Beamer slides - University of Illinois at Urbana-Champaign

Apolarity, steampunk canonical forms and the
obvious inner product
Bruce Reznick
University of Illinois at Urbana-Champaign
Geometric Commutative Algebra and Applications
2011 Spring Central Section Meeting of AMS
Iowa City, March 19, 2011
Algebra, Geometry and Combinatorics Seminar
UIUC, April 20, 2011
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

1. Introduction
Webpage for this talk: www.math.uiuc.edu/∼reznick/iowa.html
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

1. Introduction
Webpage for this talk: www.math.uiuc.edu/∼reznick/iowa.html
What is steampunk? It is a style based on combining 19th century
Victorian culture with a few elements of modern life, such as
computers.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

1. Introduction
Webpage for this talk: www.math.uiuc.edu/∼reznick/iowa.html
What is steampunk? It is a style based on combining 19th century
Victorian culture with a few elements of modern life, such as
computers.
What are canonical forms? Before I give a definition, here’s the
Basic Example for this talk.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

1. Introduction
Webpage for this talk: www.math.uiuc.edu/∼reznick/iowa.html
What is steampunk? It is a style based on combining 19th century
Victorian culture with a few elements of modern life, such as
computers.
What are canonical forms? Before I give a definition, here’s the
Basic Example for this talk.
Any binary quadratic form over C can be written as
There also exist αj ∈ C so that
at least unless a = 0 and b �= 0.
p(x, y) = ax 2 + 2bxy + cy 2 . (1)
p(x, y) = (α1x + α2y) 2 + (α3y) 2 , (2)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

1. Introduction
Webpage for this talk: www.math.uiuc.edu/∼reznick/iowa.html
What is steampunk? It is a style based on combining 19th century
Victorian culture with a few elements of modern life, such as
computers.
What are canonical forms? Before I give a definition, here’s the
Basic Example for this talk.
Any binary quadratic form over C can be written as
There also exist αj ∈ C so that
p(x, y) = ax 2 + 2bxy + cy 2 . (1)
p(x, y) = (α1x + α2y) 2 + (α3y) 2 , (2)
at least unless a = 0 and b �= 0.
That is, a “general” binary quadratic form can be written with the
square completed.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

1. Introduction
Webpage for this talk: www.math.uiuc.edu/∼reznick/iowa.html
What is steampunk? It is a style based on combining 19th century
Victorian culture with a few elements of modern life, such as
computers.
What are canonical forms? Before I give a definition, here’s the
Basic Example for this talk.
Any binary quadratic form over C can be written as
There also exist αj ∈ C so that
p(x, y) = ax 2 + 2bxy + cy 2 . (1)
p(x, y) = (α1x + α2y) 2 + (α3y) 2 , (2)
at least unless a = 0 and b �= 0.
That is, a “general” binary quadratic form can be written with the
square completed.
You don’t need a proof of this, but I’ll give you several anyway.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Note also that there are three parameters in each of these
representations.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Note also that there are three parameters in each of these
representations.
This equality is necessary, but not sufficient. Obviously,
� 3
i=1 (αix) 2 won’t work. There are three parameters in
(α1x + α2y) 2 + (iα1x + α3y) 2 , (3)
but the coefficient of x 2 is 0, so the sum (3) must have a factor of
y, which a general binary quadratic form does not. This example
may look silly, but it isn’t.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Note also that there are three parameters in each of these
representations.
This equality is necessary, but not sufficient. Obviously,
� 3
i=1 (αix) 2 won’t work. There are three parameters in
(α1x + α2y) 2 + (iα1x + α3y) 2 , (3)
but the coefficient of x 2 is 0, so the sum (3) must have a factor of
y, which a general binary quadratic form does not. This example
may look silly, but it isn’t.
What is the as-yet-nonexistent subject of steampunk canonical
forms? First approximation: 19th century algebra plus the concept
of vector spaces plus the belief that there is still something of
interest in the algebraic geometry of binary forms.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

I freely acknowledge that any result not specifically credited to
someone else may nevetheless be old. (References from the
audience welcome.)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

I freely acknowledge that any result not specifically credited to
someone else may nevetheless be old. (References from the
audience welcome.)
There is very serious mathematics in this area being done today.
There is a strong connection between this talk and the wonderful
1993 paper of Richard Ehrenborg and Gian-Carlo Rota connecting
apolarity and canonical forms using matroids. There is also recent
work on these questions from a more sophisticated point of view by
Tony Geramita, Tony Iarrobino, J. M. Landsberg, Zach Teitler,
Enrico Carlini, Giorgio Ottaviani and others I might have missed.
And of course the deep theorem of Alexander-Hirschowitz hovers
over all.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

I freely acknowledge that any result not specifically credited to
someone else may nevetheless be old. (References from the
audience welcome.)
There is very serious mathematics in this area being done today.
There is a strong connection between this talk and the wonderful
1993 paper of Richard Ehrenborg and Gian-Carlo Rota connecting
apolarity and canonical forms using matroids. There is also recent
work on these questions from a more sophisticated point of view by
Tony Geramita, Tony Iarrobino, J. M. Landsberg, Zach Teitler,
Enrico Carlini, Giorgio Ottaviani and others I might have missed.
And of course the deep theorem of Alexander-Hirschowitz hovers
over all.
This talk has been constructed to avoid profundity, especially in
the proofs.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here are two examples of what I’ll be talking about and proving
(the proofs are quite simple):
Theorem
A general binary sextic form is the sum of the cube of a quadratic
form and the square of a cubic form.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here are two examples of what I’ll be talking about and proving
(the proofs are quite simple):
Theorem
A general binary sextic form is the sum of the cube of a quadratic
form and the square of a cubic form.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here are two examples of what I’ll be talking about and proving
(the proofs are quite simple):
Theorem
A general binary sextic form is the sum of the cube of a quadratic
form and the square of a cubic form.
Theorem
A general cubic form p(x1, . . . , xn) has a unique representation in
the form
p(x1, . . . , xn) = �
(α i,j
1≤i≤j≤n
i xi + · · · + α i,j
j xj) 3 . (4)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here are two examples of what I’ll be talking about and proving
(the proofs are quite simple):
Theorem
A general binary sextic form is the sum of the cube of a quadratic
form and the square of a cubic form.
Theorem
A general cubic form p(x1, . . . , xn) has a unique representation in
the form
p(x1, . . . , xn) = �
(α i,j
1≤i≤j≤n
i xi + · · · + α i,j
j xj) 3 . (4)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here are two examples of what I’ll be talking about and proving
(the proofs are quite simple):
Theorem
A general binary sextic form is the sum of the cube of a quadratic
form and the square of a cubic form.
Theorem
A general cubic form p(x1, . . . , xn) has a unique representation in
the form
p(x1, . . . , xn) = �
(α i,j
1≤i≤j≤n
i xi + · · · + α i,j
j xj) 3 . (4)
Even though (4) is actually a canonical form, it is not a
representation of p as a minimal number of cubes, so its
application to, for example, the rank of tensors is unclear.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The canonical results given later for binary forms generalize a
couple of classical theorems of Sylvester from 1851. Sylvester also
gave algorithms in the process of giving his proofs.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The canonical results given later for binary forms generalize a
couple of classical theorems of Sylvester from 1851. Sylvester also
gave algorithms in the process of giving his proofs.
The cubic theorem was suggested by a much more profound result
of Boris Reichstein from 1987:
Theorem (Reichstein)
A general cubic p(x1, . . . , xn) can be written as
n�
k=1
(αk1x1 + · · · + αknxn) 3 + q(x1, . . . , xn−2).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The canonical results given later for binary forms generalize a
couple of classical theorems of Sylvester from 1851. Sylvester also
gave algorithms in the process of giving his proofs.
The cubic theorem was suggested by a much more profound result
of Boris Reichstein from 1987:
Theorem (Reichstein)
A general cubic p(x1, . . . , xn) can be written as
n�
k=1
(αk1x1 + · · · + αknxn) 3 + q(x1, . . . , xn−2).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The canonical results given later for binary forms generalize a
couple of classical theorems of Sylvester from 1851. Sylvester also
gave algorithms in the process of giving his proofs.
The cubic theorem was suggested by a much more profound result
of Boris Reichstein from 1987:
Theorem (Reichstein)
A general cubic p(x1, . . . , xn) can be written as
n�
k=1
Anybody here know this?
(αk1x1 + · · · + αknxn) 3 + q(x1, . . . , xn−2).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

