SECANT VARIETIES OF SEGRE-VERONESE VARIETIES

Transcription

1 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES THANH VU 1. Introduction Secant varieties of algebraic varieties is a classical topic dated back to 19th century nourishing around the Italian school. It emerged again about three decades ago with fundamental work of Zak. Recently, because of many of its applications to algebraic statistics, biology, imaging, computer science, complexity theory, it becomes one of the central topics in algebraic geometry. Especially, people are interested in the very concrete problems, like the salmon s problem, namely the defining equations of secant varieties of Segre-Veronese varieties. We will go through certain techniques and results in the field and work out as many examples as possible. The main theme of the course is on dimensions of the secant varieties, and its defining equations. For the dimension problem, we will study three techniques/approaches: the first one is the classical approach reducing the problem to finding (multi-graded) Hilbert function of fat points on (multi-graded) projective spaces. Then using the so-called simultaneous homogenization, we then reduce to finding usual Hilbert function of some fat scheme following Catalisano, Geramita, and Gimigliano. The second one is the induction approach introduced by Abo, Ottaviani, and Peterson. The third one is the tropical approach introduced by Draisma. Note that the tropical approach is closely related to the degeneration techniques introduced and studied by a number of authors (Ran, Ciliberto,...). It depends on how long it would take us through these problems, and how well we use it to apply to our problem of interest. After that, we will go over some technique to find equations of the secant varieties where representation theory is almost indispensable. I hope together we will be able to find some new equations by the end of the semester. Throughout, we will work over an algebraically closed field of characteristic zero, namely k = C. P N denotes the N-dimensional projective space over k. Varieties mean reduced irreducible subvarieties of some P N, i.e zero set of certain homogeneous polynomials in k[x 0,..., x N ]. Let X be a variety in P N, with the defining ideal I = (f 1,..., f n ). The tangent space to X at a point p is the set of all tangent lines to X at p, which is also the solution set of the linear equations f p (x i p i ) = 0 for all f = f 1,..., f n. x i The dimension of X is defined to be the dimension of the tangent space to X at a general point of X. Definition 1.1. The (s 1)-secant variety of X is the closure of all the s 1-planes spanned by s points of X. Namely, σ s (X) = p 1,..., p s. From the definition, we see that X σ 2 (X) σ 3 (X) σ s (X) = P t. Remark: If X is a linear subspace, then X = σ i (X) for all i 2. Conversely, if X = σ 2 (X), then X is linear. Thus, if X is non-degenerate, i.e, not contained in any hyperplane, then there exists r such that X σ 2 (X) σ 3 (X) σ r 1 (X) σ r (X) = P N. Example 1: ν 2 (P 1 ) P 2. It is easy to see that σ 2 (X) = P 2. Expected dimension: The secant variety can be thought of as join of varieties. For affine varieties X, Y A N+1, the sum of X and Y is defined to be the image of the map X Y A N+1 by sending (x, y) to x + y. Clearly, dim(x + Y ) dim X + dim Y. For projective varieties X, Y, the join of X and Y, is defined to be the one whose its affine cone is given by XY = ˆX + Ŷ. From the definition, we see that σ 2 (X) = X 2 the join of X with itself, and so on for higher secant varieties. Therefore, dim join(x, Y ) dim X + dim Y

2 2 THANH VU For any secant variety, we thus have dim σ t (X) min(t dim X + t 1, N). The value min(t dim X + t 1, N) is called the expected dimension of σ t (X), denoted ed t (X). Definition 1.2. When dim σ t (X) ed t (X), X is said to be t-defective or simply defective. From the definition of secant varieties as a join, we can think of secant varieties as the closure of the set of points that can be expressed as the sum of points from X. This property is closed under taking a linear projection, thus we have Lemma 1.1. Given a projective variety X P N and a linear projection π : P N P M, then we have π(x k ) = (πx) k. Example 2: Segre variety. Consider the Segre variety X = P m P n P mn+m+n. It is easy to see that it is defined by the 2-minors of a generic matrix. Thus X is the variety of rank 1 matrices on the space of all (m + 1) (n + 1) matrices. Since the sum of k rank 1 matrices has rank at most k, and conversely, there is an open set of rank 1 matrices that can be expressed as sum of k rank 1 matrices, we see that Proposition 1.2. σ k (R 1 (M m,n ) = R k (M m,n ). where R k (M m,n ) is the variety of all rank at most k m n-matrices. Thus we have a complete description of the secant varieties of X in this case. From that, one can deduce that X is defective. In a similar manner, we also have a description of the secant varieties of the second veronese embeddings of projective spaces. Example 3: Let X = ν 2 (P n ) P n(n+1)/2 1 be the second Veronese embedding of projective space. It can be check that X is the variety of rank 1 symmetric (n + 1) (n + 1) matrices. Again since a rank k symmetric matrix can be written as a sum of k rank 1 symmetric matrix, we deduce that σ k (X) is the set of rank k symmetric (n + 1) (n + 1) matrices, and is thus defined by the set of (k + 1)-minors of that matrix. 2. Secant varieties of rational normal curves and catalecticant matrices The rational normal curves, ie the Veronese embeddings of P 1 can be thought of as certain section of the varieties of rank 1 matrices. Precisely, the following is a generic catalecticant matrix x 1 x 2 x 3 x m+1 x 2 x 3 x m+2 Cat(m + 1, n + 1) = x 3.. x n+1 x m+n+1 Let d = m+n, then for any m, n 1, we have C d = ν d (P 1 ) P d is defined by the 2-minors of Cat(m+1, n+1). Thus, we see that for any k, and m, n k, we have σ k (C d ) is contained in R k (Cat(m + 1, n + 1). Indeed, they are equal. Theorem 2.1. σ k (C d ) = R k (C m+1,n+1 ) for all m, n k such that m + n = d. For detail proof, see Harris s book. 3. Terracini s Lemma and Inverse systems Terracini s Lemma: Let P and Q be general points of X and Y and R (P Q) be a general point of Z = join(x, Y ). Then T R (Z) is the linear span of T P (X) and T Q (Y ). This follows from the affine version, where we see that t X,x + t Y,y t X+Y,x+y for all x, y and the equalities hol true for generic x, y. Let s see what it tell us in the case of Veronese embedding of projective spaces. Let S = k[x 0,..., x n ]. The Veronese embeddings can be thought of as a map Let X = ν d (P n ) and P = [L d ] X, then we have ν d : P(S 1 ) P(S d ) : [L] [L d ]. T P (X) = [L d 1 M] : M S 1.

