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. PN denotes
the N -dimensional projective space over k. Varieties mean reduced irreducible subvarieties of some PN , i.e zero
set of certain homogeneous polynomials in k[x0 , ..., xN ].
Let X be a variety in PN , with the defining ideal I = (f1 , . . . , fn ). 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
X ∂f
|p (xi − pi ) = 0 for all f = f1 , . . . , fn .
∂xi
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,
σs (X) = hp1 , . . . , ps i.
From the definition, we see that
X ⊆ σ2 (X) ⊆ σ3 (X) ⊆ · · · ⊆ σs (X) = Pt .
Remark: If X is a linear subspace, then X = σi (X) for all i ≥ 2. Conversely, if X = σ2 (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 ( σ2 (X) ( σ3 (X) ( · · · ( σr−1 (X) ( σr (X) = PN .
Example 1: ν2 (P1 ) ⊂ P2 . It is easy to see that σ2 (X) = P2 .
Expected dimension: The secant variety can be thought of as join of varieties. For affine varieties X, Y ⊆
AN +1 , the sum of X and Y is defined to be the image of the map X × Y → AN +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
d = X̂ + Ŷ .
XY
From the definition, we see that σ2 (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 + 1.
1
2
THANH VU
For any secant variety, we thus have
dim σt (X) ≤ min(t dim X + t − 1, N ).
The value min(t dim X + t − 1, N ) is called the expected dimension of σt (X), denoted edt (X).
Definition 1.2. When dim σt (X) 6= edt (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 ⊆ PN and a linear projection π : PN → PM , then we have π(X k ) =
(πX)k .
Example 2: Segre variety. Consider the Segre variety X = Pm × Pn ⊆ Pmn+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. σk (R1 (Mm,n ) = Rk (Mm,n ).
where Rk (Mm,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 = ν2 (Pn ) ⊆ Pn(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 σk (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 P1 can be thought of as certain section of the
varieties of rank 1 matrices. Precisely, the following is a generic catalecticant matrix


x1
x2 x3 · · ·
xm+1
 x2
x3
xm+2 




..

x
.
Cat(m + 1, n + 1) = 
 3

 .

 ..

xn+1 · · ·
xm+n+1
Let d = m + n, then for any m, n ≥ 1, we have Cd = νd (P1 ) ⊆ Pd 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 σk (Cd ) is contained in Rk (Cat(m + 1, n + 1). Indeed, they
are equal.
Theorem 2.1. σk (Cd ) = Rk (Cm+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 TR (Z) is the linear span of TP (X) and TQ (Y ).
This follows from the affine version, where we see that tX,x + tY,y ⊆ tX+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[x0 , ..., xn ]. The
Veronese embeddings can be thought of as a map
νd : P(S1 ) → P(Sd ) : [L] → [Ld ].
Let X = νd (Pn ) and P = [Ld ] ∈ X, then we have
TP (X) = h[Ld−1 M ] : M ∈ S1 i.
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
3
If H ⊆ PN is a hyperplane, then νd−1 (H) is a degree d hypersurface. To see this, notice that H has an
equation of the form a0 z0 + · · · + aN zN , where zi are the coordinates of PN . The equation for νd−1 (H) is
obtained by substitute each zi with the corresponding degree d monomial in the x0 , . . . , xn .
Since νd−1 ([Ld ]) = [L], if H ⊆ PN is a hyperplane and [Ld ] ∈ H, then νd−1 (H) is a degree d hypersurface
passing through the point [L] ∈ PN .
If H ⊂ PN is a hyperplane such that T[Ld ] (X) ⊆ H, then νd−1 (H) is a degree d hypersurface singular at the
point [L] ∈ Pn . (We may assume that L = x0 , and it follows by direct calculation.)
Let P1 , . . . , Ps ∈ Pn be points with defining ideals p1 , . . . , ps respectively. The scheme defined by the ideal
2
p1 ∩ · · · ∩ p2s is called a 2-fat point scheme or a double point scheme.
Let X = νd (Pn ) ⊂ PN . There is a bijection between
{H ⊂ PN a hyperplane:H ⊇ hTP1 (X), . . . , TPs (X)i},
and
{degree d hypersurface of Pn singular at P1 , ..., Ps } = (p21 ∩ · · · ∩ p2s )d .
Thus we have
dim σs (X) = N − dim(p21 ∩ · · · ∩ p2s )d .
Theorem 3.1 (Alexander and Hirschowitz). The Veronese variety νd (Pn ) 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 = ν4 (P2 ) ⊂ P14 . The expected dimension of σ5 (X) is 14. In other
words, we expect that there is no quartic singular at 5 generic points of P2 . Nevertheless, assume that p1 , ..., p5
is 5 general points in P2 , we may assume that they are [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1], [1 : y1 : z1 ], [1 : y2 : z2 ]. 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 = νa1 ,...,at (Pn1 × · · · × Pnt )
be the Segre-Veronese embedding of Pn1 × · · · × Pnt by O(a1 , ..., at ). Let p1 , ..., pk be the defining ideal of k
generic points of Pn1 × · · · × Pnt , then
dim σk (X) = N − dim[p21 ∩ · · · ∩ p2k ]a1 ,...,at .
Example: Let X = ν1,2 (P1 × P2 ) → P11 . We will show that X is non-defective. Indeed, for σ2 (X), choose
2 generic points of P1 × P2 which we may assume to be [1 : 0] × [1 : 0 : 0] and [0 : 1] × [0 : 1 : 0]. Then we
have the defining ideals are p1 = (x1 , y1 , y2 ), p2 = (x0 , y0 , y2 ). Thus [p21 ∩ p22 ]1,2 = C{x0 y1 y2 , x0 y22 , x1 y0 y2 , x1 y22 }.
Furthermore, for σ3 (X), if we choose another point [1 : a] × [0 : 0 : 1] with defining ideal p3 = (x1 − ax0 , y0 , y1 ),
we see that non of the forms in [p21 ∩ p22 ]1,2 is in p23 , thus [p21 ∩ p22 ∩ p23 ]1,2 = 0 as expected.
Conjecture:
secant
variety σs (ν1,2 (Pm × Pn )) is defective if and only if
(Abo-Brambilla) Then+2
n+2
(a) 2 − n < s < min(m + 1, 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[x0 , ..., xm ; y0 , ..., yn ].
(a) For the unbalanced case: we see
that there are at least m + 1 − s linear form in S1,0 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
n
least (m + 1 − s)( n+2
−
s)
forms
in
S
vanishing twice at s generic points. Thus, we have
1,2
n
n+2
n+2
n+2
dim σs (X) ≤ (m + 1)
− 1 − (m + 1 − s)(
− s) = s(
+ m − s) − 1
n
n
2
which is smaller than the expected values in the range.