2. Preliminaries
Let Hd(Cn ) denote the set of forms p(x1, . . . , xn) of degree d with
coefficients in C. This is a vector space of dimension
N(n, d) := � � n+d−1
d . Let I(n, d) denote the index set of
monomials:
�
I(n, d) = (i1, . . . , in) : 0 ≤ ik ∈ Z, �
�
ik = d .
Let x i = x i1
1 denote the multinomial
coefficient. If p ∈ Hd(Cn ), then we can write
· · · x in
n and c(i) = d!
Q ik!
p(x1, . . . , xn) = �
i∈I(n,d)
k
c(i)a(p; i)x i .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here is an alternative that is amazing to non-algebraic geometers,
but not, apparently, to the algebraic geometers I’ve talked to. The
only accessible proof I know is in Ehrenborg-Rota.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here is an alternative that is amazing to non-algebraic geometers,
but not, apparently, to the algebraic geometers I’ve talked to. The
only accessible proof I know is in Ehrenborg-Rota.
Theorem
Suppose F : C N → C N is a polynomial map; that is,
F (t1, . . . , tN) = (f1(t1, . . . , tN), . . . , fN(t1, . . . , tN))
where each fj ∈ C[t1, . . . , tN]. Then either (i) or (ii) holds:
(i) The N polynomials {fj : 1 ≤ j ≤ N} are algebraically dependent
and F (C N ) lies in some non-trivial {P = 0} in C N .
(ii) The N polynomials {fj : 1 ≤ j ≤ N} are algebraically
independent and F (C N ) is (at least) dense in C N .
Furthermore, the second case occurs if and only if there is a point
a ∈ CN �
at which the Jacobian matrix has full rank.
� ∂fi
∂tj (a)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

When N = N(n, d), we may interpret such an F as a map from
C N to Hd(C n ) by indexing I(n, d) as {ik : 1 ≤ k ≤ N} and
making the interpretation in an abuse of notation that
F (t1, . . . , tN) =
N�
c(ik)fk(t1 . . . , tN)x ik
k=1
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

When N = N(n, d), we may interpret such an F as a map from
C N to Hd(C n ) by indexing I(n, d) as {ik : 1 ≤ k ≤ N} and
making the interpretation in an abuse of notation that
F (t1, . . . , tN) =
N�
c(ik)fk(t1 . . . , tN)x ik
k=1
To retrieve our Main Example, let i1 = (2, 0), i2 = (1, 1),
i3 = (0, 2), and suppose the original polynomial map is
F (t1, t2, t3) = (t2 1 , t1t2, t2 2 + t2 3 ). Then
F (t1, t2, t3) = t 2 1x 2 1 + 2t1t2x1x2 + (t 2 2 + t 2 3)x 2 2
= (t1x1 + t2x2) 2 + (t3x2) 2 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Definition
A canonical form for Hd(C n ) is any polynomial map F from C N
to Hd(C n ) so that almost every p ∈ Hd(C n ) is in the range of F .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Definition
A canonical form for Hd(C n ) is any polynomial map F from C N
to Hd(C n ) so that almost every p ∈ Hd(C n ) is in the range of F .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Definition
A canonical form for Hd(C n ) is any polynomial map F from C N
to Hd(C n ) so that almost every p ∈ Hd(C n ) is in the range of F .
I am somewhat puzzled that the simplest canonical forms were
often not considered so by nineteenth century mathematicians.
Suppose {pj : 1 ≤ j ≤ N} is a basis for Hd(C n ). Then for every
p ∈ Hd(C n ), there exist tj ∈ C so that p = � tjpj. This canonical
form must not have been considered “sporting” to the Victorians.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Definition
A canonical form for Hd(C n ) is any polynomial map F from C N
to Hd(C n ) so that almost every p ∈ Hd(C n ) is in the range of F .
I am somewhat puzzled that the simplest canonical forms were
often not considered so by nineteenth century mathematicians.
Suppose {pj : 1 ≤ j ≤ N} is a basis for Hd(C n ). Then for every
p ∈ Hd(C n ), there exist tj ∈ C so that p = � tjpj. This canonical
form must not have been considered “sporting” to the Victorians.
The interpretation of the Jacobian having full rank is that, for
some a ∈ C n , the forms { ∂F
∂tj (a)} span Hd(C n ).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Let’s see how this works in the Main Example.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Let’s see how this works in the Main Example.
The partials of (t1x1 + t2x2) 2 + (t3x2) 2 with respect to the tj’s are:
2x1(t1x1 + t2x2), 2x2(t1x1 + t2x2), 2x2(t3x2). (5)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Let’s see how this works in the Main Example.
The partials of (t1x1 + t2x2) 2 + (t3x2) 2 with respect to the tj’s are:
2x1(t1x1 + t2x2), 2x2(t1x1 + t2x2), 2x2(t3x2). (5)
If we specialize at (t1, t2, t3) = (1, 0, 1), so t1x1 + t2x2 = x1 and
t3x2 = x2, then (5) becomes 2x 2 1 , 2x1x2, 2x 2 2 and these do span
H2(C2 ), providing an abstract existential proof that you can
complete the square for binary quadratic forms!
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

