridional disk is to be glued in, and once this is done, all that is left is to glue in a $B^3$ along its boundary $S^2$. By a theorem of Alexander, there is essentially only one (orientation preserving) way to do this, so the Dehn surgery is specified uniquely, up to isotopy. \item{Give a surgery presentation for a homology $3$-sphere} One easy example is the empty knot. Many other examples are given by $(1,n)$ surgery on any knot. \item{Give a surgery on a knot which gives a reducible manifold} A $+6$ surgery on the right hand trefoil gives $L(3,1) + L(2,1)$. \item{Give an example of a Haken manifold with a separating incompressible surface} Let $X$ and $Y$ be two Haken manifolds with incompressible boundary of the same genus $g$. Then glue $X$ and $Y$ together along any orientation-reversing homeomorphism of their boundaries. By the Seifert-VanKampen theorem, the inclusion of the boundary in the resulting manifold will be $\pi_1$-injective, and therefore incompressible. Moreover, it is obviously separating. A concrete example is given by two non-trivial knot complements. Their boundaries are incompressible, since otherwise by compressing along a disk, we would have that the knots themselves spanned a disk and therefore would be trivial. Moreover, these manifolds are irreducible, since any PL $S^2$ embedded in a knot complement is embedded in $S^3$, and therefore bounds a $B^3$ on either side; consequently, any such $S^2$ bounds a $B^3$ on one side of the knot complement. \item{Show that a manifold $M$ is Haken iff it is incompressible and its fundamental group can be written non-trivially as $\pi_1(M) = A*_B C$ or as $\pi_1(M) = A*_B$.} We construct a smooth manifold from the mapping cone or mapping torus giving these decompositions. If necessary, we add cells to kill off sufficient higher homotopy groups so that there is a map from $M$ to this manifold realising an isomorphism on fundamental groups. By homotopy surgery on this map, therefore, we can ensure that the preimage of some suitably chosen codimension one submanifold is incompressible and two-sided, as required. \item{Give an example of a non-Haken manifold} Any manifold with finite $\pi_1$ and no boundary is not Haken, for its fundamental group cannot contain the fundamental group of any closed surface. For instance, any Lens space is non-Haken. Moreover, any connect sum is non-Haken, since Haken manifolds are required to be irreducible. Finally, examples of irreducible closed non-Haken manifolds with infinite fundamental group are given by Seifert fibered manifolds over hyperbolic triangle orbifolds whose $H_1$ is finite. \item{Show that if $M^3$ is prime and orientable then $\pi_2(M)=0$ or $M^3$ is an $S^2$ bundle over $S^1$.} If $\pi_2(M) \ne 0$ there exists an embedded sphere which does not bound a homotopy $3$-ball, by the sphere theorem. (Note: the hypothesis of orientability is essential here, or ${\Bbb RP}^2 \times S^1$ would be a counterexample). If this were separating, $M$ would not be prime, so it must be separating. But then $M$ has a connect-sum decomposition $K + M'$ where $K$ is an $S^2$ bundle over $S^1$. We must have therefore that $M' = S^3$, for $M$ is prime by hypothesis, and the statement is proved. \item{Show that any Solv manifold is virtually Haken} A Solv manifold is finitely covered by a torus bundle over $S^1$. But any fiber in a surface-bundle over $S^1$ is incompressible and two-sided, since its fundamental group lifts injectively to $surface \times {\Bbb R}$, and is therefore injective. Hence such a finite cover is Haken, as required. \item{Show that any knot has a Seifert surface} This follows by pulling back the image of a transverse point under some characteristic map $S^3 - K \to S^1$ given by the generating element of $H^1(S^3 - K) = {\Bbb Z}$. \item{Give an example of two manifolds which are homotopy equivalent but not homeomorphic} Easy example: thrice-punctured sphere and once-punctured torus. There are examples of Lens spaces with this property too. \item{If two $3$-manifolds have the same $\pi_1$ and $\pi_2$ is zero, are they homotopy equivalent?