(b) Expectedly σ3k+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
THANH VU
5. Products of P1
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 = νa1 ,...,at (Pn1 × · · · × Pnt ). Let
S = k[x1,0 , ..., x1,n1 ; · · · ; xt,0 , ..., xt,nt ] be the coordinate ring of Pn1 × · · · × Pnt .
Consider the birational map f : Pn1 ,...,nt − − → Pn defined by
(1, x1,1 ..., x1,n1 ), ..., (1, xt,1 ..., xt,nt ) → (1, x1,1 , ..., x1,n1 , ..., xt,nt )
Let Z be a zero-dimensional scheme in Pn1 ,...,nt . Let Z 0 = f (Z). Let Q0 , Q1,1 , ..., Q1,n1 , ..., Qt,1 , ..., Qt,nt be
the coordinate points of Pn . Consider the linear space Πi = hQi,1 , ..., Qi,ni i ∼
= Pni −1 with defining ideal
Ii = (z0 , z1,1 , ..., z1,n1 , · · · , zi,1
ˆ , ..., ẑi,ni , ..., zt,1 , ..., zt,nt ).
Let W = Z 0 + (a − ai )Πi , then by Catalisano-Geramita-Gimigliano
dim(IW )a = dim(IZ )a1 ,...,at .
Let’s come back to the case of two factors for a moment: Let X = νa,b (Pm ×Pn ). Let S = k[z0 , z1 , ..., zm , zm+1 , ..., zm+n ].
Let p1 , ..., pk be the defining ideal of k generic points of Pm+n . Let a = (z0 , zm+1 , ..., zm+n ) and b = (z0 , z1 , ..., zm ).
Then
dim σk (X) = N − dim(ab ∩ ba ∩ p21 ∩ · · · ∩ p2k )a+b .
1
Now for the case of P × · · · × P1 , all the ideal Ii defines a point in Pn , and we can rephrase as: Let
p1 , .., pn , q1 , ..., qk be generic points in Pn . Then
dim σk (X) = (2n − 1) − [pn−1
∩ · · · pnn−1 ∩ q12 ∩ · · · ∩ qk2 ]n .
1
Now, we introduce some preliminary result:
Let f be a form of degree d. From the exact sequence
I +f
= I/I ∩ (f ) = I/f (I : f ) → 0
0 → (I : f )(−d) → I →
f
we deduce that
dim Id ≤ dim(I : f )d + dim(I/f (I : f ))d .
For example: Let consider σ2 (ν1,3 (P1 × P1 )). In this case, we are interested in finding:
dimk [(z0 , z2 )3 ∩ (z0 , z1 ) ∩ q12 ∩ q22 ]4
where q1 , q2 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 I4 ≤ dim J3 + dim K4
where J = (z0 , z2 )3 ∩ (z0 , z1 ) ∩ q1 ∩ q2 = z0 ((z0 , z2 )2 ∩ q1 ∩ q2 ). Thus J3 has dimension one, as there is only
one quadrics in (z0 , z2 )2 vanishes simultaneously at q1 and q2 (note that these two points impose 2 equation on
these space of quarics).
Now K view as an ideal in k[z0 , z1 ], is of the form (z0 , z1 )3 ∩ r12 ∩ r22 , where r1 , and r2 are defining ideals of
two restricted points. Now, since r1 , r2 are principal, and they are relatively prime as the points q1 and q2 are
generics, we deduce that r12 ∩ r22 = r12 r22 ⊂ (z0 , z1 )3 , thus K4 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 σ4 (X) = 23.
We need the Alexander-Hirschowitz differential Horace lemma:
Let H ⊂ Pn be a hyperplane, p1 , ..., pr be generic points of Pn . Consider a zero dimension scheme Y ⊂ Pn ,
00
Y = X + 2p1 + · · · + 2pr . Let X 0 = ResH (X), X = T rH (X). Let p01 , ..., p0r be generic points in H, set
D2,H (p0i ) = 2p0i ∩ H ( a subscheme of Pn ). Consider the two schemes
Y 0 = X 0 + D2,H (p01 ) + · · · + D2,H (p0r ) ⊂ Pn ; Y
00
00
= X + p01 + ... + p0r ⊂ Pn−1 = H.
If dim(IY 0 )t−1 = 0 and dim(IY 00 )t = 0 then dim(IY )t = 0.
Now, back to our case where X = (P1 )5 ⊂ P31 , and we want to find the dimension of σ4 (X). Let W =
4q1 + · · · + 4q5 + 2p1 + · · · + 2p4 where qi are the coordinate points of P5 and qi are generic points. We need to
show that dim(IW )5 = 31 − 23 = 8.
Let H be a hyperplane which contains q2 , ..., q5 but not q1 . We choose the less generic scheme by taking p1 ,
p2 be generic points of H and p3 , p4 generic in P5 . Now add 8 points to W , call them t1 , ..., t8 with the first
four chosen generically on H and the last four chosen generically in P5 and call the resulting scheme Z. It will
be enough to prove that dim(IZ )5 = 0 (because 8 points impose at most 8 equations). We want to apply the
differential Horace lemma to this scheme.
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
5
Write Z = X + 2p3 + 2p4 , with
X = 4q1 + · · · + 4q5 + 2p1 + 2p2 + t1 + · · · + t8 .
We have
ResH (X) = 4q1 + 3q2 + · · · + 3q5 + p1 + p2 + t5 + · · · + t8 ⊂ P5
T rH (X) = 4q2 + · · · + 4q5 + 2p1 + 2p2 + t1 + · · · + t4 ⊂ P4 = H
By the Lemma, we choose two generic points p03 , p04 in H and consider the two schemes
00
Z 0 = ResH (X) + D2,H (p03 ) + D2,H (p04 ) ⊂ P5 , Z = T rH (X) + p03 + p04 ⊂ P4 .
We want to show that (IZ 0 )4 = 0 and (IZ 00 )5 = 0.
Now, note that Z 0 = 4q1 +3q2 +· · ·+3q5 +D2,H (p03 )+D2,H (p04 )+p1 +p2 +t5 +· · ·+t8 . Let W = 4q1 +3q2 +· · ·+
3q5 +D2,H (p03 )+D2,H (p04 )+p1 +p2 . It suffices to show that dim(IW )4 = 4 (then 4 generic points t5 , ..., t8 imposes
another 4 conditions). Now, let f be the defining ideal of H. We see that (IW : f )3 = 0, thus it suffices to show
that the dimension of IW |H defining the scheme W 0 = 3q2 +· · ·+3q5 +2p03 +2p04 +p1 +p2 ⊂ H = P4 is 4. Now, since
p1 , p2 are generic points of H, it suffices to show that dim(IW 0 )4 = 6 where W 0 = 3q2 +· · ·+3q5 +2p03 +2p04 ⊂ P4 .