3 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES 3 If H P N is a hyperplane, then ν 1 d (H) is a degree d hypersurface. To see this, notice that H has an equation of the form a 0 z a N z N, where z i are the coordinates of P N. The equation for ν 1 d (H) is obtained by substitute each z i with the corresponding degree d monomial in the x 0,..., x n. Since ν 1 d ([Ld ]) = [L], if H P N is a hyperplane and [L d ] H, then ν 1 d (H) is a degree d hypersurface passing through the point [L] P N. If H P N is a hyperplane such that T [L ](X) H, then ν 1 d d (H) is a degree d hypersurface singular at the point [L] P n. (We may assume that L = x 0, and it follows by direct calculation.) Let P 1,..., P s P n be points with defining ideals p 1,..., p s respectively. The scheme defined by the ideal p 2 1 p 2 s is called a 2-fat point scheme or a double point scheme. Let X = ν d (P n ) P N. There is a bijection between {H P N a hyperplane:h T P1 (X),..., T Ps (X) }, and Thus we have {degree d hypersurface of P n singular at P 1,..., P s } = (p 2 1 p 2 s) d. dim σ s (X) = N dim(p 2 1 p 2 s) d. Theorem 3.1 (Alexander and Hirschowitz). The Veronese variety ν d (P n ) is non-defective unless d = 2, n 2; d = 4, n = 2, 3, 4, or d = 3, n = 4. Example: Consider n = 2, d = 4, we have X = ν 4 (P 2 ) P 14. The expected dimension of σ 5 (X) is 14. In other words, we expect that there is no quartic singular at 5 generic points of P 2. Nevertheless, assume that p 1,..., p 5 is 5 general points in P 2, we may assume that they are [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : y 1 : z 1 ], [1 : y 2 : z 2 ]. It is easy to see that we can find a, b, c so that the quadric f = axy + bxz + cyz pass through these 5 points. The quartic f 2 is singular at these 5 points, and we get that X is defective. 4. Dimension of secant varieties of Segre-Veronese varieties Similar to the case of Veronese embedding, the dimension of the Segre-Veronese varieties can be expressed as the Hilbert function of certain ideal of fat scheme. More precisely, let X = ν a1,...,a t (P n1 P nt ) be the Segre-Veronese embedding of P n1 P nt generic points of P n1 P nt, then by O(a 1,..., a t ). Let p 1,..., p k be the defining ideal of k dim σ k (X) = N dim[p 2 1 p 2 k] a1,...,a t. Example: Let X = ν 1,2 (P 1 P 2 ) P 11. We will show that X is non-defective. Indeed, for σ 2 (X), choose 2 generic points of P 1 P 2 which we may assume to be [1 : 0] [1 : 0 : 0] and [0 : 1] [0 : 1 : 0]. Then we have the defining ideals are p 1 = (x 1, y 1, y 2 ), p 2 = (x 0, y 0, y 2 ). Thus [p 2 1 p 2 2] 1,2 = C{x 0 y 1 y 2, x 0 y2, 2 x 1 y 0 y 2, x 1 y2}. 2 Furthermore, for σ 3 (X), if we choose another point [1 : a] [0 : 0 : 1] with defining ideal p 3 = (x 1 ax 0, y 0, y 1 ), we see that non of the forms in [p 2 1 p 2 2] 1,2 is in p 2 3, thus [p 2 1 p 2 2 p 2 3] 1,2 = 0 as expected. Conjecture: (Abo-Brambilla) The secant variety σ s (ν 1,2 (P m P n )) is defective if and only if (a) ( ) ( n+2 2 n < s < min(m + 1, n+2 ) 2 ). (b) (m, n, s) = (2, 2k + 1, 3k + 2) for some k 1. (c) (m, n, s) = (4, 3, 6). Explanation of the defective case: Let S = k[x 0,..., x m ; y 0,..., y n ]. (a) For the unbalanced case: we see that there are at least m + 1 s linear form in S 1,0 vanishing at s generic points. There are at least ( ) n+2 s quadrics in S0,2 vanishing at s generic points. Thus there are at least (m + 1 s)( ( n+2 n ) n s) forms in S1,2 vanishing twice at s generic points. Thus, we have ( n + 2 dim σ s (X) (m + 1) n ) 1 (m + 1 s)( ( n + 2 n ) s) = s( ( n ) + m s) 1 which is smaller than the expected values in the range. (b) Expectedly σ 3k+2 (X) fills the space, but indeed there is one equation vanishing on it, namely the Strassen s equation. (c) Can be verified by computer.

4 4 THANH VU 5. Products of P 1 Now we will consider a dehomogenization method to translate the problem of computing the multi-graded hilbert function to that of the usual Hilbert function of fat scheme. Let X = ν a1,...,a t (P n1 P nt ). Let S = k[x 1,0,..., x 1,n1 ; ; x t,0,..., x t,nt ] be the coordinate ring of P n1 P nt. Consider the birational map f : P n1,...,nt P n defined by (1, x 1,1..., x 1,n1 ),..., (1, x t,1..., x t,nt ) (1, x 1,1,..., x 1,n1,..., x t,nt ) Let Z be a zero-dimensional scheme in P n1,...,nt. Let Z = f(z). Let Q 0, Q 1,1,..., Q 1,n1,..., Q t,1,..., Q t,nt be the coordinate points of P n. Consider the linear space Π i = Q i,1,..., Q i,ni = P ni 1 with defining ideal I i = (z 0, z 1,1,..., z 1,n1,, z i,1 ˆ,..., ẑ i,ni,..., z t,1,..., z t,nt ). Let W = Z + (a a i )Π i, then by Catalisano-Geramita-Gimigliano dim(i W ) a = dim(i Z ) a1,...,a t. Let s come back to the case of two factors for a moment: Let X = ν a,b (P m P n ). Let S = k[z 0, z 1,..., z m, z m+1,..., z m+n ]. Let p 1,..., p k be the defining ideal of k generic points of P m+n. Let a = (z 0, z m+1,..., z m+n ) and b = (z 0, z 1,..., z m ). Then dim σ k (X) = N dim(a b b a p 2 1 p 2 k) a+b. Now for the case of P 1 P 1, all the ideal I i defines a point in P n, and we can rephrase as: Let p 1,.., p n, q 1,..., q k be generic points in P n. Then dim σ k (X) = (2 n 1) [p n 1 1 pn n 1 q1 2 qk] 2 n. Now, we introduce some preliminary result: Let f be a form of degree d. From the exact sequence 0 (I : f)( d) I I + f = I/I (f) = I/f(I : f) 0 f we deduce that dim I d dim(i : f) d + dim(i/f(i : f)) d. For example: Let consider σ 2 (ν 1,3 (P 1 P 1 )). In this case, we are interested in finding: dim k [(z 0, z 2 ) 3 (z 0, z 1 ) q 2 1 q 2 2] 4 where q 1, q 2 are defining ideals of two generic points. Let f be a linear form that vanishes on these two generic points, then we see that we have an inequality dim I 4 dim J 3 + dim K 4 where J = (z 0, z 2 ) 3 (z 0, z 1 ) q 1 q 2 = z 0 ((z 0, z 2 ) 2 q 1 q 2 ). Thus J 3 has dimension one, as there is only one quadrics in (z 0, z 2 ) 2 vanishes simultaneously at q 1 and q 2 (note that these two points impose 2 equation on these space of quarics). Now K view as an ideal in k[z 0, z 1 ], is of the form (z 0, z 1 ) 3 r 2 1 r 2 2, where r 1, and r 2 are defining ideals of two restricted points. Now, since r 1, r 2 are principal, and they are relatively prime as the points q 1 and q 2 are generics, we deduce that r 2 1 r 2 2 = r 2 1r 2 2 (z 0, z 1 ) 3, thus K 4 has dimension one also, and we are done. In general, the problem with this method is to understand what is the restricted scheme. If we get a better understanding of that, we should be able to tell something about our problem in general. For example, consider the case n = 5, and k = 4. We want to show that dim σ 4 (X) = 23. We need the Alexander-Hirschowitz differential Horace lemma: Let H P n be a hyperplane, p 1,..., p r be generic points of P n. Consider a zero dimension scheme Y P n, Y = X + 2p p r. Let X = Res H (X), X = T r H (X). Let p 1,..., p r be generic points in H, set D 2,H (p i ) = 2p i H ( a subscheme of Pn ). Consider the two schemes Y = X + D 2,H (p 1) + + D 2,H (p r) P n ; Y = X + p p r P n 1 = H. If dim(i Y ) t 1 = 0 and dim(i Y ) t = 0 then dim(i Y ) t = 0. Now, back to our case where X = (P 1 ) 5 P 31, and we want to find the dimension of σ 4 (X). Let W = 4q q 5 + 2p p 4 where q i are the coordinate points of P 5 and q i are generic points. We need to show that dim(i W ) 5 = = 8. Let H be a hyperplane which contains q 2,..., q 5 but not q 1. We choose the less generic scheme by taking p 1, p 2 be generic points of H and p 3, p 4 generic in P 5. Now add 8 points to W, call them t 1,..., t 8 with the first four chosen generically on H and the last four chosen generically in P 5 and call the resulting scheme Z. It will be enough to prove that dim(i Z ) 5 = 0 (because 8 points impose at most 8 equations). We want to apply the differential Horace lemma to this scheme.

