Have Fun with computations:hilbert polynomials, Hodge diamonds, Sheaf cohomology...

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1 Have Fun with computations:hilbert polynomials, Hodge diamonds, Sheaf cohomology... Shuai Wang June 206 Introduction Basic knowledge about Hilbert schemes. 2 Computing Hibert Polynomials Example 2.. Let R = k[x 0, x, x 2, x 3 ], and I =< x 0 x 3 x x 2, x 0 x 2 x 2, x x 3 x 2 2 > We want to compute the Hilbert polynomial of S/I. For ω = (, 0, 0, ), the given generating set is a Gröbner basis. Then the initial ideal is A Stanley decomposition for S/J is J = in ω (I) =< x 0 x 3, x 0 x 2, x x 3 > {(, {2, 3}), (x, {, 2}), (x 0, {0, })} So the Hilbert polynomial is given by H S/I (t) = H k[x2,x 3](t) + H k[x,x 2](t ) + H k[x0,x ](t ) = (t + ) + ((t ) + ) + ((t ) + ) Then we have many questions like What is a Grönber basis? What is the initial ideal? = 3t + What is a Stanley decomposition?

2 Why this procedure is legal? Definition 2. (Initial ideals and Grönber basis ). Fix ω N n+, I f = t N n+ c tx t. Let T = {t c t 0, t ω maximal}, then in ω (f) = t T c t x t ; The initial ideal of I is in ω (I) =< in ω (f) f I > A set {f,..., f n } I of polynomials is a Gröbner basis for I with respect to ω if in ω (I) =< in ω (f,..., in ω (f n )) >. Remark. in ω (I) is almost never generated by the initial terms of a minimal generating set of I. Although our example above does. A Gröbner basis is not necessarily a generating set for I, not to say minimal generating set. The lexicographic order cannot be obtained by a single vector ω, however we need a chain of weight vectors: (, 0,..., 0), (0,, 0..., 0),..., (0, 0,..., 0, ) That means use a single vector, we cannot get a strict total order, to get a total order on monomials, generally, we have two ways, either use a sequence of weight vectors, or use a rational independent weight vector. We also have two identical geometric description of the initial ideal with respect to a weight vector ω. First define f = c t x t y d ω t, where d = max{ω t c t 0, t ω}. Then f S[t] and at least one monomial of f is in S. Let I y =< f f I > Then S[y]/I y is a flat k[t]-module(why?).i.e we have a flat family of schemes over A = Spec(k[y]). Spec(S[y]/I y ) Spec(S[y]) = A n A Spec(k[y]) = A If I is a homogeneous ideal, then we have a flat family of projective schemes over A P roj(s[y]/i y ) P roj(s[y]) = P n A Spec(k[y]) = A 2

3 Note that here P n is a weighted projective space with deg(y) = 0, instead of the most common projective space P n. The significance of this construction is that the firbre over y = is just P roj(s/i), and the firbre over t = 0 is P roj(s/in ω (I)), this flat degeneration froman ideal to its initial ideal is called Gröbner degeneration. A second way to understand the gemetry of the initial ideal is the following, consider the torus G n+ m -action on A n by scaling. We have a one-parameter subgroup φ(a) : G m G n+ m, k a (a ω,..., a ωn ) Then if X = Spec(S/I), φ(a) X = Spec(S/I a ), and we have Spec(S/in ω(i) ) = lim a 0 φ(a)x. Definition 2.2 (Standard monomials of I). The monomials not lying in in ω (I) are called standard monomials. Proposition 2.2. The standard monomials form a basis for S/I as k-vector space. Proof. Simply because [f] S/I, and it s represented by f, then we may assume f is a linear combinations of standard monomials, otherwise, consider f in ω (g) for some g I. They re linear independent because if F = k i f i I, consider in ω (F ) must be one of them, contradict to the fact that f i / in ω (I). Corollary H S/I = H S/inω(I) Thus to compute the Hilbert polynomial for I, we only need to compute it for monomila ideal in ω (I). The mean idea is that the standard monomials for a monomial ideal can be partitioned into translates of the monomials in smaller polynomial rings, i.e a stanley decomposition. Definition 2.3 (Stanley decomposition for S/I). Let I be a monomial ideal. A Stanley decomposition for S/I is a finite decomposition of the standard monomials of I into disjoint sets of the form where (x u, σ) = {x u+v : supp(x v ) σ} σ {0,..., n}; supp(x v ) = {i : v i > 0} Don t remember this definition, just check the following example Example 2.3 (Stanley decomposition of I =< x 2 0x >). Actually all standard monomials are {x i 0 i 0} {x 0 x k k 0} {x k k > 0}, use the language above is simply because we want to view them as translates of smaller polynomial rings. i.e {x i 0 i 0} is just k[x 0 ], {x 0 x k k 0} is a translate of k[x ], {x k k > 0} is a translate of k[x ] by x. Then the definition just use a more concise language to state this procedure. (, 0) (x 0 x, ) (x, ) 3

