Past exam of the geometry course of the University of Oxford / 2010 (B3, DMAB 2642)

The classification of compact surfaces says that every compact connected surface without boundary is homeomorphic to exactly one of the following: the sphere, an orientable surface obtained as the connected sum of $g\ge 1$ tori, or a nonorientable surface obtained as the connected sum of $k\ge 1$ real projective planes. The integers $g$ and $k$ are unique, and the orientable and nonorientable families do not overlap. Including $g=0$ makes the sphere the genus-zero member of the orientable family.

Their Euler characteristics are $2-2g$ and $2-k$, respectively. Thus orientability and the Euler characteristic determine the closed connected surface. If boundaries are allowed, one additionally specifies the number of boundary circles and removes that many open discs from a closed model.

For a finite polygonal CW complex decomposition of a compact surface, with $V$ vertices, $E$ edges and $F$ faces after all identifications,This is the Euler characteristic, independent of the chosen subdivision. In a polygonal presentation, count equivalence classes of vertices and paired edges, rather than their number before gluing.

Use a square with boundary word $ab{a}^{-1}{b}^{-1}$, identifying opposite sides with matching labels and arrows. All four vertices become one, the four sides form two edge classes, and there is one face. Therefore the torus has

A polygonal model of the real projective plane is a bigon with boundary word $aa$, with the equally labelled arrows identified. Its two vertices become one, its sides form one edge, and it has one face. ThusEquivalently, glue antipodal boundary points of a disc.

A square with boundary word $ab{a}^{-1}b$ gives the Klein bottle. The side identifications produce one vertex, two edges and one face. ConsequentlyUnlike the torus, this surface is nonorientable; its equal Euler characteristic alone does not distinguish them.

Glue two triangular discs along their boundaries to obtain a sphere. The resulting CW complex decomposition has three vertices, three edges and two faces, givingThe two triangles represent the two closed hemispheres, with an equator subdivided into three edges.

For $g\ge 1$, the standard commutator-word polygon has one face, $2g$ paired edge classes and a single vertex class. ThusAn orientable surface admits continuously compatible local orientations; for a smooth surface, this is equivalently a continuous choice of unit normal vector in an oriented ambient three-dimensional space when such an embedding is available. In a polygonal model, orientability means the faces can be oriented so that induced boundary directions on each glued edge oppose one another. Here each edge label occurs once positively and once negatively. The orientation of the polygon therefore descends consistently across every seam. Hence $X$ is orientable, and it is the closed orientable surface of genus $g$.

For a regular parametric surface $r(u,v)$, setThe first fundamental form is the induced Riemannian metricThe second fundamental form, for this choice of unit normal vector, isThe Gaussian curvature is the product of the principal curvatures, equivalently the determinant of the shape operator:Reversing the normal reverses the second fundamental form and both principal curvatures, while leaving $K$ unchanged.

Work on a regular patch with $f(u)>0$. Differentiating the surface of revolution givesThe arc-length condition gives $E=1$, $F=0$, $G=f^2$. Also $\Vert r_u\times r_v\Vert =f$, so takeThe second derivatives yieldThus the first fundamental form and second fundamental form areThe opposite normal reverses the second displayed form. At a smooth pole the revolution coordinates degenerate, so curvature is understood by a regular local chart or by continuity.

Differentiating the arc-length identity gives $f^{\prime }{f}^{\prime \prime }+h^{\prime }{h}^{\prime \prime }=0$. Multiply by $f^{\prime }$ and use $(f^{\prime }{)}^2+(h^{\prime }{)}^2=1$ to obtainThis derivation does not divide by $h^{\prime }$, so it remains valid where $h^{\prime }=0$. Inserting the first fundamental form and second fundamental form computed above gives the Gaussian curvature of a surface of revolution:

Choose $f(u)=\sin u$ and $h(u)=\cos u$ for $0\le u\le \pi$. Adding the smooth poles at the endpoints gives the unit sphere. Its meridian is parametrized by arc length, and $-f^{\prime \prime }/f=1$ away from the poles, hence everywhere by continuity. Thus the unit sphere has constant Gaussian curvature one.