5 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES 5 Write Z = X + 2p 3 + 2p 4, with We have X = 4q q 5 + 2p 1 + 2p 2 + t t 8. Res H (X) = 4q 1 + 3q q 5 + p 1 + p 2 + t t 8 P 5 T r H (X) = 4q q 5 + 2p 1 + 2p 2 + t t 4 P 4 = H By the Lemma, we choose two generic points p 3, p 4 in H and consider the two schemes Z = Res H (X) + D 2,H (p 3) + D 2,H (p 4) P 5, Z = T r H (X) + p 3 + p 4 P 4. We want to show that (I Z ) 4 = 0 and (I Z ) 5 = 0. Now, note that Z = 4q 1 +3q q 5 +D 2,H (p 3)+D 2,H (p 4)+p 1 +p 2 +t 5 + +t 8. Let W = 4q 1 +3q q 5 +D 2,H (p 3)+D 2,H (p 4)+p 1 +p 2. It suffices to show that dim(i W ) 4 = 4 (then 4 generic points t 5,..., t 8 imposes another 4 conditions). Now, let f be the defining ideal of H. We see that (I W : f) 3 = 0, thus it suffices to show that the dimension of I W H defining the scheme W = 3q q 5 +2p 3+2p 4+p 1 +p 2 H = P 4 is 4. Now, since p 1, p 2 are generic points of H, it suffices to show that dim(i W ) 4 = 6 where W = 3q q 5 +2p 3 +2p 4 P 4. But that follows from the fact that σ 2 ((P 1 ) 4 ) is non-defective. Finally, we need to show that the dimension of (I Z ) 5 = 0 where Z P 4. Note that Z = 4q q 5 + 2p 1 + 2p 2 + t t 4 + p 3 + p 4 P 4. Let W = 4q q 5 + 2p 1 + 2p 2 + p 3 + p 4 P 4. It suffices to show that dim(i W ) 5 = 4. Now, note that there is a unique hyperplane H with defining equation g in P 4 passing through q 2,..., q 5. And for any form f of degree 5 to vanishes 4 times at q 2,..., q 5, f must have g as a factor. In other words, ((I W + (g))/(g)) 5 = 0. Thus dim(i W ) 5 = dim(i V ) 4 where V = Res H (W ) = 3q q 5 + 2p 1 + 2p 2 + p 3 + p 4. Now, we are back in the previous situation, as p 3, p 4 are generic points, it suffices to show that dim(i V ) 4 = 6 where V = 3q q 5 + 2p 1 + 2p 2, and this follows from the non-defectivity of σ 2 ((P 1 ) 4 ). 6. Introduction to tropical geometry Let k be an algebraically closed field, and v : k R be a non-archimedean valuation. Let X (k ) n be a subvariety. The tropicalization of X is defined by When f = c u x u k[x ± 1,..., x± n ], we have Trop(X) = {(v(x 1 ),..., v(x n )) (x 1,..., x n ) X(k)}. [Kapranov] Trop(V (f)) = {(w 1,..., w n ) R n min(v(c u ) + u w) is achieved at least twice}. Example: Consider the field of Puiseux series C{{t}} = n 1 C((t 1/n )). For c(t) = c 1 t a1 +, where a 1 < a 2 <... has a common denominator, then v(c(t)) = a 1. The valuation of c(t) indicates the order of zero or pole at t = 0. Theorem: Assume that k is an algebraically closed field of characteristic 0, then k{{t}} is algebraically closed. Taking a trivial valuation on k, let h k[x 1,..., x n ], its tropicalization is Trop(h) : R n R sending w wt w h. More generally, let X k n be an affine variety with defining ideal I. Then Trop(X) = {w R n : in w fis not a monomial for any f I}. Using this definition, we have: suppose that X is irreducible of dimension d, then Trop(X) is a polyhedral complex in R n of dimension d. For a polynomial map f = (f 1,..., f n ) : k m k n, the tropicalization of f is Trop(f) = (Trop(f 1 ),..., Trop(f n )) : R m R n. Note that Trop(f) is a continuous map when we give R the usual topology of a half-open interval. Lemma: Let f : k m k n be a polynomial map, let X = Imf. Then Trop(f) maps R m into Trop(X). Now let C 1,..., C k be cone in V, where C i is the image of polynomial map f i : V i V. Then we define f : V i V, by (p 1,..., p k ) f 1 (p 1 ) + + f k (p k ) V. Applying the tropical results, we deduce that dim(c C k ) dim Trop(f)( V i ).

