Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 13 (Algebraic Topology)

Let $M$ be the given closed three-dimensional manifold. It is connected by the meaning of simply connected. An orientation chosen at one point can be transported along paths, and its sign change around loops defines a homomorphism ${\pi }_1(M)\to \{\pm 1\}$. Since the fundamental group is trivial, this monodromy vanishes; thus $M$ is orientable.

The first homology group is the abelianization of the fundamental group, giving $H_1(M;\mathbb{Z})=0$. Also $H_0(M;\mathbb{Z})=\mathbb{Z}$. The universal coefficient theorem for cohomology in degree one givesBoth outer groups vanish: the first because $\mathbb{Z}$ is free and the second because $H_1=0$. Hence $H^1=0$. Integral Poincare duality now gives $H_2(M;\mathbb{Z})\cong H^1(M;\mathbb{Z})=0$ and $H_3(M;\mathbb{Z})\cong H^0(M;\mathbb{Z})=\mathbb{Z}$. Homology in degrees greater than three vanishes by the same duality with negative-degree cohomology. ConsequentlyThese are precisely the homology of a sphere for $S^3$, proving that simply connected closed three-manifolds are homology spheres. No homeomorphism classification of three-manifolds is needed.

The universal cover of $S^1\times S^2$ isBecause $S^3$ is simply connected, any continuous map $f:S^3\to S^1\times S^2$ lifts to $\tilde{f}:S^3\to \mathbb{R}\times S^2$ after choosing a lift of one base point. The lifting criterion for a covering space applies because $f_{\ast }{\pi }_1(S^3)=0$. Equivalently, path lifting defines the lift and simple connectedness makes its value independent of the chosen path.

The cover deformation retracts onto $\{0\}\times S^2$, so $H_3(\mathbb{R}\times S^2;\mathbb{Z})=0$. Since $f=p\circ \tilde{f}$, functoriality of homology givesTherefore $\boxed{f_{\ast }=0\text{ and }\mathrm{deg}f=0}$, with the last equality using the oriented three-dimensional fundamental classes.

Put $\Gamma ={\pi }_1(X,x)$ and $R=\mathbb{Z}[\Gamma ]$. Give $\mathbb{Z}$ its trivial left $R$-module structure, through the augmentation $\epsilon :R\to \mathbb{Z}$, $\sum n_{g}g\mapsto \sum n_g$. The universal cover $\tilde{X}$ has the lifted cell structure, with $\Gamma$ acting by deck transformations.

For each cell of $X$, choose one lifted cell and orient every translate compatibly with its projection. The translates form one copy of the regular $R$-module. Thusis a free left $R$-module, and its boundary is $R$-linear. The augmentation $C_0(\tilde{X})\to \mathbb{Z}$ sends every vertex to one. The contractibility of $\tilde{X}$ says that its reduced homology vanishes, so the augmented complexis exact. It is therefore a free resolution of the trivial module. By the definition of the Ext functor,

An $R$-linear cochain with values in the trivial module has the same value on every translate of a lifted cell. It is therefore exactly an ordinary integral cellular cochain on $X$. The correspondence does not depend on which lift was initially chosen. It also respects the coboundary: applying a constant-on-orbits cochain to the lifted cellular boundary sums precisely the incidence coefficients of the boundary in the base. Henceas cochain complexes, not just as graded groups. Cellular cohomology computes the cohomology of a cell complex, givingAt $i=0$ both sides are $\mathbb{Z}$, by connectedness and ${\mathrm{Hom}}_R(\mathbb{Z},\mathbb{Z})=\mathbb{Z}$. This is the cellular free resolution from a contractible universal cover description of group cohomology.

Work over $\mathbb{Q}$ and put $b_i=\dim H^i(M;\mathbb{Q})$. For a closed orientable six-dimensional manifold, Poincare duality gives $b_i=b_{6-i}$, soThe remaining Betti number is even. Indeed, the cup product pairingis nondegenerate by duality and skew-symmetric by graded commutativity, since $(-1{)}^{3\cdot 3}=-1$. If its dimension were odd, a matrix $A$ for it would satisfy $\det A=\det (-A^T)=-\det A$, forcing $\det A=0$ in characteristic zero, a contradiction. Thus $b_3$ is even andThis is the Euler characteristic parity in dimensions congruent to two modulo four. If the manifold has several components, apply the argument to each component and add.

The orientability assumption is essential. The Real projective space ${\mathbb{RP}}^6$ has one cell in each dimension from zero to six, givingIt is nonorientable: its double cover $S^6$ has the orientation-reversing antipodal deck transformation, of degree $(-1{)}^7=-1$.

For the construction, ${\mathbb{CP}}^3$ is a closed orientable real six-dimensional manifold with cells in dimensions $0,2,4,6$, hence Euler characteristic four. The product $S^3\times S^3$ is also closed and orientable, with Euler characteristic $\chi (S^3{)}^2=0$. The six-sphere has Euler characteristic two.