} If $\pi_1$ is infinite, they are both $K(\pi,1)$'s, by an argument as above, so they are homotopy equivalent. If $\pi_2$ is non-zero, then $S^2\times S^1$ and $D^2 \times S^1$ are not homotopy equivalent. If $\pi_1$ is finite, then Lens spaces furnish many examples. In particular, two Lens spaces $L(p,q)$ and $L(p,q')$ have the same fundamental group, but they are homotopy equivalent iff $\pm qq'$ is a quadratic residue mod $p$. In particular, $L(5,1)$ and $L(5,2)$ are not homotopy equivalent. \item{Give an example of two $3$-manifolds that are homotopy equivalent but not homeomorphic} We know that two Lens spaces $L(p,q)$ and $L(p,q')$ are homeomorphic iff $q'= \pm q$ or $q' = \pm q^{-1}$ mod $p$. But by Whitehead's criterion, (above), this furnishes us with examples such as $L(7,1), L(7,2)$ are homotopy equivalent but not homeomorphic. \item{Let $M$ = $S^3-K$, a knot complement. Show that the splitting surfaces are all tori and that they are ``nested" - i.e. that the manifold can be decomposed as $M_1 \subset M_2 \subset \dots M_n = M$} Since the boundary is an incompressible torus, if we split along an incompressible and boundary-incompressible annulus, it would have to have both sides on the torus. But then the two boundary circles of the annulus would be parallel, and therefore by boundary incompressibility, the transverse arcs of the annulus would have to separate components of the knot in each cross-section. But this is absurd, since then the knot would have at least two components - an "inside" and an "outside" one, contradicting connectedness. Hence the outermost splitting surface is a torus. Since every embedded torus in $S^3$ bounds a solid torus on at least one side, an incompressible torus in $M$ must contain $\partial M$ in the solid-torus side. Split off this side to produce $M_{n-1}$, which is again a knot complement in $S^3$. Either it is trivial, in which case we are done, or the boundary is again incompressible so we repeat. By induction, we are done. \item{Show that any closed compact hyperbolic $3$-manifold contains no ${\Bbb Z} \oplus {\Bbb Z}$ subgroup.} If two hyperbolic isometries $a,b$ commute, then they must have the same fixed points. If $M = {\Bbb H}^3/\Gamma$ with $a,b \in \Gamma$ then $a,b$ are not elliptic, since $M$ is a manifold. Therefore they have the same fixed points at infinity, and are therefore either both parabolic or both hyperbolic. If they are both hyperbolic, they fix the same axis, and therefore by discreteness of $\Gamma$ and compactness of $S^1$, generate a ${\Bbb Z} \oplus {\Bbb Z}/n{\Bbb Z}$ for some $n$. Therefore they are both parabolic and translate elements arbitrarily small distances. But $M$ is a closed compact manifold, so it has a shortest geodesic of positive length. This contradiction proves the statement. \item{What is the Euler number of the $S^1$-bundle determined by the natural Seifert fibered structure on the Poincar\' e homology sphere?} Since the Poincar\' e homology sphere is given as the unit tangent bundle to the $(2,3,5)$ spherical triangle orbifold, the Euler number of the bundle is equal to the Euler characteristic of the base orbifold, which is $$2 - 1/2 - 2/3 - 4/5 = 1/30$$ \item{Let $K \subset S^3$ be a fibered knot. What surgeries on $K$ give a $3$-manifold which fibers over $S^1$?} Since a $3$-manifold which fibers over $S^1$ has rank of $H_1$ positive, only the $0$ surgery is a possible candidate. Let $F$ be the fiber of $S^3 - K$. Then $F$ is a Seifert surface for $K$, so $\partial F$ is a longitude of $K$. Hence $0$ surgery completes $F$ to a closed surface, and we see that the resulting manifold fibers over $S^1$. \item{Give an example of a non-prime $3$-manifold admitting a geometric structure} By the sphere theorem, any such manifold must admit a $S^2 \times S^1$ structure, since this is the only geometry with $\pi_2$ non-trivial. $P^3 + P^3$ is such an example, since it is covered by $S^2 \times S^1$ with covering translations fibre-preserving isometries. \item{Give an example of a homeomorphism $\Psi$ from the punctured torus to itself whose mapping torus does not admit a hyperbolic structure} An example is given by the identity homeomorphism, whose mapping torus is a punctured torus $\times S^1$, which admits an obvious ${\Bbb H}^2 \times {\Bbb R}$ structure. \item{What is a Dehn surgery presentation for the quotient of $S^3$ by the group of order $48$ giving the truncated cube space} By an argument similar to that which establishes that the Poincar\' e homology sphere is given by a $+1$ surgery on a right-hand trefoil, the manifold above is given by a $+2$ surgery on a right-hand trefoil. \item{Why does the Mostow Rigidity theorem fail for infinite volume manifolds?