6 6 THANH VU Let l i,b = Trop(f i,b ) for i = 1,..., k and b = 1,..., n. For v = (v 1,..., v k ) k i=1 Rmi, b {1,..., n} and i {1,..., k}, we say that i wins b at v provided that (1) l i,b (v i ) < l j,b (v j ) for all j i, and (2) l i,b is linear near v i. In this case, we denote d vi l i,b the differential R mi R of l i,b at v i. Let W i (v) = {b {1,..., n} i wins b at v} and D i (v) = {d vi l i,b b W i (v)}. Then dim Trop(f)( k V i ) = max ( dim R D i (v) ). v Here is another reformulation of the problem: For each b = 1,..., n, let A b = supp f b. For v = (v 1,..., v k ) (R m ) k, we see that i wins b at v if and only if there exists α A b such that v i, α < v j, β for all (j, β) (i, α). Let D i (v) = n b=1{α A b v i, α < v j, β for all (j, β) (i, α)}. dim Trop(f)( V i ) = max v i=1 k dim D i (v). The principle is that, for each v V i, Trop(f) is a linear function in a neighborhood of v. To find the dimension of the image of Trop(f) it suffices to find maximal of the rank of the corresponding linear map at this point v. So to prove non-defectivity, it suffices to find a good certificate vector v. For example, consider the case of Veronese embedding ν 3 (P 1 ) P 3. In this case, the map f i is defined by i=1 [x, y] [x 3, x 2 y, xy 2, y 3 ] thus if we want to find a lower bound on the dimension of σ 2 (X), we need to find the dimension of I Trop(f) = [min(3x, 3u), min(2x + y, 2u + v), min(x + 2y, u + 2v), min(3y, 3v)] Let s choose the certificate point [0, 3], [2, 1], then we see that in a neighborhood of v, the map Trop(f) is given by [3x, 2x + y, u + 2v, 3v] which has rank 4 which is the expected value, and we are done. In the case of Segre-Veronese embedding, where each A b is a singleton, the problem can also be formulated as a Voronoi problem as follows. Let S = A b. For each v = (v 1,..., v k ) (R m ) k. Let V or i (v) = {α S : v i α 2 < v j α 2 for all j i}. Then dim(c C k ) k i=1 (1 + dim Aff RV or i (v)) over all v (R m ) k. 7. Segre embeddings Let V 1, V 2, V 3 be vector spaces. The Segre embedding can be thought of as PV 1 PV 2 PV 3 P(V 1 V 2 V 3 ), (v 1, v 2, v 3 ) v 1 v 2 v 3. We want to know which Segre embedding is non-defective. The known defective Segre varieties are: (1) P n1 P n2 P n3 unbalanced: n 3 1 > (n 1 + 1)(n 2 + 1) n 1 n 2. (2) P 2 P 2n P 2n, σ 3k+1 (X) is a hypersurface defined by the Strassen s equation. (3) P 2 P 3 P 3 : The reason is as follows. First a general fact. If C is an irreducible subvariety of X passing through s generic points of X, then dim σ s (X) dim σ s (C) + s(dim X dim C). This follows from the Terracini s lemma: as picking s generic points p 1,..., p s of X, we have dim σ s (X) = dim(t p1 X + + T ps X) dim(t p1 C + + T ps C) + dim(t p1 C) + + dim(t ps C). Now for X = Seg(P 2 P 3 P 3 ), we show that there exists a rational normal curve C of degree 8 passing through 5 general points of X. This implies that dim σ 5 (X) (8 1) = 43 < 44. Now to show the existence of the rational normal curve C, first note that, through any 5 generic points on P 2, there exists a unique quadric curve g : P 1 P 2 such that g(0) = Q 1, g(1) = Q 2, g( ) = Q 3, g(x 1 ) = Q 4, g(x 2 ) = Q 5.

7 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES 7 Through 5 generic points in P 3, there exists a family a twisted cubic curve f s,t so that f s,t (0) = P 1, f s,t (1) = P 2, f s,t = P 3. And there exists parameter s and t so that f s,t (x 1 ) = P 4 and f s,t (x 2 ) = P 5. Combining these pieces together we have a rational normal curve P 1 P 2 P 3 P 3 P 47 of degree 8 passing through 5 generic points of X. Recall that a twisted cubic curve is isomorphic to the third Veronese embedding of P 1 and can be defined as follows. For any choice of linear forms L 1, L 2, L 3, M 1, M 2, M 3 so that for any [λ, µ] P 1, the linear forms λl 1 + µm 1, λl 2 + µm 2, λl 3 + µm 3 are linearly independent. Then the twisted cubic curve corresponds to these choices of linear form is given by f([λ, µ]) is the intersection point of the hyperplanes defined by the three linearly independent form above. Now, let us introduce the induction method of Abo-Ottaviani-Peterson. First of all, if p = v 1 v 2 v 3, then T p X = V 1 v 2 v 3 + v 1 V 2 v 3 + v 1 v 2 V 3. Denote G 1 px = V 1 v 2 v 3, G 2 px = v 1 V 2 v 3, and G 3 px = v 1 v 2 V 3. Then dim G i px = dim V i. For s generic points p 1,..., p s of X, let T s X = T p1 X + + T ps X. For t generic points q 1,..., q t of X, let G i tx = G i q 1 X + + G i q t X. Our goal is to find the dimension of T s X. Putting the G i tx in get us more flexible induction process. We say that T ((n 1, n 2, n 3 ); s, (a 1, a 2, a 3 ) true if T s X + G 1 a 1 X + G 2 a 2 X + G 3 a 3 X has the expected dimension. When that expected dimension is smaller than the ambient dimension, we say that T is subabundant. Similarly for the superabundant and equiabundant. The idea is that, we specify a certain set of points on a subspace H to do induction. First we have the following lemma: Lemma 7.1. With the notation as above, we have if v 1 H and if v 1 / H, where [p] denote the class of p in V 1 /H. 0 (G 1 px ) T p X (T p X ) 0 0 (T [p] X ) T p X (G 1 px) 0 Proof. Let s proof the first part, the second part is done in the similar manner. Let f : V1 V2 V3 H V2 V3 denote the canonical projection. Now, assume that p = v 1 v 2 v 3, and that v 1 H. We have f(t p X ) = f((v 1 + v 2 ) (v 1 + v 3 ) (v 2 + v 3 )) = ([v 1 ] H + v 2 ) ([v 1 ] H + v 3 ) (v 2 + v 3 ) = (T p X ) and the kernel of f restricted to T p X is equal to (V 1 /H) V2 V3 T p X = (v2 + v3 ) which is nothing but (G 1 px ). Theorem 7.2 (Abo-Ottaviani-Peterson). Let n 1 = n 1 +n 1 +1, s = s +s, a 2 = a 2 +a 2, a 3 = a 3 +a 3. Suppose (1) T ((n 1, n 2, n 3 ); s, a 1 + s, a 2, a 3) is true and subabundant. (2) T ((n 1, n 2, n 3 ; s ); a 1 + s, a 2, a 3 ) is true and subabundant. Then T ((n 1, n 2, n 3 ); s, (a 1, a 2, a 3 )) is true and subabundant. Proof. Let H V 1 be a subspace of dimension n Let X = P(H) P(V 2 ) P(V 3 ) P(H V 2 V 3 ). Similarly, let X = P(V 1 /H) P(V 2 ) PV 3 P(V 1 /H V 2 V 3 ). Specialize s points p i = v 1,i v 2,i v 3,i X, so that v 1,i H. Let f : V 1 V 2 V 3 H V 2 V 3 be the natural projection. Then we have two exact sequences for i = 1,..., s and 0 (G 1 p i X ) (T pi X) (T pi X ) 0 0 (T [pi]x ) (T pi X) (G 1 p i X) 0 where [p i ] is the class of [p i ] in X, for i = s + 1,..., s. Taking the intersection of the exact sequence give us an exact sequence 0 (G 1 p i X ) (T ) [pi]x T pi X T pi X ) G 1 p i X. i s i>s i s i s i>s