4 For example (x 0 x, ) means it s the translate of k[x ] by x 0 x, just our previous {x 0 x k k 0}. If we can always find a stanley decomposition for a monomial ideal I, then we know H S/I (t) = n H k[xj:j σ j](t u i ); u = (u) k i= Lemma 2.4 (Existence of a Stanley decomposition for a monomial ideal). Let I be a monomial ideal in S = k[x 0, x,..., x n ]. Then a Stanley decomposition exists. Proof. Let k i = min{k < I : x k i >=< I : x i >}(By Hilbert nullstellensatz, this minimal exists), and k = k i. We prove it by induction on k and n. If n = 0, S = k[x 0 ] which is a PID, thus I =< x l 0 >, l i=0 {(xi 0, )} is a Stanley decomposition of S/I, where (x i 0, ) means the translate of k by x i 0. If k = 0, I =< I : x i > for all i. Since I is a monomial ideal, we have I =< x i i / σ {0,,..., n} >, thus all standard monomials are monomials in k[x j : j σ], thus (, σ) is a Stanley decomposition of S/I. In general, consider the following short exact sequence of k-modules(i.e k-vector spaces ). 0 S/ < I : x i > S/I S/ < I, x i > 0 [f] [x 0 f]; [g] [g] Note that this is NOT a exact sequence of rings, it s exact as sequence of vector spaces because we know all monomials not lying in in ω (I) = I(since I is a monomial ideal) is a basis for S/I. If f = f j ker(s/i S/ < I, x i >), where {f j } s are all monomials, then f j < I, x i >, f j / I, wo f j = x 0 g i / I, thus f is the image of g j, and all g j s are not in < I : x i >, which simply means x i g j / I. Then by induction, a Stanley decomposition for S/ < I, x i > exists since it s a quotient in a smaller polynomial ring S/ < x i >. For S/ < I : x i >, k i < k i, for j i k j k j. Thus a Stanley decomposition for S/ < I : x i > exists. Since short exact sequence of vector spaces splits(for example, theyall are free modules, thus projective..). Then we know a Stanley decomposition of S/I exists. Careful readers may already find to really solve the question of computing the Hilbert polynomial of a given ideal I, we really have to find a way to compute a Gröbner basis for I. Luckily, we have the following 4

5 Lemma 2.5 (Buchberger s Algorithm: Computing a Gröbner basis, Wikipedia). Input: A set of polynomials F that generates I; Output: A Gröbner basis G for I; Step: G := F ; Step2: For every fi, fj in G, denote by g i the leading term of f i with respect to the given ordering, and by a ij the least common multiple of g i and g j. Step3: Choose two polynomials in G and let S ij = aij g i f i aij g j f j.(note that the leading terms here will cancel by construction). Step4: Reduce S ij, with the multivariate division algorithm relative to the set G until the result is not further reducible. If the result is non-zero, add it to G. Step5: Step Repeat steps -4 until all possible pairs are considered, including those involving the new polynomials added in step 4. Output G We don t want to prove it, instead, let s see some examples Example 2.6 (Buchberger s Algorithm). I =< f = x 0 x 3 x x 2, f 2 = x 0 x 2 x 2, f 3 = x x 3 x 2 2 >, G = {f, f 2, f 3 } We use the lexicographic order. Then g = x 0 x 3, g 2 = x 0 x 2, g 3 = x x 3, the least common multiplier l(g, g 2 ) = x 0 x 2 x 3, S 2 = x 2 f x 3 f 2 = x 2 x 3 x x 2 2. And S 2 reduces to 0 with respect to G, since S 2 = x f 3. Similar for S 23 and S 3. Then we know, G is alreday a Gröbner basis for I. Then we can fully understand the first example at the beginning of this notes. Example 2.7 (Buchberger s Algorithm). I =< f = xy x, f 2 = x 2 y >; G = {f, f 2 } Then g = xy, g 2 = x 2 a 2 = x 2 y. S 2 = x2 y xy (xy x) x2 y x 2 (x2 y) = x 2 + y 2 Then reduce S 2, since S 2 = f 2 y + y 2, let y 2 y = f 3 < min{f, f 2 }, add it to G, now we get a new G = {f, f 2, f 3 }. Then use the method above, we can check {f, f 2, f 3 } is a Gröbner basis for I. Example 2.8 (Hilbert Polynomial: A boring example). I =< x 3 0, x 2 0x 3, x 0 x x 2, x 0 x 4 2, x 4 x 3 2, x 6, x 7 2 >; S = k[x 0, x, x 2 ] 5