There is no compact smooth surface of revolution without boundary of constant Gaussian curvature minus one in this setting. On a compact such surface, the distance $f$ from the rotation axis attains a positive maximum. The maximum lies on a regular revolution patch, where the meridian arc length coordinate gives $f^{\prime \prime }\le 0$. Hence $K=-f^{\prime \prime }/f\ge 0$ there, contradicting $K=-1$. Equivalently, constant negative curvature would require $f^{\prime \prime }=f>0$, excluding any positive radial maximum. Compact patches with boundary would not be excluded by this argument, but they are not closed surfaces.

For an affinely parametrized geodesic $s\mapsto (x(s),y(s))$, the energy Lagrangian is $L=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. The coordinate $x$ is cyclic, so the Euler-Lagrange equation gives $\dot{x}/y^2=c$, a constant. Parametrize a nonconstant geodesic by hyperbolic arc length; thenIf $c=0$, $x$ is constant: these are vertical lines. If $c\ne 0$, on intervals with $\dot{y}\ne 0$,Integration gives $(x-a{)}^2+y^2=c^{-2}$. The two branches join smoothly at the highest point. Conversely these curves satisfy the geodesic equations, or are images of vertical lines under half-plane Möbius transformations that are isometries. Thus the complete nonconstant geodesics arenamely vertical lines and upper semicircles orthogonal to the real axis. Their subarcs give geodesic segments.

The Cayley transform $w=(z-i)/(z+i)$ maps the upper half-plane conformally onto the unit disc. Its inverse is $z=i(1+w)/(1-w)$, withTransporting the given hyperbolic metric therefore gives the Poincare disc modelThe Cayley transform is an isometry for this metric, so it carries the half-plane geodesics to disc geodesics. A Möbius transformation sends generalized circles to generalized circles and preserves angles. Hence these are exactly diameters and arcs of circles meeting the unit circle orthogonally. Multiplying this Riemannian metric by any positive constant leaves the unparametrized geodesics unchanged.

For distinct points $a,b$ on a geodesic, the given disc isometry $z\mapsto e^{i\theta }(z-a)/(1-\bar{a}z)$ sends $a$ to zero. Choose $\theta$ so that the image of $b$ is the positive numberAn orthogonal-circle geodesic through zero is a diameter, so this normalizes the geodesic as required.

There are two hyperbolic metric normalizations to distinguish. For the metric transferred from the stated half-plane, the radial length isThe diameter is minimizing. Indeed, any path $w(t)$ from zero to $x$ has $|\dot{w}|\ge |d|w|/dt|$, so its length is at least $2\mathrm{artanh}x$, with equality on the radial segment. Therefore the distance consistent with the given half-plane metric isFor the alternative disc metric $d{\tilde{s}}_D^2=|dw{|}^2/(1-|w{|}^2{)}^2$, every length is halved and the same isometries and geodesics remain. In that convention,The printed distance corresponds to the latter convention. It is obtained from one quarter of the given half-plane metric, rather than its direct transfer. The equal-point case gives distance zero in both conventions. This is the normalization of the Poincare disc metric.

First consider the unlettered projective frame assertion. A list of $n+2$ points in ${\mathbb{CP}}^n$ is in general position when every $n+1$ of their representative vectors are linearly independent, equivalently no $n+1$ of the points lie in a projective hyperplane. Choose representatives $v_1,\dots ,v_{n+1}$ of the first $n+1$ points as a basis. Write $v_{n+2}=\sum c_i{v}_i$. Each $c_i$ is nonzero, since otherwise $v_{n+2}$ and the other $n$ basis vectors would be dependent. Rescaling the representatives makes $v_{n+2}=v_1+\cdots +v_{n+1}$.

Do the same for the second frame with representatives $w_i$. The invertible linear map $v_i\mapsto w_i$ induces a projective transformation carrying all $n+2$ points as required. A transformation fixing a normalized frame is represented by a diagonal matrix, since it fixes the first $n+1$ points. Fixing the sum point forces every diagonal entry to coincide, so the matrix is scalar and the projective transformation is the identity. This proves existence and uniqueness.

Return to the four chosen points on the two lines. Since the six points are distinct, at most one is $K\cap L$; put its index third if necessary. No three of $p_1,p_2,q_1,q_2$ are collinear: every triple contains two on one line and a point on the other line away from the intersection. Thus they form a projective frame in ${\mathbb{CP}}^2$. The four target points also have no collinear triple, as their three-by-three determinants are nonzero. The frame result gives a unique projective transformation with the prescribed four images.