But that follows from the fact that σ2 ((P1 )4 ) is non-defective.
00
Finally, we need to show that the dimension of (IZ 00 )5 = 0 where Z ⊂ P4 . Note that
00
Z = 4q2 + · · · + 4q5 + 2p1 + 2p2 + t1 + · · · + t4 + p03 + p04 ⊂ P4 .
Let W 0 = 4q2 + · · · + 4q5 + 2p1 + 2p2 + p03 + p04 ⊂ P4 . It suffices to show that dim(IW 0 )5 = 4. Now, note that
there is a unique hyperplane H 0 with defining equation g in P4 passing through q2 , ..., q5 . And for any form f
of degree 5 to vanishes 4 times at q2 , ..., q5 , f must have g as a factor. In other words,
((IW 0 + (g))/(g))5 = 0.
Thus
dim(IW 0 )5 = dim(IV )4
where V = ResH 0 (W ) = 3q2 + · · · + 3q5 + 2p1 + 2p2 + p03 + p04 . Now, we are back in the previous situation, as
p03 , p04 are generic points, it suffices to show that dim(IV 0 )4 = 6 where V 0 = 3q2 + · · · + 3q5 + 2p1 + 2p2 , and this
follows from the non-defectivity of σ2 ((P1 )4 ).
0
6. Introduction to tropical geometry
Let k be an algebraically closed field, and v : k → R̄ be a non-Archimedean valuation. Let X ⊆ (k ∗ )n be a
subvariety. The tropicalization of X is defined by
Trop(X) = {(v(x1 ), ..., v(xn ))|(x1 , ..., xn ) ∈ X(k)}.
When f =
P
cu x u ∈
±
k[x±
1 , . . . , xn ],
we have
[Kapranov] Trop(V (f )) = {(w1 , ..., wn ) ∈ Rn | min(v(cu ) + u · w) is achieved at least twice}.
Example: Consider the field of Puiseux series C{{t}} = ∪n≥1 C((t1/n )). For c(t) = c1 ta1 + · · · , where
a1 < a2 < ... has a common denominator, then v(c(t)) = a1 . 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[x1 , ..., xn ], its tropicalization is
Trop(h) : R̄n → R̄ sending w → wtw h.
More generally, let X ⊂ k n be an affine variety with defining ideal I. Then
Trop(X) = {w ∈ R̄n : inw f is 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 R̄n of dimension d.
For a polynomial map f = (f1 , ..., fn ) : k m → k n , the tropicalization of f is Trop(f ) = (Trop(f1 ), ..., Trop(fn )) :
m
R̄ → R̄n . Note that Trop(f ) is a continuous map when we give R̄ 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 R̄m into Trop(X).
Now let C1 , ..., Ck be cone in V , where Ci is the image of polynomial map fi : Vi → V . Then we define
Q
f : Vi → V , by (p1 , ..., pk ) → f1 (p1 ) + · · · + fk (pk ) ∈ V . Applying the tropical results, we deduce that
Y
dim(C1 + · · · + Ck ) ≥ dim Trop(f )( Vi ).
6
THANH VU
Qk
Let li,b = Trop(fi,b ) for i = 1, ..., k and b = 1, ..., n. For v = (v1 , ..., vk ) ∈ i=1 Rmi , b ∈ {1, ..., n} and
i ∈ {1, ..., k}, we say that i wins b at v provided that
(1) li,b (vi ) < lj,b (vj ) for all j 6= i, and
(2) li,b is linear near vi .
In this case, we denote dvi li,b the differential Rmi → R of li,b at vi .
Let Wi (v) = {b ∈ {1, ..., n}|i wins b at v} and Di (v) = {dvi li,b |b ∈ Wi (v)}. Then
dim Trop(f )(
Y
Vi ) = max(
v
k
X
dimR hDi (v)i).
i=1
Here is another reformulation of the problem:
For each b = 1, ..., n, let Ab = supp fb . For v = (v1 , ..., vk ) ∈ (Rm )k , we see that i wins b at v if and only if
there exists α ∈ Ab such that
hvi , αi < hvj , βi
for all (j, β) 6= (i, α). Let
Di (v) = ∪nb=1 {α ∈ Ab |hvi , αi < hvj , βi for all (j, β) 6= (i, α)}.
dim Trop(f )(
Y
Vi ) = max
v
k
X
dim hDi (v)i.
i=1
Q
The principle is that, for each v ∈ Vi , 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 ν3 (P1 ) → P3 . In this case, the map fi is defined by
[x, y] → [x3 , x2 y, xy 2 , y 3 ]
thus if we want to find a lower bound on the dimension of σ2 (X), we need to find the dimension of
= 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 Ab is a singleton, the problem can also be formulated
as a Voronoi problem as follows. Let S = ∪Ab . For each v = (v1 , ..., vk ) ∈ (Rm )k . Let
V ori (v) = {α ∈ S : kvi − α|2 < kvj − αk2 for all j 6= i}.
Pk
Then dim(C1 + · · · + Ck ) ≥ i=1 (1 + dim Af fR V ori (v)) over all v ∈ (Rm )k .
7. Segre embeddings
Let V1 , V2 , V3 be vector spaces. The Segre embedding can be thought of as
PV1 × PV2 × PV3 → P(V1 ⊗ V2 ⊗ V3 ), (v1 , v2 , v3 ) → v1 ⊗ v2 ⊗ v3 .
We want to know which Segre embedding is non-defective. The known defective Segre varieties are:
(1) Pn1 × Pn2 × Pn3 unbalanced: n3 − 1 > (n1 + 1)(n2 + 1) − n1 − n2 .
(2) P2 × P2n × P2n , σ3k+1 (X) is a hypersurface defined by the Strassen’s equation.
(3) P2 × P3 × P3 : 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 σs (X) ≤ dim σs (C) + s(dim X − dim C).
This follows from the Terracini’s lemma: as picking s generic points p1 , ..., ps of X, we have
dim σs (X) = dim(Tp1 X + · · · + Tps X) ≤ dim(Tp1 C + · · · + Tps C) + dim(Tp1 C)⊥ + · · · + dim(Tps C)⊥ .
Now for X = Seg(P2 × P3 × P3 ), we show that there exists a rational normal curve C of degree 8 passing through
5 general points of X. This implies that
dim σ5 (X) ≤ 8 + 5 · (8 − 1) = 43 < 44.