6 Since all generators are monomials, we know they form a Gröbner basis for I already. And in(i) = I. For this example, we don t need to consider Stanley decomposition, afterall, standard monomials are finite, thus P S/I (t) = 0 And we can also fire up a compute algebra package such as Macaulay2 to compute Hilbert polynomials for more complicated homogeneous ideals. For example Example 2.9. I =< 3x 0 x 2x 2 2 3x 0 x 2 + x x 2 + x 2 2 9x 2 0 4x 2 8x 0 x 2 + 8x x 2 + 5x 2 2, x 3 3x x x 3 2 > After understanding all above, let s see some interesting examples. Note in Macaulay2, the default ordering is the Graded reverse lexicographic order. Example 2.0 (A rational quartic curve in P 3 ). I =< xz yw, xy 2 z 3, x 2 y z 2 w, x 3 zw 2 > Then a Gröbner basis for I is given by [xz yw, xy 2 z 3, x 2 y z 2 w, x 3 zw 2, z 4 y 3 w] The corresponding initial ideal is A Stanley decomposition is given by in(i) =< xz, xy 2, x 2 y, x 3, z 4 > {y i w k } {zy i w k } {z 2 y i w k } {z 3 y i w k } {xw k } {xyw k } {x 2 w k } Thus the Hilbert polynomial for S/I is H S/I (t) = H P (t) + H P (t ) + H P (t 2) + H P (t 3) + 3H P 0(t) = (t + ) + t + (t ) + (t 2) + 3 = 4t + Thus the dimension of this variety is, and the degree of this variety is! 4 = 4, these re the reasons we call it a quartic curve in P 3. Example 2. (Double lines in P 3 ). I =< x 2, xy, y 2, u d x v d y >; S = C[u, v, x, y] Let consider d = 7. Then a Gröbner basis is given by [y 2, xy, x 2, u 7 x v 7 y] 6

7 The corresponding initial ideal is given by in(i) =< y 2, xy, x 2, u 7 x > A Standley decomposition is given by Thus the Hilbert polynomial for S/I is {u i v j } {yu i v j } 7 i=0 {xu i v k } H S/I (t) = (t + ) + t + 7 = 2t + 8 Thus this is a dimensional projective scheme of degree 2. And (P roj(s/i)) red = P 2 There s a better way to memorize this example, consider a matrix a b 2 bd c b ac c 2 d Then among all six monors of this matrix,2 of them are generated by the other 4. So the ideal defined by all the 2 2 minors of the matrix is actually I =< ad bc, ca 2 b 3, c 2 a b 2 d, c 3 bd 2 > Example 2.2 (Bezout s Theorem fails in general). We claim, let X P 4, be the cone over this quartic curve, and Y P 4 be a copy of P 2 which is defined by < a, d >, then however codim(x Y ) = codim(x) + codim(y ) deg(x Y ) > deg(x)deg(y ) Proof. Actually, since the Hilbert polynomial of I is 4t +, so codim(x) = 2 = codim(y ). We also know X Y id defined by the ideal J =< ad bc, ca 2 b 3, c 2 a b 2 d, c 3 bd 2 > + < a, d >, it s Hilbert polynomial is 5. So it s just a zero-dimensional scheme of degree 5. Meanwhile, deg(x) = 4, deg(y ) =, Bezout s Theorem fails. To see why this happens note that, a minimal primary decomposition of I + ideal(a) is given by < a, bc, b 2, c 3 bd 2 > < a, d, bc, c 3, b 3 >, that is we have an embedded point in the intersection which caused the failure. 7