} The Gromov norm of such manifolds is infinite, so in particular, a quasi-isometric lift to ${\Bbb H}^3$ of a map between them need not preserve regular ideal simplices (after straightening). \item{Give an explicit construction of a closed hyperbolic $3$-manifold} The Seifert-Weber dodecahedral space is an example. It is obtained topologically from a dodecahedron by identifying opposite faces after a twist of $\pi/5$. By Poincar\' e's theorem, one need only check that there exists a regular hyperbolic dodecahedron with all dihedral angles equal to $2 \pi/5$. But an ideal hyperbolic dodecahedron has dihedral angles equal to $2 \pi/6$ and a Euclidean dodecahedron has dihedral angles greater thatn $2 \pi/5$ so some such hyperbolic dodecahedron must exist. \item{Suppose $\alpha$ and $\beta$ are simple closed non-intersecting curves on a hyperbolic surface $S$. Show that their geodesic representatives $\alpha'$ and $\beta'$ do not intersect.} Lift $\alpha$ and $\beta$ to $\tilde \alpha$ and $\tilde \beta$ in the universal cover. Then $\tilde \alpha'$ and $\tilde \beta'$ will have the same endpoints at infinity. Then $\alpha'$ and $\beta'$ intersect iff some lift $\tilde \alpha'$ and $\tilde \beta'$ intersect. But then some lift $\tilde \alpha$ and $\tilde \beta$ intersect, since their endpoints at infinity are linked. This contradicts our assumption; therefore $\alpha'$ and $\beta'$ do not intersect. \item{Let an isometry of hyperbolic $3$-space be represented by $A \in PSL(2,{\Bbb C})$. Show that if $tr(A) \in {\Bbb R}, |tr(A)|<2$ then the isometry is elliptic, that if $tr(A)= \pm 2$ then the isometry is parabolic, and that otherwise it is hyperbolic.} We know that the equation $$\frac {az+b} {cz+d} = z$$ has a solution in the extended complex plane. Since trace is invariant under conjugation, we can place this fixed point at infinity. Then we have $c=0,d=1/a$. If $|a|=1$ then either $a= \pm 1$ in which case we have a parabolic isometry, or $a = e^{1\theta}$ in which case we have an elliptic isometry. The only other possibility is that $|a| \ne 1$, in which case we have a hyperbolic isometry, as required. \item{Show that no closed compact surface in ${\Bbb R}^3$ can have strictly negative curvature everywhere.} We take a height function on the surface which is a Morse function. This must have some critical point of index $0$, since $H_0$ is not zero, by the Morse inequalities (for instance). But at such a point, the curvature is clearly non-negative. \item{Show that there are ${\Bbb R}^3$ hyperbolic structures on a pair of pants with geodesic boundary.} By the standard construction, there is exactly one hyperbolic structure on a right angled hexagon such that the three ``right-angled" sides have lengths $a,b,c$. We double this to produce a pair of pants with prescribed geodesic boundary lengths. Conversely, given any such pair of pants, double it to produce a hyperbolic structure on the closed surface of genus $2$. Then rotation in the obvious axis of symmetry fixes three closed geodesics which must intersect the boundary arcs at right angles. Moreover, by the obvious symmetry which interchanges opposite components of the arcs, these geodesics cut the boundary geodesics into two equal pieces. This completes the proof. \item{Let $A$ be a regular ideal hyperbolic cube and $B$ a regular ideal hyperbolic tetrahedron. What is the ratio $vol(A)/vol(B)$?