Now to show the existence of the rational normal curve C, first note that, through any 5 generic points on P2 ,
there exists a unique quadric curve g : P1 → P2 such that
g(0) = Q1 , g(1) = Q2 , g(∞) = Q3 , g(x1 ) = Q4 , g(x2 ) = Q5 .
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
7
Through 5 generic points in P3 , there exists a family a twisted cubic curve fs,t so that fs,t (0) = P1 , fs,t (1) =
P2 , fs,t = P3 . And there exists parameter s and t so that fs,t (x1 ) = P4 and fs,t (x2 ) = P5 . Combining these
pieces together we have a rational normal curve P1 → P2 × P3 × P3 → P47 of degree 8 passing through 5 generic
points of X. Recall that a twisted cubic curve is isomorphic to the third Veronese embedding of P1 and can be
defined as follows. For any choice of linear forms L1 , L2 , L3 , M1 , M2 , M3 so that for any [λ, µ] ∈ P1 , the linear
forms λL1 + µM1 , λL2 + µM2 , λL3 + µM3 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 = v1 ⊗ v2 ⊗ v3 , then
Tp X = V1 ⊗ v2 ⊗ v3 + v1 ⊗ V2 ⊗ v3 + v1 ⊗ v2 ⊗ V3 .
G1p X
Denote
= V1 ⊗ v2 ⊗ v3 , G2p X = v1 ⊗ V2 ⊗ v3 , and G3p X = v1 ⊗ v2 ⊗ V3 . Then dim Gip X = dim Vi . For s
generic points p1 , ..., ps of X, let
Ts X = Tp1 X + · · · + Tps X.
For t generic points q1 , ..., qt of X, let
Git X = Giq1 X + · · · + Giqt X.
Our goal is to find the dimension of Ts X. Putting the Git X in get us more flexible induction process. We say
that T ((n1 , n2 , n3 ); s, (a1 , a2 , a3 ) true if
Ts X + G1a1 X + G2a2 X + G3a3 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
00
0 → (G1p X )⊥ → Tp X ⊥ → (Tp X 0 )⊥ → 0
if v1 ∈ H and
00
0 → (T[p] X )⊥ → Tp X ⊥ → (G1p X)⊥ → 0
if v1 ∈
/ H, where [p] denote the class of p in V1 /H.
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 = v1 ⊗ v2 ⊗ v3 , and that v1 ∈ H. We have
f (Tp X ⊥ ) = f ((v1⊥ + v2⊥ ) ∩ (v1⊥ + v3⊥ ) ∩ (v2⊥ + v3⊥ ))
⊥
⊥
⊥
⊥
⊥
0 ⊥
= ([v1 ]⊥
H + v2 ) ∩ ([v1 ]H + v3 ) ∩ (v2 + v3 ) = (Tp X )
and the kernel of f restricted to Tp X ⊥ is equal to (V1 /H)∗ ⊗ V2∗ ⊗ V3∗ ∩ Tp X ⊥ = (v2⊥ + v3⊥ ) which is nothing
00
but (G1p X )⊥ .
00
00
00
00
Theorem 7.2 (Abo-Ottaviani-Peterson). Let n1 = n01 + n1 + 1, s = s0 + s , a2 = a02 + a2 , a3 = a03 + a3 . Suppose
00
(1) T ((n01 , n2 , n3 ); s0 , a1 + s , a02 , a03 ) is true and subabundant.
00
00
00
00
(2) T ((n1 , n2 , n3 ; s ); a1 + s0 , a2 , a3 ) is true and subabundant.
Then T ((n1 , n2 , n3 ); s, (a1 , a2 , a3 )) is true and subabundant.
Proof. Let H ⊆ V1 be a subspace of dimension n01 + 1. Let X 0 = P(H) × P(V2 ) × P(V3 ) → P(H ⊗ V2 ⊗ V3 ).
00
Similarly, let X = P(V1 /H) × P(V2 ) × PV3 → P(V1 /H ⊗ V2 ⊗ V3 ). Specialize s0 points
pi = v1,i ⊗ v2,i ⊗ v3,i ∈ X, so that v1,i ∈ H.
Let f : V1∗ ⊗ V2∗ ⊗ V3∗ → H ∗ ⊗ V2∗ ⊗ V3∗ be the natural projection. Then we have two exact sequences
00
0 → (G1pi X )⊥ → (Tpi X)⊥ → (Tpi X 0 )⊥ → 0
for i = 1, ..., s0 and
00
0 → (T[pi ] X )⊥ → (Tpi X)⊥ → (G1pi X)⊥ → 0
00
where [pi ] is the class of [pi ] in X , for i = s0 + 1, ..., s.
Taking the intersection of the exact sequence give us an exact sequence
\
\
\
\
\
00
00
0→
(G1pi X )⊥
(T[pi ] X )⊥ →
Tpi X ⊥ →
Tpi X 0 )⊥
G1pi X ⊥ .
i≤s0
i>s0
i≤s
i≤s0
i>s0
8
THANH VU
Now this sequence can be written as
00
00
0 → (G1s0 X )⊥ ∩ (Ts00 X )⊥ → (Ts X)⊥ → (Ts0 X 0 )⊥ ∩ (G1s00 X 0 )⊥ .
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
00
0 → (G1q X )⊥ → G1q X ⊥ → (G1q X 0 )⊥ → 0
00
for generic point q ∈ X. Furthermore, if q = v1 ⊗ v2 ⊗ v3 , then dim Giq X = dim Giq X 0 = dim Giq X = dim Vi
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
00
00
00
00
00
T ((n01 , n2 , n3 ); s0 ; (a1 + s , a02 , a03 )) and T ((n1 , n2 , n3 ); s ; (a1 + s0 , a2 , a3 ))
00
00
00
00
implies T ((n01 + n2 + 1, n2 , n3 ); s0 + s ; (a1 , a02 + a2 , a03 + a3 )).
To deal with the Segre embedding of P1 × Pm × Pn , 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, P1 × Pm × Pm+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 P1 × Pm × Pm .
Proposition 7.4. T ((1, m, m); m + 1; (0, 0, 0)) holds. In particualr, P1 × Pm × Pm 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 P2 × Pm × Pn . 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 ∈ V1 ⊗ V2 ⊗ V3 , then T corresponds to map V1∗ → V2 ⊗ V3 , and so on. These are called the
flattening of T . Let r1 , r2 , r3 be ranks of these matrices. The multi-linear rank of T is defined to be (r1 , r2 , r3 ).