We can now prove the first theorem. Consider the representation
p(x, y) = f 2 (x, y) + g 3 (x, y) =
(t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ) 2 + (t5x 2 + t6xy + t7y 2 ) 3 :
f (x, y) = t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ,
g(x, y) = t5x 2 + t6xy + t7y 2 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

We can now prove the first theorem. Consider the representation
p(x, y) = f 2 (x, y) + g 3 (x, y) =
(t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ) 2 + (t5x 2 + t6xy + t7y 2 ) 3 :
f (x, y) = t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ,
g(x, y) = t5x 2 + t6xy + t7y 2 .
Then the partials with respect to the tj’s are:
2x 3 f , 2x 2 yf , 2xy 2 f , 2y 3 f ; 3x 2 g 2 , 3xyg 2 , 3y 2 g 2 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

We can now prove the first theorem. Consider the representation
p(x, y) = f 2 (x, y) + g 3 (x, y) =
(t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ) 2 + (t5x 2 + t6xy + t7y 2 ) 3 :
f (x, y) = t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ,
g(x, y) = t5x 2 + t6xy + t7y 2 .
Then the partials with respect to the tj’s are:
2x 3 f , 2x 2 yf , 2xy 2 f , 2y 3 f ; 3x 2 g 2 , 3xyg 2 , 3y 2 g 2 .
If we specialize at distinct linear powers for {f , g}, say
f = x 3 , g = y 2 , then these partials become:
2x 6 , 2x 5 y, 2x 4 y 2 , 2x 3 y 3 ; 3x 2 y 4 , 3xy 5 , 3y 6 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

We can now prove the first theorem. Consider the representation
p(x, y) = f 2 (x, y) + g 3 (x, y) =
(t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ) 2 + (t5x 2 + t6xy + t7y 2 ) 3 :
f (x, y) = t1x 3 + t2x 2 y + t3xy 2 + t4y 3 ,
g(x, y) = t5x 2 + t6xy + t7y 2 .
Then the partials with respect to the tj’s are:
2x 3 f , 2x 2 yf , 2xy 2 f , 2y 3 f ; 3x 2 g 2 , 3xyg 2 , 3y 2 g 2 .
If we specialize at distinct linear powers for {f , g}, say
f = x 3 , g = y 2 , then these partials become:
2x 6 , 2x 5 y, 2x 4 y 2 , 2x 3 y 3 ; 3x 2 y 4 , 3xy 5 , 3y 6 .
These trivially span H6(C 2 ). No algorithm, though.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The classical version of the Jacobian argument is called the
Lasker-Wakeford Theorem. It’s worthwhile to spend a few minutes
on who these people are.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The classical version of the Jacobian argument is called the
Lasker-Wakeford Theorem. It’s worthwhile to spend a few minutes
on who these people are.
Emanuel Lasker (1868-1941) received his Ph.D. under Max
Noether at Göttingen in 1902. He first developed the concept of a
primary ideal and proved the primary decomposition theorem for an
ideal of a polynomial ring in terms of primary ideals.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The classical version of the Jacobian argument is called the
Lasker-Wakeford Theorem. It’s worthwhile to spend a few minutes
on who these people are.
Emanuel Lasker (1868-1941) received his Ph.D. under Max
Noether at Göttingen in 1902. He first developed the concept of a
primary ideal and proved the primary decomposition theorem for an
ideal of a polynomial ring in terms of primary ideals.
He is probably better known for being the world chess champion
for 27 years (1894-1921), which spanned the life of ...
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Edward Kingsley Wakeford (1894-1916). I quote from the
memorial article by J. H. Grace in the Proceedings of the London
Mathematical Society. It may be the angriest obituary I’ve ever
read, and it can be found in its entirety on the Iowa webpage.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Edward Kingsley Wakeford (1894-1916). I quote from the
memorial article by J. H. Grace in the Proceedings of the London
Mathematical Society. It may be the angriest obituary I’ve ever
read, and it can be found in its entirety on the Iowa webpage.
“He [EKW] was slightly wounded early in 1916, and soon after
coming home was busy again with Canonical Forms.... [H]e
discovered a paper of Hilbert’s which contained the very theorem
he had long been in want of – first vaguely, and later quite
definitely. This was in March; April found him, full of the most
joyous and reverential admiration for the great German master,
working away in fearful haste to finish the dissertation ...
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Edward Kingsley Wakeford (1894-1916). I quote from the
memorial article by J. H. Grace in the Proceedings of the London
Mathematical Society. It may be the angriest obituary I’ve ever
read, and it can be found in its entirety on the Iowa webpage.
“He [EKW] was slightly wounded early in 1916, and soon after
coming home was busy again with Canonical Forms.... [H]e
discovered a paper of Hilbert’s which contained the very theorem
he had long been in want of – first vaguely, and later quite
definitely. This was in March; April found him, full of the most
joyous and reverential admiration for the great German master,
working away in fearful haste to finish the dissertation ...
He returned to the front in June and was killed in July.... He only
needed a chance, and he never got it.”
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

I found the Lasker-Wakeford Theorem in The theory of
determinants, matrices and invariants by H. W. Turnbull, who was
by some accounts the last old-style invariant theorist. Turnbull was
a good mathematician and his book is a real Rosetta Stone for
understanding 19th century algebra. He described the theorem as
“paradoxical and very curious”. I will rephrase it for simplicity.
Theorem (Lasker-Wakeford)
If F : C M → Hd(C n ) (M ≥ N(n, d)) then F is a canonical form if
and only if there is a point a ∈ C M so that there is no non-zero
form q which is apolar to all N forms { ∂F
∂F (a), . . . , ∂t1 ∂tN (a)}.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

I found the Lasker-Wakeford Theorem in The theory of
determinants, matrices and invariants by H. W. Turnbull, who was
by some accounts the last old-style invariant theorist. Turnbull was
a good mathematician and his book is a real Rosetta Stone for
understanding 19th century algebra. He described the theorem as
“paradoxical and very curious”. I will rephrase it for simplicity.
Theorem (Lasker-Wakeford)
If F : C M → Hd(C n ) (M ≥ N(n, d)) then F is a canonical form if
and only if there is a point a ∈ C M so that there is no non-zero
form q which is apolar to all N forms { ∂F
∂F (a), . . . , ∂t1 ∂tN (a)}.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