8 8 THANH VU Now this sequence can be written as 0 (G 1 s X ) (T s X ) (T s X) (T s X ) (G 1 s X ). Thus we see that if the outer terms has the expected dimension, then so is the middle term. Thus, the for the induction step to work, we need to include the G s stuff in. That is why we need a more general form as stated in the Theorem. To get that general form, simply add on the desired number of points, and remark that we have the exact sequence 0 (G 1 qx ) G 1 qx (G 1 qx ) 0 for generic point q X. Furthermore, if q = v 1 v 2 v 3, then dim G i qx = dim G i qx = dim G i qx = dim V i for i 2. Thus we can put either of the factor on the other side of the intersection depending on the way we specilize the set of points. We thus get the general statement. Here is some application of the result. For simplicity, I m gonna write the statement of the theorem as T ((n 1, n 2, n 3 ); s ; (a 1 + s, a 2, a 3)) and T ((n 1, n 2, n 3 ); s ; (a 1 + s, a 2, a 3 )) implies T ((n 1 + n 2 + 1, n 2, n 3 ); s + s ; (a 1, a 2 + a 2, a 3 + a 3 )). To deal with the Segre embedding of P 1 P m P n, we know that for it to be balanced, then either n = m or n = m + 1. In the following, we will show that they are non-defective if they are balance. First, we have some prelimilary remark. For the triple (1, 1, 1), we know T ( ; 2; (0, 0, 0)), T ( ; 1; (1, 1, 0)); T ( ; 0; (2, 1, 1)); T ( ; 0; (2, 2, 0)); T ( ; 0; (3, 0, 0)). Thus, we have T ((3, 1, 1); 1; (1, 3, 0)). More generally, we have Proposition 7.3. T ((1, m, m + 1); m + 1; (0, 1, 0)) holds. In particular, P 1 P m P m+1 is non-defective. Proof. We prove by induction on m, with the case m = 1 is known. Now, by induction, we know T ((m, 1, m + 1); m + 1; (1, 0, 0)). Thus if we know T ((1, 1, m + 1); 1; (m + 1, 0, 1)), then by the Abo-Ottaviani-Peterson Theorem, we deduce that T ((m + 2, 1, m + 1); m + 2; (0, 0, 1) holds which is the claim. Thus it remain to show that T ((1, 1, m + 1); 1; (m + 1, 0, 1)) holds. For the later, I also do by induction on m, assuming that the statement holds for m = 1 and 2. Now to verify T ((m + 1, 1, 1); 1; (1, 0, m + 1)), I know by induction T ((m 1, 1, 1); 1; (1, 0, m 1)). Furthermore, I know T ((1, 1, 1); 0; (2, 0, 2)). Thus another application of the Theorem gives me the desire conclusion. Also, as a consequence, we can complete the case P 1 P m P m. Proposition 7.4. T ((1, m, m); m + 1; (0, 0, 0)) holds. In particualr, P 1 P m P m is non-defective. Proof. By previous proposition, we know that T ((m, 1, m+1); m+1; (1, 0, 0)) holds. Thus, it suffices to note that T ((0, 1, m+1); 1; (m+1, 0, 0)) holds. The Theorem then applies to show that T ((m+1, 1, m+1); m+2; (0, 0, 0)) holds which is what we want. From that we deduce for the triple (3, 1, 1) T ( ; 2; (0, 1, 1)); T ( ; 2; (0, 2, 0)); T ( ; 1; (1, 2, 1)); T ( ; 1; (1, 3, 0)); T ( ; 1; (0, 3, 2)); T ( ; 0; (2, 3, 1)); T ( ; 0; (2, 2, 2)); T ( ; 0; (2, 4, 0)); T ( ; 0; (3, 0, 0)) It s worth trying the next case of P 2 P m P n. In this case, for it to be balanced, we must have n 3(m + 1) m 2 = 2m + 1. In this case, we want the statements of the form T ((2, 2n, 2n+2l); 3n+l; ()) T ((2, 2n, 2n+2l+1); 3n+l+1; ()) T ((2; 2n+1; 2n+2l+1); 3n+l+2; ()) T ((2, 2n+1, 2n+2l); 3n+l+2; ()) 8. Flattenings Assume that T V 1 V 2 V 3, then T corresponds to map V 1 V 2 V 3, and so on. These are called the flattening of T. Let r 1, r 2, r 3 be ranks of these matrices. The multi-linear rank of T is defined to be (r 1, r 2, r 3 ). The space of tensor of multirank (r 1, r 2, r 3 ) is a subvariety, denoted Sub r1,r 2,r 3. If r r i for all i, then σ r (X) Sub r1,r 2,r 3. Thus we have some equations that vanishes on the secant varieties. In particular, σ 2 (X) Sub 2,2,2 where the later is defined by the set of 3 3 minors of the flattening. Indeed, in the case of first secant variety, they are equal (Claudiu Raicu). The case of higher secant varieties is unknown.

9 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES 9 For example, if we consider flattening, then we get no equation for σ 4 (P 2 P 2 P 2 ) P 26, as the flattenings are 3 9 matrices. Nevertheless, it is not the whole space, it is defined by the Strassen s equation. Now, assume that V 1 = C 3 and consider the flattening V1 V 2 V 3. Choose a basis {v 1, v 2, v 3 } for V 1 and write T as a linear combination of matrices T = v 1 T 1 + v 2 T 2 + v 3 T 3. The 3 3 matrices T i are called the slices of T in the V 1 -direction with respect to the chosen basis. Now consider the following matrix 0 T 1 T 2 ϕ T = T 0 T 3 T 2 T 3 0 Then if T has rank 1 then ϕ T has rank 2 and ϕ T +T = ϕ T + ϕ T. Thus we see that if T has rank r, then ϕ T has rank 2r. Therefore, det(ϕ T ) vanishes on σ 4 (P 2 P 2 P 2 ) which is the Strassen s equation. For the Veronese varieties, and so are the Segre-Veronese varieties, one have the notion of symmetric flattenings. For any vector space V, one has the natural map S e V S d e V S d V which gives rise to a map S e V S d e V whose matrix corresponding to the multiplication table of S d S e. Since there are quite a number of way to break, Geramita conjectures that they all give the same ideal. 9. Prolongation Let S = k[x 1...x n ], and A S d be a vector space of forms of degree d. The rth prolongation of A, denoted by A (r) is {f S d+r r f x α A for all α Nn with α = r}. The connection to secant variety is given by [Landsberg-Manivel; Sidman-Sullivant] Theorem 9.1. Let I be the defining ideal of X P n 1, assume that I m d. Let A = I d. Then A (d 1)(r 1) = I(σ r (X)) r(d 1)+1. Pick a basis {x 1,..., x n } for V, then Sym(V ) is identified with k[x 1,..., x n ]. In this case, the co-multiplication map d,r : S d+r V S d V S r V can be identified as d,r (F ) = 1 r F α! x α xα. α =r Thus the prolongation can be computed as follows: Let A S d V, then A (r) = (A S r V ) S d+r V. The idea of polarization helps to understand prolongation, and bridge it to invariant theory and secant varieties Suppose F is a homogeneous polynomial of degree d in k[x 1,..., x n ]. For each i = 1,..., d, we introduce a new set of n variables x i = (x i 1,..., x i n). We also introduce an auxiliary set of variables t = (t 1,..., t d ). The polarization of F, denoted P F (x 1,..., x d ), is the coefficient of t 1 in the expansion of F (t 1 x t d x d ). One has the following properties (1) P F (x 1,..., x d ) is linear in each set of variables x i. (2) P F (x 1,..., x d ) is symmetric in the x i in the sense that if σ is a permutation of d elements, then P F (x 1,..., x d ) = P F (x σ(1),..., x σ(d) ). (3) P F (x,..., x) = d!f (x). Here is the relation to partial differentiation. Let F be a homogeneous polynomial of degree d + r. (4) Assume that F = x α, then (5) If β = r, then P F (x,..., x, y,..., y) = α! β =r ( d α β d!r! r F x β = r P F (x,..., x, y,..., y) y β. )( ) r x α β y β. β Lemma 9.2. Let A S d V, and F S d+r V be a homogeneous polynomial with polarization P F (x,..., x, y,.., y) of degree d in the x-variables. The following are equivalent F is in A (r).