The space of tensor of multirank ≤ (r1 , r2 , r3 ) is a subvariety, denoted Subr1 ,r2 ,r3 . If r ≤ ri for all i, then
σr (X) ⊂ Subr1 ,r2 ,r3 .
Thus we have some equations that vanishes on the secant varieties. In particular, σ2 (X) ⊂ Sub2,2,2 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.
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
9
For example, if we consider flattening, then we get no equation for σ4 (P2 × P2 × P2 ) ⊆ P26 , 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 V1 ∼
= C3 and consider the flattening V1∗ → V2 ⊗ V3 . Choose a basis {v1 , v2 , v3 } for V1 and write T as a
linear combination of matrices
T = v1 ⊗ T 1 + v2 ⊗ T 2 + v 3 ⊗ T 3 .
i
The 3×3 matrices T are called the slices of T in the V1 -direction with respect to the chosen basis. Now consider
the following matrix


0
T 1 −T 2
0
T3 
ϕT = −T 1
2
3
T
−T
0
Then if T has rank 1 then ϕT has rank 2 and ϕT +T 0 = ϕT + ϕT 0 . Thus we see that if T has rank r, then ϕT
has rank ≤ 2r. Therefore, det(ϕT ) vanishes on σ4 (P2 × P2 × P2 ) 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[x1 ...xn ], and A ⊂ Sd be a vector space of forms of degree d. The rth prolongation of A, denoted
by A(r) is
∂rf
{f ∈ Sd+r | α ∈ A for all α ∈ Nn with |α| = r}.
∂x
The connection to secant variety is given by [Landsberg-Manivel; Sidman-Sullivant]
Theorem 9.1. Let I be the defining ideal of X ⊂ Pn−1 , assume that I ⊂ md . Let A = Id . Then
A(d−1)(r−1) = I(σr (X))r(d−1)+1 .
Pick a basis {x1 , ..., xn } for V ∗, then Sym(V ∗ ) is identified with k[x1 , ..., xn ]. In this case, the co-multiplication
map ∆d,r : S d+r V ∗ → S d V ∗ ⊗ S r V ∗ can be identified as
X 1 ∂rF
∆d,r (F ) =
⊗ xα .
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[x1 , ..., xn ]. For each i = 1, ..., d, we introduce a
new set of n variables xi = (xi1 , ..., xin ). We also introduce an auxiliary set of variables t = (t1 , ..., td ). The
polarization of F , denoted P F (x1 , ..., xd ), is the coefficient of t1 in the expansion of F (t1 x1 + · · · + td xd ).
One has the following properties
(1) P F (x1 , ..., xd ) is linear in each set of variables xi .
(2) P F (x1 , ..., xd ) is symmetric in the xi in the sense that if σ is a permutation of d elements, then
P F (x1 , ..., xd ) = P F (xσ(1) , ..., xσ(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 = xα , then
X d r P F (x, ..., x, y, ..., y) = α!
xα−β y β .
α−β
β
|β|=r
(5) If |β| = r, then
d!r!
∂rF
∂ r P F (x, ..., x, y, ..., y)
=
.
β
∂x
∂y 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
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 .
r
Proof. The equivalence of (1) and (2) follows from the fact that ∂∂xFβ is just β! times the coefficient of y β in
P F . Similar for the equivalent of (2) and (3). The equivalence of (2) and (4) follows
from the fact that if we
P
choose sufficiently generic points vi , the vectors in indeterminates Fα of the form |α|=r Fα (v i )α will be linearly
independent. Hence, they all lie in A if and only if every Fα ∈ 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 = t1 v 1 + · · · + tr v r where ti and v i are indeterminates.
We will show that for any specialization of the v i to points of X and ti ∈ K, F (v) = 0. By polarization, it
suffices to prove
P F (v, ..., v) = 0.
The point now is that the polarization P F (x1 , ..., xl ) is linear in each set of variables xi , l = r(d − 1) + 1. Thus,
we deduce that
X
r(d − 1) + 1 β
P F (v, ..., v) =
t P F ((v 1 )β1 , ..., (v r )βr )
β
|β|=r(d−1)+1
where v i is repeated βi 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 ) = 0.
10. 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
s : X̂ × · · · X̂ → W
which corresponds to map of rings
s# : Sym(W ∗ ) → k[X̂ × · · · × X] = k[X̂] ⊗ · · · ⊗ k[X̂].
The defining ideal of σk (X) is the kernel of this map, and the coordinate ring of σk (X) is the image of s# .
In the case X = SVd1 ,d2 ,d3 (P V1∗ ×P V2∗ ×P V3∗ ) the homogeneous coordinate ring of X has the decomposition
M
k[X] =
(S rd1 V1 ⊗ S rd2 V2 ⊗ S rd3 V3 ).
r≥0
The homogeneous coordinate ring Sym(W ∗ ) decomposes into GL(V1 ) × GL(V2 ) × GL(V3 )-irreducible representations lead us to consider the following map. For each partition µ = (µi11 ...µiss ) of r, we consider the
map
s
O
πµ : Sr (Sd1 V1 ⊗ Sd2 V2 ⊗ Sd3 V3 ) →
Sij (Sµj d1 V1 ⊗ Sµj d2 V2 ⊗ Sµj d3 V3 )
j=1
which is the composition of the natural inclusion
Sr (Sd1 V1 ⊗ Sd2 V2 ⊗ Sd3 V3 ) → (Sd1 V1 ⊗ Sd2 V2 ⊗ Sd3 V3 )⊗r
and the tensor product of the natural multiplication maps
m : (Sdi Vi )⊗µj → Sµj di Vi .
Here is the description on monomials. Each variable zα is identified with an 1 × 3 block with entries the
multi-sets αi of cardinality di . A monomial zα1 · · · zαr of degree r is represented as an r × 3 block M , whose
rows correspond to the variables zαi . The way πµ 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
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
11
1, 2 1
1, 1 3 π(2,2) 1, 1, 1, 2 1, 3 1, 1, 1, 2 1, 1 1, 2, 2, 2 1, 2
−→
+
+
1, 1, 2, 2 1, 2 1, 1, 2, 2 2, 2 1, 1, 1, 1 1, 3
1, 1 1
2, 2 2
Multi-prolongation[Landsberg] For a positive integer r, the polynomials of degree r vanishing on σk (SVd1 ,d2 ,d3 (P V1∗ ×
P V2∗ × P V3∗ ) are precisely the elements of Sr (Sd1 V1 ⊗ Sd2 V2 ⊗ Sd3 V3 ) in the intersection of the kernels of the
maps πµ , where µ ranges over all partitions of r with exactly k parts.
