ON THE DIMENSION OF THE LOCUS OF DETERMINANTAL
HYPERSURFACES
ZINOVY REICHSTEIN† AND ANGELO VISTOLI‡
Abstract. The characteristic polynomial PA (x0 , . . . , xr ) of an r-tuple A := (A1 , . . . , Ar )
of n × n-matrices is defined as
PA (x0 , . . . , xr ) := det(x0 I + x1 A1 + · · · + xr Ar ) .
We show that if r > 3 and A := (A1 , . . . , Ar ) is an r-tuple of n × n-matrices in general
position, then up to conjugacy, there are only finitely many r-tuples A0 := (A01 , . . . , A0r )
such that pA = pA0 . Equivalently, the locus of determinantal hypersurfaces of degree n
in Pr is irreducible of dimension (r − 1)n2 + 1.
1. Introduction
Let r, n > 2 be integers, and k be a base field of characteristic not dividing n. Given an
r-tuple A := (A1 , . . . , Ar ) ∈ Mrn of n × n-matrices, we define the characteristic polynomial
of A as
PA (x0 , . . . , xr ) := det(x0 I + x1 A1 + · · · + xr Ar ) .
The purpose of this paper is to answer the following question, due to B. Reichstein.
Question 1.1. For (A1 , . . . , Ar ) in general position in Mrn , are there finitely many or
infinitely many conjugacy classes of r-tuples A0 := (A01 , . . . , A0r ) such that pA = pA0 ?
To restate this question in geometric terms, consider the following diagram
(1.1)
Mrn
π
Qr,n = Mn /P GLn
P
P
/
(
DHypr,n 
/
Hypersurf r,n ,
where
r+n
• Hypersurf r,n ' P( n )−1 denotes the space of degree n hypersurfaces in Pr ,
• Qr,n := Mrn //PGLn = Spec k[Mrn ]PGLn denotes the categorical quotient space for
the conjugation action of PGLn on r-tuples of n × n-matrices, and
• π denotes the natural projection induced by the inclusion k[Mrn ]PGLn ,→ k[Mrn ].
• P takes an r-tuple A = (A1 , . . . , Ar ) of n × n matrices to the hypersurface in Pr
cut out by the homogeneous polynomial PA (x0 , . . . , xr ) of degree n. Hypersurfaces
of this form are called “determinantal”.
2000 Mathematics Subject Classification. 14M12, 15A22, 05A10.
Key words and phrases. Determinantal hypersurfaces, matrix invariants, q-binomial coefficients.
†
Supported in part by an NSERC Discovery grant.
‡
Supported in part by research funds from the Scuola Normale Superiore.
1
2
REICHSTEIN AND VISTOLI
• DHypr,n denotes the closure of the image of P in Hypersurf r,n This is the “locus
of determinantal hypersurfaces” of degree n in Pr .
Question 1.2. What is the dimenson of DHypr,n ?
Questions 1.1 and 1.2 are closely related. Indeed, Question 1.1 asks whether or not
fibers of P in general position are finite, or equivalently, whether or not
dim(DHypr,n ) = dim(Qr,n ) ,
where
dim(Qr,n ) = dim(Mrn ) − dim(PGLn ) = (r − 1)n2 + 1 .
Our main result answers Questions 1.1 and 1.2 for r > 3.
Theorem 1.3. Assume r > 3. Then the general fiber of the map P is finite. Equivalently,
dim(DHypr,n ) = (r − 1)n2 + 1, for any n > 2.
Several remarks are in order.
(1) Theorem 1.3 was previously known in the following special cases.
If r > n2 − 1, then Theorem 1.3 is an easy consequence of a classical theorem of
Frobenius [F1897] which asserts that the only linear transformations T of the matrix
algebra preserving the determinant function are of the form A → P XQ or A → P X t Q,
where X t denotes the transpose of X, and P and Q are fixed n × n matrices, such that
det(P ) det(Q) = 1. For a modern proof of Frobenius’ theorem, further references, and
generalizations, see [Wat87].
In the case, where n = r = 3, Theorem 1.3 is equivalent to the following assertion: a
general hypersurface of degree 3 in P3 is determinantal. Equivalently, the map P : M33 →
Hypersurf 3,3 ' P19 is dominant. This result goes back to (at least) H. Grassmann [G1855];
for a modern proof, see [Bou00, Corollary 6.4].
In the case, where r = 3 and n = 4, Theorem 1.3 is equivalent to the assertion of that
determinantal quartic hypersurfaces in P3 form a codimension 1 locus in Hypersurf 3,4 '
P34 . This is proved in [Dol12, Example 4.2.23].
(2) It follows from the above-mentioned theorem of Frobenius that for r > n2 −1 the general fiber of P consists of two points, corresponding to the similarity classes (A1 , . . . , Ar )
and (At1 , . . . , Atr ). Consequently, for r > n2 − 1, deg(P ) = 2. Here as usual, deg(P )
denotes the number of points in a general fiber of P . In the case, where r = n = 3,
deg(P ) = 72; see [Bou00, Corollary 6.4] or [Dol12, Theorem 9.3.6]. We do not know what
deg(P ) is in general; our proof of Theorem 1.3 sheds no light on this question.
(3) Theorem 1.3 fails for r = 2, as long as n > 3. Indeed, in the case
n+2
2
dim(Q2,n ) = n + 1 >
− 1 = dim(Hypersurf 2,n ),
2
so the fibers of P cannot be finite. In fact, this setting has been much studied, both
from the theoretical point of view and in connection to applications to control theory.
