ENUMERATIVE GEOMETRY FOR THE REAL GRASSMANNIAN OF LINES IN PROJECTIVE SPACE FRANK SOTTILE Abstract. Given Schubert conditions on lines in projective space which gener-
ically determine a nite number of lines, we show there exist general real conditions determining the expected number of real lines. This extends the classical Schubert calculus of enumerative geometry for the Grassmann variety of lines in projective space from the complex realm to the real. Our main tool is an explicit geometric description of rational equivalences, which also constitutes a novel determination of the Chow rings of these Grassmann varieties of lines. The combinatorics of these rational equivalences suggests a non-commutative, associative product on the free Abelian group on Young tableaux. We conclude by considering some of these enumerative problems over nite elds.
1. Introduction Describing the common zeroes of a set of polynomials is more problematic over non-algebraically closed elds. For systems of polynomials with few monomials on a complex torus (\fewnomials"), Khovanskii [8] showed that the number of real zeroes are at most a small fraction of the the number of complex zeroes. Fulton ([5], x7.2) asked how many solutions to a problem of enumerative geometry can be real; for example, how many of the 3264 conics tangent to ve general real conics can be real. He later showed that all, in fact, can be real. This was rediscovered by Ronga, Tognoli, and Vust [13]. Robert Speiser suggested the classical Schubert calculus of enumerative geometry would be a good testing ground for this question. For problems of enumerating lines in Pn incident to real linear subspaces in general position, we show that all solutions can be real. Let G Pn be the Grassmannian of lines in Pn. A ag and a partition = ( ; ) determine a Schubert subvariety of type , which has codimension jj = + . The automorphism group PGLn of Pn acts transitively on G Pn and on the set of Schubert varieties of a xed type. Over elds of characteristic zero, Kleiman's Transversality Theorem [9] shows that a general collection of Schubert subvarieties of G Pn will intersect generically transversally. In x5, we extend this to 1
+1
1
1
Date : September 6, 1996. 1991 Mathematics Subject Classi cation. 14M15, 14N10, 14P99, 05E10. Key words and phrases. Grassmannian, real enumerative geometry, Young tableaux. Research supported in part by NSERC grant # OGP0170279. To appear in Duke Mathematical Journal. 1
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FRANK SOTTILE
elds of positive characteristic. A consequence is that for Schubert-type enumerative problems in G Pn, the basic principle of the Schubert calculus remains valid in positive characteristic: each component of a general intersection of Schubert varieties appears with multiplicity one. For partitions ; : : : ; m, let G ( ; : : : ; m) be the (non-empty) set of points of the Chow variety of G Pn representing cycles arising as generically transverse intersections of Schubert varieties of types ; : : : ; m. Any generically transverse intersection of Schubert varieties is rationally equivalent to a sum of Schubert varieties; the Schubert calculus gives algorithms for determining how many of each type. In x4, we prove 1
1
1
1
1
Theorem A. Let ; : : : ; m be partitions. Then there is a cycle (depending upon ; : : : ; m) whose components are explicitly described Schubert varieties, such that is in the Zariski closure of G ( ; : : : ; m). Moreover, for each cycle X in G ( ; : : : ; m), there is an explicit chain of rational curves between X and with each curve lying in the Zariski closure of G ( ; : : : ; m). 1
1
1
1
1
The proof of Theorem A constitutes an explicitly geometric determination of the Schubert calculus of enumerative geometry for lines in Pn. In fact, Theorem A shows these `Schubert-type' enumerative problems may be solved without reference to the Chow ring, a traditional tool in enumerative geometry. As well, it determines products in the Chow ring: Let be the rational equivalence class of a Schubert variety of type . Equating the rational equivalence class of cycles in G ( ; : : : ; m) to that of yields a formula for products in AG Pn, the Chow ring of G Pn. Corollary B. Let c be the number of components of of type . Then 1
1
1
m Y i
=1
i =
X
c :
Thus the structure of these Chow rings is determined in a strong sense: All products among classes from the Schubert basis are expressed as linear combinations of basis elements and these expressions are obtained by exhibiting rational equivalences between a generically transverse intersection of Schubert varieties and the cycle . Moreover, these expressions require only a `set-theoretic' understanding, as the cycle is free of multiplicities. Let Fl be the manifold of real ags which parameterizes real Schubert varieties of xed type. When k = R, Theorem A has the following consequence.
REAL ENUMERATIVE GEOMETRY
3
Theorem C. Let n; : : : ; m be partitions with j j + + jmj = 2n ? 2, the dimension of G P . Then there exists a non-empty classically open subset of (Fl)m consisting of m-tuples of ags whose corresponding Schubert varieties 1
1
1
(of respective types ; : : : ; m) meet transversally, with all points of intersection real. To the best of the author's knowledge, this is the rst result showing that a large class of non-trivial enumerative problems can have all of their solutions real. Theorem C may have applications outside of geometry. For example, some problems involving real matrices, such as the pole assignment problem in systems control theory [1], may be expressed as intersection problems on a real Grassmannian. The construction of the cycle and rational curves of Theorem A uses the combinatorics of Young tableaux and suggests a non-commutative, associative algebra with additive basis the set of Young tableaux, described in x7. This algebra has surjections to the Chow rings of Grassmann varieties and the algebra of symmetric functions. However, it di ers fundamentally from the plactic algebra of Lascoux and Schutzenberger [11], which is also non-commutative, associative, constructed from Young tableaux, and related to symmetric functions. In x8, we ask which enumerative problems may be solved over which ( nite) elds and give the answer for two classes of Schubert-type enumerative problems. We also show how some of our constructions may be carried out over nite elds. The rational equivalences of Theorem A arise from a sequence of deformations which transform generically transverse intersections of Schubert varieties into , a sum of distinct Schubert varieties. We regard this as the classical method of degeneration, with a shift in focus. Traditionally, a product of cycles is computed by moving the cycles into special position and then studying the resulting intersection cycle. This may fail when applied to more than a few cycles; an intersection typically becomes improper before the positions are special enough for the intersection to be easily recognized. Rather than considering deformations of cycles to be intersected, we use an idea of Chaivacci and Escamilla-Castillo [2] and study deformations of intersection cycles. In particular, the cycle of Theorem A is usually not an intersection of Schubert varieties. Theorem C is deduced from Theorem A and `real' conservation of number; the number of real points in a real zero-cycle is constant under small real deformations. 1
2. Preliminaries Let k be an in nite eld. Varieties will be quasi-projective, reduced (not necessarily irreducible), and de ned over k. All subvarieties (except some curves) are assumed to be closed. When k = R, let X (R) be the R-valued points of X , equipped with the classical topology. Note that X (R) need not be topologically connected, even when X is irreducible and projective.
4
FRANK SOTTILE
Let X be a smooth variety, U and W subvarieties of X , and set Z = U \ W . We say U and W meet properly if either Z is empty or the codimension of Z in X is the sum of the codimensions of U and W . We say U and W meet generically transversally if each irreducible component of Z has an open subset along which U and W are nonsingular and meet transversally. In this case, Z is generically reduced (reduced at the generic point of each component), the fundamental cycle [Z ] of Z is multiplicity free, and in the Chow ring AX of X [U ] [W ] = [U \ W ] = [Z ] =
r X i
[Zi ];
=1
where Z ; : : : ; Zr are the irreducible components of Z . 2.1. Chow varieties. Suppose X is projective. Then positive cycles of a xed dimension and degree on X are parameterized by Chow X , a Chow variety of X . We write Chow X for any Chow variety of X ; context will indicate the dimension and degree intended. Let U be a normal variety and a subvariety of X U with generically reduced equidimensional bres over U . The association of a point u of U to the fundamental cycle of the bre u determines a function from U to Chow X . Since U is normal, this function is algebraic ([10], [4]). Moreover, if X , U , and are de ned over k, then so are Chow X and the map : U ! Chow X ([15], xI.9). Cycles represented by points on a rational curve in Chow X are rationally equivalent; the converse to this statement is not true. If Y and Z are rationally equivalent, then there is a third cycle W such that Y + W and Z + W are connected by a chain of rational curves, as points in some Chow variety of X. 2.2. Grassmannians and Schubert varieties. For S Pn, let hS i denote the linear span of S . For a vector space V , let PV be the projective space of all one dimensional subspaces of V . Suppose K = PU and M = PW . De ne Hom(K; M ) to be Hom(U; W ), the vector space of linear maps from U to W . If K M , set M=K = P(W=U ). A complete ag F is a collection of linear subspaces Fn F F = Pn, where dim Fi = n ? i. We adopt the convention that Fp = ; for p > n. Let G Pn be the Grassmannian of lines in Pn, a (2n ? 2)-dimensional variety. For a partition = ( ; ), let Fl denote the variety of partial ags of type ; those K M with K an (n ? ? 1)-plane and M an (n ? )-plane. If Fl is non-empty, then necessarily 0 < + 1 n, so + 2n ? 2. A partial
ag K M determines a Schubert variety (K; M ) comprising lines contained in M which also meet K . The type of (K; M ) is the type, = ( ; ), of its de ning partial ag K M , and its codimension is jj = + . If = , then (K; M ) = G M , the Grassmannian of lines in M . If M = Pn, so = 0, then we write K for this Schubert variety. The tangent space to ` 2 G Pn is naturally identi ed with the linear space Hom(`; Pn=`). In fact, ` has an ane 1
q
1
0
1
1
1
REAL ENUMERATIVE GEOMETRY
5
neighbourhood isomorphic to Hom(`; Pn=`). It is not hard to verify the following lemma, whose proof we omit. 2.3. Lemma. Let K; M be subspaces of Pn 1. If K M , then the smooth locus of (K; M ) consists of those lines ` with ` 6 K . For such `,
T` (K; M ) = f 2 Hom(`; Pn=`) j (`) M=` and (` \ K ) 2 hK; `i=`g:
T
2. We have K G M = (K \ M; M ). This intersection is transverse at the smooth points of (K \ M; M ) if and only if K and M meet properly in Pn . T 3. Let Ki Mi, for i = 1; 2. If the intersection (K ; M ) (K ; M ) is proper, then Mi meets Kj properly for i 6= j and M \ M is proper. 1
1
1
1
2
2
2
An intersection of two Schubert varieties may be generically transverse and reducible. In fact, this observation is at the heart of our methods. 2.4. Lemma. Let H Pn be a hyperplane, P 6 H a linear subspace, and F P \ H a proper linear subspace. Let N 6 H be a linear subspace meeting F |and hence P |properly, and set L = N \ H . Then (F; P ) and L meet generically transversally in G Pn,
(F; P )
\
1
L = (N \ F; P ) + (F; P \ H )
\
N ;
and the second component is itself a generically transverse intersection. Proof: The rightT hand side is a subset of the left; we show the other inclusion. Let ` 2 (F; P ) L . If ` meets L\F = N \F , then ` 2 (N \F; P ). Otherwise, ` is spanned by its intersections with F and L, hence ` P \ hF; Li P \ H T and so ` 2 (F; P \ H ) N . We use (1) of Lemma 2.3 to verify these intersections are generically transverse. T Let ` 2 (F; P \H ) N and suppose that p = `\F is distinct from q = `\N , so that each Schubert variety is smooth at ` = hp; qi. Then hF; qi = hF; `i P \ H and hN; pi = hN; `i, so that
T` (F; P \ H ) = f 2 Hom(`; Pn=`) j (q) 2 P \ H=` and (p) 2 hF; qi=`g T` N = f 2 Hom(`; Pn=`) j (q) 2 hN; pi=`g: The assumptions on F; P; N , and H ensure these tangent spaces meet properly, proving (F; P \ H ) and N meet transversally at `. Similar arguments may be used to verify the other transversality assertions.