Proof. We have
M
(k[X]⊗k )r =
(Sµ1 d1 V1 ⊗ Sµ1 d2 V2 ⊗ Sµ1 d3 V3 ) ⊗ · · · ⊗ (Sµk d1 V1 ⊗ Sµk d2 V2 ⊗ Sµk d3 V3 ).
µ1 +···+µk =r
The degree r component of π is then a direct sum of maps πµ where µ ranges over all partitions of r with k
parts.
11. Introduction to representation theory of finite groups
Definition 11.1. 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 Λn (V ) and symmetric powers Symn (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 ResH 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 11.2. 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 0 such that V = W ⊕ W 0 .
As a consequence, every finite-dimensional representation of G is completely reducible.
Proof. Let W
00
00
be a subspace such that V = W ⊕ W . Let P : V → V be the projector onto W . Let
1 X
ρg ◦ P ◦ ρ−1
P̄ =
g .
|G|
g∈G
Then P̄ is an intertwining projectors, so if we let W 0 = ker P̄ , then W 0 is invariant and V = W ⊕ W 0 .
Lemma 11.2. 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 {ρs 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 EndG (V ) ∼
= C.
Proof. (1) Since ker ϑ and =ϑ are invariant subspaces.
(2) Since C is algebraically closed, so ϑ has at least one eigenvector.
Lemma 11.4. If U is a CG−module, then
dim(HomCG (CG, U )) = dim U.
Theorem 11.5. Write CG = U1 ⊕· · ·⊕Ur , a direct sum of irreducible CG−submodules. If U is any irreducible
CG−module, then the number of CG−modules Ui with Ui ∼
= U is equal to dim U .
Proof. Follows from lemma.
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
ai 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 ∗ , Λk (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)) 6= 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 11.6. 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 = S3 . We have, as with any symmetric group, two one dimensional representations, the trivial representation, denoted U , and the alternating representation U 0 , defined by setting gv = sgn(g)v. The permutation representation, in which G acts on C3 by permuting the coordinates, g(z1 , z2 , z3 ) = (zg−1 (1) , zg−1 (2) , zg−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 = {(z1 , z2 , z3 ) : z1 + z2 + z3 = 0}. This is an irreducible representation, called
the standard representation of G.
In the following, we will describe all irreducible representations of S3 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 A3 ⊂ S3 , the space W is spanned by eigenvectors vi for the action of τ , whose eigenvalues are
the powers of a cube root of unity ω = e2πi/3 . Thus W = ⊕Vi , where Vi = Cvi , and τ vi = ω ai vi . Next, we ask
how the remaining elements of S3 act on W in terms of this decomposition. Let σ be any transposition, so that
τ and σ together generate S3 , with the relation στ σ = 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 τ a = ωa, τ β = ω 2 β, σa = β, σβ = a.
Now, let v be an eigenvector for τ . If the eigenvalue of v is ω i 6= 1, then σ(v) is an eigenvector with eigenvalue
2i
ω 6= ω i , and so is independent of v, and v, σ(v) together span a two - dimensional subspace V 0 of W invariant
under S3 . In fact, V 0 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 S3 are the
trivial, alternating, and standard repsentation. In the decomposition W = aU ⊕ bU 0 ⊕ 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 0 ⊕V .
12. Characters
Definition 12.1. Suppose that V is a representation of G. Then the character of V is the function χ : G → C
defined by χ(g) = trace(gV ).
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
13
Theorem 12.1. 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) = λIn 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
1 X
ϕ=
g,
|G|
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
1 X
χV (g)
m = dim V G = trace(ϕ) =
|G|
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 χV is zero. Note that, with V and W are irreducible representations of G, then
dim HomG (V, W ) is one if V ∼
= W and 0 otherwise. Moreover, the character of the representation Hom(V, W ) =
V ∗ ⊗ W is χV · χW . Thus, we have
(
1, if V ∼
1 X
=W
χV (g)χW (g) =
|G|
0, if V 6∼
= 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
P
1
hϑ, φi = |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 hχV , χV i = 1.
Proposition
12.2. Let a : G → C be any function on the group G, and for any representation V of G set,
P
ϕa,V =
a(g) · g : V → V . Then ϕa,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
ϕa,V (hv) =
X
a(g) · g(hv)
= a(hgh−1 )hgh−1 (hv)
X
= h(
a(hgh−1 )g(v))
= hϕa,V (v).
Thus ϕa,V is a class function. Conversely,
Theorem 12.3. The number of irreducible representations of G is equal to the number of conjugacy classes of
G. Equivalently, their characters {χV } form an orthonormal basis for Cclass (G).
Proof. Suppose that a is a class function, and ha, χV i = 0 for all irreducible representations V , we must show
that a = 0. Consider ϕa,V , by Schur lemma, ϕa,V = λid, and if n = dim V , then
1
|G|
· trace(ϕa,V ) =
ha, χV ∗ i = 0.
n
n
P
Thus, ϕa,V = 0, or
a(g) · g = 0 for any representation V of G. In particular, for the regular representation
R, ϕa,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.
λ=
13. Representations of symmetric groups
For each partition λ of n, we will construct a representation of Sn , called Vλ . We will show that Vλ is
irreducible for each partition λ, and Vλ is not isomorphic to Vµ if λ and µ are different partitions. Then since
the number of conjugacy classes of Sn is the same as the number of partitions, they are all the irreducible
representation of Sn .
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 Sn
P = {g ∈ Sn : g preserves each row} and
14
THANH VU
Q = {g ∈ Sn : g preserves each column}.
P
In the group algebra CSn , we introduce two elements corresponding to these subgroups, we set aλ = g∈P eg
P
and bλ = g∈Q sgn(g)eg . Finally we set cλ = aλ · bλ ∈ CSn , this is called the Young symmetrizer.
The following lemma is the basic tool need for representation of Sn .
Lemma 13.1. Let T and T 0 be tableaux on shapes l and l0 . Assume that l does not strictly dominate l0 . Then
exactly one of the following occurs:
(i) There are two distinct integers that occur in the same row of T 0 and the same column of T .
(ii) λ0 = λ and there is some p0 in Pλ0 and some q in Qλ such that p0 · T 0 = q · T .
Proof. Suppose (i) is false. The entries of the first row of T 0 must occur in different columns of T , so there is
a q1 ∈ Qλ so that these entries occur in the first row of q1 · T . The entries of the second row of T 0 occur in
different columns of T , so also of q1 · T , so there is a q2 in Qλ , not moving the entries equal to those in the
first row of T 0 , so that these entries all occur in the first two rows of q2 · q1 · T . Continueing in this way, we
get q1 , . . . , qk ∈ Qλ such that the entries in the first k rows of T 0 occurs in the first k rows of qk · · · q1 · T . In
particular, since T and qk · · · q1 · T have the same shape, it follows that λ01 + · · · + λ0k ≤ λ1 + · · · + λk . Since this
is true for all k, this means that l0 l.