In particular, it is well known that the map P : Q2,n → Hypersurf 2,n is dominant, and
the points of the fiber of P over a general plane curve C of degreee n are in a natural
DETERMINENTAL HYPERSURFACES
3
n(n − 1)
on C. For details and
2
further references, see [CT79], [Vin86], [Bou00, Section 3], [Dol12, Section 4.1], [Ne11].
(4) On the other hand, Theorem 1.3 remains true for r = n = 2. Indeed, in this case
the k-algebra k[Q2,n ] = k[M22 ]PGLn is generated by five algebraically independent elements,
Tr(A1 ), Tr(A2 ), det(A1 ), det(A2 ) and Tr(A1 A2 ); see, [P67, Theorem 2.1], [H71, p. 20]
or [FHL81, Lemma 1(1)]. One easily checks that these five elements lie in the k-algebra
generated by the coefficients of det(x0 I + x1 A1 + x2 A2 ). We conclude that for r = n = 2
the map P : M22 //PGL2 → Hypersurf 2,2 ' P5 is, in fact, a birational isomorphism, i.e.,
deg(P ) = 1. If r, n > 2 but (n, r) 6= (2, 2), then (A1 , . . . , Ar ) and (At1 , . . . , Atr ) are not
conjugate, for (A1 , . . . , Ar ) ∈ Mrn in general position (see, e.g., [R93, Remark 1 on p. 73])
and hence, deg(P ) ≥ 2.
(5) The fact that P : Mrn → Hypersurf r,n is dominant if and only if r = 2 or r = n = 3
was known to L. E. Dickson; see [Di21].
bijective correspondance with line bundles of degree
2. The general strategy
The rest of this paper will be devoted to proving Theorem 1.3. We will begin by
reducing to the case, where r = 3; see Section 3.
Clearly dim(DHyp3,n ) 6 dim(Q3,n ) = 2n2 + 1, since the morphism P : Q3,n → DHyp3,n
is dominant, by definition. Our strategy for proving the opposite inequality will be to find
a triple of n × n matrices A = (A1 , A2 , A3 ) ∈ M3n with the following property: the tangent
space TA (FA ) to the fiber FA of P passing through A, is of dimension n2 −1. If we can find
an A with this property, then by the Fiber Dimension Theorem, dim(DHyp3,n )+(n2 −1) >
dim(M3n ) = 3n2 , and the proof of Theorem 1.3 will be complete.
To implement this strategy, we will exhibit a homogeneous system of linear equations
cutting out the tangent space TA (FA ) to FA at A inside the tangent space TA (M3n ) (which
we identify with M3n ) in Section 4. We will do this for A = (A1 , A2 , A3 ) ∈ M3n in general position; see Lemma 4.1(b). However, for a general triple A = (A1 , A2 , A3 ) ∈ Mn3
the resulting system of linear equations is rather complicated (in particular, it is badly
overdetermined), and we have not been able to compute the dimension of the solution
space in this generality. However, we will choose A = (A1 , A2 , A3 ) in a particular way
(see (5.1)), so that the space TA (FA ) carries a (Z/nZ)2 -grading. In other words, for this
choice of A, TA (FA ) remains invariant under a certain linear action of the finite abelian
group G := (Z/nZ)2 on Mn3 ; see Section 6. We will then decompose TA (FA ) as a direct
sum of n2 three-dimensional character spaces and verify that TA (FA ) has the desired dimension, n2 − 1, by finding solutions of our linear system in each character space. This
computation is carried out in Section 7. It relies on properties of q-binomial (and trinomial) coefficients, which are recalled in Section 5.
3. Reduction to the case, where r = 3
For notational convenience, let us denote the maps
P : Mrn → Hypersurf r,n and P : Qr,n → Hypersurf r,n
in diagram (1.1) by P r,n and P
r,n
, respectively (in this section only).
4
REICHSTEIN AND VISTOLI
r,n
Proposition 3.1. Assume r > 3. If the general fiber of P is finite, then the general
r+1,n
fiber of P
is also finite. In particular, in the course of proving Theorem 1.3 we may
assume without loss of generality that r = 3.
Let us say that an r-tuple A = (A1 , . . . , Ar ) of n × n-matrices is saturated if Ai and Aj
generate the full matrix algebra Mn (k) for every 1 6 1 < j 6 r.
Our proof of Proposition 3.1 will rely on the following lemma.
r,n
Lemma 3.2. Assume that r > 2 and the general fiber of P is finite. Then there exists
a dense open PGLn -invariant subset U of Mrn such that for A = (A1 , . . . , Ar ) ∈ U
(i)
(i)
(a) the fiber of P containing A is the union of the conjugacy classes of A(i) = (A1 , . . . , Ar ),
for i = 1, . . . , d. Here A(1) = A, and d is the degree of P .
(b) The r-tuple A(i) is saturated for every i = 1, . . . , d.
Proof. Let Z be the subvariety of Mrn consisting of r-tuples of n × n-matrices A =
(A1 , . . . , Ar ) which are not saturated. The lemma amounts to showing that P r,n does
r,n
not map Z dominantly onto DHypr,n . Since P r,n = P ◦ π, and P r,n is generically
finite-to-one, this is equivalent to showing that π does not map Z dominantly onto
Qr,n = Mrn /PGLn .
To prove the last assertion, it suffices to show that if C = (C1 , . . . , Cr ) is saturated, then
π(C) 6∈ π(Z). Indeed, Z and the PGLn -orbit of C are both closed and PGLn -invariant in
Mrn . Hence, they can be separated by PGLn -invariant polynomial functions on Mrn ; see,
e.g., [MFK94, p. 29, Corollary 1.2]. In other words, π(C) 6∈ π(Z), as desired.