8 3 Computation of sheaf cohomology Example 3.. Consider a smooth (rational) quartic curve X in P 3 given by the image of P P 3 [x, y] [x 4, x 3 y, xy 3, y 4 ] Let S = k[a, b, c, d] denote the coordinate ring of P 3, then we have the following free resolution where 0 S( 5) S 4 ( 4) S( 2) S 3 ( 3) I(X) 0 I(X) =< bc ad, c 3 bd 2, ac 2 b 2 d, b 3 a 2 c > and the last map is given by < a, b, c, d > Say, we want to compute H (O X ), since we have 0 I X O P 3 O X 0 We also know, H (O P 3) = H 2 (O P 3) = 0,thus Then we also have H (O X ) = H 2 (I X ) 0 K O P 3( 2) O 3 P 3( 3) I X 0 Here K denotes the first syzygy sheaf of X, by Serre duality and the fact that ω P 3 = O P 3( 4), we know H 2 (O P 3( 2) O 3 P 3 ( 3)) = 0, H 3 (O P 3( 2) O 3 P 3 ( 3)) = 0, therefore Finally, we also have H 2 (I X ) = H 3 (K) 0 O P 3( 5) O 4 P 3( 4) K 0 together with the fact that H 4 (O P 3( 5)) = 0, we know by Serre duality again, H 3 (K) = coker(h 3 (O P 3( 5)) H 3 (O 4 P 3( 4))) H 3 (K) = coker(h 0 (O P 3()) H 0 (O 4 P 3)) which is just the dual of S 4 S() induced by < a, b, c, d >, and it s an injection in degree 0 part. Thus we know the dual is an epimorphism in degree 0, thus we conclude H (O X ) = H 3 (K) = coker(h 3 (O P 3( 5)) H 3 (O 4 P 3( 4))) = 0 Actually, X is a rational curve in P 3, the result is not surprising at all. 8

9 Proposition 3.2 (Review of cotangent,conormal sequence). If we have ring maps R S T, then we have a relative cotangent sequence T S Ω S/R Ω T/R Ω T/S 0 the name relative cotangent sequence comes from the geometric intuition,if we have Z Y X T X/Y Z T X/Z T Y/Z 0 And if S T is surjective, and I is the kernel, then Ω T/S = 0, I/I 2 in the conormal sequence is actually the first cotangent functor the geometric intuition behind is I/I 2 T S Ω S/R Ω T/R 0 N X/Y T X Y T Y 0 Specially, in the case where S = R[x, x..., x r ]/I, I = (f,..., f n ), then we have S R Ω R[x,...,x r]/r = n i= Sdx i, by the conormal sequence, Ω S/R = coker(i/i 2 n i=sdx i ) together with the epimorphisms n i= Se i I/I 2, e i f i, we conclude Ω S/R = coker( n i=se i r i=sdx i ) which is a quotient module of the free S module r i= Sdx i), hence it has a natural S module structure, and the map is given by the Jacobian of I. Example 3.3 (Hodge diamond of Fermat quartic in P 3 ). Actually, here I want to explain my somewhat weird method to compute the Hodge diamond of the Fermat quartic in P 3, but it s also natural in some other sense. X = P roj(k[x, y, z, w]/(x 4 + y 4 + z 4 + w 4 ) Technical point here is that we can find a minimal free resolution of the cotangent sheaf Ω X 0 O P 3( 8) O P 3( 4) OP 4 3( 5) O4 P 3( 3) O6 P 3( 3) Ω X 0 which we would like to view it as two short exact sequences 0 K OP 6 3( 3) Ω X 0 0 O P 3( 8) O P 3( 4) OP 4 3( 5) O4 P 3( 3) K 0 shere K is the first syzygy sheaf of Ω X. 9