I found the Lasker-Wakeford Theorem in The theory of
determinants, matrices and invariants by H. W. Turnbull, who was
by some accounts the last old-style invariant theorist. Turnbull was
a good mathematician and his book is a real Rosetta Stone for
understanding 19th century algebra. He described the theorem as
“paradoxical and very curious”. I will rephrase it for simplicity.
Theorem (Lasker-Wakeford)
If F : C M → Hd(C n ) (M ≥ N(n, d)) then F is a canonical form if
and only if there is a point a ∈ C M so that there is no non-zero
form q which is apolar to all N forms { ∂F
∂F (a), . . . , ∂t1 ∂tN (a)}.
Most writers restrict themselves to M = N(n, d) as do I.
The question for us now is, what does “apolarity” mean in this
context?
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Apolarity
I call the following the “obvious inner product”; it can be found in
ancient invariant theory, and analysts call it the “Fisher inner
product”. Either way, it’s at least 120 years old.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Apolarity
I call the following the “obvious inner product”; it can be found in
ancient invariant theory, and analysts call it the “Fisher inner
product”. Either way, it’s at least 120 years old.
Recall that
p(x1 . . . , xn) = �
c(i)a(p; i)x i .
We now define, for p, q ∈ Hd(C n ):
[p, q] = �
i∈I(n,d)
i∈I(n,d)
c(i)a(p; i)a(q; i).
(Ehrenborg-Rota use this extensively, but never explicitly.)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Apolarity
I call the following the “obvious inner product”; it can be found in
ancient invariant theory, and analysts call it the “Fisher inner
product”. Either way, it’s at least 120 years old.
Recall that
p(x1 . . . , xn) = �
c(i)a(p; i)x i .
We now define, for p, q ∈ Hd(C n ):
[p, q] = �
i∈I(n,d)
i∈I(n,d)
c(i)a(p; i)a(q; i).
(Ehrenborg-Rota use this extensively, but never explicitly.)
This is only an inner product for real forms; for complex forms you
need a(q; i). The conjugate actually only makes our expressions
more complicated, [p, q] is really just a bilinear form on Hd(C n ).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Definition
p, q ∈ Hd(C n ) are apolar if [p, q] = 0.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

For α ∈ C n , define (α·) d ∈ Hd(C n ) by
(α·) d (x) = (α · x) d =
⎛
⎝
n�
j=1
αjxj
⎞
⎠
d
= �
i∈I(n,d)
c(i)α i x i ,
where the usual multinomial conventions apply. We define the
differential operator q(D) for q ∈ He(C n ) in the usual way by
q(D) = �
i∈I(n,e)
� �i1 � �in ∂
∂
c(i)a(q; i) · · · .
∂x1 ∂xn
The reason the obvious inner product is so useful is that it has
many nice properties; all can be verified formally. Proofs are
available in many of the papers at Iowa.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The inner product satisfies these properties:
[p, q] = [q, p].
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The inner product satisfies these properties:
[p, q] = [q, p].
p(α) = [p, (α·) d ].
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The inner product satisfies these properties:
[p, q] = [q, p].
p(α) = [p, (α·) d ].
If e = d, then p(D)q = d![p, q] = q(D)p.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The inner product satisfies these properties:
[p, q] = [q, p].
p(α) = [p, (α·) d ].
If e = d, then p(D)q = d![p, q] = q(D)p.
If f ∈ He(C n ) and g ∈ Hd−e(C n ), then
d![fg, p] = (fg)(D)p = f (D)g(D)p = e![f , g(D)p]
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The inner product satisfies these properties:
[p, q] = [q, p].
p(α) = [p, (α·) d ].
If e = d, then p(D)q = d![p, q] = q(D)p.
If f ∈ He(C n ) and g ∈ Hd−e(C n ), then
d![fg, p] = (fg)(D)p = f (D)g(D)p = e![f , g(D)p]
1 ∂p
d ∂xj (α) = [p, xj(α·) d−1 ], etc.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The inner product satisfies these properties:
[p, q] = [q, p].
p(α) = [p, (α·) d ].
If e = d, then p(D)q = d![p, q] = q(D)p.
If f ∈ He(C n ) and g ∈ Hd−e(C n ), then
d![fg, p] = (fg)(D)p = f (D)g(D)p = e![f , g(D)p]
1 ∂p
d ∂xj (α) = [p, xj(α·) d−1 ], etc.
If e ≤ d and g ∈ Hd−e(Cn ), then
g(D)(α·) d = d!
e! g(α)(α·)e .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The consequences of this definition include:
p is apolar to (α·) d iff p(α) = 0.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The consequences of this definition include:
p is apolar to (α·) d iff p(α) = 0.
p is apolar to g(x)(α·) d−e for every g ∈ He(C n ) iff all e-th
order derivatives of p vanish at α iff p vanishes to e-th order
at α.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The consequences of this definition include:
p is apolar to (α·) d iff p(α) = 0.
p is apolar to g(x)(α·) d−e for every g ∈ He(C n ) iff all e-th
order derivatives of p vanish at α iff p vanishes to e-th order
at α.
If deg q ≤ deg p and q(D)p = 0, then all multiples of q in
Hd(C n ) are apolar to p.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The consequences of this definition include:
p is apolar to (α·) d iff p(α) = 0.
p is apolar to g(x)(α·) d−e for every g ∈ He(C n ) iff all e-th
order derivatives of p vanish at α iff p vanishes to e-th order
at α.
If deg q ≤ deg p and q(D)p = 0, then all multiples of q in
Hd(C n ) are apolar to p.
In Ehrenborg-Rota, if p and q are forms of possibly different
degree, p is apolar to q if p(D)q = 0. The definitions coincide
when the degrees are equal, but not otherwise. In that case,
the definition is not symmetric: if deg p > deg q, then p(D)q
will always equal 0.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