10 10 THANH VU Every coefficient of F as a polynomial in the y-variables is in A. Every coefficient of the form P F (x,..., x, y 1,..., y r ) viewed as a polynomial in all of the y-variables, is in A. P F (x,..., x, v,..., v) A for every choice of v V. Proof. The equivalence of (1) and (2) follows from the fact that r F is just β! times the coefficient of y β in x β P F. Similar for the equivalent of (2) and (3). The equivalence of (2) and (4) follows from the fact that if we choose sufficiently generic points v i, the vectors in indeterminates F α of the form α =r F α(v i ) α will be linearly independent. Hence, they all lie in A if and only if every F α A. Proof of the theorem: Suppose that F A (r 1)(d 1), thus deg F = r(d 1) + 1. A general point on the rth secant variety of X is the span of r points of X. So let v = t 1 v t r v r where t i and v i are indeterminates. We will show that for any specialization of the v i to points of X and t i K, F (v) = 0. By polarization, it suffices to prove P F (v,..., v) = 0. The point now is that the polarization P F (x 1,..., x l ) is linear in each set of variables x i, l = r(d 1) + 1. Thus, we deduce that ( ) r(d 1) + 1 P F (v,..., v) = t β P F ((v 1 ) β1,..., (v r ) βr ) β β =r(d 1)+1 where v i is repeated β i times. Since F A (r 1)(d 1), every coefficient of a monomial in the y-variables in the polynomial P F (x,..., x, y 1,..., y (r 1)(d 1) ) is in A by part (3) of the preceding lemma. Therefore, each of these degree d coefficients is in A (written in the v i -variables. Thus if we specialize all of the v i to points of X, P F ((v 1 ) β1,..., (v r ) βr ) = Multi-prolongation For simplicity, we will describe multi-prolongation for the secant varieties of Segre-Veronese in three factor case. Note that, if X is a projective varieties, then the kth secant variety of X is the image of the addition map which corresponds to map of rings s : ˆX ˆX W s # : Sym(W ) k[ ˆX X] = k[ ˆX] k[ ˆX]. The defining ideal of σ k (X) is the kernel of this map, and the coordinate ring of σ k (X) is the image of s #. In the case X = SV d1,d 2,d 3 (P V1 P V2 P V3 ) the homogeneous coordinate ring of X has the decomposition k[x] = r 0(S rd1 V 1 S rd2 V 2 S rd3 V 3 ). The homogeneous coordinate ring Sym(W ) decomposes into GL(V 1 ) GL(V 2 ) GL(V 3 )-irreducible representations lead us to consider the following map. For each partition µ = (µ i1 1...µis s ) of r, we consider the map s π µ : S r (S d1 V 1 S d2 V 2 S d3 V 3 ) S ij (S µjd 1 V 1 S µjd 2 V 2 S µjd 3 V 3 ) which is the composition of the natural inclusion j=1 S r (S d1 V 1 S d2 V 2 S d3 V 3 ) (S d1 V 1 S d2 V 2 S d3 V 3 ) r and the tensor product of the natural multiplication maps m : (S di V i ) µj S µjd i V i. Here is the description on monomials. Each variable z α is identified with an 1 3 block with entries the multi-sets α i of cardinality d i. A monomial z α1 z αr of degree r is represented as an r 3 block M, whose rows correspond to the variables z αi. The way π µ acts on an r 3 block M is as follows. it partitions in all possible ways the set of rows of M into subsets of sizes equal to the parts of µ, collapses the elements of each subset into a single row, and takes the sum of all blocks obtained in this way. Here by collapsing a set of rows we mean taking the columnwise union of the entries of the rows. For example