♠
We are now ready to complete the proof of Proposition 3.1. Suppose deg(P
Let U ⊂ Mrn be as in Lemma 3.2 and
r,n
) = d.
A = (A1 , . . . , Ar ) ∈ (U × Mn ) ∩ (Mn ×U ) ⊂ Mrn .
r+1,n
We claim that the fiber of P
containing A has at most d2 points. Proposition 3.1 is
an immediate consequence of this claim.
To prove the claim, assume the contrary: there are d2 + 1 non-conjugate (r + 1)-tuples
(i)
(i)
A(i) = (A1 , . . . , Ar+1 ) such that the generalized characteristic polynomial PA(i) (t0 , . . . , tr+1 )
is the same as PA (t0 , . . . , tr+1 ) for each i = 1, . . . , d2 + 1.
Let C := (A1 , . . . , Ar ) and D := (A2 , . . . , Ar+1 ) be the r-tuples of matrices obtained
from A by dropping the last matrix and the first matrix, respectively. Similarly, set
(i)
(i)
(i)
(i)
C (i) := (A1 , . . . , Ar ) and D(i) := (A2 , . . . , Ar+1 ). By our choice of A, both C and D
lie in U . Note that the generalized characteristic polynomial PC is obtained from PA by
setting t1 = 0. Since PA = PA(i) , we see that
PC (t2 , . . . , tr+1 ) = PC (i) (t2 , . . . , tr+1 ).
Similarly PD (t1 , . . . , tr ) = PD(i) (t1 , . . . , tr ).
Since there are exactly d non-conjugate r-tuples of n × n matrices with the same char2
acteristic polynomial as C, at least d + 1 r-tuples among C (1) , . . . , C (d +1) lie in the same
conjugacy class. After relabeling, we may assume that C (1) , . . . , C (d+1) are pairwise conjugate. Similarly two of the r-tuples D(1) , . . . , D(d+1) have to be conjugate. Let us assume
that D(1) and D(2) are conjugate.
DETERMINENTAL HYPERSURFACES
5
In summary, we have found that two non-conjugate (r + 1)-tuples of n × n-matrices,
(1)
(1)
(2)
(2)
A = (A1 , . . . , Ar+1 ) and A(2) = (A1 , . . . , Ar+1 ), with the following properties:
(1)
(i) A(1) and A(2) are saturated,
(1)
(1)
(2)
(2)
(ii) (A1 , . . . , Ar ) and (A1 , . . . , Ar ) are conjugate, and
(1)
(1)
(2)
(2)
(iii) (A2 , . . . , Ar+1 ) and (A2 , . . . , Ar+1 ) are conjugate.
We will now show that (i), (ii) and (iii) imply that A(1) and A(2) are, in fact, conjugate.
This contradiction will complete the proof of the claim and thus of Proposition 3.1.
(1)
(1)
After replacing A(1) and A(2) by conjugate r-tuples, we may assume that (A1 , . . . , Ar ) =
(2)
(2)
(A1 , . . . , Ar ) = (A1 , . . . , Ar ). In other words, we want to show that if (A2 , . . . , Ar , X)
is conjugate to (A2 , . . . , Ar , Y ), for some n × n-matrices X and Y , then X = Y .
Indeed, suppose g(A2 , . . . , Ar , X)g −1 = (A2 , . . . , Ar , Y ), for some g ∈ PGLn . Then g
commutes with A2 , . . . , Ar . Now recall that we are assuming r ≥ 3, and by our construction A is saturated. In particular, A2 and A3 generate Mn as a k-algebra. Schur’s lemma
now tells us that g is a scalar matrix, i.e., g = 1 in PGLn . Thus Y = gXg −1 = X, as
desired.
♠
4. The tangent space to a fiber of P
Given an n × n matrix X, we will denote the classical adjoint of X by X ad . Recall that
X is, by definition, the n × n matrix whose (i, j)-component is (−1)i+j det(Xji ), where
Xji is the (n − 1) × (n − 1) matrix obtained from X by deleting row j and column i. If
X is invertible, then X ad = det(X)X −1 .
ad
Lemma 4.1. Let A = (A1 , . . . , Ar ) be an r-tuple of n × n-matrices and let FA ⊂ Mrn be
the fiber of P over PA . Then the tangent space TA (FA ) to FA at A is the linear subspace
of TA (Mrn ) ' Mrn consisting on the r-tuples (B1 , . . . , Br ) of n × n-matrices satisfying
(a) Tr((x0 I + x1 A1 + · · · + xr Ar )ad (x1 B1 + · · · + xr Br )) = 0.
(b) Suppose some matrix in the linear span of A1 , . . . , Ar has distinct eigenvalues. Then
the tangent space TA (FA ) admits the following alternative description: TA (FA ) is the linear
subspace of Mrn consisting on the r-tuples (B1 , . . . , Br ) which satisfy
Tr((x1 A1 + · · · + xr Ar )d (x1 B1 + · · · + xr Br )) = 0
for every d = 0, 1, . . . , n − 1.
Here we view (B1 , . . . , Br ) as an element of the tangent space to the affine space Mrn at
A. We identify this tangent space with Mrn .
Note also that the left hand side of the formulas in part (a) is a polynomial in x0 , x1 , . . . , xr .
We require all coefficients of this polynomial to vanish. This gives rise to system of linear
equations on B1 , . . . , Br , which cut out the tangent space to FA in Mrn . Similarly, in part
(b) we require that for every d = 0, 1, . . . , n − 1 the left hand side of the formula should
be identically zero as a polynomial in x1 , . . . , xr .