Since we have assumed that l does not strictly dominate λ0 , we must have λ = λ0 . Taking k to be the number
of rows in l, and q = qk · · · q1 , we see that q · T and T 0 have the same entries in each row. This means that there
is a p0 in Pλ0 such that p0 · T 0 = q · T .
Define a linear ordering on the set of all tableaux with n boxes, by saying that T 0 > T if either 1) the shape
of T 0 is larger than the shape of T in the lexicographic ordering, or 2) T 0 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 0 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 ∈ Pλ , and q ∈ Qλ , p · T ≥ T and q · T ≤ T .
Corollary 13.2. If T and T 0 are standard tableaux with T 0 > T , then there is a pair of integers in the same
row of T 0 and the same column of T .
Proof. Since T 0 > T , the shape of T cannot dominate the shape of T 0 . If there is no such pair, we are in case
(ii) of the lemma: p0 · T 0 = q · T . Since T and T 0 are standard, we have qT ≤ T and p0 T 0 ≥ T 0 , this contradict
the assumption T 0 > T .
Let A = CSn . For a partition l, let Vλ = Acλ be the corresponding representation. Note that P ∩ Q = {1}
so
an
element of Sn can be written in at most one way as a product p · q, p ∈ P , q ∈ Q. Thus c is the sum
P
±eg , the sum over all g that can be written as p · q, with coefficient ± being sgn(q).
Lemma 13.3. (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.
P
Proof. Only the last assertion is not obvious. If
ng eg satisfies the condition in (3), then npgq = sgn(q)ng for
all g, p, q, in particular, npq = sgn(q)n1 . Thus, it suffices to verify that ng = 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 ng = −ng . If T 0 = 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 0 , 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 p1 ∈ P , and q10 ∈ Q0 = gQg −1 so that p1 T and q10 T 0 have the same first row, repeating on the
rest of the tableau, one gets p ∈ P and q 0 ∈ Q0 so that pT = q 0 T 0 . Then pT = q 0 gT , so p = q 0 g, and therefore
g = pq, where q = g −1 (q 0 )−1 g ∈ Q, as required.
Lemma 13.4. (1) If λ > µ lexicographically, then for all x ∈ A, aλ xbµ = 0. In particular, if λ > µ, then
cλ · cµ = 0.
(2) For all x ∈ A, cλ xcλ is a scalar multiple of cλ . In particular, cλ cλ = nλ cλ , for some nλ ∈ C.
Proof. For (1), we may take x = g ∈ Sn . Since gbµ g −1 is the element constructed from gT 0 , where T 0 is the
tableau used to construct bµ , it suffices to show that aλ bµ = 0. One verifies that λ > µ implies that there are
two integers in the same row of T and the same column of T 0 . If t is the transposition of these integers, then
aλ t = aλ , tbµ = −bµ , so aλ bµ = aλ ttbµ = −aλ bµ , as required. Part (2) follows from the previous lemma.
Lemma 13.5. (1) Each Vλ is an irreducible representation of Sn .
(2) If λ 6= µ, then Vλ and Vµ are not isomorphic.
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
15
Proof. For (1) note that cλ Vλ ⊂ Ccλ by lemma 13.4. If W ⊂ Vλ is a subrepresentation, then cλ W is either Ccλ
or 0. If the first is true, then Vλ = Acλ ⊂ W . Otherwise, W · W ⊂ A · cλ W = 0, but this implies W = 0. Indeed,
a projection from A onto W is given by right multiplication by an element ϕ ∈ A with ϕ = ϕ2 ∈ W · W = 0.
This argument also shows that cλ Vλ 6= 0.
For (2), we may assume λ > µ. Then cλ Vλ = Ccλ 6= 0, but cλ Vµ = 0, so thay cannot be isomorphic A−
modules.
Proposition 13.6. Vλ0 = Vλ ⊗ sgn.
as sgn(s)s. Then V σ = V ⊗ sgn. Moreover
σ(Acλ ) = σ(A)σ(aλ )σ(bλ ) = Abλ0 aλ0 ∼
= Aaλ0 bλ0 .
Proof. Define σ : k[Sn ] → k[Sn ] by σ(
P
as s) =
P
14. Representations of general linear groups
k
Let G = GL(C ) and C[G] denote the ring of regular functions on G. A representation ρ : G → GL(V ) is
polynomial if ρ is a regular function. The whole space C[G] has a natural structure of a representation if we
put Rg f (x) = f (xg). The space C[G]n of homogeneous polynomials of degree n is invariant. Thus there is a
decomposition C[G] = ⊕∞
0 C[G]n .
Let ρ : G → GL(V ) be a polynomial representation. For any ϕ ∈ V ∗ define a map ρ0ϕ : V → C[G] by the
formula
ρ0ϕ (v) = fv,ϕ , where fv,ϕ (s) = hϕ, ρs (v)i.
By definition, ρ0ϕ (ρg v) = Rg ρ0ϕ (v), so this map is an intertwiner. Therefore, every irreducible polynomial
representation of G is a subrepresentation in C[G].
Let E = Ck be the standard representation of G.
Lemma 14.1. For each n, the map π : (E ∗ )⊗n ⊗ E ⊗n → C[G]n given by ϕ ⊗ v → fv,ϕ is surjective.
Proof. Let e1 , . . . , ek be a basis in E and f1 , . . . , fk the dual basis in E ∗ . Then fj1 ⊗ · · · ⊗ fjn ⊗ ei1 ⊗ · · · ⊗ eik →
gi1 j1 . . . gin jn , where gij is a matrix entry of a matrix g in the basis e1 , . . . , ek . Thus the monomial basis of
C[G]n belongs to the image of π.
Therefore to classify polynomial irreducible representations of G we have to find all irreducible subrepresentations of E ⊗n . To study them, we use the general theory of dual pair.
Theorem 14.2. Let ρ : G → GL(V ), σ : H → GL(V ) be compatible representations of G and H over an
algebraically closed field k. Assume that EndG (V ) = σ(k(H)) and ρ is completely reducible. Then
V = ⊕m
1 Vi Wi
where Vi is an irreducible representation of G and Wi is an irreducible representation of H. Moreover, Vi is not
isomorphic to Vj if i 6= j and similarly, Wi is not isomorphic to Wj if i 6= j.
Proof. Since ρ is completely reducible, one can write
V = ⊕m
1 (Vi ⊗ Wi )
where the action of G is trivial on Wi . Then
EndG (V ) =
m
Y
Endk (Wi ).