The classical “Fundamental Theorem of Apolarity” can now be
easily stated and proved. I don’t know how it was understood
before the Nullstellensatz. The theorem applies even when e > d.
Theorem (FTA)
Suppose q ∈ He(C n ) is irreducible and p ∈ Hd(C n ). Then
q(D)p = 0 iff there exist αk ⊂ {α : q(α) = 0} and λk ∈ C such
that
p(x) =
n�
λk(αk · x) d .
k=1
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
Fix q. Define the two subspaces
A = {p ∈ Hd(C n ) : q(D)p = 0},
B = {p ∈ Hd(C n ) : p = � λk(αk·) d , q(αk) = 0}.
We want to show that B ⊆ A and B ⊥ ⊆ A ⊥ ; if so, then since
Hd(C n ) is finite dimensional, it will follow that A = B.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
Fix q. Define the two subspaces
A = {p ∈ Hd(C n ) : q(D)p = 0},
B = {p ∈ Hd(C n ) : p = � λk(αk·) d , q(αk) = 0}.
We want to show that B ⊆ A and B ⊥ ⊆ A ⊥ ; if so, then since
Hd(C n ) is finite dimensional, it will follow that A = B.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
Fix q. Define the two subspaces
A = {p ∈ Hd(C n ) : q(D)p = 0},
B = {p ∈ Hd(C n ) : p = � λk(αk·) d , q(αk) = 0}.
We want to show that B ⊆ A and B ⊥ ⊆ A ⊥ ; if so, then since
Hd(C n ) is finite dimensional, it will follow that A = B.
First, if e > d, then A = Hd(Cn �
), so B ⊆ A. If e ≤ d and p ∈ B,
λkq(αk)(αk·) d−e = 0, so p ∈ A.
then q(D)p = d!
(d−e)!
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
Fix q. Define the two subspaces
A = {p ∈ Hd(C n ) : q(D)p = 0},
B = {p ∈ Hd(C n ) : p = � λk(αk·) d , q(αk) = 0}.
We want to show that B ⊆ A and B ⊥ ⊆ A ⊥ ; if so, then since
Hd(C n ) is finite dimensional, it will follow that A = B.
First, if e > d, then A = Hd(Cn �
), so B ⊆ A. If e ≤ d and p ∈ B,
λkq(αk)(αk·) d−e = 0, so p ∈ A.
then q(D)p = d!
(d−e)!
Observe that f ∈ B ⊥ iff q(α) = 0 implies [f , (α·) d ] = f (α) = 0.
Since q is irreducible, the Nullstellensatz implies that q | f . If
e > d, this is impossible unless f = 0, so B ⊥ = {0} ⊆ A ⊥ . If
e ≤ d, then f = gq where g ∈ Hd−e(Cn ). But p ∈ A =⇒
[q(D)p, g] = 0. It follows
q(D)p = 0 =⇒ [p, f ] = [p, gq] = d!
(d−e)!
that f ∈ A ⊥ , completing the proof.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

About 15 years ago, I proved a fairly obvious generalization of this
theorem when q is not irreducible. The proof is, in spirit, the same
as the one given above.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

About 15 years ago, I proved a fairly obvious generalization of this
theorem when q is not irreducible. The proof is, in spirit, the same
as the one given above.
Theorem
Suppose q ∈ He(Cn ) factors as �r j=1 qmj
j into a product of distinct
irreducible factors and suppose p ∈ Hd(Cn ). Then q(D)p = 0 iff
there exist αjk ⊂ {qj(α) = 0}, and φjk ∈ Hmj −1(Cn ) such that
p(x) =
r�
j=1
� nj
�
k=1
φjk(x)(αkj · x) d−(mj −1)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
�
.

About 15 years ago, I proved a fairly obvious generalization of this
theorem when q is not irreducible. The proof is, in spirit, the same
as the one given above.
Theorem
Suppose q ∈ He(Cn ) factors as �r j=1 qmj
j into a product of distinct
irreducible factors and suppose p ∈ Hd(Cn ). Then q(D)p = 0 iff
there exist αjk ⊂ {qj(α) = 0}, and φjk ∈ Hmj −1(Cn ) such that
p(x) =
r�
j=1
� nj
�
k=1
φjk(x)(αkj · x) d−(mj −1)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
�
.

About 15 years ago, I proved a fairly obvious generalization of this
theorem when q is not irreducible. The proof is, in spirit, the same
as the one given above.
Theorem
Suppose q ∈ He(Cn ) factors as �r j=1 qmj
j into a product of distinct
irreducible factors and suppose p ∈ Hd(Cn ). Then q(D)p = 0 iff
there exist αjk ⊂ {qj(α) = 0}, and φjk ∈ Hmj −1(Cn ) such that
p(x) =
r�
j=1
� nj
�
k=1
φjk(x)(αkj · x) d−(mj −1)
In the case of binary forms, where zeros correspond to linear
factors, the theorem is even simpler, and can be made equivalent
to Gundelfinger’s generalization of Sylvester’s canonical forms.
This next theorem also appears in Ehrenborg-Rota.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
�
.

Theorem
Suppose � mj = d + 1 and let ℓi(x, y) = αix + βiy. Suppose
further that ℓi and ℓj are pairwise linearly independent for i �= j.
Then every p ∈ Hd(C 2 ) can be written as
�
j φj(x, y)(βjx − αjy) d−(mj −1) , where φj ∈ H d−(mj −1)(C 2 ).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Theorem
Suppose � mj = d + 1 and let ℓi(x, y) = αix + βiy. Suppose
further that ℓi and ℓj are pairwise linearly independent for i �= j.
Then every p ∈ Hd(C 2 ) can be written as
�
j φj(x, y)(βjx − αjy) d−(mj −1) , where φj ∈ H d−(mj −1)(C 2 ).
Proof.
This is an easy corollary of the last theorem. More directly,
�
�
A = φj(x, y)(βjx − αjy) d−(mj −1)
: φj ∈ Hmj −1(C 2 �
) .
j
Observe that p ∈ A ⊥ ⇐⇒ p is apolar to each possible
x r y mj −1−r (βjx − αjy) d−(mj −1) ⇐⇒ p vanishes to mj-th order at
(βj, −αj) ⇐⇒ p is divisible by each ℓ mj
j . Since �
j mj = d + 1,
this implies p = 0, hence A = {0} ⊥ = Hd(C 2 ).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Remember canonical forms? Suppose α1, . . . , αn appear in a
canonical form as
(α1x1 + · · · + αnxn) d = ℓ d .
Then ∂F
∂αj = dxjℓ d−1 , and in applying Lasker-Wakeford, note that a
form is apolar to each of these if and only if it is singular at
(α1, . . . , αn). Start thinking about general forms which are singular
at general sets of points and you enter the context in which
Alexander-Hirschowitz comes in, and we change topics.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Some canonical forms for binary forms
In 1869, J. J. Sylvester (1814-1897) reflected on the discovery of
some of his most famous research in 1851, done while he was
working as an actuary.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Some canonical forms for binary forms
In 1869, J. J. Sylvester (1814-1897) reflected on the discovery of
some of his most famous research in 1851, done while he was
working as an actuary.
“I discovered and developed the whole theory of canonical binary
forms for odd degrees, and, as far as yet made out, for even
degrees too, at one evening sitting, with a decanter of port wine to
sustain nature’s flagging energies, in a back office in Lincoln’s Inn
Fields. The work was done, and well done, but at the usual cost of
racking thought — a brain on fire, and feet feeling, or feelingless,
as if plunged in an ice-pail. That night we slept no more.”
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Theorem (Sylvester)
(i) A general binary form of degree d = 2k − 1 can be written as
k�
(αjx + βy) 2k−1 .
j=1
(ii) A general binary form of degree d = 2k can be written as
λx 2k +
k�
(αjx + βy) 2k .
j=1
Note that the constant counts work out: 2 · k = (2k − 1) + 1 and
1 + 2 · k = 2k + 1.
These are immediate consequences of the results I just described,
but Sylvester did even better, giving an algorithmic construction.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Theorem (Sylvester)
Suppose p(x, y) = �d � � d
j=0 j ajx d−j y j and h(x, y) =
�r t=0 ctx r−t y t = �r j=1 (βjx − αjy) is a product of pairwise
distinct linear factors. Then there exist λk ∈ C so that
if and only if
⎛
⎜
⎝
p(x, y) =
r�
λk(αkx + βky) d
k=1
a0 a1 · · · ar
a1 a2 · · · ar+1
.
.
. ..
ad−r ad−r+1 · · · ad
.
⎞
⎟
⎠ ·
⎛ ⎞
c0
⎜
⎜c1
⎟
⎜ ⎟
⎝ . ⎠ =
⎛ ⎞
0
⎜
⎜0
⎟
⎜ ⎟
⎝.
⎠
0
.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
cr