6
REICHSTEIN AND VISTOLI
r+n
Proof. (a) Observe that the image of the map P lies in the affice subspace A( n )−1 of
r+n
P( n )−1 = Hypersurf r,n consisting of hypersurfaces of the form
X
ai1 ,...,ir xi00 . . . xirr = 0 ,
i0 +···+ir =n
where ar,0,...,0 6= 0 (or equivalently, ar,0,...,0 = 1, after rescaling). Thus we may view P as a
r+n
polynomial map between the affine spaces Mrn and A( n )−1 . A vector B = (B1 , . . . , Br )
in the tangent space TA (Mrn ) (which we identify with Mrn ) lies in TA (FA ) if and only if
the directional derivative of P at A in the direction of B vanishes. Our computation of
this directional derivative, will rely on formula (4.2) below. We will now digress, briefly,
to derive this formula.
Let ∆Y = (∆yij ) be a n × n matrix; we think of the entries ∆yij of ∆Y as being
“small”. Then expanding det(I + ∆Y ), we see that
(4.1)
det(I + ∆Y ) = 1 + Tr(∆Y ) + (terms of degree > 2 in ∆yij ).
More generally, if Y = (yij ) is a fixed n × n-matrix, then we claim that
(4.2)
det(Y + ∆Y ) = det(Y ) + Tr(Y ad ∆Y ) + (terms of degree > 2 in ∆yij ).
Indeed, both sides are n × n-matrices, whose entries are polynomials in yij and ∆yij .
Hence, in order to prove formula (4.2) we may assume without loss of generality that Y
is non-singular. In this case,
det(Y + ∆Y ) = det(Y ) det(I + Y −1 ∆Y ) .
Applying formula (4.1) to the second factor, we obtain (4.2).
We are now ready to compute the directional derivative of P in the direction of
(B1 , . . . , Br ). Setting Y := x0 I + x1 A1 + · · · + xr Ar and ∆Y := (x1 B1 + · · · + xr Br )h, and
applying (4.2), we see that
P (A1 + hB1 , . . . Ar + hBr ) = det(Y + ∆Y ) = det(Y + ∆Y ) = det(Y ) + Tr(Y ad ∆Y )h + O(h2 )
= P (A1 , . . . , Ar ) + Tr((x0 I + x1 A1 + · · · + xr Ar )ad (x1 B1 + · · · + xr Br ))h + O(h2 ) .
This shows that the directional derivative of P at A in the direction of B is
Tr((x0 I + x1 A1 + · · · + xr Ar )ad (x1 B1 + · · · + xr Br )),
and part (a) follows. (Note that in the last computation h → 0 but x0 , x1 , . . . , xn remain
constant throughout.)
(b) Let A be an n × n matrix with distinct eigenvalues, over a field K. We claim that
B ∈ Mn satisfies
(i) Tr((x0 I + A)ad B) = 0 for every x0
if and only if B satisfies
(ii) Tr(Ad B) = 0 for every d = 0, . . . , n − 1.
Once this claim is established, we can deduce part (b) from part (a) by setting A :=
x1 A1 +· · ·+xr Ar and B := x1 B1 +· · ·+xr Br and working over the field K = k(x1 , . . . , xr ).
To prove the claim, we may pass to the algebraic closure of K. By our assumption A
has distinct eigenvalues, and hence, is diagonalizable. We may thus assume without loss
DETERMINENTAL HYPERSURFACES
7
of generality that A is the diagonal matrix diag(λ1 , . . . , λn ), where λ1 , . . . , λn are distinct
elements of K. Then
Π(t)
Π(t)
(tI + A)ad = diag(
,...,
),
t + λ1
t + λn
where Π(t) = (t + λ1 )(t + λ2 ) . . . (t + λn ) = det(tI + A) and each diagonal enty
a polynomial of degree n − 1 in t. Condition (i) now translates to
n
X
bii
i=1
Π(t)
is
t + λi
Π(t)
= 0,
t + λi
where b11 , . . . , bnn are the diagonal entries of B. Setting t = −λi , for i = 1, . . . , n, we
obtain b11 = b22 = · · · = bnn = 0. On the other hand, condition (ii) translates to
n
X
λdi bii = 0,
i=1
for each d = 0, 1, . . . , n − 1, which we view as a homogeneous system of n linear equations
in n unknowns b11 , . . . , bnn . The matrix of this system is the Vandermonde matrix


1
1
...
1
 λ1
λ2 . . . λn 
 .
.
..
..
.. 
 ..
.
. 
.
λn−1
λ2n−1 . . .
1
λnn−1
Since λ1 , . . . , λn are distinct, this Vandermonde matrix is non-singular, and the above
system has only the trivial solution, b11 = b22 = · · · = bnn = 0.
In summary, for A = diag(λ1 , . . . , λn ) both (i) and (ii) are equivalent to b11 = b22 =
· · · = bnn = 0. Hence, (i) and (ii) are equivalent to each other. This completes the proof
of the claim and thus of Lemma 4.1(b).
♠
5. Skew-commuting matrices and q-binomial coefficients
Recall that we are working over a base field k of characteristic not dividing n. For
the sake of proving Theorem 1.3, we may assume without loss of generality that k is
algebraically closed. In particular, we may assume that k contains a primitive nth root
of unity, which we will denote by q. For the rest of this paper we will set




1 0 0 ...
0
0 1 0 ... 0
0 q 0 . . .


0 
 , A2 := 0 0 1 . . . 0 , and A3 := A1 A2 .