} We can decompose $A$ into a disjoint union of four ideal tetrahedra by the obvious symmetry of the alternating vertices of $A$. The ``middle" tetrahedron is clearly regular. Moreover, putting one of the vertices of the ideal cube at infinity, we see that by symmetry, the simplex parameter of one of the other tetrahedra is a sixth root of unity - i.e. it too is regular. Hence $A$ can be decomposed into $4$ regular ideal tetrahedra, and the ratio is $4$, as required. \item{Show that there are no non-trivial sequences of the volume function in dimension $n>3$ for closed hyperbolic manifolds} By Margulis' Lemma, we have a decomposition of such manifolds into a ``thick" piece and a ``thin" piece, such that every thin piece is a collar neighborhood of short geodesic. By compactness and the bound below on the injectivity radius on the thick piece, for a given volume range, there are only finitely many diffeomorphism types of the thick piece (follows by covering the piece with open balls and investigating the combinatorial type of their intersections.) Now, by the Seifert-Van-Kampen theorem, excising the thin pieces from the manifolds does not affect their fundamental groups. Therefore by Mostow rigidity, there are only finitely many manifolds in a given volume range, as required. \item{Give an example of two hyperbolic link complements having the same volume} Many such examples are furnished by Colin Adams' result that incompressible thrice punctured spheres in hyperbolic manifolds are totally geodesic (and rigid), and therefore the manifolds can be ``cut" and ``reglued" along such pieces to produce new manifolds with the same volume. Other examples are given by manifolds built from the same ideal tetrahedra, like the many manifolds (for example, the Whitehead link complement) obtained by identifying faces of the regular ideal octahedron. \end{itemize} \section{Minor Topic: Complex Analysis} \begin{itemize} \item{How would you write down the power series for $tan z$?} In general for an analytic function $f$, the Taylor series can be calculated by the usual method to give $$f(z) = f(0) + f'(0)z + f^{(2)}(0)z^2/2 + f^{(3)}(0)z^3/6 + \dots$$ \item{What is the radius of convergence of a power series?} The radius of convergence of a power series of a meromorphic function corresponds to the distance between the centre of the expansion and the nearest pole of the function. This follows from the formula for the radius of convergence: if we write $$f(z) = a_0 + a_1z + a_2z^2 + \dots$$ then the radius of convergence of the power series $R$ satisfies $$1/R = \limsup_j |a_j|^{1/j}$$ For, if we choose a point with $|z|$ bigger than this value, infinitely many of the values diverge; whereas if we choose a point with $|z|$ smaller than this value, $|f(z)|$ can be bounded by a convergent geometric series. To see that there exists a pole somewhere on the radius of convergence, note that otherwise $|f(z)|$ is bounded below on some neighborhood of the disk of this radius, and by the Cauchy integral formula for $f^{(n)}$, we can estimate $\limsup_j |a_j|^{1/j}$ above. \item{Show that if all the zeroes of a polynomial lie in a half-plane, then all zeroes of the derivative lie in the same half plane} Write $$P(z) = (z-\alpha_1)(z-\alpha_2) \dots (z-\alpha_n)$$ Then $$P'(z)/P(z) = \sum_i 1/(z-\alpha_i)$$ which for $z$ on the other side of the half-plane, is a sum of non-zero complex numbers with a positive projection onto a fixed line. Hence the sum is non-zero, so $P'(z)$ is non-zero in this half-plane, and we are done. \item{What is the area of a spherical triangle?} If the angles of the triangle are $\alpha,\beta,\gamma$ then the area of the triangle is $\alpha+ \beta + \gamma - \pi$. For, continue the geodesic sides of the triangle in a great circle to dissect the sphere into five regions, two of area equal to the triangle in question, and three lunes of area $2(\pi - \alpha), 2(\pi - \beta), 2(\pi - \gamma)$ so that we have a formula $$Area = (4 \pi - 2(\pi- \alpha) - 2(\pi - \beta) - 2(\pi - \gamma))/2 = \alpha + \beta + \gamma - \pi$$ \item{What is a complex torus?} A complex torus is a quotient of ${\Bbb C}$ by the translation group of some discrete lattice. \item{What is the automorphism group of a complex torus?} In general, it will be ${\Bbb R}/{\Bbb Z} \oplus {\Bbb R} / {\Bbb Z}$. However, if the torus is square or hexagonal, its automorphism group has a central extension of order $2$ and $3$ respectively. \item{If $f_n$ is a family of holomorphic functions such that $f_n \to f$ uniformly on compact subsets in some domain $\Omega$, what can you say about $f_n'$? In fact, by Weierstrass' theorem, the same is true for $f_n'$, by Cauchy's integral theorem. Note in fact that the hypothesis that these functions be holomorphic is essential, since in the smooth case, it is not true even for functions on ${\Bbb R}$. \item{Give an example of a sequence $f_n \to f$ where every $f_n$ is holomorphic and injective, and $f$ is not. Is this the most general such example?