6
FRANK SOTTILE
2.5. Young tableaux I. The Young diagram of a partition = ( ; ) is a tworowed array j jjof kboxesj jwith k boxes in the rst row and in the second. Note j that . Here, brc denotes the greatest integer less than or equal to r. We make no distinction between a partition and its Young diagram. A Young tableau T of shape is a lling of the boxes of with the integers 1; 2; : : : ; jj. These integers increase left to right across each row and down each column. Thus the ith entry in the second row of T must be at least 2i. Call jj the degree of T , denoted jT j. If = , then T is rectangular. Here are three Young tableaux; the rst is rectangular. +1
2
2
1 2 5 3 4 6
1 2 5 7 3 4 6
1 3 4 6 7 8 2 5 9
2.6. Arrangements. A key to our proof of Theorem A is the form of the cycle as well as the cycles which are intermediate in the rational equivalences. These cycles depend upon a particular lattice of subspaces of Pn de ned by an arrangement of hyperplanes H ; : : : ; H n? with speci ed linear dependencies. Given hyperplanes H ; : : : ; H n? andT a subset A f2; 3; : : : ; 2n ? 2g, de ne HA by H; = Pn and, if A 6= ;, HA = j2A Hj . 2
2
2
2
2
2
2.7. De nition. An arrangement F in Pn is a collection of 2n ? 3 distinct hyperplanes H ; : : : ; H n? in Pn such that for m = 2; : : : ; n, the following two conditions hold: (A.1) Fm := Hf ; ;:::; m? g has codimension m, and if m 6= n, then Fm H m? but Fm 6 H m. Thus Fm = Fm \ H m , and if A f2; : : : ; 2m ? 1g, then HA 6 H m . (A.2) If A f2; : : : ; 2m ? 2g, and Fm 6= HA, then HA 6 H m? . A consequence of (A.1) is that the subspaces F ; F ; : : : ; Fn form part of a complete ag in Pn, and that if l 2, then Fl Hj if and only if l j + 1. A main point of these conditions is that if A f2; : : : ; lg, then HA meets Hl properly unless l even and HA = F l , or dropping the condition that l is even, unless HA = Fb l c (which implies l is even, by (A.1)). Arrangements can be constructed over an arbitrary in nite eld. The potential obstruction for nite elds is nonexistence of hyperplanes H m? satisfying (A.2). In x8, we describe an inductive construction of arrangements and estimate over which nite elds it is possible to have an arrangement. 2
2
2 3
2
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+1
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+1
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+1 2
+1
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1
2.8. An arrangement in P . We illustrate this de nition with an arrangement in P . Let x ; x ; : : : ; x be homogeneous coordinates for P . De ne hyperplanes 5
5
5
0
1
5
REAL ENUMERATIVE GEOMETRY
7
H ; : : : ; H by the linear forms: H : x H : x H : x H : x +x +x (1) H : x H : x + 2x + 3x + x H : x The ag F associated to this arrangement is the standard ag, where Fi is de ned by the equations 0 = x = = xi? . The hyperplanes H ; : : : ; H restricted to the P de ned by x = 0 give an arrangement in P . Figure 1 shows (part of) the Hasse diagram of the lattice of subspaces in P generated by this arrangement (also part of the Hasse diagram of the original arrangement in P ). 2
8
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H2 H23 = F 2
H3
H4
H5
H24 H34 H25 H35 H45
H234 = F 3
H6 H26
H36
H46
H56
H236 H246 H256 H346 H356 H456
F 4 = H2346
Figure 1. The Arrangement in P
4
2.9. De nition-Lemma. Let F be an arrangement in Pn. For each tableau T with jT j 2n ? 2, de ne HT to be HA , where A is the set of entries in the second row of T . If T has shape ( ; ) with 1, then the following hold. (1) dim HT = n ? . (2) HT 6= F . (3) If S is another tableau with shape ( ; ), then HS = HT only if S = T . (4) If T is rectangular with degree 2 , then HT \ H = F . Proof: Let the second row of T be the set A [ fj g, where j is the largest entry in the second row of T . We prove the rst part by induction on . Let S be the tableau obtained from T by removing the entries j; j + 1; : : : ; + . Then HS = HA has dimension n ? + 1 and so HT will have dimension n ? if HS 6 Hj , since HT = HS \ Hj . As A f2; : : : ; j ? 1g, we see by (A.1) and (A.2) that HA Hj only if j = 2m ? 1 and Fm = HA. In this case, m = ? 1, 2
+1
+1
8
FRANK SOTTILE
thus j = 2 ? 3. But this contradicts j 2 , as j is the th entry in the second row of T . To prove the second statement, note that F 6 Hj , since Fb j c 6 Hj and Fb j c F , as j 2 . Since HT Hj , it follows that F 6= HT . For (3), suppose that is minimal with respect to the existence of tableaux S 6= T of shape ( ; ) where HS = HT . Let j be the largest entry in the second row A [ fj g of T and let m be the largest entry in the second row B [ fmg of S . If m 6= j , suppose m < j . Then HS Hj , as HT Hj . Since B [ fmg f2; : : : ; j ? 1g, this implies HS = F , contradicting (2). Suppose now that j = m. By the minimality of , we have HA 6= HB and so HA[B = HA \ HB = HT Hj . Then (A.1) implies j is odd and (A.2) implies HT = F , as A [ B f2; : : : ; j ? 1g, again contradicting (2). For the last statement, suppose T is rectangular with jT j = 2 . Since the second row of T is a subset of f2; : : : ; 2 g and F = Hf ;::: ; g, we have F HT , and so F is a hyperplane in HT . Then (4) is immediate, as F H but HT 6 H , by (A.2). 2
2
+1
2
+1
2
2
+1
+1
2
+1
+1
2.10. De nition-Corollary. Let F be an arrangement and T a tableau of shape ( ; ) with + 2n ? 2. De ne (T ) := (F ; HT ). Then (1) If < n, then F HT is a partial ag of type ( ; ). (2) If n then (T ) = ;; otherwise, (T ) is a Schubert variety of type ( ; ). (3) If S is a tableau and (S ) = (T ) 6= ;, then S = T . Proof: For the rst statement, note that the second row of T is a subset of f2; : : : ; + g, so that Fb c HT , by (A.1). This shows F HT , as . Since dim HT = n ? by Lemma 2.9(1), F HT is a partial ag of type ( ; ). The second statement is a consequence of the rst. For the nal statement, rst note that if ; 6= K M , then M is the union of all lines in (K; M ). Thus (S ) = (T ) 6= ; implies that HT = HS . But then S = T , by Lemma 2.9(3). +1
+1
+ 2
+1
+1
+ 2
+1
3. The Cycles (T ) and (Ts; ; L) We describe the cycle = (T ) of Theorem A as well as the cycles (Ts; ; L) which arise intermediately in the rational equivalences between X and in Theorem A. These are de ned with respect to a xed arrangement F , though our notation suppresses the dependence on F . These cycles are also de ned with respect to a xed set T of Young tableaux, all of which have the same degree l. In other words, jT j = l for every T 2 T .
REAL ENUMERATIVE GEOMETRY
De ne (T ), a multiplicity-free cycle on G Pn with pure dimension, by X (T ) := (T ):
9
1
T 2T
By Corollary 2.10, each Schubert variety summand of (T ) has dimension 2n ? 2 ? l and they are all distinct, showing (T ) is multiplicity-free and of pure dimension. 3.1. Young tableaux II. Suppose T is a tableau with jT j = l and s a positive integer. Let T (s) be the tableau obtained from T by adjoining the consecutive integers l + 1; : : : ; l + s to the rst row of T . Let T s be the tableau obtained by adjoining l + 1; : : : ; l + s to the second row of T , if that is possible. For example, the following picture shows T , T (3) and T . Note that T is not de ned. +
+3
1 2 3 5 4
+4
1 2 3 5 6 7 8 4
1 2 3 5 4 6 7 8
With this notation, HT s = HT and HT s = HT \ Hl \ \ Hl s. These relations are the motivation for arrangements and our indexing of subspaces by tableaux. If a; b 0, let T a (b) be (T a)(b). In (T a)(b), the consecutive integers l + 1; : : : l + a + b occur in distinct columns, in order left to right. If T is a set of Young tableaux for which jT j = l for every T 2 T , then set [ T (s) := T (s): +
( )
+
Similarly de ne T by
a
+
+1
+
and T
a (b).