(5.1) A1 := 
. . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . .
n−1
0 0 0 ... q
1 0 0 ... 1
It is easy to see that
A2 A1 = qA1 A2 ,
and An1 = An2 = I ,
where I denotes that n × n-identity matrix. Hence, cojugation by A1 commutes with
conjugation by A2 ; we will denote these commuting linear operators by ConjA1 and
8
REICHSTEIN AND VISTOLI
ConjA2 : Mn → Mn , respectively. They generate a subgroup of GL(Mn ) isomorphic to
(Z/nZ)2 . One readily checks that
ConjA1 (Ae11 Ae22 ) = q −e2 Ae11 Ae22 and ConjA2 (Ae11 Ae22 ) = q e1 Ae11 Ae22 .
In particular,
(5.2)
(
n, if e1 ≡ e2 ≡ 0
Tr(Ae11 Ae22 ) =
0, otherwise.
(mod n), and
Letting e1 and e2 range over Z/nZ, we see that each of the n2 one-dimensional subspaces
Spank (Ae11 Ae22 ) is a character space for the abelian group
hConjA1 , ConjA2 i ' (Z/nZ)2 .
Since these spaces have distinct associated characters, the matrices Ae11 Ae22 form a k-basis
of Mn , as e1 and e2 range over Z/nZ. In the sequel it will often be more convenient for us
to work in this basis, than in the standard basis of Mn , consisting of elementary matrices.
We now recall that the q-factorial [d]q ! of an integer d > 0 is given by
[d]q ! := [1]q [2]q . . . [d]q ,
1 − qa
= 1 + q + · · · + q a−1 . In particular, [0]q ! = 1. (Recall that we are
1−q
assuming that n > 2 throughout, and thus q 6= 1.) If a and b are non-negative integers
and a + b = d 6 n − 1, then
d
[d]q !
.
(5.3)
:=
a, b q
[a]q ![b]q !
where [a]q :=
is called a q-binomial coefficient. If a < 0 or b < 0, we set
d
:= 0 .
a, b q
Similarly, if a + b + c = d 6 n − 1, then

[d]q !


, if a, b, c > 0, and

[a]q ! [b]q ! [c]q !
d
(5.4)
:=

a, b, c q


0, otherwise.
is called a q-trinomial coefficient. This terminology is justified by parts (a) and (b) of the
following lemma. Part (c) will play an important role in the sequel.
Lemma 5.1. Assume d = 0, . . . , n − 1.
(a) Let X and Y be matrices such that XY = qY X. Then
X d d
(X + Y ) =
X aY b .
a,
b
q
a+b=d
DETERMINENTAL HYPERSURFACES
9
(b) Let A1 and A2 be as in (5.1). Then
d
(x1 A1 + x2 A2 + x3 A1 A2 ) =
X
q
c(c−1)
2
a+b+c=d
d
a, b, c
b+c
xa1 xb2 xc3 Aa+c
1 A2 .
q
(c) For any e1 , e2 ∈ Z/nZ,
d
Tr((x1 A1 + x2 A2 + x3 A1 A2 )
Ae11 Ae22 )
=n
X
q
e1 (b+c)+
a,b,c
c(c−1)
2
d
a, b, c
xa1 xb2 xc3 ,
q
where the sum ranges over triples of non-negative integers (a, b, c), subject to the following
conditions: a + b + c = d, a + c + e1 ≡ 0 (mod n), and b + c + e2 ≡ 0 (mod n).
Proof. The binomial formula in part (a) was proved by M. P. Schützenberger [Sch53]; for
a detailed discussion of this formula and further references, see [HMS04].
(b) We apply part (a) twice. First we set X = x1 A1 + x3 A1 A2 and Y := x2 A2 to obtain
X d
d
(5.5)
(x1 A1 + x2 A2 + x3 A1 A2 ) =
(x1 A1 + x3 A1 A2 )i xj2 Aj2 .
i,
j
q
i+j=d
Next we apply part (a) with X := x1 A1 and Y := x3 A1 A2 :
X i i
(5.6)
(x1 A1 + x3 A1 A2 ) =
xa1 xc3 Aa1 (A1 A2 )c .
a, c q
a+c=i
Substituting (5.6) into (5.5), setting i := a + c and b := j, and using the identities
d
d
i
(5.7)
=
a, b, c q
i, b q a, c q
and
(5.8)
(A1 A2 )c = q
c(c−1)
2
Ac1 Ac2 ,
we obtain the formula in part (b). Note that (5.7) is an immediate consequence of the
definitions (5.3) and (5.4), and (5.8) follows from A2 A1 = qA1 A2 .
e1
To deduce part (c) from part (b), multiply both sides of (b) by Ae11 Ae22 , rewrite Ab+c
2 A1
as q e1 (b+c) Ae11 Ab+c
2 , and take the trace on both sides. The desired equality now follows
from (5.2).
♠
For future reference we record a simple identity involving q-trinomial coefficients.
Lemma 5.2. Suppose α, β, and γ are integers, 0 6 α, β, γ 6 n−1 and 1 6 α+β +γ 6 n.
Set d := α + β + γ − 1. Then
d
d
d
(
:
:
) = (1 − q α : 1 − q β : 1 − q γ )
α − 1, β, γ q
α, β − 1, γ q
α, β, γ − 1 q
as points in the projective plane P2 .
10
REICHSTEIN AND VISTOLI
Proof. If α, β, γ > 0, the lemma is obtained by multiplying each of the numbers
d
d
d
,
, and
α − 1, β, γ q
α, β − 1, γ q
α, β, γ − 1 q
[α]q ! [β]q ! [γ]q !