} If $f_n$ is injective, then in particular $f_n'$ is never zero. Hence by Cauchy's theorem, $f'$ is either never zero or identically zero. If it is never zero, it is locally injective, and therefore by continuity, it is injective. Otherwise it is constant, and certainly not injective - for instance, the sequence $f_n(z) = z/n$ converges uniformly on compact sets to $f(z)=0$. \item{Why is there no conformal automorphism from the punctured disk to an annulus?} If there were such an automorphism, we could extend it by the reflection principle to an automorphism from ${\Bbb C}^*$ to $D$ and then to an automorphism from ${\Bbb C}$ to $D$ which would be an entire bounded function and therefore constant by Liouville's theorem, contradicting our assumption. \item{Show that for a doubly periodic function $f$ that the number of zeroes of $f$ and the number of poles of $f$ (counted with multiplicity) is equal.} If $f$ is doubly periodic, then so is $f'$. Then integrating $f'/f$ around a generic fundamental domain (i.e. one without poles or zeroes on it) we get the sum of the zeroes minus the number of poles, which is also zero, since by periodicity, this integral vanishes. Alternatively, such an $f$ determines an orientation preserving map from $T$ to $\hat {\Bbb C}$ which therefore has the same number of preimages (equal to the Brouwer degree) for any point. \item{Suppose $f_i$ are harmonic functions on the unit disk $D$. Show that no linear combination of the $f_i$ can be negative on $\partial D$ and positive at some point in the interior of $D$} Since the Laplacian is a linear operator, any linear combination of the $f_i$'s will be a harmonic function. Then a harmonic function is determined by its values on the boundary of a disk by the visual extension which is a weighted mean, and therefore if the boundary values are all negative, the function is negative everywhere in the disk. Alternatively, by the maximum principle for harmonic functions, a harmonic function cannot achieve a maximum on the interior of a disk, and therefore cannot be positive if it is negative on the boundary. \item{Find the poles and residues of $1/\sin(z)$} The poles of $1/\sin(z)$ are precisely at the zeroes of $\sin(z)$, which are at $n\pi$ for $n$ an integer. This follows from the additive formula for $\sin(z)$ given by $$\sin(a+bi) = \sin(a)\cos(bi) + \sin(bi)\cos(a)$$ $$= \sin(a)\cosh(b) + i\sinh(b)\cos(a)$$ which is zero when $\sin(a)=0$ (and therefore $\cos(a) \ne 0$) and $\sinh(b)=0$, which is when $b=0$ and $a=n\pi$, as stated above. The residue at $n\pi$ is given by the constant term of $(z-n\pi)/\sin(z)$ which is equal to the reciprocal of the limit of $\sin(z)/(z-n\pi)$ at $n\pi$, which is $1$ or $-1$ according to whether $n$ is even or odd. \item{Give the formula for a conformal map from the unit disk to the inside of a polygon with angles $2\pi - \beta_i \pi$} This is just the Schwarz-Christoffel Formula, which is given by $$F(w) = C \int_0^w \prod_{k=1}^{n-1} (w-w_k)^{- \beta_k} dw + C'$$ where $C,C'$ are complex constants, and the $w_i$ are points on the unit circle. Note that in this case the polygon has $n$ sides. In the case of a triangle, this formula simplifies to $$F(w) = \int_0^w w^{- \beta_1}(w-1)^{- \beta_2} dw$$ This maps the unit disk onto a triangle with angles $\pi(1 - \beta_1)$, $\pi(1 - \beta_2)$ and $\pi(\beta_1 + \beta_2 - 1)$ \item{Show that a continuous real-valued function $u$ on some region $\Omega$ which has the mean-value property is harmonic} Since being harmonic is a local property, we restrict our attention to the case that $\Omega$ is a small disk. We construct a harmonic function $v$ on this disk by the visual extension (Poisson's formula). Then such a function agrees with $u$ on the boundary, and $u-v$ satisfies the mean-value property, since both $u$ and $v$ do. But the mean-value property implies the maximum and minimum principle, so since $u-v$ is identically zero on the boundary of the disk, it is identically zero on the interior. Hence $u$ is harmonic, as required. \end{itemize} \end{document}