+
+
+
T 2T
For integers 0 s , recursively de ne Ts;
T ; := T ( ); [ Ts; := Ts? ; T s( ? s) [ [ [ = T ( ) T ( ? 1) T s( ? s): 3.2. Intermediate cycles (Ts; ; L). The intermediate cycle (Ts; ; L) is de ned with respect to an integer s with 0 s , a set T of tableaux with jT j = l for every T 2 T , and a subspace L of codimension ? s + 1 in Pn which has a particular position with respect to the xed arrangement F . We say that a subspace L of codimension ? s + 1 in Pn meets F (l,s)-properly if the following 0
+
1
+1
+
conditions are satis ed 1. L meets the subspaces Fb l c ; : : : ; Fl properly; 2. L \ Fl = Fl ?s ; and 3. For each tableau T with jT j = l, L meets the subspaces HT properly. +1 2
+1
+
+2
+1
+1
s?1)
+(
and HT
s
+
10
FRANK SOTTILE
If Fl ?s 6= ;, then the last condition is redundant: For any b, either HT b is unde ned, or else Fl HT b , implying that L meets HT b properly, as L meets Fl properly with non-empty intersection. Note also that L is a linear subspace of Pn=Fl ?s with codimension ? s + 1. Suppose l; s; , and T are as in the previous paragraph, and that L is a subspace of Pn which meets F (l; s)-properly. Let T 2 Ts? ; . Then the rst row of T has length b l s + 1 + ? s. De ne the Schubert variety (T ) (= (F ; H )) if b l + ? s + 2 (T ; L) = (L \ F b; H ) T otherwise b? s T and set X (T ; L): (Ts? ; ; L) := +
+
+2
+1
+
+
+1
+
+2
1
+ 2
+1
+
1
T 2Ts?1;
The cycle (Ts? ; ; L) is well-de ned and multiplicity-free: 1
3.3. Lemma. For T 2 Ts? ; , (T ; L) is a Schubert variety of type equal to the shape of T . Furthermore, (Ts? ; ; L) is a multiplicity-free cycle with each component having dimension 2n ? 2 ? l ? . Proof: Let T 2 Ts? ; have shape (b; a). Then jT j = a + b = l + and no entry in the second row of T exceeds l + s ? 1. If the length of the rst row of T is at least l + ? s + 2, then (T ; L) = (T ) is a Schubert variety of type (b; a). Otherwise l s + 1 b ? + s l + 1, so L meets Fb? s properly in a space of codimension b + 1. Since no entry in the second row of T exceeds l + s ? 1, HT contains Fb l s c , hence also Fb? s \ L, so (T ; L) is a Schubert variety of type (b; a). To see that (Ts? ; ; L) is multiplicity-free, suppose (T ; L) = (S ; L) for some S; T 2 Ts? ; . Then S and T have the same shape and HS = HT , so S = T by Lemma 2.9(3). 1
1
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3.4. Young tableaux III. Let T be a tableau of shape (a + s; a), and let be a positive integer. De ne a set T of tableaux by T := fT ( ); T ( ? 1); : : : ; T g: (It is impossible to form the tableau T s , so if > s, then the last tableau listed will be T s( ? s).) If jT j = l, then T consists of all tableaux obtained from T by adjoining the consecutive integers l + 1; : : : ; l + to T in distinct columns and in order from left to right. For example, +1
+
+( +1)
+
1 2 5 6 3 4
*3 =
1 2 5 6 7 8 9 , 1 2 5 6 8 9 , 1 2 5 6 9 3 4 3 4 7 3 4 7 8
REAL ENUMERATIVE GEOMETRY
11
S Given a set T of tableaux, de ne T := T 2T T . Then, by the de nition of , [ [ [ T = T ( ) T ( ? 1) T = T ; : Let ; : : : ; m be positive integers and ; the empty tableau, the only tableau +1
+
1
of shape (0; 0). De ne the set of tableaux m := ( ((; ) ) ) m : P i Let si = j j . Then m is the set of tableaux where, for each 1 i m, the consecutive integers si? + 1; : : : ; si? + i = si occur in distinct columns, and in order from left to right. In particular 1 1 (l 1's) is the set of all tableaux of degree l, and 2 4 3 is the set of 11 tableaux shown in Figure 2. 1
1
2
1
=1
1
1
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 8 9 7
1 2 3 4 5 6 9 7 8
1 2 3 4 5 6 7 8 9
1 2 4 5 6 7 8 9 3
1 2 4 5 6 8 9 3 7
1 2 4 5 6 9 3 7 8
1 2 4 5 6 3 7 8 9
1 2 5 6 7 8 9 3 4
1 2 5 6 8 9 3 4 7
1 2 5 6 9 3 4 7 8
Figure 2. 2 4 3
3.5. Families of intermediate cycles. The cycles (Ts? ; ; L) are used to construct families of cycles which arise in the proof of Theorem A. These families, i;s ! Ui;s, have base Ui;s which is an open subset of a product of Grassmann varieties. The di erent families are compared by considering the subset Gi;s of the Chow variety whose points represent fundamental cycles of bres of the family i;s ! Ui;s. In fact, we only use the Chow variety as a convenient place to compare cycles from di erent families. Let ; : : : ; m be positive integers and F an arrangement. Fix 1 i m and set T := i? . Then every tableau T 2 T has degree l := + + i? . For 0 a n, let Ga Pn = Gn?aPn, the Grassmannian of codimension a planes in Pn. Q Let 0i; be theQsubscheme of G Pn mj i G j Pn whose bre at a point (Ki; : : : ; Km) of mj i G j Pn is \ \ \ (2) (T ) Ki Km : Q Let Ui; mj i G j Pn consist of those (m ? i + 1)-tuples for which this intersection is generically transverse, and let i; be the restriction of 0i; to Ui; , 1
1
1
1
1
1
+1
1
0
=
+1
=
+1
0
=
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12
FRANK SOTTILE
so i; ! Ui; is a family with generically reduced equidimensional bres. This family induces a map : Ui; ! Chow G Pn. Let Gi; Chow G Pn be its image, the set of cycles which are generically transverse intersections of the form (2). Similarly, for 1 s i, let V G i ?s (Pn=Fl i?s ) be the (open) subset of those L of codimension i ?sQ +1 in Pn which meet F (l; s)-properly. Let 0i;s be the subscheme of G Pn V mj i G j Pn whose bre at (L; Ki ; : : : ; Km) is h \ i\ \ \ (Ts? ; i ; L) + (T s) L Ki Km : Q De ne Ui;s V mj i G j Pn, i;s, and Gi;s analogously to Ui; , i; , and Gi; . Set Gm ; to be the singleton f ( m)g. When k = R, set Gi;s R = (Ui;s(R)), the set of fundamental cycles of bres of i;s over real points of Ui;s. 0
0
0
1
0
1
+1
+
+2
+1
1
+1
= +1
+
1
+1
+1
0
= +1
0
+1 0
0
1
;
4. Main Results Kleiman's Transversality Theorem [9] establishes the transversality of a general translate in characteristic zero. In positive characteristic, there exists a smooth subvariety of G P which does not meet a particular smooth Schubert variety, nor any PGL -translate of that Schubert variety, generically transversally ([9], x9). However, inn x5, we show (Theorem E) that general translates of Schubert varieties of G P do meet generically transversally and use this to show the sets Ui;s are non-empty, proving the following lemma. 3
1
4
1
4.1. Lemma. Let ; : : : ; m be positive integers and let F be any arrangement. Then for all 1 i m and 0 s i , Ui;s is a dense open subset of the corresponding product of Grassmannians. Thus Gi;s is a unirational subset of Chow G Pn. When k = R, Gi;s R is the image of the (non compact) real algebraic manifold Ui;s(R) under the morphism . In x6, we prove: Theorem D. Let ; : : : ; m be positive integers, F an arrangement, and 1 i m. 1. Let X be a closed point of Gi ; . Then there is an open subset U of P n f0g and a family of cycles X over U [ f0g such that X is the bre of X over 0 and bres of XjU are in Gi; i . Thus (U ) is a rational curve in Gi; i such that X 2 (U ) [ f0g. 2. Fix s with 0 s < i. Let X be a closed point of Gi;s . Then there is an open subset U of P n f0g and a family of cycles X over U [ f0g such that X is the bre of X over 0 and bres of XjU are in Gi;s. Thus (U ) is a rational curve in Gi;s such that X 2 (U ) [ f0g. 1
1
;
1
1
+1 0
+1
1
REAL ENUMERATIVE GEOMETRY
13
4.2. Proof of Theorem A using Theorem D. Suppose i = ( i ; 0) for 1 i m. Let F be an arrangement and let ( m ) be the cycle of Theorem A. Then G ; = G ( ; : : : ; m ) and Gm ; = f( m )g. Since Gi;s is unirational, any two points of Gi;s are connected by a chain of rational curves, each lying within the closure of Gi;s. Downward induction on the lexicographic order on pairs (i; s), together with and Theorem D proves the existence of a chain of rational curves between ( m) and an arbitrary cycle X 2 G ( ; : : : ; m ). Thus Theorem D implies Theorem A when each partition i is a single row. Now suppose i = ( i + i; i) for 1 i m and set = + + m. For a linear subspace M Pn of codimension , let F be an arrangement in M , and put = ( m). De ne Ui;s and Gi;s as in x3.5, with M replacing Pn. nThen G ; G ( ; : : : ; m). In fact, if PGLn is the automorphism group of P , then G ( ; : : : ; m ) is the union of all translates of G ; under the action of PGLn . Let X 2 G ( ; : : : ; m), so X is a generically transverse intersection \ \
(K ; M ) (Km ; Mm ); where Ki Mi has type i for 1 i m. Set M = M \ \ Mm . Iteration of Lemma 2.3(3) shows that if M has codimension in Pn and Li = M \ Ki has codimension i + 1 in M . Thus \ \ X = L Lm is a generically transverse intersection in G M . Let be any automorphism of Pn with M = M , so that (X ) 2 G ; , and suppose that ? is a one parameter subgroup containing . The orbit ? X is a rational curve (or a point) in G ( ; : : : ; m), and contains (X ). Since
(X ) 2 G ; , previous arguments show there exists a chain of rational curves between (X ) and ( m ), with each curve contained in the closure of G;. 1
1
1 0
+1 0
1
1
1
1
0
0
1
0
1
1 0
+1
1
1 0
1
+1
1
1
1
1
1
0
1 0
1
1 0
1
1 0
4.3. The Schubert calculus. Interpreting Theorem A in terms of products in the Chow ring of G Pn, we have: Corollary B0. Let ; : : : ; m be partitions with i = ( i + i; i) for 1 i m, n and set = + + m . Then in A G P , 1
1
1
Y m
i
=1
1
i =
X
c a;b;:::; m : (
= ( + a; + b) a+b= 1 ++ m
)
1
(3)
where c ;:::; m is the numberPof tableaux ofPshape such that for 1 i m, the i consecutive integers 1 + ij? j ; : : : ; ij j occur in distinct columns and in order from left to right. 1
1
=1
=1
14
FRANK SOTTILE