In dimension six, removing an open ball subtracts one from the Euler characteristic. To see this, write $M$ as the punctured manifold joined to a closed ball along $S^5$, whose Euler characteristic is zero, and apply the cellular inclusion-exclusion formula. Gluing two punctured manifolds along $S^5$ therefore givesA connected sum of $r$ copies of ${\mathbb{CP}}^3$ and $s$ copies of $S^3\times S^3$ haswith the empty sum interpreted as $S^6$. Given any even integer $n$, takeBoth are nonnegative integers and the displayed Euler characteristic equals $n$. Hence every even integer is the Euler characteristic of a closed oriented six-manifold, including zero and negative integers; the construction is connected.

Use integral singular chains. The inclusion $Y\subseteq X$ identifies $C_k(Y)$ with the subgroup of $C_k(X)$ generated by the singular simplices whose images lie in $Y$. It is a chain subcomplex. DefineThis boundary is well defined because $\partial C_k(Y)\subseteq C_{k-1}(Y)$, and its square is zero because ${\partial }^2=0$. Its homology is, by definition, $H_k(X,Y)$. A relative cycle has a representative $c$ with $\partial c\in C_{k-1}(Y)$; representatives of the same relative homology class differ by $\partial b+y$, where $b\in C_{k+1}(X)$ and $y\in C_k(Y)$.

Inclusion and quotient induce homomorphisms $i_{\ast }:H_k(Y)\to H_k(X)$ and $j_{\ast }:H_k(X)\to H_k(X,Y)$. Define the relative homology connecting homomorphism byHere $\partial c$ is a cycle in $Y$ since ${\partial }^{2}c=0$. If $c$ is changed to $c+\partial b+y$, its boundary changes by $\partial y$, a boundary in $Y$, proving independence of representatives. The formula is additive, so $\delta$ is a homomorphism.

The long exact sequence in relative homology isand ends inWe prove all three types of exactness directly, so no unproved abstract lemma about chain complexes is required.

At $H_k(X)$, $j_{\ast }{i}_{\ast }=0$ because a chain in $Y$ is zero in the relative chain group. Conversely, let an absolute cycle $c$ have $j_{\ast }[c]=0$. The definition of a relative boundary gives $c=\partial b+y$ with $y\in C_k(Y)$. Since $c$ is a cycle, $\partial y=0$. Thus $[c]=i_{\ast }[y]$, proving $\ker j_{\ast }=\mathrm{im}{i}_{\ast }$.

At $H_k(X,Y)$, $\delta j_{\ast }=0$ because an absolute cycle has zero boundary. Conversely, suppose $\delta [c]=0$. Then $\partial c=\partial y$ for some $y\in C_k(Y)$. Therefore $c-y$ is an absolute cycle and $j_{\ast }[c-y]=[c]$. Hence $\ker \delta =\mathrm{im}{j}_{\ast }$.

At $H_{k-1}(Y)$, $i_{\ast }\delta =0$ because $\partial c$ is a boundary in $X$. Conversely, if a cycle $y\in C_{k-1}(Y)$ becomes zero in $H_{k-1}(X)$, then $y=\partial c$ for some $c\in C_k(X)$. That $c$ is a relative cycle, and $\delta [c]=[y]$. Thus $\ker i_{\ast }=\mathrm{im}\delta$.

These arguments also prove the assertions at shifted indices. Finally, every zero-dimensional relative cycle lifts to a chain in $C_0(X)$, and every zero-chain is an absolute cycle; this proves surjectivity onto $H_0(X,Y)$. All maps, representative changes, compositions and exactness assertions have consequently been justified from the chain definitions.

We first prove that fixed-point-free sphere maps are homotopic to the antipodal map. If $f:S^d\to S^d$ has no fixed point, defineThe denominator could vanish only if $f(x)$ is a positive multiple of $x$. Both are unit vectors, so this would require $f(x)=x$ and $t=1/2$, excluded by the hypothesis. Thus $H$ is a continuous homotopy from $f$ to $x\mapsto -x$.

On $S^{2r}$, with $r\ge 1$, the antipodal map has degree $(-1{)}^{2r+1}=-1$, since it is the boundary map of the ambient linear transformation $-I$ on ${\mathbb{R}}^{2r+1}$. Homotopy invariance therefore gives degree minus one for every fixed-point-free action map. The degree of each homeomorphism is $\pm 1$, and functoriality makes degree a group homomorphismEvery nonidentity element in a free action has degree minus one, so this homomorphism has trivial kernel. HenceFor $S^0$ the same bound follows immediately from its two points. This proves that free actions on even-dimensional spheres have order at most two, without an Euler-characteristic argument requiring a cell structure on the quotient.

Identify $S^{2b-1}$ with the unit sphere in ${\mathbb{C}}^b$, and put $\zeta =e^{2\pi i/a}$. The generator of the cyclic group $\mathbb{Z}/a$ acts byIt preserves the sphere because $|\zeta |=1$. If its $k$th power fixes a point, then $({\zeta }^k-1)z_j=0$ for every $j$. Some coordinate is nonzero, so ${\zeta }^k=1$, meaning $k\equiv 0(\mathrm{mod}a)$. Thus the scalar root-of-unity action is free. The case $a=1$ is the trivial group and also satisfies freeness. The quotient is a standard lens space.