The normalized lines are $K=\{z=0\}$ and $L=\{y=0\}$. A point of $K$ with first coordinate zero is $p_1$, so $p_3\ne p_1$ permits unique normalization $p_3=[1,\alpha ,0]$. The inequality $p_3\ne p_2$ gives $\alpha \ne 1$. Similarly, $q_3=[1,0,\beta ]$ uniquely with $\beta \ne 1$.

The lines $p_1{q}_2$ and $p_2{q}_1$ are $x-z=0$ and $x-y=0$, giving $r_{12}=[1,1,1]$. The lines $p_1{q}_3$ and $p_3{q}_1$ are $\beta x-z=0$ and $\alpha x-y=0$, giving $r_{13}=[1,\alpha ,\beta ]$. Finally, the two lines defining $r_{23}$ have equationsTaking the cross product of their coefficient vectors givesThe last vector cannot vanish: if $\alpha \beta =1$, both parameters are nonzero and vanishing of its other coordinates would force $\alpha =\beta =1$, which is excluded.

Let $u=(1,1,1)$ and $v=(1,\alpha ,\beta )$. The vector representing $r_{23}$ satisfiesConsequently the three representative vectors are linearly dependent, so $r_{12},r_{13},r_{23}$ are collinear in the projective plane. Equivalently their three-by-three determinant vanishes. This is the two-line configuration of Pappus's hexagon theorem.

The strong Bézout theorem says that if two projective plane algebraic curves $C,D$ over $\mathbb{C}$ have degrees $m,n$ and no common irreducible component, their intersection is finite andEach intersection multiplicity is a positive integer. Scheme-theoretic multiplicities, rather than just distinct intersection points, are essential: tangencies and singularities can make individual terms larger than one.

For curves with no common local component at $p$,In local affine coordinates, their defining equations have linearly independent gradients exactly in this case. These independent linear terms form local coordinates, so the local intersection quotient has dimension one. Conversely, if either equation lacks a nonzero linear term or the two linear terms are dependent, the local intersection quotient cannot have dimension one. This is transverse intersection of plane curves of two smooth plane curves.

Write the irreducible curve as $F=0$, with $F$ homogeneous of degree $n$. For $n=1$ there are no singular points. Otherwise choose a nonzero partial derivative, say $F_x$, which has degree $n-1$. Such a derivative exists in characteristic zero. Irreducibility and its smaller degree show that $F$ and $F_x$ have no common component.

At every singular point $p$ of $C$, $F_x(p)=0$ and the curve multiplicity of $F$ is at least two. The standard local lower bound for intersection multiplicity gives $I_p(F,F_x)\ge 2$. Applying the Bézout theorem,ThereforeThis is the polar bound for singular points of a plane curve.

Take the union of $n$ distinct projective lines in general position, with no three concurrent. Its defining polynomial is the product of their linear equations, so its degree is $n$. Exactly two lines meet at each of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairwise intersections, producing an ordinary double point; elsewhere just one line passes through and the curve is smooth. HenceFor $n=1$, the line itself gives zero singular points, as required.

The curves have no common component, since their intersection is finite. Let $R=0$ define the irreducible degree-$m$ curve $E$. Choose $p\in E$ outside $C\cap D$, and select $(\lambda ,\mu )\ne (0,0)$ with $\lambda P(p)+\mu Q(p)=0$. The polynomial $H=\lambda P+\mu Q$ is not identically zero, since $P,Q$ are not proportional.

The degree-$n$ curve $H=0$ meets $E$ at the prescribed $nm$ base points and also at $p$. If $R$ did not divide $H$, irreducibility of $R$ and the Bézout theorem would bound the number of distinct intersections by $nm$, contradicting these $nm+1$ points. ThusAt every remaining common zero of $P,Q$ lying off $E$, $R$ is nonzero and hence $S=0$. ThereforeThis is residual intersection through a curve pencil. The existence of a point $p$ outside the finite base locus is automatic on the projective curve $E$.