Corollary B0 follows from Theorem A: The integers c ;:::; m count the number of tableaux in the set m of shape and hence the number of components of ( m ) of type . Thus Corollary B 0 asserts the equality of the rational equivalence classes of cycles shown to be rationally equivalent in Theorem A: The left hand side is the rational equivalence class of cycles in G ( ; : : : ; m) and the right hand side is the rational equivalence class of ( m ). The number c ;:::; m is in fact the Kostka number K , where = ( ; : : : ; m ) (see [14]). If = = m = 0 and + + m = 2n ? 2, then the only non-zero term n? ;n? ;n? on the right hand side of (3) is c n?;:::; n? ;n? , or c ;:::; m times the class m of a point (line). Since i ; is the rational equivalence class of L , whenever L has dimension n ? i ? 1, we deduce: 1
1
1
1
1
1
1
1
1
(
1
(
(
1)
1
(
1
1
1)
1
1)
0)
4.4. Corollary. The number of lines meeting general (n ? i ? 1)-planes for 1 i m is equal to the number of tableaux of shape (nP? 1; n ? 1) such that for P i ? 1 i m, the consecutive integers 1 + j j ; : : : ; ij j occur in distinct columns and in order from left to right. 1
=1
=1
4.5. Enumerative geometry for the real Grassmannian. Let ; : : : ; m be partitions and de ne GR to be those cycles in G ( ; : : : ; m) which arise as generically transverse intersections of Schubert varieties de ned by real ags. Theorem C0. Let ; :n: : ; m be partitions with i = ( i + i; i), and set P m = i i. Let M P be a real (n ? )-plane, F a real arrangement in M , and de ne ( m ) in terms of F . 1. ( m ) is in the closure of GR . 2. If Q j mj + + jmj = 2n ? 2, then there is a non-empty classically open subset of i Fli whose corresponding Schubert varieties meet transversally, with all points of intersection real. 1
1
1
=1
1
1
1
=1
Proof: De ne the sets Ui;s and Gi;s as in x3.5 for the arrangement F in M and the integers ; : : : ; m. Arguing as in the proof of Theorem A shows G ; R GR. Restricting to the real points of the varieties in Theorem D shows Gi;s R G ; R. 1
1 0;
;
1 0;
Assertion (1) is the case (i; s) = (m + 1; 0). ;n? . Then ( ) consists of d distinct real For (2), let d = c n?;:::; m m lines. Hence Gi;s S dG Pn, the Chow variety of e ective degree d zero cycles on G Pn. The real points S d G Pn(R) of S dG Pn are e ective degree d zero cycles stable under complex conjugation. The dense subset of S dG Pn(R) of multiplicity free cycles has a topological component M parameterizing sets of d distinct real lines and ( m) 2 M. By (1), ( m ) 2 GR , which T shows GR M 6= ;, a restatement of (2). (
1
1)
1
1
1
1
1
1
1
1
1
REAL ENUMERATIVE GEOMETRY
15
5. Generically Transverse Intersections We give a proof, valid in any characteristic, that general translates of Schubert varieties of G Pn meet generically transversally. This extension of Kleiman's characteristic zero transversality Theorem [9] in this situation relies upon his result that general translates have proper intersection. Theorem E. Let ; : : : ; m be partitions. Then the set U Qmi Fli of partial
ags (K M ; : : : ; Km Mm ) for which the intersection \ \
(K ; M ) (Km ; Mm) 1
1
=1
1
1
1
Q
1
is generically transverse is a dense open subset of mi Fli . Proof: We begin with an observation which simpli es the geometry. For 1 i m, let Ki Mi be a partial ag of type i = ( i + i; i) and suppose the corresponding Schubert varieties meet properly. By Lemma 2.3(3), =1
\ \
\ \
\
(K ; M ) (Km ; Mm) = G M N Nm ; where M = M \ \ Mm has codimension = + + m and Ki = Ni \ Mi , with Ni meeting Mi properly for 1 i m. Fix a codimension subspace M of Pn. As U is stable under the diagonal action of PGLn , it is the union of the translates of U \ X , where X consists of those m-tuples of ags with M Mi, for 1 i m. Moreover, U is open if and only if V := U \ X is open in X . Let Y X be the set of m-tuples of ags where M = M \ \ Mm and Ki meets M properly. The product of the maps (Ki ; Mi) 7! Ki \ M = Li; i = 1; : : : ; m Q i M , and V is the inverse image exhibits Y as a bre bundle with base Tmi GT Lm is generically transverse. It of the set of (L ; : : : ; Lm) for which TL T suces to prove the lemma for L Lm in G M , that is, for the case when i = 0 for 1 i m. Let be the subscheme of 1
1
1
1
1
1
+1
1
+1
=1
1
1
1
1
(Pn)m G
1
Pn
m Y i
G i Pn +1
=1
consisting of (2m + 1)-tuples (p ; : : : ; pm; `; L ; : : : ; Lm ) such that pi 2 ` \ Li for 1 i mQ . The projection of to (Pn)m G Pn exhibits as a bre bundle with bre mi G i (Pn=pi) and base f(p ; : : : ; pm; `) j each pi 2 `g. This base has dimension m + 2n ? 2, so is irreducible of dimension 1
1
1
+1
1
=1
m + 2n ? 2 +
m X
m X
i
i
=1
(n ? i ? 1)( i + 1) = 2n ? 2 ?
=1
i +
m X i
=1
(n ? i)( i + 1):
16
FRANK SOTTILE
Q The projection of to mi G i Pn has image consisting of those (L ; : : : ; Lm ) whose corresponding Schubert varieties P have non-empty intersection. This image is a proper subvariety if 2n ? 2 < mi i. In this case, let U be the complement of this image. P Suppose 2n?2 m . Let W consist of those points where ; : : : ;
+1
1
=1
=1
i
i
L1
Lm
meet transversally at `. By Lemma 2.3, W consists of those points such that 1. ` 6 Li for 1 i m; thus pi = ` \ Li and ` is a smooth point of Li . 2. The tangent spaces T` Li for 1 i m meet transversally. Thus W is an open subset of . We show W 6= ;. Fix ` 2 G Pn and x distinct points p ; : : : ; pm of `. Suppose ` = PH and for each 1 i m, x a point qi 2 H such that pi = Phqii. Suppose also that Pn=` = PM and identify Pn with P(H M ). Recall that T`G Pn = Hom(`; Pn=`) = Hom(H; M ). De ne the linear map f : Hom(H; M ) ?! M m by f ( ) = ( (q ); : : : ; (qm )). If m 2 (the case m=1 is trivial), then f ? (0) = f0g, as H = hq ; q i. For any linear subspace Mi M of dimension n ? 1 ? i , de ne Li := Phqi; Mi i. Then Li 2 G i Pn and pi = ` \ Li. Moreover, T` Li = f 2 Hom(H; M ) j (qi) 2 Mig. Fix linear subspaces M ; : : : ; Mm of M where dim Mi = n ? 1 ? i. Since (GL(M ))m acts transitively on M m n f0g, Theorem 2(i) of [9] shows there exists a dense open subset V of (GL(M ))m consisting of g such that f ? (g(M Mm )) P either is f0g or has codimension equal to mi i, the codimension of M Mm in M . Let g = (g ; : : : ; gm) 2 V and set Mi0 := giMi and L0i := hqi; Mi0 i. Then L0i 2 G i Pn, and f ? (M 0 Mm0 ) = f 2 Hom(H; M ) j (qi) 2 Mi0 ; for i = 1; : : : ; mg =1
1
1
1
1
1
1
2
+1
1
1
1
1
=1
1
+1
1
\m
1
=
i
T` L0i :
P
=1
Since i is the codimension of T` L0i in T`G Pn and mi i 2n ? 2, the varieties L0 ; : : : ; L0m meet transversally at `. Thus W 6= ;. Q Let Z = n W and let : ! mi G i Pn be the projection. T Then
L : : : ; Lm meet generically transversally if dim(? (LP; :m: : ; Lm) Z ) is less than the expected dimension of a bre, which is 2n ? 2 ? i i. Take U to be the set of m-tuples (L ; : : : ; Lm) where 1
=1
1
+1
=1
1
1
1
=1
1
dim(?
1
(L ; : : : ; Lm ) 1
\
Z ) < 2n ? 2 ?
m X i
=1
i :
Thus U is open and non-empty, for otherwise dim Z = dim , which would imply Z = and contradict W 6= ;.
REAL ENUMERATIVE GEOMETRY
17
We also need a further lemma: 5.1. Lemma. Let d; ; : : : ; m be positive integers and let Z be a subscheme of Q m n G P with dim(Z ) < d. Then the set W i G i Pn consisting of those (K ; : : : ; Km ) for which 1
+1
1
=1
1
dim(Z
\
K
\ \
Km ) < d ?
1
m X i
i
=1
is open and dense. Qm G i Pn whose bre at (K ; : : : ; K ) Proof: Let be the subscheme of Z m i T T T is Z K Km , and let Xi(`) G i Pn denote the set of Ki which meet `. Projection to Z exhibits as a bre bundle with bre X (`) Xm (`) at a point ` 2 Z . The codimension of Xi(`) in G i Pn is i, so has dimension +1
=1
1
+1
1
1
+1
dim Z ?
m X
i +
m X
(n ? i)( i + 1):
i i P Q m n i W i G P is the locus where the bre dimension is less than d? mi i. By upper semicontinuity of bre W is open. If W were empty, then Qmdimension, i Pn would have dimension at least d ? G all bres of the projection to Pm . This would imply i i i m m X X =1
=1
+1
=1
=1
+1
=1
=1
dim d ?
i
=1
i +
(n ? i)( i + 1);
i
=1
a contradiction, as d > dim Z . Thus W must be non-empty. 5.2. Proof of Lemma 4.1. We show that for each (i; s) with 1 i m and 0 s i, the sets Ui;s are open dense subsets of the corresponding products of Grassmannians. We rst show this for Ui; . Let T = i? . Recall that Ui; consists of those (Ki; : : : ; Km) such that the intersection \ \ \ (4) (T ) Ki Km P is generically transverse. The cycle (4) has dimension d = 2n ? 2 mj j . Let Z be the singular locus of (T ). Since (T ) has pure dimension 2n ? P i transverse if for every component 2 ? j? j , the intersection (4) is generically T T T (T ) of (T ),Tthe intersection T T (T ) Ki Km is generically transverse, and if dim(Z Ki Km ) < d. By Corollary 2.10(3), (T ) = (S ) 6= ; implies T = S . Thus Z is a union of intersections of components and the singular loci of components, and hence Z ) < dim((T )). By Lemma 5.1, there is an open subset W of Qm Gdim( j Pn consisting of those (K ; : : : ; K ) for which dim(Z T T T ) < i m Ki Km j i d. 0
1
1
0
=1
1
=1
+1
=
18
FRANK SOTTILE
Q For T 2 T , let UT mj i G j Pn be the set of (Ki; : : : ; Km) for which the intersection \ \ \ (T ) Ki Km ?T U \ W , it suces to show that is generically transverse. Because U = i; T 2T T Q m j U is a dense open subset of G Pn for each T 2 T . Let T 2 T and +1
=
0
T
+1
j i
suppose T has shape = ( ; ). De ne =
V Fl
m Y j i
G j Pn +1
=
to be the set of ags for which the intersection \ \ \
(F; H ) Ki Km is generically transverse. By Theorem E, V is dense and open. Note that
fF HT g UT = V
\
fF HT g +1
+1
m Y j i
!