11 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES 11 1, 2 1 1, 1 3 1, 1 1 2, 2 2 π (2,2) 1, 1, 1, 2 1, 1, 2, 2 1, 3 1, 1, 1, 2 + 1, 2 1, 1, 2, 2 1, 1 1, 2, 2, 2 + 2, 2 1, 1, 1, 1 1, 2 1, 3 Multi-prolongation[Landsberg] For a positive integer r, the polynomials of degree r vanishing on σ k (SV d1,d 2,d 3 (P V P V2 P V3 ) are precisely the elements of S r (S d1 V 1 S d2 V 2 S d3 V 3 ) in the intersection of the kernels of the maps π µ, where µ ranges over all partitions of r with exactly k parts. Proof. We have (k[x] k ) r = (S µ1d 1 V 1 S µ1d 2 V 2 S µ1d 3 V 3 ) (S µk d 1 V 1 S µk d 2 V 2 S µk d 3 V 3 ). µ 1+ +µ k =r The degree r component of π is then a direct sum of maps π µ where µ ranges over all partitions of r with k parts. 11. Introduction to representation theory of finite groups Definition A representation of a group G is a homomorphism from G into GL(V ) for some n. The integer n is then called the degree of the representation. If V and W are representations, the tensor product V W is also representation via g (v w) = g v g w. The exterior powers Λ n (V ) and symmetric powers Sym n (V ) are subrepresentation of V n. The dual V = Hom(V, C) of V is also a representation via ρ (G) = ρ(g 1 ) t : V V for all g G. Let H be a subgroup of G. For any ρ : G GL(V ), we denote by Res H ρ the restriction of ρ on H. Let p : G H be a homomorphism of groups. For every representation ρ : H GL(V ), ρ p : G GL(V ) is also a representation, called the lift of ρ to G. Exterior tensor product: Let ρ : G GL(V ) and σ : H GL(W ) be representations, then their exterior product ρ σ : G H GL(V W ) is defined by (ρ σ)(s, t)(v w) = ρ(s)(v) σ(t)w. If δ : G G G is the diagonal embedding, then ρ σ = (ρ σ) δ. Definition A representation is called irreducible if it does not contain a proper non-zero invariant subspace. A representation is called indecomposable if it can not be written as direct sum of two subrepresentations. Theorem 11.1 (Maschke). Let G be a finite group, and char k do not divide G. Let ρ : G GL(V ) be a representation and W be an invariant subspace. Then there is an invariant subspace W such that V = W W. As a consequence, every finite-dimensional representation of G is completely reducible. Proof. Let W be a subspace such that V = W W. Let P : V V be the projector onto W. Let P = 1 ρ g P ρ 1 g. G Then P is an intertwining projectors, so if we let W = ker P, then W is invariant and V = W W. g G Lemma Let G be a finite group, ρ : G GL(V ) be an irreducible representation. Then dim V G. Proof. Take any non-zero vector v V, then the set {ρ s v} s G spans an invariant subspace which must coincide with V. Hence dim V G. Theorem 11.3 (Schur s lemma). Let V and W be irreducible CG modules. (1) If ϑ : V W is a CG homomorphism, then either ϑ is an isomorphism, or ϑ = 0. (2) If ϑ : V V is a CG isomorphism, then ϑ is a scalar multiple of the identity endomorphism. As a consequence, a representation V of G is irreducible if and only if End G (V ) = C. Proof. (1) Since ker ϑ and Iϑ are invariant subspaces. (2) Since C is algebraically closed, so ϑ has at least one eigenvector. Lemma If U is a CG module, then dim(hom CG (CG, U)) = dim U. Theorem Write CG = U 1 U r, a direct sum of irreducible CG submodules. If U is any irreducible CG module, then the number of CG modules U i with U i = U is equal to dim U. Proof. Follows from lemma. 1

12 12 THANH VU Thus in analyzing the representation of G, we need: (i) Describe all the irreducible representation of G. Once we have done this, there remains the problem of carrying out in practice the description of a given representation in these terms. Thus our second goal will be: (ii) Find technique for giving the direct sum decomposition, and in particular determining the multiplicities a i of an arbitrary representation V. Finally, it is the case that the representations we will most often be concerned with are those arising from simpler ones by the short of linear or multilinear algebraic operation. We would like, therefore, to be able to describe, the representation we get when we perform these operations. This is known generally as : (iii) Plethysm: Describe the decomposition, with multiplicities of representation derived from a given representation V, such as V V, V, Λ k (V ). Note that, it suffices to work out these plethysm for irreducible representations. Similarly, if V and W are two irreducible representations, we want to decompose V W, this is usually known as the Clebsch - Gordon problem. If V is a representation of G. In general, for g, h G, we will have g(h(v)) h(g(v)). Indeed, ρ(g) : V V will be G linear if and only if g is in the center Z(G) of G. In particular, if G is abelian, and V is an irreducible representation, then by Schur lemma, every element g G acts on V by a scalar multiple of the identity. Every subspace of V is thus invariant, so V must be one dimensional. Lemma Let V be an irreducible representation of G, and let z Z(CG). Then there exists λ C such that zv = λv for all v V. Proof. Since C is algebraically closed, so it has a non-trivial eigenvector v with eigenvalue l. For any g G, we have gz v = lg v. Thus g v is also an eigenvector of z with eigenvalue l. Let W = {g v g G}. Then W is an invariant subspace of V, thus W = V. Example: Let G = S 3. We have, as with any symmetric group, two one dimensional representations, the trivial representation, denoted U, and the alternating representation U, defined by setting gv = sgn(g)v. The permutation representation, in which G acts on C 3 by permuting the coordinates, g(z 1, z 2, z 3 ) = (z g 1 (1), z g 1 (2), z g 1 (3)). This representation, is not irreducible, the line spanned by the sum (1, 1, 1) of the basis vectors is invariant, with complementary subspace V = {(z 1, z 2, z 3 ) : z 1 + z 2 + z 3 = 0}. This is an irreducible representation, called the standard representation of G. In the following, we will describe all irreducible representations of S 3 without character theory, (this is superfuous in finite group theory), though the idea would be the key in understanding representation of Lie groups. The idea is a very simple one. We look at the actions of the maximal abelian subgroup on an abitrary representation W. The action of this abelian subgroup yields a very simple decomposition: If we take τ to be any generator of A 3 S 3, the space W is spanned by eigenvectors v i for the action of τ, whose eigenvalues are the powers of a cube root of unity ω = e 2πi/3. Thus W = V i, where V i = Cv i, and τv i = ω ai v i. Next, we ask how the remaining elements of S 3 act on W in terms of this decomposition. Let σ be any transposition, so that τ and σ together generate S 3, with the relation στσ = τ 2. Thus, if v is an eigenvector for τ with eigenvalue ω i, we have τ(σ(v)) = ω 2i σ(v) so σ(v) is again an eigenvector for τ with eigenvalue ω 2i. With σ = (12), τ = (123), the standard representation has a basis a = (ω, 1, ω 2 ), β = (1, ω, ω 2 ), with τa = ωa, τβ = ω 2 β, σa = β, σβ = a. Now, let v be an eigenvector for τ. If the eigenvalue of v is ω i 1, then σ(v) is an eigenvector with eigenvalue ω 2i ω i, and so is independent of v, and v, σ(v) together span a two - dimensional subspace V of W invariant under S 3. In fact, V is isomorphic to the standard representation by the explicit computation above. If the eigenvalue of v is 1, then if σ(v) is not independent of v, then v spans a one dimensional subrepresentation of W, isomorphic to the trivial representation if σ(v) = v, and to the alternating representation if σv = v. If σ(v) and v are independent, then v + σ(v) and v σ(v) span one dimensional representations of W isomorphic to the trivial and alternating representations. Thus, the only three irreducible representations of S 3 are the trivial, alternating, and standard repsentation. In the decomposition W = au bu cv, we can determine the multiplicities a, b, c as followed: c, is the number of independent eigenvectors for τ with eigenvalue ω, where a + c is the multiplicity of 1 as an eigenvalue of σ, and b + c is the multiplicity of 1 as an eigenvalue of σ. This approach gives us the answer to our third problem as well, since if we know the eigenvalues of τ on a representation, we know the eigenvalues of τ on the various tensor powers of W. For example, V V = U U V. 12. Characters Definition Suppose that V is a representation of G. Then the character of V is the function χ : G C defined by χ(g) = trace(g V ).