10 Then with a straightforward use of the Serre duality, H 0 (Ω X ) = H (K) = 0 Thus we know h 0, (X) = h 2, (X) = 0, here we mention H 2 (Ω X ) = 0 simply for the later use. Then similar strategy tells us H (Ω X ) = H 2 (K) and H 2 (K) fits into the long exact sequence of cohomology groups associated to the second short exact sequence above, with the help of Serre duality 0 H 2 (K) H 3 (O P 3( 8) O P 3( 4)) H 3 (O 4 P 3( 5) O4 P 3( 3)) H3 (K) However the long exact sequence associated to the first short exact sequence together with Serre duality tells us Thus we have H 3 (K) = H 2 (Ω X ) = 0 H (Ω X ) = H 2 (K) = ker(h 3 (O P 3( 8) O P 3( 4)) H 3 (O 4 P 3( 5) O4 P 3( 3))) the Hodge number h, (i.e the dimension of H (Ω X )) is thus easy to compute ( ) 7 h 3 (O P 3( 8) O P 3( 4)) = h 0 (O P 3(4) O P 3) = + = 36 3 ( ) 4 h 3 (OP 4 3( 5) O4 P 3( 3)) = h0 (OP 4 3() O4 P 3( )) = 4 = 6 We conclude h, (X) = 36 6 = 20 All other terms in the Hodge diamond are pretty easy to compute, the diamond of Fermat quartic in P 3 is given by

11 My friend, Sam, showed me a picture of his friend. In that picture, his friend s holding two balloons, on one of them is the Hodge diamond of Quintic threefold in P 4 and on the other one is the Hodge diamond of its mirror. Which I think is pretty cool. So I decided to compute one more example here use the most straightforward method as above. Example 3.4 (Hodge diamond of quintic threefold in P 4 ). The technical point we want to emphasize here is that, sometimes we might not be as luck as above. However we ll always find ways out. X = P roj(k[x, y, z, w, t]/(x 5 + y 5 + z 5 + w 5 + t 5 )) Then we can find a minimal free resolution of the cotangent sheaf Ω X 0 O P 4( 5) O 5 P 4( 4) O P 4( 0) OP 5 4( 6) O P4( 5) O0 P 4( 3) O P4( 5) O0 P 4( 2) Ω X 0 Use the first and second syzygy sheaf K, K 2 to break this resolution into three short exact sequences. Here we only want to compute h 2, (X), h, = is easier to compute, and all other terms can be computed by the symmetry of the Hodge diamond. Use Serre vanishing theorem in the first two short exact sequences, we get and H 2 (Ω X ) = H 3 (K ) 0 H 3 (K ) H 4 (K 2 ) H 4 (OP 5 4( 6) O P4( 5) O0 P 4( 3)) H4 (K ) 0 First note that, by Serre duality ( ) 5 h 4 (OP 5 4( 6) O P 4( 5) O0 P 4( 3)) = h0 (OP 5 4() O P 4 O0 P 4( 2)) = 5 + = 26 H 4 (K ) also fits into the following sequence 0 H 3 (Ω X ) H 4 (K ) H 4 (O P 4( 5) O 0 P 4( 2)) H4 (Ω X ) However H 4 (Ω X ) = 0, h 3, (X) = h 0,2 (X) = h 2,0 (X) = dim(h 2 (O X )) = 0 and by Serre duality again Thus we also get dim(h 4 (O P 4( 5) O 0 P 4( 2))) = dim(h 4 (K )) =

12 That is dim(h 3 (K )) = dim(h 4 (K 2 )) 26 + = dim(h 4 (K 2 )) 25 To compute H 4 (K 2 ),notice that it fits into the following sequence 0 H 3 (K 2 ) H 4 (O P 4( 5)) H 4 (O 5 P 4( 4) O P 4( 0)) H4 (K 2 ) 0 Since h, = and dim(h 4 (O P 4( 5))) = the second arrow must be an isomorphism. So we know how to compute the dimension of H 4 (K 2 ) dim(h 4 (K 2 )) = dim(h 4 (O 5 P 4( 4) O P 4( 0))) = ( 9 4 ) = 26 Together with our argument above, we conclude h 2, (X) = dim(h 2 (Ω X )) = dim(h 3 (K )) = = 0 The Hodge diamond of the quintic threefold X and its mirror are given by Conclusion Every Street that we met, is a starting point of my brand new life [] References [] D. Adams. Notes on Hilbert Schemes. San Val,