G j Pn ; +1
=
so UT is open. Since Fl = PGLn fF HT g, and V is stable under the diagonal action of PGLn , we see that V = PGLn (fF HT g UT ). Thus UT is non-empty. The case of Ui;s for s > 0 follows by similar arguments. +1
+1
+1
+1
+1
6. Construction of Explicit Rational Equivalences We use the following lemma to parameterize the explicit rational equivalences in the proof of Theorem D. 6.1. Lemma. Let F be a complete ag in Pn. Suppose L1 is a hyperplane notScontaining Fn. Then there exists a pencil of hyperplanes Lt , for t 2 P = A f1g, such that if t 6= 0, then Lt meets the subspaces of F properly, T and, for each i n ? 1, the family of codimension i + 1 planes induced by Lt Fi , for t 6= 0, has bre Fi at 0. Proof: Let x ; : : : ; xn be coordinates for Pn such that L1 is given by xn = 0 n and Fi by x = = xi? = 0. Let e ; : : : ; en be a basis for P dual to these coordinates. For t 2 A , de ne Lt = htej + ej j 0 j n ? 1i: T For t 6= 0, Lt Fi = htej + ej j i j n ? 1i and so has codimension i + 1. The bre of this family at 0 is hej j i j n ? 1i = Fi . In the situation of Lemma 6.1, write limt! Lt \ Fi = Fi . q
1
1
q
+1
0
0
1
0
1
+1
+1
+1
+1
0
+1
REAL ENUMERATIVE GEOMETRY
19
6.2. Proof of Theorem D, Part 1. Fix an arrangement F and an integer i with 1 i m. Set T := i? and l := + + i? , the degree of all tableaux in T . Suppose X is a cycle in Gi ; . That is \ \ \ X = (T i) Ki Km ; where the intersection is generically transverse. Let L1 be any hyperplane which meets F (l; i)-properly. We use Lemma 6.1 to produce a family of cycles with base P whose special member is X , and whose general member is a cycle in Gi; i . Apply Lemma 6.1 to the ag induced by F in Pn=Fl and the hyperplane L1=Fl to obtain a pencil Lt of hyperplanes such that if t 6= 0 and i l + 1, then Lt meets Fi properly and limt! Lt \ Fi = Fi . Let X be the subscheme of P G Pn whose bre at non-zero t 2 P is h i\ \ \ Xt = (T i ? ; i ; Lt) + (T i ) Ki Km : It is shown below that X is the bre of X at 0; we rst deduce Part 1 from this fact. Let U 00 P be the open subset of those t for which Xt is generically reduced and dim Xt = dim X . As X is generically reduced, U 00 is non-empty. Since L1 meets F (l; i)-properly, the set U 0 P consisting of those t where Lt meets F (l; i)-properly is open and dense. Since Lt = G Pn, we have \ (T i ) = (T i ) Lt : Thus, for t 2 U := U 0 \ U 00, the bre Xt is a cycle in Gi? ; i . The restriction of X to U [ f0g gives a family over a smooth curve with generically reduced equidimensional bres. Thus the association of a point u of U [ f0g to the bre Xu gives a morphism : U [ f0g ! Chow G Pn with (U ) Gi? ; i and (0) = X , proving Part 1. We show X is the bre of X at 0 by examining components of X separately. For T 2 T i ? ; i , let XT be the subscheme of P G Pn whose bre at a non-zero t 2 Pn is \ \ \ (XT )t = (T ; Lt) Ki Km : Since 0 1 1
1
1
0
1
+1 0
0
+1
1
0
q
+2
+2
0
+1
1
1
1
+
1
+1
0
1
0
0
1
1
+
+
1
1
1
0
0
1
1
1
+1
X =
and
@ X XT A T 2T i ?1; i
+ P (T 1
i )
+
[
\
Ki
+1
\ \
Km ;
T i = T i ? ; i T i ; it suces to show that for each T 2 T i ? ; i , the bre of XT at 0 is \ \ \ (T ) Ki Km : 1
1
+1
+
20
FRANK SOTTILE
Let T 2 T i ? ; i . If the rst row of T has length exceeding Tl+1, thenT (TT; Lt) = (T ), so XT is the constant family over P with bre (T ) Ki Km . Now suppose the rst row of T has length b l + 1. Then, for t 6= 0, (T ; Lt) = T
(Fb \ Lt ; HT ). Since limt! Fb Lt = Fb , we see that (T ) = (Fb ; HT ) is theT bre atT0 of Tthe family over P whose bre at t 6= 0 is (T ; Lt). Since is an open subset UT (T ) Ki Km is generically transverse, there T P such that for t 2 UT n f0g, the intersection (T ; Lt ) Ki TT T TKm isT generically transverse. This shows that the bre at 0 of XT is (T ) Ki Km , which completes the proof of Part 1 of Theorem D. 1
1
+1
0
+1
+1
1
+1
1
+1
+1
6.3. Proof of Theorem D, Part 2. Let 0 s < i and suppose X 2 Gi;s . Then h \ i\ \ \ X = (Ts; i ; N ) + (T s ) N Ki Km ; 0
+1
+( +1)
0
+1
where N has codimension i ? s in Pn and meets F (l; s + 1)-properly, and this intersection is generically transverse. We make a useful calculation. 6.4. Lemma. Let L = N \ Hl s , a hyperplane in N . Then \ \ (T s( i ? s); N ) + (T s ) N = (T s) L : 0
+ +1
+
+( +1)
+
0
S
Since Ts; i = Ts? ; i T s( i ? s), this Lemma shows that
h
1
+
\ i\
\ \
Ki Km : X = (Ts? ; i ; N ) + (T s) L We use Lemma 6.4 to complete the proof of Theorem D, and prove it at the end of the section. Let N be any complete ag in N=Fl i ?s re ning the images of the partial ag Fl i ?s N \ Fl N \ Fb l s c L : +
0
1
0
+1
q
+
+
+2
+ 2
+1
+1
0
Let L1 be any hyperplane of N which meets F (l; s)-properly. Application of Lemma 6.1 to the ag N in N=Fl i ?s yields a pencil Lt of hyperplanes of N , each containing Fl i ?s , such that for l s + 1 j l and t 6= 0, Lt meets N \ Fj properly, with limt! Lt \ N \ Fj = N \ Fj . Since Lt \ N \ Fj = Lt \ Fj for j in this range, Lt meets Fj properly, as N meets F (l; s + 1)-properly. Let X be the subscheme of P G Pn whose bre at a non-zero t 2 P is h \ \ \ i\ Km : Xt = (Ts? ; i ; Lt ) + (T s) Lt Ki q
+
+2
+
+
+2
2
0
+1
1
1
1
+
1
+1
We claim that X is the bre of X at 0; Part 2 follows from this fact in much the same manner as Part 1 followed from the analogous fact in x6.2. We show X 0
0
REAL ENUMERATIVE GEOMETRY
21
is the bre of X at 0 by examining each component separately. For T 2 Ts? ; i let XT be the subscheme of P G Pn whose bre at t 6= 0 is \ \ \ (XT )t = (T ; Lt) Ki Km : T T T Arguing as at the end of x6.2, we conclude that (T ; N ) Ki Km is the bre of XT at 0. For S 2 T s, let XS be the subscheme of P G Pn whose bre at t is \ \ \ \ Km : (XS )t = (S ) Lt Ki T T T T Arguing as at the end of x6.2, we conclude that (S )P L Ki Km P is the bre of XS at 0. Since X = T 2Ts? ; i XT + S2T s XS , we see that the bre of X at 0 is X . 6.5. Proof of Lemma 6.4. Let T 2 T s. We show \ \ (T ) L = (T ( i ? s)) + (T ) N : Summing over T 2 T s will complete the proof. We treat the case when T is rectangular separately from the general case. T Suppose T is rectangular. Then (T ) = G HT , so by Lemma 2.3(2), (T ) L is equal to (HT \L ; HT ), and this intersection is generically transverse only if L meets HT properly. Since T is a rectangular tableau in T s, s l and l + s = 2k l s is even, so = k l. Then N meets Fk properly, as N meets F (l; s + 1)properly. Since T is rectangular, Lemma 2.9(4) implies HT \ Hl s = Fk . It follows that HT meets L = N \ Hl s properly, and HT \ L = N \ Fk . Since HT = HT i ?s and k + i ? s = l s + i ? s is the length of the rst row of T ( i ? s), we have \ (T ) L = (N \ Fb l s c ; HT i?s ) = (T ( i ? s); N ): Now suppose T is not rectangular. Let b be the length of the rst row of T , so b l s . Then HT \ Hl s = HT . Since T 2 T s , N meets HT properly, so HT meets L properly. Lemma 2.4 with F = Fb , P = HT , and H = Hl s implies \ \
(Fb ; HT ) L = (Fb \ N; HT ) + (Fb ; HT \ Hl s ) N : T But this is (T ( i ? s); N ) + (T ) N , which completes the proof. 1
1
1
+1
+1 1
+
1
+1
0
+1
+
1
0
+
+1
0
+
1
0
0
+
+ +1 2
+ +1
0
+ +1
0
+1
+1
+
(
)
2
+ 2
0
+ +1
+1
+ +1
2
(
+1
)
+1
0
+( +1)
+1
+1
+ +1
+1
0
+1
+1
+ +1
+1
7. An algebra of tableaux The Schubert classes of a Grassmann variety form an integral basis for its Chow ring. Thus there exist integral constants c de ned by the identity: X = c :
0
22
FRANK SOTTILE
In 1934, Littlewood and Richardson [12] gave a conjectural formula for these constants, which was proven in 1978 by Thomas [16]. Lascoux and Schutzenberger [11] showed how to construct the ring of symmetric functions as a subalgebra of a non-commutative associative ring called the plactic algebra whose additive group is the free abelian group with basis the set of (non-standard) Young tableaux. Each tableau T of shape determines a monomial summand of the Schur function s . Evaluating s at Chern roots of the dual to the tautological bundle of the Grassmannian gives the Schubert class . Non-symmetric monomials in these Chern roots are not de ned, so individual Young tableaux are not expected to appear in the geometry of Grassmannians. In this context, the crucial use we made of the Schubert varieties (T ) is surprising. A feature of our methods is the correspondence between an iterative construction of the set m and the rational curves in the proof of Theorem D. This suggests an alternate non-commutative, associative product on . The resulting algebra has surjections to the ring of symmetric functions and to Chow rings of Grassmannians. Additional combinatorial preliminaries for this section may be found in [14]. Here, partitions , , and may have any number of rows. Suppose T andSU are, respectively, a tableau of shape and a skew tableau of shape =. Let T U be the tableau of shape whose rst jj entries comprise T , and remaining entries comprise U , with each increased by jj. For tableaux S and T where the shape of S is , de ne X [ S T = S U; the sum taken over all and all skew tableaux U of shape = Knuth equivalent to T . Figure 3 shows an example. This product is related to the operation of x2.5: Let Y be the unique standard tableau of shape ( ; 0). If T has at most 2 rows, then T is the set of summands of T Y with at most two rows. 1
1 2 5 6 3 4
1 2 3
=
1 2 5 6 7 8 9 1 2 5 6 8 9 1 2 5 6 9 + + + 3 4 3 4 7 3 4 7 8 1 2 5 6 8 9 3 4 7 1 2 5 6 9 3 4 7 8
+
+
1 2 5 6 9 3 4 8 7 1 2 5 6 3 4 9 7 8
Figure 3. The composition
+
.
