Oral exam practice problems: Algebraic Geometry

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1 Oral exam practice problems: Algebraic Geometry Alberto García Raboso TP1. Let Q 1 and Q 2 be the quadric hypersurfaces in P n given by the equations f 1 x x 2 n = 0 f 2 a 0 x a n x 2 n = 0 respectively, where all the coefficients a i are nonzero and different from one another. How many singular elements are there in the pencil generated by Q 1 and Q 2? Denote by Q [λ:µ] P n the element of the pencil over [λ : µ] P 1, given by the equation n λ + µa i x 2 i = 0 i=0 A quadric hypersurface is singular if and only if it is not of maximal rank, so the singular elements are located at λ + µa i = 0, i = 0,... n; since the coefficients a i are distinct, there are n + 1 of them. Moreover, all of the singular fibers are quadric hypersurfaces of rank one less than the maximal, and thus cones over a nonsingular quadric hypersurface in P n 1 with vertex a point. RD1. a Calculate the Betti numbers of the intersection of the quadrics Q 1 and Q 2 of problem TP1. b In the case of odd n, calculate the Hodge numbers of Q 1 Q 2. a First of all, notice that Q 1 Q 2 is a complete intersection. By an iteration of Krull s principal ideal theorem, a subscheme defined by two equations has codimension at most two. It is clearly not of codimension zero, for f 1 and f 2 are nonzerodivisors. To see that it is not of codimension one, notice that f 1 and f 2 are irreducible and of the same degree; since they do not differ by a unit, they define different prime ideals. Q 1 Q 2 is smooth: the projective tangent plane at a point p = [x 0 :... : x n ] Q 1 Q 2 is the intersection of the tangent planes of Q 1 and Q 2 at p, i.e., it is given by the kernel of the matrix f 1 / x 0... f 1 / x n 2x x n = f 2 / x 0... f 2 / x n 2a 0 x a n x n Hence p is a singular point if the 2 2 minors of this matrix vanish at p. The equations of this singular locus are a i a j x i x j = 0 for 0 i < j n, whose only solutions are the coordinate points [1 : 0 :... : 0], [0 : 1 :... : 0],... [0 : 0 :... : 1], which are not contained in Q 1 nor in Q 2. 1

2 Using the Lefschetz hyperplane theorem and Poincaré duality, we have 1 if i even, 0 i 2n 2 b i = b n 2 if i = n 2 χ = n 1 b n 2 for n odd, and 1 if i even, 0 i 2n 2, i n 2 b i = b n 2 if i = n 2 χ = n 2 + b n 2 for n even. For another way of calculating the Euler characteristic of Q 1 Q 2, consider the exact sequence of vector bundles 0 T Q1 Q 2 T P n Q1 Q 2 N Q1 Q 2 0 with N Q1 Q 2 = OP n2 2 Q1 Q 2, which yields an equation on total Chern classes: c T P n Q1 Q 2 = c TQ1 Q 2 c NQ1 Q ω n+1 = 1 + c 1 + c n ω1 + 2ω where c i are the Chern classes of the tangent bundle of Q 1 Q 2 and ω = c 1 O P n1 Q1 Q 2. The top Chern class c n 2 is then the coefficient of ω n 2 in 1 + ω n+1 /1 + 2ω 2 ; equivalently, [ ] 1 + ω n+1 c n 2 = Res ω=0 ω n ω 2 dω [ ] [ ] 1 + ω n ω n+1 = Res ω= 1/2 ω n ω 2 dω Res ω= ω n ω 2 dω [ ] [ ] = 1n 1 + z n z n+1 Res z=0 8 z 2 dz + Res 1 z n 1 z=0 z z 2 dz = 1n n 4 + n 4 = { 0 if n odd n/2 if n even Hence { 0 if n odd χ = c n 2 = Q 1 Q 2 2n if n even { n 1 if n odd = b n 2 = n + 2 if n even 2