1
Thus σ : k(H) → Endk (Wi ) is surjective, that implies that each Wi is irreducible representation of K and
Wi 6∼
= Wj if i 6= j.
Let ρ : Sn × GLk → GL(E ⊗n ) be a representation of Sn × GLk given by
ρs (v1 ⊗ vn ) = vs(1) ⊗ · · · ⊗ vs(n) and ρg (v1 ⊗ · · · ⊗ vn ) = gv1 ⊗ · · · ⊗ gvn .
Let Mk = EndC (E) be the matrix algebra, then EndC (E ⊗n ) = Mk⊗n . Therefore
EndSn (E ⊗n ) = (Mk⊗n )Sn = Symn (Mk ).
Moreover, by the formula
2n x1 . . . xn =
X
(−1)i2 +···+in (x1 + (−1)i2 x2 + · · · + (−1)in xn )n
ij ∈{0,1}
the span of g ⊗ · · · ⊗ g, g ∈ GLk in Symn (Mk ) is Symn (Mk ), since GLk is dense in Mk . This is by definition is
ρ(C(GLk )), therefore we have the situation of dual pair between Sn and GLk .
16
THANH VU
Theorem 14.3. Let Γn,k denote the set of all Young diagrams with n boxes such that the number of rows is
not bigger than k. Then
E ⊗n = ⊕λ∈Γn,k Vλ ⊗ Wλ ,
where Wλ is an irreducible representation of GLk . Moreover Wλ and Wµ are not isomorphic if λ 6= µ.
Proof. We know that
E ⊗n = ⊕λ∈Γ0 Vλ ⊗ Wλ ,
for some set Γ0 of Young diagrams with n boxes. It is left to show that Γ0 = Γn,k . By construction Γ0 consists
of all diagrams λ for which Wλ = cλ (E ⊗n ) 6= 0. Let λ be such a diagram, an consider the canonical standard
tableau on λ and consider cλ as defined by this tableau. Consider an element v = ei1 ⊗ · · · ⊗ ein of the standard
basis in E ⊗n . If λ has more than k rows, then one can find ej which appears twice in the same column. Then
bλ (v) = 0, and therefore cλ v = 0. Since this holds for any basis vector, we have cλ (E ⊗n ) = 0. On the other
hand, if λ has k or less row, one can check that
k
1
⊗ · · · ⊗ e⊗λ
) 6= 0.
cλ (e⊗λ
1
k
Therefore Γ0 = Γn,k .
15. Weight vectors
16. First secant varieties of Segre-Veronese varieties
My research is in algebraic geometry, commutative algebra. I use combinatorial methods in many of my
current research problems. One such example is the problem of classifying defective Segre-Veronese varieties.
With Brian Harbourne, we are running a working seminar on the tropical approach to secant dimensions and
Nagata’s conjecture. Currently, as I am teaching a topic course on secant varieties of Segre-Veronese varieties,
we focus on the case of Segre-Veronese embeddings of P m xP n by O(1,2). We have verified that many of them
are non-defective using the tropical approach. We think that a complete solution for this case should be done
shortly. Another example is the problem of finding the Betti tables of the Segre-Veronese varieties. There I focus
on the Ottaviani-Paoletti conjecture for Veronese embeddings, and the Betti tables for Veronese embeddings of
P 2 . Furthermore, I am also interested in problems related to Hilbert schemes, toric varieties, among others.
Given that, my research is tightly related to the theme of the program. Thus the program will be very beneficial
to me as it will allow me to learn more about combinatorial methods in algebraic geometry, and talk to experts
in the field to enhance my knowledge. That opportunity will help me expand my research interests. I also
believe that I can make good contribution to the program, given my background in the field.
17. Salmon’s problem
18. Secant varieties of P2 × Pn
Project 1: Defectiveness of Segre product Pa × Pb × Pc : Determine all triples (a, b, c) so that the Segre variety
P × Pb × Pc is defective.
During lectures, we will study two approaches to this problem. One is using the theory of inverse system
to reduce the problem to compute Hilbert function of certain ideal. The other approach is by induction and a
reduction to the case of Veronese varieties.
Project 2: Equations of secant varieties of Segre-Veronese varieties ν1,2 (Pa × Pb ).
During lectures, we will study the general approach by Claudiu Raicu, and a specific result of Cartwright,
Erman and Oeding.
Project 3: Secant varieties of the Segre product X = P3 × P3 × P3 . Note that the salmon’s problem is to
determine σ4 (X).
a
References
[1] H. Abo, and M. Brambilla, Secant varieties of Segre-Veronse varieties Pm × Pn embedded by O(1, 2)., Exper. Math. 18 (2009),
369–384.
[2] H. Abo, and M. Brambilla, New examples of defective secant varieties of Segre-Veronse varieties., Collect. Math. 63 (2012),
287–297.
[3] H. Abo, G. Ottaviani, and C. Peterson, Induction for secant varieties of Segre varieties., Trans. Amer. Math. Soc. 361 (2009),
767–792.
[4] E. Carlini, N. Grieve and L. Oeding, Four lectures on secant varieties. Connections between algebra, combinatorics and
geometry, 101–146. Springer Proc. Math. Stat. 76. Springer, New York 2014.
[5] D. Cartwright, D. Erman, and L. Oeding, Secant varieties of P2 × Pn embedded by O(1, 2)., J. Lond. Math. Soc. 85 (2012),
121–141.
[6] M. V. Catalisano, A.V. Geramita and A. Gimigliano, Ranks of tensors, secant varieties of Segre varieties and fat points.,
Linear Alg. Appl. 355 (2002), 263–285.
SECANT VARIETIES OF SEGRE-VERONESE VARIETIES
17
[7] M. V. Catalisano, A.V. Geramita and A. Gimigliano, Higher secant varieties of Segre-Veronese varieties., In Projective
Varieties with unexpected properties.
[8] M. V. Catalisano, A.V. Geramita and A. Gimigliano, Higher secant varieties of the Segre varieties P1 × · · · × P1 ., J. Pure
Appl. Algebra 201 (2005), 367–380.
[9] M. V. Catalisano, A.V. Geramita and A. Gimigliano, Secant varieties of P1 × · · · × P1 (n-times) are not defective for n ≥ 5.,
J. Alg. Geometry 20 (2011), 295–327.
[10] C. Raicu, Secant varieties of Segre-Veronese varieties. Algebra Number Theory 6 (2012), 1817–1868.
[11] C. Raicu, 3 × 3 minors of catalecticants. Math. Res. Lett. 20 (2013), 745–756.