∈ k. If one of the integers α, β, γ is 0, say,
[d]q !
d
= 1 − qα = 0 ,
α − 1, β, γ q
♠
by the non-zero scalar (1 − q)
α = 0, then
and the lemma follows.
6. A grading of the tangent space
Let A1 , A2 and A3 = A1 A2 be as in (5.1). Let V ⊂ M3n be the tangent space TA (FA ),
where FA is the fiber of the map P : M3n → Hypersurf 3,n containing A. Since A1 has
distinct eigenvalues, Lemma 4.1 tells us that V ⊂ M3n consists of triples (B1 , B2 , B3 )
satisfying
Tr((x1 A1 + x2 A2 + x3 A1 A2 )d (x1 B1 + x2 B2 + x3 B3 )) = 0
for d = 0, 1, . . . , n − 1. Here the left hand side is required to be zero as a polynomial in
x1 , x2 , x3 , for every d = 0, 1, . . . , n − 1.
Following the strategy outlined in Section 2, in order to complete the proof of Theorem 1.3 it suffices to show that dim(V ) = n2 − 1.
Lemma 6.1. V is invariant under the linear action of the finite abelian group (Z/nZ)2 =
hτ, σi on M3n given by
(6.1)
σ : (B1 , B2 , B3 ) 7→ (ConjA1 (B1 ), q ConjA1 (B2 ), q ConjA1 (B3 ))
(6.2)
τ : (B1 , B2 , B3 ) 7→ (q −1 ConjA2 (B1 ), ConjA2 (B2 ), q −1 ConjA2 (B3 )) .
Proof. Suppose (B1 , B2 , B3 ) ∈ V , i.e.,
fB1 ,B2 ,B3 ,d (x1 , x2 , x3 ) := Tr((x1 A1 + x2 A2 + x3 A1 A2 )d (x1 B1 + x2 B2 + x3 B3 )) = 0
for every d = 0, . . . , n − 1. Here fB1 ,B2 ,B3 ,d is a polynomial in x1 , x2 , x3 with coefficients in
k, and fB1 ,B2 ,B3 ,d (x1 , x2 , x3 ) = 0 means that fB1 ,B2 ,B3 ,d is the zero polynomial, i.e., every
coefficient vanishes. Let
(C1 , C2 , C3 ) := σ(B1 , B2 , B3 ) = (ConjA1 (B1 ), q ConjA1 (B2 ), q ConjA1 (B3 )) ,
as above. To prove that V is invariant under σ, we need to show that (C1 , C2 , C3 ) ∈ V ,
i.e., fC1 ,C2 ,C3 ,d is identically 0 for every d = 0, 1, . . . , n − 1. Keeping in mind that
A1 := Conj A1 (A1 ),
A2 := q ConjA1 (A2 ),
and A1 A2 := q ConjA1 (A1 A2 ),
we see that
0 = fB1 ,B2 ,B3 ,d (x1 , x2 , x3 ) = Tr(ConjA1 ((x1 A1 + x2 A2 + x3 A1 A2 )d (x1 B1 + x2 B2 + x3 B3 ))
= Tr((x1 A1 + x2 q −1 A2 + x3 q −1 A1 A2 )d (x1 C1 + x2 q −1 C2 + x3 q −1 C3 ))
= fC1 ,C2 ,C3 ,d (x1 , q −1 x2 , q −1 x3 ).
DETERMINENTAL HYPERSURFACES
11
This shows that fC1 ,C2 ,C3 ,d (x1 , q −1 x2 , q −1 x3 ) is identically zero as a polynomial in x1 , x2 , x3 .
Hence, so is fC1 ,C2 ,C3 ,d (x1 , x2 , x3 ), as desired.
A similar argument shows that V is invariant under τ . (Here we conjugate by A2 ,
rather than A1 .) This completes the proof of Lemma 6.1.
♠
Since we are working over an algebraically closed base field k of characteristic not
dividing n, Lemma 6.1 tells us that V is a direct sum of character spaces for the action
of (Z/nZ)2 on M3n . There are n2 character spaces, each of dimension 3 (one for each
character of (Z/nZ)2 ). They are defined as follows
We1 ,e2 := {(t1 Ae11 +1 Ae22 , t2 Ae11 Ae22 +1 , t3 Ae11 +1 Ae22 +1 ) | t1 , t2 , t3 ∈ k},
where (e1 , e2 ) ∈ (Z/nZ)2 . Here σ multiplies every vector in We1 ,e2 by q −e2 and τ by q e1 .
In other words, (Z/nZ)2 acts on We1 ,e2 by the character
χ : σ a τ b 7→ q −e2 a+e1 b .
In summary, V =
Ln−1
e1 ,e2 =0
Ve1 ,e2 , where
Ve1 ,e2 := V ∩ We1 ,e2 .
Recall that our goal is to show that dim(V ) = n2 −1. Thus in order to prove Theorem 1.3,
it suffices to establish the following proposition.
Proposition 6.2. (a) V0,0 = (0).
(b) dim(Ve1 ,e2 ) = 1 for any (0, 0) 6= (e1 , e2 ) ∈ (Z/nZ)2 .
Proposition 6.2 will be proved in the next section.