Here is an example of Sylvester’s algorithm in action. Let
p(x, y) = 3x 5 − 20x 3 y 2 + 10xy 4 =
� �
5
· 3 x
0
5 +
� �
5
+
3
� 5
1
· 0 x 2 y 3 +
�
· 0 x 4 � �
5
y +
2
� �
5
· 2 xy
4
4 +
· (−2) x 3 y 2
� 5
5
�
· 0 y 5 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here is an example of Sylvester’s algorithm in action. Let
Since
p(x, y) = 3x 5 − 20x 3 y 2 + 10xy 4 =
� �
5
· 3 x
0
5 +
� �
5
+
3
� 5
1
· 0 x 2 y 3 +
�
· 0 x 4 � �
5
y +
2
� �
5
· 2 xy
4
4 +
· (−2) x 3 y 2
� 5
5
�
· 0 y 5 .
⎛ ⎞
⎛
⎞ 0
3 0 −2 0 ⎜
⎝ 0 −2 0 2⎠
· ⎜1
⎟
⎝0⎠
−2 0 2 0
1
=
⎛ ⎞
0
⎝0⎠
,
0
we have h(x, y) = y(x 2 + y 2 ) = y(y − ix)(y + ix).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Here is an example of Sylvester’s algorithm in action. Let
Since
p(x, y) = 3x 5 − 20x 3 y 2 + 10xy 4 =
� �
5
· 3 x
0
5 +
� �
5
+
3
� 5
1
· 0 x 2 y 3 +
�
· 0 x 4 � �
5
y +
2
� �
5
· 2 xy
4
4 +
· (−2) x 3 y 2
� 5
5
�
· 0 y 5 .
⎛ ⎞
⎛
⎞ 0
3 0 −2 0 ⎜
⎝ 0 −2 0 2⎠
· ⎜1
⎟
⎝0⎠
−2 0 2 0
1
=
⎛ ⎞
0
⎝0⎠
,
0
we have h(x, y) = y(x 2 + y 2 ) = y(y − ix)(y + ix).
Accordingly, there exist λk ∈ C so that
p(x, y) = λ1x 5 + λ2(x + iy) 5 + λ3(x − iy) 5 .
Indeed, λ1 = λ2 = λ3 = 1, as may be checked.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks about Sylvester’s algorithm
If h(D) = �r ∂ ∂
j=1 (βj ∂x − αj ∂y ) = �r t=0 ct
h(D)p =
�d−r
m=0
d!
(d − r − m)!m!
� d−r
�
i=0
∂r ∂x r−t ∂y t , then
ai+mci
�
x d−r−m y m
The coefficients of h(D)p are, up to multiple, the rows in the
matrix product, so the matrix condition is h(D)p = 0. This is
FTA configured for products of linear factors.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks about Sylvester’s algorithm
If h(D) = �r ∂ ∂
j=1 (βj ∂x − αj ∂y ) = �r t=0 ct
h(D)p =
�d−r
m=0
d!
(d − r − m)!m!
� d−r
�
i=0
∂r ∂x r−t ∂y t , then
ai+mci
�
x d−r−m y m
The coefficients of h(D)p are, up to multiple, the rows in the
matrix product, so the matrix condition is h(D)p = 0. This is
FTA configured for products of linear factors.
An alternate proof of Sylvester’s Theorem is basically
equivalent to computing the solution of constant-coefficient
linear recurrence sequences.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks about Sylvester’s algorithm
If h(D) = �r ∂ ∂
j=1 (βj ∂x − αj ∂y ) = �r t=0 ct
h(D)p =
�d−r
m=0
d!
(d − r − m)!m!
� d−r
�
i=0
∂r ∂x r−t ∂y t , then
ai+mci
�
x d−r−m y m
The coefficients of h(D)p are, up to multiple, the rows in the
matrix product, so the matrix condition is h(D)p = 0. This is
FTA configured for products of linear factors.
An alternate proof of Sylvester’s Theorem is basically
equivalent to computing the solution of constant-coefficient
linear recurrence sequences.
If d = 2s − 1 and r = s, then the matrix is s × (s + 1) and
has a non-trivial null-vector. The corresponding h (which can
be given in terms of the coefficients of p) has distinct factors
unless its discriminant vanishes. This gives the canonical form
in odd degree.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

If d = 2s and r = s, then the matrix is square, and in general,
there exists λ so that p(x, y) − λx 2s has a matrix with a
non-trivial null-vector as above. This gives the canonical form
in even degree. That extra wobble is what Sylvester must
have meant by “as far as yet made out”.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

If d = 2s and r = s, then the matrix is square, and in general,
there exists λ so that p(x, y) − λx 2s has a matrix with a
non-trivial null-vector as above. This gives the canonical form
in even degree. That extra wobble is what Sylvester must
have meant by “as far as yet made out”.
In this even case, the determinant of the square matrix is the
catalecticant. Sylvester apologized for introducing this term:
“Meicatalecticizant would more completely express the
meaning of that which, for the sake of brevity, I denominate
the catalecticant.” Sylvester was very interested in the
technical aspects of poetry and a “catalectic” verse is one in
which the last line is missing a foot.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

If d = 2s and r = s, then the matrix is square, and in general,
there exists λ so that p(x, y) − λx 2s has a matrix with a
non-trivial null-vector as above. This gives the canonical form
in even degree. That extra wobble is what Sylvester must
have meant by “as far as yet made out”.
In this even case, the determinant of the square matrix is the
catalecticant. Sylvester apologized for introducing this term:
“Meicatalecticizant would more completely express the
meaning of that which, for the sake of brevity, I denominate
the catalecticant.” Sylvester was very interested in the
technical aspects of poetry and a “catalectic” verse is one in
which the last line is missing a foot.
To his credit, in the same paper, Sylvester introduced the
term “unimodular” in its current meaning.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Possibly new results
It is easy to see that if ℓj(x, y) = αjx + βjy, 1 ≤ j ≤ d + 1 is a set
of d + 1 pairwise distinct linear forms, then {ℓd j } is a basis for
Hd(C2 ): the representation of {ℓd j } with respect to the basis
{ � d
j
� x d−j y j } has Vandermonde determinant
�
1≤i