3 b Since we want to consider the case of n odd, let n = 2g + 1, so that b n 2 = b 2g 1 = 2g. Denote by P 2g+1 P 1 X π P 1 the family of quadrics given by the pencil in problem TP1; we have seen that it is the vanishing locus of the bihomogeneous polynomial n n n λ + µa i x 2 i = λ x 2 i + µ a i x 2 i i=0 i=0 The expression on the right hand side exhibits X as the blowup of P 2g+1 along the base locus of the pencil, B := Q 1 Q 2 ; in particular, since B is smooth, X is too. By the Lefschetz hyperplane theorem we know then that h p,q X = h p,q P 2g+1 P 1 for p+q 2g. Together with Poincaré duality, this gives all the Hodge numbers of X outside of the middle dimension p+q = 2g+1: the only nonzero ones are h 0,0 X = h 2g+1,2g+1 X = 1 and h p,p X = 2 for 1 p 2g. We can also find the Betti number b 2g+1 X from the topological Euler characteristic of X, which in turn can be calculated from the knowledge of the projection X π P 1. The general fiber of this map is a smooth quadric Q 2g in P 2g+1, whose Euler characteristic can be calculated as above to be χ Q 2g = 2g + 2. Using once more the Lefschetz hyperplane theorem and Poincaré duality we thus have the Betti numbers of Q 2g : b k Q 2g 2 if k = 2g = 1 if k even, 0 k 4g, k 2g The 2g + 2 singular elements of the pencil are cones over a nonsingular quadric Q 2g 1 in P 2g with vertex a point, hence isomorphic to the Thom space of a trivial real vector bundle of rank 2 over Q 2g 1, which in turn is isomorphic to Σ 2 Q 2g 1 +. Proceeding as with Q2g, we obtain b k Q 2g 1 = i=0 { 1 if k even, 0 k 4g 2 { b k Σ 2 Q 2g 1 1 if k even, 0 k 4g = + i.e., the topological Euler characteristic of the singular elements of the pencil is one less than that of the general ones. Since there are 2g + 2 of them, we arrive at χx = χp 1 χ Q 2g [ + 2g + 2 χ Σ 2 Q 2g 1 χ Q 2g] + = 22g + 2 2g + 2 = 2g + 2 Collecting all the previous results, we have b 2g+1 X = 2g and hence p+q=2g+1 hp,q X = 2g. 3

4 We now consider the Leray spectral sequences associated to the map X π P 1 and the sheaves Ω p X : E r,s 2 Ωp X = Hr P 1, R s π Ω p X Hr+s X, Ω p X Notice that, since dim P 1 = 1, these spectral sequences have only two columns, and they actually degenerate at the E 2 -step. Hence, for p, q 1, h p,q X = h q X, Ω p X = h0 P 1, R q p Ω p X + h1 P 1, R q 1 p Ω p X Moreover h 0,q X = h q X, O X = 0, the last equality coming from the long exact sequence in cohomology associated to the exact sequence of sheaves 0 O P 2g+1 P 1 2, 1 O P 2g+1 P 1 O X 0 We now investigate the sheaves R s π Ω p X for p > 0. Let U P1 the complement of the 2g + 2 points over which the singular fibers are located; then π 1 U = U Q 2g, so that R s π Ω p X U = H s U Q 2g, Ω p U Q 2g = a+b=s i+j=p = H 0 U, O U H a U, Ω i U H b Q 2g, Ω j Q 2g [H s Q 2g, Ω pq 2g H s Q 2g, Ω p 1 Q 2g ] From the Hodge numbers of Q 2g, we see that R s π Ω p X = 0 unless p = s or p = s + 1. For the middle dimension p + q = 2g + 1, we have h 0,2g+1 X = h 2g+1,0 X; if 1 p 2g, then h p,2g+1 p X = h 0 P 1, R 2g+1 p p Ω p X + h1 P 1, R 2g p p Ω p X The first term vanishes unless p = g + 1, while the second is nonzero only if p = g, that is, the only nonzero Hodge numbers in the middle dimension are h g,g+1 X and h g+1,g X, which have to be equal because of complex conjugation. In sum, 1 if p = q = 0 or p = q = 2g + 1 h p,q 2 if 1 p = q 2g X = g if p = g, q = g + 1 or p = g + 1, q = g We can finally compute the Hodge numbers of B: since X is the blowup of P 2g+1 along B, we have 1 if 0 p = q 2g 1 h p,q B = h p+1,q+1 X h p+1,q+1 P 2g+1 = g if p = g 1, q = g or p = g, q = g 1 4