Remark 6.3. If X and Y are n × n-matrices, then clearly Tr(X d [X, Y ]) = 0 for every
d > 0. Setting X = x1 A1 + x2 A2 + x3 A1 A2 , Y = Ae11 Ae22 , and thus
[X, Y ] = x1 (1 − q e2 )Ae11 +1 A2 + x2 (q e1 − 1)Ae11 Ae22 +1 + x3 (q e1 − q e2 )Ae11 +1 Ae22 +1 ,
we see that the triple
(B1 , B2 , B3 ) = ((1 − q e2 )Ae11 +1 Ae22 , (q e1 − 1)Ae11 Ae22 +1 , (q e1 − q e2 )Ae11 +1 Ae22 +1 )
lies in Ve1 ,e2 . Here (B1 , B2 , B3 ) = (0, 0, 0) if (e1 , e2 ) = (0, 0) in (Z/nZ)2 and (B1 , B2 , B3 ) 6=
(0, 0, 0) otherwise. Proposition 6.2 tells us that, in fact, (B1 , B2 , B3 ) spans Ve1 ,e2 for every
(e1 , e2 ) ∈ (Z/nZ)2 .
7. Conclusion of the proof of Theorem 1.3
It remains to prove Proposition 6.2. Given t1 , t2 , t3 ∈ k, recall that an element
w := (t1 Ae11 +1 Ae22 , t2 Ae11 Ae22 +1 , t3 Ae11 +1 Ae22 +1 )
of We1 ,e2 lies in Ve1 ,e2 if and only if
Tr((x1 A1 + x2 A2 + x3 A1 A2 )d (t1 x1 Ae11 +1 Ae22 + t2 x2 Ae11 Ae22 +1 + t3 x3 Ae11 +1 Ae22 +1 ))
12
REICHSTEIN AND VISTOLI
is identically 0 as a polynomial in x1 , x2 , x3 , for every d = 0, . . . , n − 1. Rewriting this
polynomial as
t1 x1 Tr((x1 A1 + x2 A2 + x3 A1 A2 )d Ae11 +1 Ae22 )
+t2 x2 Tr((x1 A1 + x2 A2 + x3 A1 A2 )d Ae11 Ae22 +1 )
+t3 x3 Tr((x1 A1 + x2 A2 + x3 A1 A2 )d Ae11 +1 Ae22 +1 )
and applying Lemma 5.1(c) to each term, we obtain
X
t1
nq
(e1 +1)(b+c)+
c(c−1)
2
d
a, b, c
(a,b,c)
(7.1)
c0 (c0 −1)
2
00 00
00 +c00 )+ c (c −1)
2
0
X
+t2
0
nq e1 (b +c )+
d
a0 , b0 , c0
(a0 ,b0 ,c0 )
+t3
X
nq (e1 +1)(b
(a00 ,b00 ,c00 )
b c
xa+1
1 x2 x3
q
0
0
0
xa1 +1 xb2 xc3
q
d
a00 b00 c00 +1
x
x x
= 0,
a00 , b00 , c00 1 2 3
where the sums are takes over triples of non-negative integers (a, b, c), (a0 , b0 , c0 ) and
(a00 , b00 , c00 ) satisfying
a00 + b00 + c00 = d
a0 + b 0 + c 0 = d
a+b+c=d
0
0
a00 + c00 + e1 + 1 ≡ 0 (mod n)
a + c + e1 + 1 ≡ 0 (mod n) a + c + e1 ≡ 0 (mod n)
b0 + c0 + e2 + 1 ≡ 0 (mod n), b00 + c00 + e2 + 1 ≡ 0 (mod n).
b + c + e2 ≡ 0 (mod n),
The expression on the left hand side of (7.1) is a homogeneous polynomial of degree d + 1.
e2
e1 e2 +1
Our element w = (t1 Ae+1
, t3 Ae11 +1 Ae22 +1 ) of We1 ,e2 lies in Ve1 ,e2 if and only
1 A2 , t2 A1 A2
if this polynomial is identically zero.
To make the conditions the vanishing of this polynomial imposes on t1 , t2 , t3 more
explicit, let us examine the coefficient of xα1 xβ2 xγ3 (with d + 1 = α + β + γ). This coefficient
is zero unless α, β and γ are chosen so that
(7.2)
α+β+γ 6n
α + γ + e1 ≡ 0 (mod n)
β + γ + e2 ≡ 0 (mod n).
On the other hand, if α, β and γ satisfy conditions (7.2), then setting
d := α + β + γ − 1
a = α − 1, b = β, c = γ
a0 = α,
a00 = α,
b0 = β − 1,
b00 = β,
c=γ
c00 = γ − 1,
DETERMINENTAL HYPERSURFACES
13
we see that the coefficient of xα1 xβ2 xγ3 is
γ(γ−1)
γ(γ−1)
d
d
(e1 +1)(β+γ)+ 2
e1 (β−1+γ)+ 2
t1 nq
+ t2 nq
α − 1, β, γ q
α, β − 1, γ q
(γ−2)(γ−1)
d
2
+t3 nq (e1 +1)(β+γ−1)+
.
α, β, γ − 1 q
γ(γ−1)
Equating this coefficient to 0 and dividing through by nq e1 (β+γ)+ 2 , we obtain
d
d
d
β−e1
−e1
β+γ
=0
+ t3 q
+ t2 q
(7.3)
t1 q
α, β − 1, γ q
α − 1, β, γ q
α, β, γ − 1 q
e2
e1 e2 +1
In summary, w = (t1 Ae+1
, t3 Ae11 +1 Ae22 +1 ) lies in Ve1 ,e2 if and only if (7.3)
1 A2 , t2 A1 A2
holds for every α, β, γ satisfying conditions (7.2).