Possibly new results
It is easy to see that if ℓj(x, y) = αjx + βjy, 1 ≤ j ≤ d + 1 is a set
of d + 1 pairwise distinct linear forms, then {ℓd j } is a basis for
Hd(C2 ): the representation of {ℓd j } with respect to the basis
{ � d
j
� x d−j y j } has Vandermonde determinant
�
1≤i

Theorem
Suppose m, n ≥ 0 and d ≥ 1 are integers so that m + n = d + 1.
Suppose ℓj(x, y) = βjx + γjy, 1 ≤ j ≤ m, are fixed pairwise
linearly independent linear forms and suppose ek | d, 1 ≤ k ≤ r
and �r k=1 (ek + 1) = n. Then a general binary form of degree d
can be written as
p(x, y) =
m�
j=1
cjℓ d j (x, y) +
r�
k=1
where cj ∈ C and fk is a form of degree ek.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
f d/ek
k (x, y), (6)

Theorem
Suppose m, n ≥ 0 and d ≥ 1 are integers so that m + n = d + 1.
Suppose ℓj(x, y) = βjx + γjy, 1 ≤ j ≤ m, are fixed pairwise
linearly independent linear forms and suppose ek | d, 1 ≤ k ≤ r
and �r k=1 (ek + 1) = n. Then a general binary form of degree d
can be written as
p(x, y) =
m�
j=1
cjℓ d j (x, y) +
r�
k=1
where cj ∈ C and fk is a form of degree ek.
f d/ek
k (x, y), (6)
The novelty here is the existence of forms of intermediate degree
taken to higher powers.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
The parameters are the m constants cj and, for each k, the ek + 1
coefficients of
ek�
fk(x, y) = αkux ek−u u
y
u=0
The partials with respect to the cj’s are simply {ℓ d 1 , · · · , ℓd m}, and
the partial with respect to αku is
(d/ek)x ek−u u d/ek−1
y fk .
Now evaluate the Jacobian at a choice of parameters so that
fk(x, y) = ˜ℓ ek
k , where the linear forms ˜ℓk are chosen so that the
combined set {ℓj, ˜ℓk} is pairwise linearly independent. Then
f d/ek−1
k = ˜ℓ d−ek
k , and it is taken times a basis of Hek (C2 ). By
earlier theorems, this set, taken all together, is a basis for Hd(C2 )
and so (6) is a canonical form.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks:
This result follows easily from the end of the Ehrenborg-Rota
paper, but does not seem to appear there explicitly.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks:
This result follows easily from the end of the Ehrenborg-Rota
paper, but does not seem to appear there explicitly.
If ek ≡ 1, m = d + 1 and n = 0, this is the “Vandermonde
basis” canonical form.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks:
This result follows easily from the end of the Ehrenborg-Rota
paper, but does not seem to appear there explicitly.
If ek ≡ 1, m = d + 1 and n = 0, this is the “Vandermonde
basis” canonical form.
If ek ≡ 1, d = 2r − 1, m = 0 and n = r, this recovers
Sylvester’s canonical form for binary forms of odd degree.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks:
This result follows easily from the end of the Ehrenborg-Rota
paper, but does not seem to appear there explicitly.
If ek ≡ 1, m = d + 1 and n = 0, this is the “Vandermonde
basis” canonical form.
If ek ≡ 1, d = 2r − 1, m = 0 and n = r, this recovers
Sylvester’s canonical form for binary forms of odd degree.
If ek ≡ 1, d = 2r, m = 1, ℓ1(x, y) = x and n = r, this
recovers Sylvester’s canonical form for binary forms of even
degree.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

A few remarks:
This result follows easily from the end of the Ehrenborg-Rota
paper, but does not seem to appear there explicitly.
If ek ≡ 1, m = d + 1 and n = 0, this is the “Vandermonde
basis” canonical form.
If ek ≡ 1, d = 2r − 1, m = 0 and n = r, this recovers
Sylvester’s canonical form for binary forms of odd degree.
If ek ≡ 1, d = 2r, m = 1, ℓ1(x, y) = x and n = r, this
recovers Sylvester’s canonical form for binary forms of even
degree.
If ek ≡ 1, so that n = 2r and m = n − 2r, this interpolates
between the first two examples, and Sylvester’s algorithm can
still be used. Let q = �m j=1 (βjx − αjy). Then q(D) kills the
fixed d-th powers and q(D)p has degree d − m = 2r − 1.
Write q(D)p as a sum of r 2r − 1-st powers and then
“integrate” to find an explicit representation for p.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

If ek ≡ 2, then an analogue to Sylvester’s canonical forms
occurs for general forms of even degree d = 2k: they are the
sum of the k-th power of ⌊(d + 1)/3⌋ quadratics plus a linear
combination of the d − 3⌊(d + 1)/3⌋ ∈ {0, 1, 2} 2k-th powers
of specified linear forms, say, chosen from {x 2k , y 2k }. We
don’t have an algorithm for this. We want one.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

If ek ≡ 2, then an analogue to Sylvester’s canonical forms
occurs for general forms of even degree d = 2k: they are the
sum of the k-th power of ⌊(d + 1)/3⌋ quadratics plus a linear
combination of the d − 3⌊(d + 1)/3⌋ ∈ {0, 1, 2} 2k-th powers
of specified linear forms, say, chosen from {x 2k , y 2k }. We
don’t have an algorithm for this. We want one.
If d = 4, m = 0, e1 = 2 and e2 = 1, a general binary quartic
can be written as the sum of the square of a quadratic and
the fourth power of a linear form. (We have an algorithm for
this which shows that it can be done in six different ways.)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

If ek ≡ 2, then an analogue to Sylvester’s canonical forms
occurs for general forms of even degree d = 2k: they are the
sum of the k-th power of ⌊(d + 1)/3⌋ quadratics plus a linear
combination of the d − 3⌊(d + 1)/3⌋ ∈ {0, 1, 2} 2k-th powers
of specified linear forms, say, chosen from {x 2k , y 2k }. We
don’t have an algorithm for this. We want one.
If d = 4, m = 0, e1 = 2 and e2 = 1, a general binary quartic
can be written as the sum of the square of a quadratic and
the fourth power of a linear form. (We have an algorithm for
this which shows that it can be done in six different ways.)
If d = 6, m = 0, e1 = 3 and e2 = 2, then ( 6
6
3 + 1) + ( 2 + 1) =
6 + 1 implies that a general binary sextic can be written as the
sum of the square of a cubic and the cube of a quadratic
form. We don’t have an algorithm for this. We want one.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

More variables
Why are canonical forms even an issue? The main reason is that
maps which one would think have full range don’t. Apart from
sums of squares, where the orthogonal group plays a role, the
simplest example occurs in H4(C3 ). Since N(3, 4) = � � 6
2 = 15, one
expects that a general ternary quartic could be written as
p(x1, x2, x3) =
5�
k=1
(αk1x1 + αk2x2 + αk3x3) 4
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
(7)