1 2 5 6 3 4 8 9 7
+
REAL ENUMERATIVE GEOMETRY
23
Theorem F. The product determines an associative non-commutative Z-algebra structure on with unit the empty tableau ;. Moreover, is not the plactic product.
Proof: In the plactic algebra, the product of two tableaux is always a third, showing is not the plactic product. For any tableau T , we have ;T = T ; = T . Note 1
1 2
1 3 2
1 2 3
1 2 3
1 2 3
1 2
1 ,
so is non-commutative. To show associativity, let R, S , and T be tableaux. Then X [ R (S T ) = R W; S the sum taken over W Knuth equivalent to S V , where V is Knuth equivalent to T . Let U 0 be the rst jS j entries in W , and V 0 the last jT j entries, each decreased by jS j, thus,
R (S T ) =
X [ 0[ R
U
V 0;
the sum taken over U 0 Knuth equivalent to S and V 0 to T , which is (R S ) T . Let m < n. For a tableau T of shape , let (T ) be the Schur function s. De ne m; n(T ) to be 0 if + m n or m 6= 0 and otherwise. Then and m; n are additive surjections from to, respectively, the algebra of symmetric functions and AGm Pn. Theorem G. The maps and m; n are Z-algebra homomorphisms. Proof: For any tableaux S and T of shape , and arbitrary partitions and , there is a natural bijection (given by Haiman's dual equivalence [6]) between the set of tableaux with shape = Knuth equivalent to S and those Knuth equivalent to T , and this common number is c . Thus is an algebra homomorphism. It follows that m; n is as well. 1
+1
8. Enumerative Geometry and Arrangements Over Finite Fields A main result of this paper, Theorem C, shows that any Schubert-type enumerative problem concerning lines in projective space may be solved over R. By `solved' over a eld k, we mean there are ags in Pnk determining Schubert varieties which meet transversally in nitely many points, all of which are de ned over k. We describe two families of enumerative problems and, for each problem in each family, we determine the elds over which it may be solved. We also consider the problem of nding arrangements over nite elds.
24
FRANK SOTTILE
8.1. The n lines meeting four (n ? 1)-planes in P n? . Given three pairwise non-intersecting (n ? 1)-planes L ; L ; and L in P n? , there are coordinates x ; : : : ; x n for P n? such that L , L , and L are de ned by the equations: L : x = x = = xn = 0 L : xn = = x n = 0 L : x ? xn = = x n ? x n = 0 T T One may check that L L L is a transverse intersection, and that if ;n? P n? is the union of the lines meeting each of L ; L ; and L , then ;n? is the image of the standard Segre embedding of P Pn? into P n? ([7]): : [a; b] [y ; : : : ; yn] 7?! [ay ; : : : ; ayn; by ; : : : ; byn]: The lines meeting L , L , and L are the images of P fpg, for p 2 Pn? . The Segre variety ;n? has degree n, so a general (n ? 1)-plane L meets ;n? in n distinct points, each determining a line meeting L ; : : : ; L . These lines, ` ; : : : ; `n, meet L in distinct points which span L . If these lines are de ned over k, then we may change coordinates so that each `j is the span of the standard basis elements ej and en j . For 1 j n, let pj = [ j ; j ] 2 Pk be the rst coordinate of ? (`j \ L ). Then L : x ? xn = = nxn ? nx n = 0: Also, p ; : : : ; pn are distinct; otherwise L \ ;n? contains a line. Thus, if this enumerative problem may be solved over k, then k has at least n?1 elements. Conversely, suppose k has at least n?1 elements and let p ; : : : ; pn be distinct elements in Pk and de ne L ; : : : ; L and ` ; : : : ; `n as above. Then ` ; : : : ; `n are precisely the lines which meet each of L ; : : : ; L . 8.2. The n lines meeting a xed line and n + 1 (n ? 1)-planes in Pn . A line ` and (n ? 1)-planes K ; : : : ; Kn in Pn are in linear general position if for every p 2 `, the hyperplanes ?i (p) = hp; Kii, for 1 i n, meet in a line. In this case, the union [ ? (p) \ \ ?n(p) S ;n? = 2
1
2
1
2
1
1
2
1
3
2
2
1
+1
2
3
2
1
1
2
+1
3
2
2
3
1
1
1
2
1
1
1
1
2
3
1
1
1
1
1
1
2
1
1
1
1
3
1
1
4
1
1
2
1
1
1
4
1
+
1
4
1
1
1
1
+1
4
2
1
4
1
1
1
1
1
4
1
1
1
4
+1
+1
1
1
1
p2`
1
is a rational normal surface scroll. Moreover, the lines meeting each of `, K ; : : : ; Kn are precisely those lines (p) = ? (p) \ \ ?n(p) for p 2 `. Since S ;n? has degree n, a general (n ? 1)-plane Kn meets S ;n? in n distinct points, each determining a line (p) which meets `; K ; : : : ; Kn . If k is nite with q elements, there are only q + 1 lines (p) de ned over k. Thus it is necessary that q n ? 1 to solve this problem over k. We show this condition suces. 1
1
1
1
+1
1
1
1
+1
REAL ENUMERATIVE GEOMETRY
25
All rational normal surface scrolls are projectively equivalent, ([7], x9), so we may assume that S ;n? has the following standard form. Let x ; x ; y ; : : : ; yn be coordinates for Pn where ` has equation y = = yn = 0. Then for p = [a; b; 0; : : : ; 0] 2 `, we have that (p) is the linear span of p and the point [0; 0; an? ; an? b; : : : ; abn? ; bn? ]. P Let p ; : : : ; pn 2 P be distinct, and let F = ni Aibi an?i be a form on P vanishing at p ; : : : ; pn. De ne Kn by the vanishing of the two linear forms : x ?y : A x + A y + + A n yn : The intersection of S ;n? and the hyperplane de ned by is the rational normal curve : [a; b] 7?! [an; an? b; an? b; an? b ; : : : ; abn? ; bn]: Since ( ) is the form F , we see that (p ); : : : ; (pn) are the lines meeting each of `; K ; : : : ; Kn . Thus, if k has at least n ? 1 elements, this enumerative problem may be solved over k. These two families are the only non-trivial examples of Schubert-type enumerative problems for which we know an explicit description of their solutions. Each of these problems can be solved over any eld k where #Pk exceeds the number of solutions. In particular, they may be solved over Q. It would be interesting to nd exact solutions to other enumerative problems in order to test whether these observations hold more generally. 1
1
1
+1
1
2
2
1
1
2
1
1
1
1
=0
1
+1
1
2
1
1
2
0
1
1
1
1
1
1
1
2
2
2
1
1
1
+1
1
8.3. Arrangements over nite elds. In x2.6 we remarked it is possible to construct arrangements over some nite elds. Here we show how. Recall that an arrangement F in Pn is a collection of 2n ? 3 distinct hyperplanes H ; : : : ; H n? in Pn such that for m = 2; : : : ; n, the following two conditions hold: (A.1) Fm := Hf ; ;:::; m? g has codimension m, and if m 6= n, then Fm H m? but Fm 6 H m. Thus Fm = Fm \ H m , and if A f2; : : : ; 2m ? 1g, then HA 6 H m . (A.2) If A f2; : : : ; 2m ? 2g, and Fm 6= HA, then HA 6 H m? . We characterize subsets A of f2; : : : ; 2n ? 2g which give distinct subspaces HA. This is used to estimate the order of a eld k sucient to construct an arrangement. 2
2 3
2
2
2
2
2
+1
2
1
2
2
2
1
8.4. Lemma. Let F be an arrangement and A a subset of f2; : : : ; 2n ? 2g with dim HA = n ? m. Then there exists B = fb < < bm g f2; : : : ; 2n ? 2g with HB = HA, where B satis es (A.3) Either bi 2i, for i = 1; : : : ; m, or else b = 2, b = 3, and there is some 2i ? 2 for 2 < i j . j = 2; : : : ; m with bbi = 2i for j < i 1
1
i
2
26
FRANK SOTTILE
Moreover, if B 0 is another subset satisfying (A.3), then dim HB0 = n ? m and HB = HB0 if and only if B = B 0. Proof: Let A f2; : : : ; 2n ? 2g and suppose A = fa < < ak g. Discarding, if necessary, elements ai of A such that Hfa ;:::;ai? g Hai , we may assume that m = k and, for 1 i m, dim Hfa ;:::;ai g = n ? i. If ai 2i for all i, set B = A. Otherwise, let j be the largest index with aj < 2j . Since fa ; : : : ; aj g is a subset of f2; : : : ; 2j ? 1g, (A.1) implies Fj = Hf ;::: ; j? g Hfa ;:::;aj g . But dim Fj = n ? j = dim Hfa ;:::;aj g , which shows Fj = Hfa ;:::;aj g . De ne b ; : : : ; bm by b = 2, b = 3, and for i > 2, 2i ? 2 if 2 < i j bi = a : if j < i i The HB = HA, as an easy induction and (A.1) show Hfb ;::: ;bj g = Fj . If B 0 f2; : : : ; 2n?2g satis es (A.3), then arguing as in the proof of Lemma 2.9(3) shows HB = HB0 ) B = B 0. We estimate the size of a nite eld k sucient to construct an arrangement. Theorem H. There exists an arrangement in Pnk if the order of k is at least n? X (2n ? 4)! (2i)! : + (n ? 2)!(n ? 1)! i i!(i + 1)! 1
1
1
1
1
2
2
1
1
1
1
1
1
2
1
4
=0
Proof: Consider the problem of inductively constructing an arrangement in Pnk. Let 2 m n ? 2 and suppose we have found H ; : : : ; H m? satisfying (A.1) and (A.2). This de nes Fm = H \ \ H m? . Let H m be any hyperplane in Fm . Set Fm = Fm \ H m. Then, we must nd a hyperplane Pnk not containing H m Pnk=Fm with HB 6 H m , for all B f2; : : : ; 2m ? 2g satisfying the condition (A.3) of Lemma 8.4. We investigate when this is possible. Let Sm be the set of all B satisfying (A.3), with B f2; : : : ; 2m ? 2g. Let P m be the set of hyperplanes in Pn de ned over k which contain Fm . 2