Let the six distinct conic points be $p_1,\dots ,p_6$, and write ${\ell }_{ij}$ for their joining line. Form two reducible cubicsNo three distinct points of an irreducible conic are collinear, by the Bézout theorem applied to a line and the conic. Thus all six side lines are distinct, the cubics have no common component, and six of their intersections are precisely the vertices. The remaining pairings giveThese three points are off the conic and distinct. For example, equality of $q_1$ and $q_2$ would put that point on both ${\ell }_{34}$ and ${\ell }_{45}$, making it $p_4$ and then placing $p_4$ on ${\ell }_{12}$, impossible. The other cases are identical. Hence the two cubics meet in nine distinct points.

Apply residual intersection through a curve pencil with $n=3$, $m=2$ and $E$ the conic: six intersections lie on $E$, so the other three lie on a curve of degree at most one. Therefore $q_1,q_2,q_3$ are collinear, proving Pascal's theorem.

First record the unlettered definitions. A divisor on an algebraic curve on a compact Riemann surface is a finite formal sum $D={\sum }_p{n}_p[p]$ with $n_p\in \mathbb{Z}$. Its degree is $\mathrm{deg}D={\sum }_p{n}_p$. A principal divisor is $(f)={\sum }_p{\mathrm{ord}}_p(f)[p]$ for a nonzero meromorphic function, with positive order at a zero and negative order at a pole. Two divisors are linearly equivalent if their difference is principal. Principal divisors form a subgroup because $(fg)=(f)+(g)$ and $(1/f)=-(f)$.

For the doubly periodic function, put $h=f^{\prime }/f$. It is periodic with the same period lattice, and its residue at $p_i$ equals $n_i$. Opposite sides of the fundamental parallelogram are translates by a period with opposite directions, so their integrals of $h$ cancel. The residue theorem givesThusIn particular an elliptic function has equally many zeros and poles counted with multiplicity in a period cell.

Order the period lattice basis so the indicated contour is positively oriented; reversing its orientation only changes the signs of the resulting integers. SetPairing the translated sides while retaining the factor $w$ givesFor example, the right side is the translate by ${\lambda }_1$ of the left side with reversed orientation; their combined contribution is ${\lambda }_{1}B$. The bottom and top similarly give $-{\lambda }_{2}A$.

Along either path there is a continuously continued logarithm of the nonzero finite function $f$. Since its endpoint values agree by periodicity, each logarithmic increment is an integer multiple of $2\pi i$. Therefore $A=2\pi ir$, $B=2\pi is$ for integers $r,s$. The residue theorem applied to $wh$ now yieldsThis is the elliptic divisor-sum identity.

For a divisor on an algebraic curve $D={\sum }_j{n}_j[p_j+\Lambda ]$, defineChanging the representatives $p_j$ changes the second coordinate only by an element of $\Lambda$. The two preceding contour arguments show that a principal divisor has degree zero and period lattice sum zero. Thus the formula is unchanged by linear equivalence of divisors and is a well-defined group homomorphism on the quotient.

It is nontrivial, since $[0]$ maps to $(1,0+\Lambda )$. In fact it is surjective: for any $d\in \mathbb{Z}$ and $z+\Lambda \in \mathbb{C}/\Lambda$, the divisor on an algebraic curve $(d-1)[0]+[z+\Lambda ]$ maps to $(d,z+\Lambda )$. Only well-definedness and a nontrivial homomorphism are required here; identifying its entire kernel would be an additional theorem.

Set $g(w)=a\wp (w)+b{\wp }^{\prime }(w)+c$. The Weierstrass elliptic function is doubly periodic, with a double pole at the origin modulo $\Lambda$ and no other poles. Its derivative has leading term $-2w^{-3}$. Since $b\ne 0$, no lower-order term cancels this, so $g$ has a pole of order exactly three at zero and no others.

The point $\Phi (0+\Lambda )=[0,1,0]$ does not lie on the line, because its line equation evaluates to $b\ne 0$. At every other point of the image, lying on the line is equivalent to $g(w)=0$. The three distinct image points therefore yield three distinct zeros modulo $\Lambda$. By the zero-pole count, these are all the zeros and each is simple. Hence its principal divisor on the torus isApplying the elliptic divisor-sum identity givesThis is the period lattice form of the chord-and-tangent group law on an elliptic cubic.