Proof of Proposition 6.2(a). Our goal is to show that w = (t1 A1 , t2 A2 , t3 A1 A2 ) lies in
V0,0 if and only if t1 = t2 = t3 = 0. Note that here e1 = e2 = 0, and (α, β, γ) =
(n, 0, 0), (0, n, 0), (0, 0, n) satisfy conditions (7.2). Substituting these values into (7.3),
d
and remembering that a,b,c = 0 whenever a, b or c is < 0, we obtain
n−1
t1
= 0,
n − 1, 0, 0 q
or equivalently, t1 = 0. Similarly, setting (α, β, γ) = (0, n, 0) yields t2 = 0, and setting
(α, β, γ) = (0, 0, n) yields t3 = 0. This proves part (a).
♠
Proof of Proposition 6.2(b). Here (e1 , e2 ) 6= (0, 0), and we can use Lemma 5.2 to simplify
formula (7.3) as follows
t1 q β+γ (1 − q α ) + t2 q −e1 (1 − q β ) + t3 q β−e1 (1 − q γ ) = 0 .
Using (7.2), we can rewrite this in a more symmetric way, as
(7.4)
t1 (q −e2 − q d+1 ) + t2 (q −e1 − q d+1 ) + t3 (q d+1 − q −e1 −e2 ) = 0 ,
where d + 1 = α + β + γ, as before.
Claim. Suppose e1 , e2 = 0, . . . , n − 1 and (e1 , e2 ) 6= (0, 0). Then there exist two triples
of non-negative integers, (α1 , β1 , γ1 ) and (α2 , β2 , γ2 ) satisfying conditions (7.2) such that
d1 6≡ d2 (mod n). Here d1 = α1 + β1 + γ1 − 1 and d2 = α2 + β2 + γ2 − 1.
We will now deduce Proposition 6.2(b) from this claim. The proof of the claim will be
deferred to the end of this section. Assuming the claim is established, formula (7.4) tells
us that if (t1 Ae11 +1 Ae22 , t2 Ae11 Ae22 +1 , t3 Ae11 +1 Ae22 +1 ) lies in Ve1 ,e2 , then t1 , t2 and t3 satisfy the
linear equations
t1 (q −e2 − q d1 +1 ) + t2 (q −e1 − q d1 +1 ) + t3 (q d1 +1 − q −e1 −e2 ) = 0,
(7.5)
t1 (q −e2 − q d2 +1 ) + t2 (q −e1 − q d2 +1 ) + t3 (q d2 +1 − q −e1 −e2 ) = 0.
14
REICHSTEIN AND VISTOLI
The matrix of this system
e
q 2 − q d1 +1 q −e1 − q d1 +1 q d1 +1 − q −e1 −e2
q e2 − q d2 +1 q −e1 − q d2 +1 q d2 +1 − q −e1 −e2
is easily seen to have rank 2. Indeed, the determinants of the 2 × 2 minors are
±(q d1 +1 − q d2 +1 )(q −e2 − q −e1 ),
±(q d1 +1 − q d2 +1 )(q −e1 −e2 − q −e1 ),
and
±(q d1 +1 − q d2 +1 )(q −e1 −e2 − q −e1 ) .
Since q d1 +1 6= q d2 +1 , all three of these determinants can only be zero if q −e1 = q −e2 =
q −e1 −e2 or equivalently, e1 ≡ e2 ≡ e1 + e2 (mod n), i.e., (e1 , e2 ) = (0, 0) (mod n), contradicting our assumption that (e1 , e2 ) 6= (0, 0). We conclude that the solution space to
system (7.5) is of dimension ≤ 1 and consequently, dim(Ve1 ,e2 ) 6 1 On the other hand, by
Remark 6.3. dim(Ve1 ,e2 ) > 1. This shows that dim(Ve1 ,e2 ) = 1, thus completing the proof
of Proposition 6.2(b).
We now turn to the proof of the claim. The statement of the claim is clearly symmetric
with respect to e1 and e2 . That is, if the triples
(α1 , β1 , γ1 ) and (α2 , β2 , γ2 )
satisfy the claim for (e1 , e2 ), then the triples (β1 , α1 , γ1 ), (β2 , α2 , γ2 ) will satisfy the claim
for (e2 , e1 ). Thus for the purpose of proving this claim, we may assume without loss of
generality that 0 6 e2 6 e1 6 n − 1.
Case 1: e2 > 1. Here the triples
(α1 , β1 , γ1 ) = (0, e1 − e2 , n − e1 ) and (α, β, γ) = (1, e1 − e2 + 1, n − e1 − 1)
satisfy conditions (7.2) and yield distinct sums d1 + 1 = α1 + β1 + γ1 = n − e2 and
d2 + 1 = α2 + β2 + γ2 = n − e2 + 1. Note that d2 + 1 6 n, because we are assuming that
e2 > 1.
Case 2: e2 = 0 but 1 6 e1 6 n − 1. Set (α1 , β1 , γ1 ) = (0, e1 , n − e1 ), as in Case 1, and
(α2 , β2 , γ2 ) = (n − e1 , 0, 0). Then d1 + 1 = n and d2 = n − e1 are, once again, distinct
modulo n. This completes the proof of the claim and hence, of Proposition 6.2 and of
Theorem 1.3
♠
Acknowledgments. We are grateful to Boris Reichstein for bringing Question 1.1 to
our attention and for helpful discussions.
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(Reichstein) Department of Mathematics, University of British Columbia, Vancouver,
B.C., Canada V6T 1Z2
E-mail address: reichst@math.ubc.ca
(Vistoli) Scuola Normale Superiore, Piazza dei Cavalieri 7, 56126 Pisa, Italy
E-mail address: angelo.vistoli@sns.it