13 SECANT VARIETIES OF SEGRE-VERONESE VARIETIES 13 Theorem Let ρ : G GL(n, C) be a representation of G, and let χ be the character of ρ. (1) For g G, χ(g) = χ(1) ρ(g) = λi n for some λ C. (2) Ker ρ = {g G : χ(g) = χ(1)}. For any representation V of a group G, set V G = {v V : gv = v : for all g G}. Let ϕ = 1 g, G then ϕ is a G module homomorphism. The map ϕ is a projection of V onto V G. We thus have the number of copies of the trivial representation appearing in the decomposition of V, which is m = dim V G = trace(ϕ) = 1 χ V (g) G In particular, for an irreducible representation V other than the trivial representation, the sum over all g G of the values of the character χ V is zero. Note that, with V and W are irreducible representations of G, then dim Hom G (V, W ) is one if V = W and 0 otherwise. Moreover, the character of the representation Hom(V, W ) = V W is χ V χ W. Thus, we have { 1 1, if V χv (g)χ W (g) = = W G 0, if V = W. A class function on G is a function ψ : G C such that ψ(x) = ψ(y) whenever x and y are conjugate elements of G. Define an Hermitian inner product on the space of class functions on G, by ϑ, φ = 1 G ϑ(g)φ(g). Then, the characters of the irreducible representation of G are orthonormal. As a corollary, the number of irreducible representations of G is less than or equal to the number of conjugacy classes. Also, V is irreducible if and only if χ V, χ V = 1. Proposition Let a : G C be any function on the group G, and for any representation V of G set, ϕ a,v = a(g) g : V V. Then ϕ a,v is a homomorphism of G modules for all V if and only if a is a class function. Proof. When a is a class function, we have g G ϕ a,v (hv) = a(g) g(hv) g G = a(hgh 1 )hgh 1 (hv) = h( a(hgh 1 )g(v)) = hϕ a,v (v). Thus ϕ a,v is a class function. Conversely, Theorem The number of irreducible representations of G is equal to the number of conjugacy classes of G. Equivalently, their characters {χ V } form an orthonormal basis for C class (G). Proof. Suppose that a is a class function, and a, χ V = 0 for all irreducible representations V, we must show that a = 0. Consider ϕ a,v, by Schur lemma, ϕ a,v = λid, and if n = dim V, then λ = 1 n trace(ϕ a,v ) = G n a, χ V = 0. Thus, ϕ a,v = 0, or a(g) g = 0 for any representation V of G. In particular, for the regular representation R, ϕ a,r = 0. But in R the elements {g G}, thought of as elements of End(R), are linearly independent. Thus a(g) = 0 for all g. 13. Representations of symmetric groups For each partition λ of n, we will construct a representation of S n, called V λ. We will show that V λ is irreducible for each partition λ, and V λ is not isomorphic to V µ if λ and µ are different partitions. Then since the number of conjugacy classes of S n is the same as the number of partitions, they are all the irreducible representation of S n. To start, let λ be a partition of n. We label the boxes of λ in the standard way. Define two subgroups of the symmetric group S n P = {g S n : g preserves each row} and

14 14 THANH VU Q = {g S n : g preserves each column}. In the group algebra CS n, we introduce two elements corresponding to these subgroups, we set a λ = g P e g and b λ = g Q sgn(g)e g. Finally we set c λ = a λ b λ CS n, this is called the Young symmetrizer. The following lemma is the basic tool need for representation of S n. Lemma Let T and T be tableaux on shapes l and l. Assume that l does not strictly dominate l. Then exactly one of the following occurs: (i) There are two distinct integers that occur in the same row of T and the same column of T. (ii) λ = λ and there is some p in P λ and some q in Q λ such that p T = q T. Proof. Suppose (i) is false. The entries of the first row of T must occur in different columns of T, so there is a q 1 Q λ so that these entries occur in the first row of q 1 T. The entries of the second row of T occur in different columns of T, so also of q 1 T, so there is a q 2 in Q λ, not moving the entries equal to those in the first row of T, so that these entries all occur in the first two rows of q 2 q 1 T. Continueing in this way, we get q 1,..., q k Q λ such that the entries in the first k rows of T occurs in the first k rows of q k q 1 T. In particular, since T and q k q 1 T have the same shape, it follows that λ λ k λ λ k. Since this is true for all k, this means that l l. Since we have assumed that l does not strictly dominate λ, we must have λ = λ. Taking k to be the number of rows in l, and q = q k q 1, we see that q T and T have the same entries in each row. This means that there is a p in P λ such that p T = q T. Define a linear ordering on the set of all tableaux with n boxes, by saying that T > T if either 1) the shape of T is larger than the shape of T in the lexicographic ordering, or 2) T and T have the same shape, and the largest entry that is in a different box in the two tableaux occurs earlier in the column word of T than in the column word of T. An important property of this ordering is that if T is a standard tableau, then for any p P λ, and q Q λ, p T T and q T T. Corollary If T and T are standard tableaux with T > T, then there is a pair of integers in the same row of T and the same column of T. Proof. Since T > T, the shape of T cannot dominate the shape of T. If there is no such pair, we are in case (ii) of the lemma: p T = q T. Since T and T are standard, we have qt T and p T T, this contradict the assumption T > T. Let A = CS n. For a partition l, let V λ = Ac λ be the corresponding representation. Note that P Q = {1} so an element of S n can be written in at most one way as a product p q, p P, q Q. Thus c is the sum ±eg, the sum over all g that can be written as p q, with coefficient ± being sgn(q). Lemma (1) For p P, p a = a p = a. (2) For q Q, (sgn(q)q)b = b(sgn(q)q) = b. (3) For all p P, q Q, pc(sgn(q)q) = c, and up to multiplication by a scalar, c is the only such element in A. Proof. Only the last assertion is not obvious. If n g e g satisfies the condition in (3), then n pgq = sgn(q)n g for all g, p, q, in particular, n pq = sgn(q)n 1. Thus, it suffices to verify that n g = 0 if g / P Q. For such g it suffices to find a transposition t such that p = t P and q = g 1 tg Q, for then g = pgq, so n g = n g. If T = gt is the tableau obtained by replacing each entry i of T by g(i), the claim is that there are two distinct integers that appear in the same row of T and in the same column of T, t is then the transposition of these two integers. We must verify that if there were no such pair of integers, then one could write g = pq for some p P, q Q. To do this, first take p 1 P, and q 1 Q = gqg 1 so that p 1 T and q 1T have the same first row, repeating on the rest of the tableau, one gets p P and q Q so that pt = q T. Then pt = q gt, so p = q g, and therefore g = pq, where q = g 1 (q ) 1 g Q, as required. Lemma (1) If λ > µ lexicographically, then for all x A, a λ xb µ = 0. In particular, if λ > µ, then c λ c µ = 0. (2) For all x A, c λ xc λ is a scalar multiple of c λ. In particular, c λ c λ = n λ c λ, for some n λ C. Proof. For (1), we may take x = g S n. Since gb µ g 1 is the element constructed from gt, where T is the tableau used to construct b µ, it suffices to show that a λ b µ = 0. One verifies that λ > µ implies that there are two integers in the same row of T and the same column of T. If t is the transposition of these integers, then a λ t = a λ, tb µ = b µ, so a λ b µ = a λ ttb µ = a λ b µ, as required. Part (2) follows from the previous lemma. Lemma (1) Each V λ is an irreducible representation of S n. (2) If λ µ, then V λ and V µ are not isomorphic.