5 TP2. a Let Q be the singular quadric hypersurface in P 4 given by the equation x 2 0 +x2 1 +x2 2 +x2 3 = 0. Calculate its divisor class group, its Picard group and its cohomology with coefficients in O and C. b Compute the Picard group of Spec O p,q, where p Q is the singular point. Use this to show that p is not a quotient singularity. c Show that Q contains a plane P through p. Prove that the blow-up Q of Q along P is smooth and answer the same questions. d Show that the blow-up of Q at p is a P 1 -bundle over P 1 P 1 and identify which one it is. a First of all, change coordinates in P 4 : y 0 = 1 2 x 0 + ix 1, y 1 = 1 2 x 0 ix 1, y 2 = 1 2 x 2 + ix 3, y 3 = 1 2 x 2 + ix 3, y 4 = x 4 Now, Q is the cone over X := V + y 0 y 1 y 2 y 3 P 3 with vertex the point p = [0 : 0 : 0 : 0 : 1]. The singular point is of codimension 3 in Q, so that ClQ = ClQ {p}. But Q {p} = X A 1, which yields ClQ = ClQ {p} = ClX A 1 = ClX = Z Z since X = P 1 P 1. Following the chain of isomorphisms, we find generators for ClQ: P 1 := V + y 1, y 2 Q P 2 := V + y 1, y 3 Q Neither of these Weil divisors is Cartier, for Q is normal and they are not locally principal at p. Indeed, m p,q /m 2 p,q = m p,p 4/m 2 p,p 4 is a vector space of dimension 4 with basis the classes of y 0, y 1, y 2, y 3 ; the image of the prime ideal associated to P 1 contains the classes of both y 1 and y 2, so it could not be generated by a single element, and similarly for P 2. However, P 1 P 2 = V + y 1, so that Z P 1 + P 2 CaClQ. Suppose there were another Weil divisor in CaClQ\Z P 1 + P 2. Then, by taking an appropriate linear combination, we would have kp 1 CaClQ for some k 1. To show that kp 1 is not locally principal for any such k, look at m k p,q m k+1 p,q = mk p,p + q/q 4 = m k+1 + q/q p,p 4 m k p,p 4 m k p,p 4 m k+1 p,p 4 + q = m k p,p 4 m k+1 p,p 4 + m k p,p 4 q where q O p,p 4 is the ideal generated by y 0 y 1 y 2 y 3. It is clear that this is a vector space with basis {y0y a 2y b 3 c a, b, c 0, a + b + c = k} {y1y a 2y b 3 c a, b, c 0, a + b + c = k} Again the image of the ideal associated to kp 1 is of dimension strictly greater than one, proving that CaClQ = Z P 1 + P 2. Finally, since Q is integral, we have PicQ = CaClQ = Z. The cohomology with coefficients in O can be easily deduced from the exact sequence 0 O P 4 2 O P 4 O Q 0 5