2
2
+1
2
+1
+1
2
1
1
2
2
2
+1
Every codimension m subspace HB containing Fm determines a hyperplane H B in P m consisting of those hyperplanes in Pn containing HB . Moreover, H B is de ned over k whenever HB is de ned over k. Thus there exists a hyperplane H m in Pnk=Fm with HB 6 H m for all B 2 Sm, if, as sets of k-valued points, [ HB 6= ;: X = P m n +1
+1
2
+1
+1
S We estimate #
2
+1
B2Sm
to show that the hypotheses imply X 6= ;. We claim that for each m = 0; : : : ; n ? 2, m X? (2i)! (2 m )! : sm := #Sm = m!(m + 1)! + i !( i + 1)! i B2Sm HB ,
2
=0
REAL ENUMERATIVE GEOMETRY
27
Granting this, suppose k has q sn? elements. Since P m has (qm ? 1)=(q ? 1) elements and each H B has (qm ? 1)=(q ? 1) elements, X is non-empty if qm ? 1 > (qm ? 1)sm. This holds because qm ? 1 qm ? 1 q sn? sm; by our assumption on q = #k. To enumerate Sm , let B = fb < < bm g 2 Sm . Then there is some j 2 f0; 2; : : : ; mg such that bi 2i for i > j , but bi < 2i for i j . If we set ci = bj i ? 2j for i = 1; : : : ; m ? j , then c ; : : : ; cm?j is the second row of a Young tableau of shape (m ? j; m ? j ). This is because for each i, 2i ci 2(m ? j ). Conversely, if c ; : : : ; cm?j is the second row of a tableau of shape (m ? j; m ? j ), then f2 < 3 < < 2j ? 2 < c + 2j < < cm?j + 2j g 2 Sm : Let Ts be the set of tableaux of shape (s; s). These arguments show there is a bijection Sm ! Tm [ Tm? [ Tm? [ : : : [ T : By the hook length formula of Frame, Robinson, and Thrall [3], #Ts = s ss . P Thus #Sm = m mm + im? i i i . This result is not the best possible, as we used a crude estimate to count the set X . For P , Theorem H gives #k 3 and for P , #k 7. However, the arrangements in x2.8 are de ned over smaller elds. The arrangement in P is de ned over the eld with two elements, while the arrangement in P may be de ned over the eld with four elements: If F is the eld with 2 elements so that F = F [x]=(x + x + 1) is the eld with 4 elements, then changing the coecients 2 and 3 in (1) of x2.8 to x and x + 1, gives an arrangement in P de ned over F . +1
2
+1
+1
2
1
+
1
1
1
2
3
0
(2 )!
!( +1)!
(2
!(
2
)!
+1)!
=0
(2 )!
!( +1)!
5
4
4
5
2
4
2
2
5
4
Acknowledgments
I am grateful to the anonymous referee and to Andy Hwang for their detailed and helpful comments on earlier versions of this paper. Theorem C was proven in my 1994 Ph.D. Thesis from the University of Chicago, supervised by William Fulton. I would like to thank Professor Fulton for suggesting these problems, for thoughtful advice, and for introducing me to algebraic geometry. References [1] C. I. Byrnes, Pole assignment by output feedback, in Three Decades of Mathematical Systems Theory, H. Nijmeijer and J. M. Schumacher, eds., vol. 135 of Lecture Notes in Control and Information Sciences, Springer-Verlag, 1989, pp. 31{78. [2] R. Chiavacci and J. Escamilla-Castillo, Schubert calculus and enumerative problems, Bollettino U.M.I., 7 (1988), pp. 119{126.
28
FRANK SOTTILE
[3] J. S. Frame, G. de B. Robinson, and R. M. Thrall, The hook graphs of the symmetric group, Canad. J. Math., 6 (1954), pp. 316{325. [4] E. Freidlander and B. Mazur, Filtrations on the Homology of Algebraic Varieties, no. 529 in Memoirs of the American Mathematical Society, American Mathematical Society, July 1994. [5] W. Fulton, Introduction to Intersection Theory in Algebraic Geometry, CBMS 54, AMS, 1984. [6] M. Haiman, Dual equivalence with applications, including a conjecture of Proctor, Discrete Math, 99 (1992), pp. 79{113. [7] J. Harris, Algebraic Geometry: A First Course, Graduate Texts in Mathematics 133, Springer-Verlag, 1992. [8] A. G. Khovanskii, Fewnomials, Trans. of Math. Monographs, 88, AMS, 1991. [9] S. Kleiman, The transversality of a general translate, Comp. Math., 28 (1974), pp. 287{ 297. [10] J. Kollar, Rational Curves on Algebraic Varieties, no. 32 in Ergebnisse der Math., Springer-Verlag, 1996. [11] A. Lascoux and M.-P. Schutzenberger, Le monod plaxique, in Non-Commutative Structures in Algebra and Geometric Combinatorics, Quaderni de \la ricerca scienti ca," 109 Roma, CNR, 1981, pp. 129{156. [12] D. E. Littlewood and A. R. Richardson, Group characters and algebra, Phil. trans. Roy. Soc. London., 233 (1934), pp. 99{142. [13] F. Ronga, A. Tognoli, and T. Vust, The number of conics tangent to 5 given conics: the real case. manuscript, http://www.unige.ch/math/bibl/preprint/liste.html, 1995. [14] B. Sagan, The Symmetric Group; Representations, Combinatorics, Algorithms & Symmetric Functions, Wadsworth & Brooks/Cole, 1991. [15] P. Samuel, Methodes d'Algebre Abstraite en Geometrie Algebrique, Ergebnisse der Math., Springer-Verlag, seconde ed., 1967. [16] G. Thomas, On Schensted's construction and the multiplication of Schur functions, Adv. in Math., 30 (1978), pp. 8{32.
REAL ENUMERATIVE GEOMETRY
29
9. Appendix: The 5 lines meeting 6 P 's in P . We illustrate our results in the rst case which is not covered by the exact solutions given in x8. By Corollary 3.3, the number of lines meeting 6 generic P 's in P equals the number of tableaux of shape (3; 3). This number is 5 as 2
4
2
1 3 5 2 4 6
1 3 4 2 5 6
1 2 5 3 4 6
1 2 4 3 5 6
4
1 2 3 4 5 6
are all tableaux of shape (3; 3). By Theorem B, there are six real P 's with ve real lines incident on all six. Here, we use our methods to construct a family of sextuples of P 's in P such that for small real values of some parameters, we obtain six real P 's with ve real lines incident on all six. 2
2
4
2
9.1. Arrangements. We wish to construct arrangements as simply as possible over R. To that end, we use the ideas of x7.3 to nd arrangements whose hyperplanes are given by linear forms with small integral coecients. Arrangements may be constrctucted inductively; to nd an arrangement in Pn, it suces to begin with one in Pn? , and then select the last two hyperplanes. We illustrate this by inductively nding an arrangement in P . Let x ; : : : ; x be coordinates for P and F be the ag with Fi : x = = xi = 0. 1
5
1
5
6
q
1
P : We rst select H , a hyperplane containing F . Thus H has equation 2
2
2
2
Ax + Bx for some A, B . Let A = 0 and B = 1, a choice with small integral coecients, gives H : x = 0. P : H contains F and is distinct from H . It suces to take H : x = 0. T Since H F = F , it suces to take H : x = 0. P : H must contain F , but none of the three codimension 2 planes: 1
2
2
3
3
2
4
4
2
2
2
3
5
3
4
1
3
3
H :x =x =0 23
1
H :x =x =0
2
24
2
H : x = x = 0:
3
34
1
3
Thus H has an equation Ax + Bx + Cx with A B C 6= 0. Among all possible choices, T H : x + x + x = 0 has the smallest integral coecients. Since H F = F , it suces to take H : x = 0. P : H must contain F , but none of the seven 2-planes HA for 5
1
5
6
5
7
3
1
4
2
2
3
3
6
4
4
A 2 f234; 236; 246; 256; 346; 356; 456g:
30
FRANK SOTTILE
Letting [a b c] be coordinates for P , these 7 P 's have coordinates 2
H H H H H H H
234
236
246
256
346
356
456
= = = = = = =
2
[ a b [ a b [a b [ a ?a b [ a b [ a ?a b [ a ?a b
c] c] c] c] c] c] c ];
where has been used in place of 0. Since H must contain F , it has equation Ax + Bx + Cx + Dx = 0. The conditions on H ammount to the conditions that A B C D 6= 0 and A; B; C are distinct. We can solve this system in Z=5Z: If we normalize so that D = 1, then there are 24 choices for A; B; C , and 6 if we want A; B; C (as integers)Tto be as small as possible. We select H : x +2x +3x +x = 0. Since H F = F , it suces to take H : x = 0. 7
4
1
2
3
4
7
7
8
4
5
8
1
2
3
4
5
Question: Find a nice way to iterate this procedure; or an algorithm (involving tableaux?) to inductively construct arrangements in Pn. 9.2. The 5 lines meeting 6 P 's in P . We list the hyperplanes in the arrangement constructed in x1.1 for P : 2
4
4
H H H H H
2
3
4
5 6
= = = = =
[ [ [ [ ?a ?b a + b [
] ] ] ] ]
x x x x +x +x x
2
1
3
1
2
3 4
=0 =0 =0 =0 = 0;
where ; a; b are arbitrary numbers and is 0. In Theorem C', we showed that (T ) = fH ; H ; H ; H ; H g, a degree 5 reduced zero cycle, is in the closure of the set of degree 5 zero cycles on G P which meet six real P 's in general position, considered as a subset of S G P (R). We construct a 7-parameter family of sextuples of real P 's such that (T ) is in the closure of the set of degree 5 zero cycles on G P which meet sextuples in this family. 6