Suppose a nontrivial finite group acts freely on ${\mathbb{R}}^n$ by homeomorphisms. Choose a nonidentity element of finite order $m$, a prime $p$ dividing $m$, and the power of that element with exponent $m/p$. It has order $p$, so a subgroup $C_p$ also acts freely.

Let $M={\mathbb{R}}^n/C_p$. A finite free action is properly discontinuous: around any point choose mutually disjoint neighborhoods of the finitely many orbit points, then intersect their inverse translates to obtain an evenly covered neighborhood. Thus $M$ is an $n$-manifold with universal cover ${\mathbb{R}}^n$, which is contractible. The group-cohomology argument of Question 2 applies even if no cell structure has been chosen: singular simplices in $M$ lift to the cover, their lifts form free deck-group orbits, and the augmented singular chains of the contractible cover form an exact free resolution. Therefore

We calculate the right-hand side explicitly. Write $R=\mathbb{Z}[C_p]$, let $t$ generate $C_p$, and set $N=1+t+\cdots +t^{p-1}$. The periodic resolution of a finite cyclic group isIts exactness can be seen in coefficients. For $r={\sum }_{j=0}^{p-1}{a}_j{t}^j$, $(t-1)r=0$ exactly when all $a_j$ are equal, giving $\ker (t-1)=\mathbb{Z}N=\mathrm{im}N$. Also $Nr=\epsilon (r)N$, so $\ker N=\ker \epsilon$. If $\epsilon (r)=0$, thenHence $\ker N=\ker \epsilon =(t-1)R$, proving exactness everywhere.

Apply ${\mathrm{Hom}}_R(-,\mathbb{Z})$ with the trivial action. Multiplication by $t-1$ induces zero, and multiplication by $N$ induces multiplication by $p$. The cochain complex is thereforeIts positive even-degree cohomology iswhile its positive odd-degree cohomology is zero.

This contradicts the dimension of the quotient manifold. The noncompact form of Poincaré duality for noncompact manifolds, with the orientation local system when needed, isHere Borel-Moore homology is the homology of locally finite chains. Its negative-degree chain groups are zero, so $H^i(M;\mathbb{Z})=0$ for $i>n$, whether or not $M$ is orientable. Taking any $2j>n$ contradicts the displayed cyclic group calculation. ThusThis proves finite groups cannot act freely on Euclidean space for general homeomorphisms; it does not assume that the action consists of affine maps or isometries.

Let $\gamma$ be the real tautological line bundle over ${\mathbb{RP}}^2$. A homogeneous polynomial of degree $a$ defines a section of the real line bundleIndeed, on a projective line $\ell \subset {\mathbb{R}}^3$ its restriction satisfies $F(rv)=r^{a}F(v)$, exactly the transformation law for a linear functional on ${\ell }^{\otimes a}$. Its zero set is well defined without treating an odd-degree polynomial as a global real-valued function on projective space.

Put $u=w_1(\gamma )\in H^1({\mathbb{RP}}^2;{\mathbb{F}}_2)$. This is the nonzero generator: the tautological line changes sign along the generating projective loop. Dualization preserves this sign, while tensoring multiplies transition signs, soFor generic coefficients both polynomial sections are transverse to zero and their zero curves meet transversely. This genericity follows from parametric transversality: at each projective point some coordinate is nonzero, and varying the coefficient of its $a$th power varies the section's value freely; the two polynomial coefficient sets vary the two values independently. The common zero set is consequently a zero-dimensional submanifold of a compact space, and is finite.

We justify the needed parity of zeros of a real line-bundle section. Along a closed loop transverse to the zero curve, trivialize the pulled-back line bundle on the cut interval. At a simple zero the real component function changes sign. The product of these sign changes equals the sign of the bundle's transition around the loop, which is the evaluation of its first Stiefel–Whitney class. Therefore the intersection parity of the zero curve with every loop equals the evaluation of $w_1$ on that loop. This says that its mod-two Poincare dual is $w_1$, so

The mod-two Poincare duality intersection rule givesThere are no orientation signs in these coefficients, so every transverse intersection contributes one. Two distinct projective lines meet in exactly one point; both have dual class $u$ by the preceding degree-one calculation. Hence $⟨u^2,[{\mathbb{RP}}^2{]}_2⟩=1$, andThis proves the parity of real projective plane-curve intersections.

Equality with $ab$ need not hold. Choose $a=1$, $b=2$, with the line and conicOn the line, the conic equation becomes $x^2+y^2=0$, which has no nonzero real solution. Thus $\boxed{|A\cap B|=0\ne 2=ab}$. Both real curves are smooth and nonempty, and they are disjoint, so their real intersection is transverse vacuously. Even over the complex numbers they meet in two distinct transverse points $[1,i,0]$ and $[1,-i,0]$; the example is not a tangency or shared-component exception. On the unit sphere $F^2+g^2$ has a positive minimum, so disjointness also persists under sufficiently small real coefficient perturbations, so it is compatible with the generic real situation.