6 We obtain H 0 Q, O Q = C and H i Q, O Q = 0 for i > 0. Since Q {p} = X A 1, we can identify Q with the Thom space of a trivial real bundle of rank 2 over P 1 P 1, so that Q Σ 2 P 1 P 1 +. From this the Betti numbers of Q follow: b 0 = 1, b 1 = 0, b 2 = 1, b 3 = 0, b 4 = 2, b 5 = 0 and b 6 = 1. b First of all, observe that a prime divisor in Spec O p,q, i.e., a height one prime ideal of O p,q, corresponds to a prime divisor on Q that passes through p. Let ι : Spec O p,q Q be the canonical inclusion. If P is a prime divisor in Q, then define ι P := { ι 1 P if p P 0 if p P and extend linearly to a map ι : DivQ DivSpec O p,q. Moreover, if f KQ, then ι f = ι v P f P = v P f ι 1 P P prime P prime p P is the principal divisor associated to f considered as an element of KSpec O p,q = KQ. Hence ι descends to a map of divisor class groups, which we still denote ι. The surjectivity of the these maps is clear: if D = n P ι 1 P then ι D = D for D = P prime p P P prime p P n P P Since P 4 = V + y 4 Q DivQ does not pass trough the singular point, we have that ι y 1 /y 4 = ι P 1 + P 2 P 4 = ι 1 P 1 + ι 1 P 2 is a principal divisor in DivSpec O p,q. Furthermore, if f = k ι 1 P 1 then f O Q U for some open neighborhood U of p in Q, in contradiction with the fact that kp 1 is not locally principal at the singular point. In conclusion, ClSpec O p,q = Z P 1. We now study the divisor class group of a quotient singularity. Let G be a finite subgroup of SU3 acting on C 3 ; the C-algebra homomorphism C[x, y, z] G C[x, y, z] corresponds to the canonical quotient map π : C 3 C 3 /G. We can mimic the above construction to define a map π : DivC 3 /G DivC 3 which also descends to a homomorphism of divisor class groups: if P be a prime divisor in C 3 /G passing through the origin, then π P is a Weil divisor which is invariant under the action of G. Since ClC 3 = 0 C[x, y, z] is a UFD, there exists f KC 3 such that 6

7 f = π P. But then G f := g G g f is G-invariant and descends to a rational function on the quotient; hence G f = G P DivC 3 /G and ClC 3 /G is torsion. Reasoning as above, we can also show that the divisor class group of the local scheme at the singularity is a quotient of the latter group, hence also torsion. This proves that the singular point of Q is not a quotient singularity. c We have already seen that Q contains two planes through the singular point, namely P 1 and P 2. The blow-up Q of Q along P 1 can be constructed as the proper transform of Q with respect to the blow-up of P 4 along P 1. The latter can be constructed as follows: there is a surjection of O P 4[u, v] onto the blow-up algebra k 0 Ik, where I = y 1 O P y 2 O P 4 1 and the map takes u and v to y 1 and y 2, respectively, considered as elements of degree one; the kernel of this map is generated by the element uy 2 vy 1, so that Bl P1 P 4 := Proj I k = Proj O P 4[u, v]/uy 2 vy 1 k 0 = V + uy 2 vy 1 P 4 P 1 The total transform of Q is then V + y 0 y 1 y 2 y 3, uy 2 vy 1 P 4 P 1. Looking at the inverse image of D + y 0 Q P 4 under the blow-up map π we can differentiate the component along the exceptional divisor E = P 2 P 1 from the proper transform Q of Q: π 1 D + y 0 Q = V y 10 y 20 y 30, uy 20 vy 10 = V y 10 y 20 y 30, y 20 u vy 30 D + y 0 P 1 Thus, Q = V + y 0 y 1 y 2 y 3, uy 2 vy 1, uy 0 vy 3 P 4 P 1 The smoothness can be checked affine-locally. In order to calculate the cohomology of Q with coefficients in O, notice that E Q is a complete intersection in P 4 P 1, so that we have the exact sequence 0 O P 4 P 1 3, 1 O P 4 P 1 2, 0 O P 4 P 1 1, 1 O P 4 P 1 O E Q 0 The cohomology of P 4 P 1 with coefficients in O P 4 P1a, b follows immediately from that of P4 and P 1 and the Künneth formula. A short calculation yields H 0 E Q, O E Q = C and H i E Q, O E Q = 0. Similarly, E Q is the vanishing locus of uy 0 vy 3 in P 2 P 1 with coordinates [y 0 : y 3 : y 4 ] [u : v], so that 0 O P 2 P 1 1, 1 O P 2 P 1 O E Q 0 and H 0 E Q, O E Q = C and H i E Q, O E Q = 0. Finally E = P 2 P 1 gives H 0 E, O E = C and H i E, O E = 0. Plugging this information in the exact sequence we obtain H 0 Q, O Q = C and H i Q, O Q = 0. 0 O E Q O E O Q O E Q 0 7