4
246
256
346
356
456
2
1
5
1
4
2
1
4
6
REAL ENUMERATIVE GEOMETRY
31
Let K |K be P 's depending upon parameters ; ; ; ; ; ; with coordinates and de ning ideals: K = [ a b c] (x ; x ) K = [ a b b c] (x ; x ? x ) K = [ ? a b c a ? c ?a ] (x ? x ; x + x + x ) K = [ a b c c ?c ] (x ? x ; x + x ) K = [ ?a b a ? b c ? b ?c ] (x + x + x ; x + x + x ) K = [ a b ?c c ] (x + x ; x ): Let Ki be the Schubert variety of lines meeting Ki. We claim: 1. For general values of the parameters ; ; : : : ; , the intersection \ \ \ \ \
K K K K K K is a generically transverse intersection. 2. For suciently small values of ; ; and , and all values of ; ; , this intersection consists of 5 real lines in P . 9.3. De ning Ideals. The Grassmannian G P is embedded into P ( C ) = P via the Plucker embedding. The coordinates x ; x ; x ; x ; x of P give rise to Plucker coordinates f[i j ] j 1 i < j 5g for P , where [i j ] $ xi ^ xj . The ideal of G P in P is given by the Plucker equations [12] [34] ? [13] [24] + [14] [23] [12] [35] ? [13] [25] + [15] [23] [12] [45] ? [14] [25] + [15] [24] [13] [45] ? [14] [35] + [15] [34] [23] [45] ? [24] [35] + [25] [34] Let K be P de ned by two linear forms A and B . Then the set of lines meeting K in P is a hyperplane section of G P in P , the hyperplane de ned by the form A ^ B . Thus K |K induce the following hyperplanes in P : K : [12] K : [23] + [34] K : [13] + [35] + [45] + [14] + [15] K : [34] + [45] + [35] K : [12] + [14] ? [23] + [24] + [34] + ([15] + [25] + [35]) K : [45] + [35] 1
2
6
1
1
2
3
1
4
3
2
5
3
3
5
1
5
2
6
3
2
4
5
5
4
4
5
5
6
5
4
3
2
4
4
2
3
1
2
4
1
4
9
2
1
2
3
4
5
5
4
9
1
4
9
2
4
1
1
4
9
9
6
1
2
2
3
4
5
6
Let I ( ; ; ; ; ; ; ) be the ideal in P generated by the Plucker equations and these six linear forms. In x8.4 we use geometry to show that for a generic 9
32
FRANK SOTTILE
choice of the parameters ; : : : ; , the ideal I ( ; ; ; ; ; ; ) de nes a reduced zero-dimensional scheme of degree 5. For these parameters, the intersection \ \ \ \ \
K K K K K K is generically transverse, proving assertion 1 from the end of x8.2. These methods will also prove assertion 2, that for certain real values of the parameters, all points of intersection are real. 9.4. Geometry. If F ( H is a partial ag in P , let (F; H ) be those lines in H which also meet F . If H = P , write F for this Schubert variety and, if F has codimension 1 in H , then (F; H ) is all the lines in H , G H . If T is a tableau whose rst row has length a 3, then T = (Fa ; HT ), where Fa is in the ag F and HT is the intersection of those Hj where j is in the second row of T . Let T be the set of all Young tableaux of degree i whose rst row has length at most 3. For example, T is 6
5
4
3
2
1
4
4
1
+1
q
+1
1
4
1 3 2 4
1 3 4 2
1 2 3 4
T
T
T
1 2 4 3
T
T
1 2 3 4
We now study the cycles K K K K K K as the parameters ; : : : ; vary. Our aim is to use geometric arguments to verify the assertions 1 and 2 from the end of x8.2. : If 6= 0, then K and K ( ) meet properly, so K T K is generically transverse and irreducible.T When = 0, then K K ( ) = F and hK ; K ( )i = H . This and other such statements in this section are easily veri ed by examining the de ning equations of K {K from x8.2. Thus, by Lemma 2.4, \
K K = F + G H = + = (T ): 1
1
3
4
5
2
1
2( )
6
2( )
1
2
1
1
2
3
1
2
2
6
1
3
2
2
1 2
1 2
Thus 1. If ; : : : ; are chosen so that (T ), K , : : : , K meet generically transversally, then for general , the Schubert varieties K , K , : : : ,
K meet generically transversally. T T T 2. If ; : : : ; are chosen so that (T ) T K T T K is 5 real lines, then for suciently close to 0, K K K is 5 realTlines. : For any ( ; ), K ( ; ) meets H properly, and if 6= 0, then K F = ;. Thus if 6= 0, K meets (T ) generically transversally. If = 0 and 6= 0, then K (0; ) meets F in a point, F ( ), and hK ; F i = H . Let F ( ) = K T H . Then \ \ \
(T ) K ; = K G H + K F = (F ( ); H ) + F + (F ; H ): 2
3
6
1
2
6
2
1
3
3
6
2
6
2
3
2
3
3
3
3
3
2
3
3(
)
3
4
3
3
1
3
3
2
2
3
4( )
3
3
3
3
REAL ENUMERATIVE GEOMETRY
33
Thus 1. If ; : : : ; are chosen so that (T ), K ; , : : : , K meet generically transversally, then for general , then the cycles (T ), K ; , : : : ,
K meet generically transversally. T T T 2. If ; : : : ; are chosen so that (T ) K ; T KT is 5Treal lines, then for suciently close to 0, then (T ) K ; K is 5 real lines.
: Since F K (0; 0), (T ) T TK ; is an improper intersection. However, when 6= 0, let X ( ) = (T ) K ; , which is (F ( ); H ) + F +
(F ; H ). Coordinates and de ning ideals for F ( ) and F ( ) are: F ( ) = [ a b ? a ?b ] (x ; x ; x + x + x ) F ( ) = [ b ?b ] (x ; x ; x ; x + x ) Thus F (0) = F and F (0) = F , and so the family X ( ) over R ? f0g extends to a family over R, whose bre over 0 is X (0) = (F ; H ) + F + (F ; H ) = +
+ = (T ): 2
3 (0
)
6
2
3(
)
6
2
3 (0
)
6
2
3
3
2
)
6
3 (0 0)
2
3
3(
3 (0
3
)
3
2
3
3
1
2
4
3
3
4
4
3
1
4( )
2
2
4
3
4
5
5
4
3
2
3
4
1 2 3
1 3 2
3
3
1 2 3
Thus 1. If ; : : : ; are chosen so that (T ), K , K , and K meet generically transversally, then for general 6= 0, the cycles (T ), K ; , : : : , K meet generically transversally. T T T 5 realTlines, 2. If ; : : : ; are chosen so that (T ) K K T K is T then for 6= 0 suciently close to 0, then (T ) K ; K is 5 real lines. : For any , K () meets both HJ and H properly, and for 6= 0, K () \ F = ;. Thus if 6= 0, (T ) meets K ; generically transversally. If = 0, then K (0; ) H and K (0; ) \ F = F (). Thus by Lemma 2.4, (T ) meets K ; generically transversally and \ \
(T ) K ; = ( (F ; H ) + F + (F ; H )) K ; = (F ; H ) + (F (); H ) + (F ; H ) + (F ; H ) + (F (); H ) = + (F (); H ) +
+ + (F (); H ): 3
4
5
6
2
3
3 (0
2
4
2
3
3
3
4 (0
4(
4
4 (0
3
)
3
1 3 2 4
3 (0
)
6
3
3
4
6
6
5
4
)
4
)
4
3
4
)
2
3
4
24
4
4
3
4 (0
2
2
4
1 2 3 4
)
4
3
34
4
1 2 3 4
4
3
Thus 1. If ; , and are chosen so that (T ), K ; , K , and K meet generically transversally, then for general , the cycles (T ), K ; ,
K , and K meet generically transversally. T T T 2. If ; , and are chosen so that (T ) K ;T K T K isT ve real lines, then for suciently close to 0, (T ) K ; K K is ve real lines. 3
4 (0
)
5
6
3
5
4(
)
6
3
4 (0 3
)
6
5
4(
)
5
6
3
34
FRANK SOTTILE
; ; : Let M; L ; L be any three P 's whichTare inTlinear general position, so that M \ L is a point p = 6 p = M \ L . Then G M L L = fhp ; p ig, a single line. This is real 5
2
6
5
6
5
6
1
5
6
5
6
if M; L and L are all real. Likewise, if H ' P and T q 2TH is a point so that q 2 H , L Tand L Tare in linear general position, then (q; H ) L L is the single line hq; L i hq; L i H . This is real if q; H; L and L are all real. For any ; ; , the subspaces K () and K ( ) are in linear general position with each of F 2 H , F () 2 H , F () 2 H , H , and H . Thus, for any ; ; , the cycles
(T ),T K ; , TK , and T K meet generically transversally, and
(T ) K ; K K = fH (); H (; ); H (); H (; ); H g; (1) where these ve lines have coordinates, equations and arise as: H () = [ a b b ] x ; x ; x + x H (; ) = [ a ?a a + b b ] x ; x + x + x ; x + x + x H () = [ a b b ] x ; x ; x + x H (; ) = [ a ?a a + b b ] x ; x + x + x ; x + x + x H = [ a ?a b] x +x +x ;x ;x : Thus for any , , and , the cycles (T ), K ; , K , and K meet generically transversally. Then the observations 1. in ; ; , and above together imply that 1. For general ; ; ; ; ; , and , the Schubert subvarieties K , K , K , K , K , and K meet gererically transversally. T T T If ; , and are real numbers, then (T ) K ; K K is ve real lines. Then the observations 2. in ; ; , and above together imply that 2. Given any T real ;T, andT , for suciently small real numbers ; ; , and with T
6= 0, K K K K K K is ve real lines. Lastly, consider the cycle (1) when (; ; ) = (0; 0; 0). We see that it equals
(T ) = fH ; H ; H ; H ; H g: Thus (T ) is in the closure of the set of degree 5 zero cycles on G P which meet sextuples (K ; K ; K ; K ; K ; K ) in this family. 5
6
3
5
5
6
5
5
6
6
6
5
4
4
3
4
4 (0
3
3
)
4 (0
4
2
5( )
)
6
24
34
6( )
5( )
6( )
246
256
346
246
356
2
256
2
1
3
2
346
1
1
456
3
4 (0
)
3
2
1
4
5
3
1
356
456
3
4
3
2
4
5
4
5
5
3
3
5( )
3
4
6( )
1
2
3
6
3
4
3
2
1
6
246
6
4 (0
)
5( )
6( )
6
5
256
346
356
456
1
1
2
3
4
5
4
6
Mathematical Sciences Research Institute, 1000 Centennial Drive, Berkeley, California 94720, USA Department of Mathematics, University of Toronto, 100 St. George Street, Toronto, Ontario M5S 3G3, Canada (on leave)
E-mail address :
[email protected]
4
5