8 Since P 4 P 1, Bl P1 P 4 and Q are all smooth, we have the following diagram of vector bundles on Q: T Q T Q 0 0 T BlP1 P 4 Q T P 4 P 1 Q N BlP1 P 4 \P 4 P 1 Q 0 0 N Q\BlP1 P 4 N Q\P 4 P 1 N BlP1 P 4 \P 4 P 1 Q We know N BlP1 P 4 \P 4 P 1 = O P 4 P 11, 1 Bl P1 P 4 and N Q\BlP1 P 4 = O P 4 P 12, 0 Q, so that N Q\P 4 P 1 = O P 4 P 12, 0 O P 4 P 11, 1 Q From the middle column of the diagram above, we obtain c T P 4 P 1 Q = c T Q c N Q\P 4 P ω η 2 = 1 + c 1 + c 2 + c ω1 + ω + η where c i are the Chern classes of the tangent bundle of Q and ω = c 1 OP 4 P 11, 0 Q, η = c 1 OP 4 P 10, 1 Q We calculate c 1 = 2ω + η, c 2 = ω2ω + 3η, c 3 = 3ω 2 η and χ = Q c 3 = 6. The Lefschetz hyperplane theorem yields the Betti numbers b 0 E Q = 1, b 1 E Q = 0 and b 2 E Q = 2. Moreover, we also know all of the Betti numbers of E and E Q. We will use a Mayer-Vietoris argument to compute the Betti numbers of Q. The following table contains all the information that we have so far. 8

9 i b i E Q b i E Q b i E Q b b b x 0 + b b b b 0 0 Notice that we have also made use of Poincaré duality for Q. From the first two rows, it is immediate that b 0 = 1 and b 1 = 0. Moreover, from the third and fourth rows we have b 2 = 2 x+b 3 and x b 3. Equating the value of the Euler characteristic that we calculated before with the one obtained from the table, we arrive at the condition b 3 = 2x. The only possible solution is then x = 0, b 2 = 2 and b 3 = 0. More precisely, the Lefschetz hyperplane theorem gives H 2 E Q, Z = Z Z; since we also have H 2 E Q, Q = Z Z, it is clear from the table that H 2 Q, Z is also torsion-free. We then deduce from the exponential sequence that Pic Q = Z Z. Moreover, since Q is smooth, its divisor class group is isomorphic to its Picard group. d First consider the blow-up of C 4 at the origin: it can be described as the set {p, l C 4 P 3 : p l}, i.e., as the total space of the tautological line bundle O P 3 1 over P 3 : Bl 0 C 4 = Tot OP 3 1 := Spec Sym O P 31 The blow-up of P 4 at the origin is then a fiberwise compactification of the latter, hence a P 1 -bundle over P 3. Any such bundle is the projectivization of a rank two vector bundle E on P 3, which always splits as a sum of line bundles. To prove the latter fact, we may assume without loss of generality that h 0 E > 0: otherwise take a big enough twist, which can be undone at the end. If s : O P 3 E is a non-zero global section, the image of its dual s : E O P 3 is an ideal sheaf O P 3 D for some effective divisor D and its kernel L is locally free of rank one; dualizing we obtain a short exact sequence 0 O P 3D E L 0 Since Ext 1 L, O P 3D = H 1 P 3, LD = 0, the extension is split. Write then Bl 0 P 4 = PL1 L 2 := Proj SymL 1 L 2 π P 3 For 0 i 3, let f 1 1,i and f 1 2,i be local generators of L 1 and L 2, respectively, as O P 3-modules over the affine open D + z i P 3. We then have isomorphisms O P 3 D+ z i [u i, v i ] = SymL 1 L 2 D + z i 9

10 sending u i f 1,i SymL 1 L 2 D + z i v i f 2,i SymL 1 L 2 D + z i Suppose that the section of π given by the proper transform of the hyperplane at infinity V + y 4 is given by the canonical projection L 1 L 2 L 1. In the notation just introduced, this corresponds to the vanishing locus of the v i over each D + z i. The complement of this section is given locally by Spec O P 3 D+ z i [u i /v i ]. These pieces glue together to Spec SymL 1 L 2, so that L 1 L 2 = O P 31. Since PE = PE L 2, we conclude that Bl 0P 4 = PO P 3 1 O P 3. The blow-up of Q at the origin is the proper transform of Q under π, i.e., Bl 0 Q = PO P 1 P 1 1, 1 O P 1 P RD2. Analyze the plane curve C given by y 2 = fx = 10 i=1 x ii calculate its geometric and arithmetic genus, singular points,.... Let C be the projective closure of C, with equation Y 2 Z i=1 X izi = 0. Its arithmetic genus follows from the degree-genus formula: p a C = = The singular points of C are the solutions of the equations 0 = y 2 fx, 0 = x y 2 fx = f x, 0 = y 2 fx = 2y y These conditions are satisfied at the points m, 0 such that f has a multiple root x = m; this happens when m = 2, In an infinitesimal neighborhood of each of these points, we can redefine the x coordinate to put the equation in the form y 2 = x m. Resolving these singularities, we see that the normalization N of C has two points over each of the singular points for which m is even, and only one for those for which m is odd. The intersection of C with the hyperplane at infinity Z = 0 is a highly singular point. Instead of trying to resolve that singularity, we perform a clever change of coordinates in the affine plane, valid for x 0: we define z = x 1 and w = x 28 y; the equation of C then takes the form w 2 z 10 i=0 1 iz i = 0 Adding the point z = w = 0 to C compactifies it; notice, in particular, that the point added is nonsingular. In summary, N is a 2-to-1 cover of P 1 ramified over the points x = 1, 3, 5, 7, 9,. By the Riemann-Hurwitz formula, 2gN 2 = = gn = 2 10

11 TP3. Classify all singular quadric hypersurfaces in P n. Describe the singularities, resolve them and describe the corresponding resolutions. Since the only invariant of a quadratic form on C n+1 is its rank, there are n + 1 isomorphism classes of quadric hypersurfaces in P n. A complete set of class representatives is given by Q k,n = V + x x 2 k 1 Pn where k = 1,..., n + 1 is the rank of the associated quadratic form. It is clear that a quadric hypersurface is smooth if and only if its rank is maximal in this case, if k = n + 1. A singular quadric of rank 1 is a double hyperplane; in this case, all points are singular and a resolution is provided by that same hyperplane with its reduced scheme structure. If k = 2, we have the union of two hyperplanes intersecting transversely in a linear subvariety of dimension n 2, whose resolution consists of two disjoint copies of P n 1. For 3 k n, Q k,n is a cone over a smooth quadric hypersurface in V + x k,..., x n = P k 1 with vertex S := V + x 0,..., x k 1 = P n k. To perform a resolution of singularities, we blow up P n along S: Bl S P n = V + {xi u j x j u i } 0 i<j k 1 P n P k 1 TO DO: prove smoothness and describe these resolutions. RD3. Look at the stratification of the theta divisor of a curve C by singularity type, and define a stratification of the moduli space M g by this data. Work this out for g 6. TP4. Consider a generic pencil of plane curves of degree d. Show that its singular elements have a unique singular point, and that it is an ordinary double point. Count how many of them there are. 11