More variables
Why are canonical forms even an issue? The main reason is that
maps which one would think have full range don’t. Apart from
sums of squares, where the orthogonal group plays a role, the
simplest example occurs in H4(C3 ). Since N(3, 4) = � � 6
2 = 15, one
expects that a general ternary quartic could be written as
p(x1, x2, x3) =
5�
k=1
(αk1x1 + αk2x2 + αk3x3) 4
This would be a canonical form, if the partials with respect to the
αkj’s at some chosen value would span H4(C 3 ). By apolarity, this
means that there should be no non-singular quartic which is
singular at the five points αk = (αk1, αk2, αk3).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
(7)

More variables
Why are canonical forms even an issue? The main reason is that
maps which one would think have full range don’t. Apart from
sums of squares, where the orthogonal group plays a role, the
simplest example occurs in H4(C3 ). Since N(3, 4) = � � 6
2 = 15, one
expects that a general ternary quartic could be written as
p(x1, x2, x3) =
5�
k=1
(αk1x1 + αk2x2 + αk3x3) 4
This would be a canonical form, if the partials with respect to the
αkj’s at some chosen value would span H4(C 3 ). By apolarity, this
means that there should be no non-singular quartic which is
singular at the five points αk = (αk1, αk2, αk3).
However, as Clebsch argued in the 1860’s, since N(3.2) = 6, any
choice of five αk’s pass through a non-zero quadratic h(x1, x2, x3),
and so h 2 will be apolar to all the partials and so (7) is not a
canonical form.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
(7)

A few years later, Sylvester gave another proof. Given
p(x1, x2, x3) = �
r+s+t=4
4!
r!s!t! arstx r 1x s 2x t 3,
define the catalecticant Hp as a quadratic form in 6 variables (or a
6 × 6 symmetric matrix defined linearly in terms of p.
⎛
⎜
Hp = ⎜
⎝
a400 a220 a202 a310 a301 a211
a220 a040 a022 a130 a121 a031
a202 a022 a004 a112 a103 a013
a310 a130 a112 a220 a211 a121
a301 a121 a103 a211 a202 a112
a211 a031 a013 a121 a112 a022
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
⎞
⎟
⎠

Under this definition, H (α·) 4 is a perfect square. Thus if p is a sum
of five fourth powers, then rank(Hp) ≤ 5, so Hp is singular. This
can’t happen for a general ternary quartic, for which the
determinant is non-zero. This gives the algebraic relation of the
coefficients in (7).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Under this definition, H (α·) 4 is a perfect square. Thus if p is a sum
of five fourth powers, then rank(Hp) ≤ 5, so Hp is singular. This
can’t happen for a general ternary quartic, for which the
determinant is non-zero. This gives the algebraic relation of the
coefficients in (7).
Clebsch’s proof and Sylvester’s proof are really the same, because
Hp(t1, . . . , t6) =
[(t1x 2 1 + t2x 2 2 + t3x 2 3 + t4x1x2 + t5x1x3 + t6x2x3) 2 (D)]p.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Under this definition, H (α·) 4 is a perfect square. Thus if p is a sum
of five fourth powers, then rank(Hp) ≤ 5, so Hp is singular. This
can’t happen for a general ternary quartic, for which the
determinant is non-zero. This gives the algebraic relation of the
coefficients in (7).
Clebsch’s proof and Sylvester’s proof are really the same, because
Hp(t1, . . . , t6) =
[(t1x 2 1 + t2x 2 2 + t3x 2 3 + t4x1x2 + t5x1x3 + t6x2x3) 2 (D)]p.
This becomes a serious topic in algebraic geometry, so I won’t talk
about it here. Our 19th century ancestors saw that funny things
happen when (n, d) = (3, 4), (4, 4), (5, 4), (5, 3). In the early
1990s, Alexander and Hirschowitz proved that these are the only
cases these funny things can happen.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Sums of two squares
Suppose p ∈ H2r (C2 ), has even degree. Then the FTA (algebra
this time) says that p can be written as a product of 2r linear
forms, which in general are pairwise linearly independent, so by
pairing them off, we have p = gh, where f , g ∈ Hr (C2 ) in
essentially � � 2r−1
different ways. Further, by the old algebraist’s
trick:
r
�
g + h
p = gh =
2
� 2
� �2 g − h
+ .
2i
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Sums of two squares
Suppose p ∈ H2r (C2 ), has even degree. Then the FTA (algebra
this time) says that p can be written as a product of 2r linear
forms, which in general are pairwise linearly independent, so by
pairing them off, we have p = gh, where f , g ∈ Hr (C2 ) in
essentially � � 2r−1
different ways. Further, by the old algebraist’s
trick:
r
�
g + h
p = gh =
2
� 2
� �2 g − h
+ .
2i
A sum of two squares from Hr (C 2 ) is not a canonical form because
it uses 2(r + 1) = (2r + 1) + 1 parameters. But, if (u, v) ∈ C 2 and
u 2 + v 2 = 1, then
a 2 + b 2 = (ua + vb) 2 + (va − ub) 2
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
(8)

Sums of two squares
Suppose p ∈ H2r (C2 ), has even degree. Then the FTA (algebra
this time) says that p can be written as a product of 2r linear
forms, which in general are pairwise linearly independent, so by
pairing them off, we have p = gh, where f , g ∈ Hr (C2 ) in
essentially � � 2r−1
different ways. Further, by the old algebraist’s
trick:
r
�
g + h
p = gh =
2
� 2
� �2 g − h
+ .
2i
A sum of two squares from Hr (C 2 ) is not a canonical form because
it uses 2(r + 1) = (2r + 1) + 1 parameters. But, if (u, v) ∈ C 2 and
u 2 + v 2 = 1, then
a 2 + b 2 = (ua + vb) 2 + (va − ub) 2
(These correspond to gh = (ζg)(ζ −1 h) where ζ = u + iv; in terms
of sos forms, when u, v ∈ R, all the representations in (8) have the
same Gram matrix.)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
(8)

Accordingly, for a general p, we can choose (u, v) so that the
coefficient of x r , say, disappears in the second sum, and
p(x, y) =
� r�
k=0
αkx r−k y k
� 2
+
� r�
k=1
βkx r−k y k
is another old canonical form for binary forms of even degree.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
� 2
(9)

Accordingly, for a general p, we can choose (u, v) so that the
coefficient of x r , say, disappears in the second sum, and
p(x, y) =
� r�
k=0
αkx r−k y k
� 2
+
� r�
k=1
βkx r−k y k
is another old canonical form for binary forms of even degree.
This also gives another proof of the Basic Example.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
� 2
(9)

Accordingly, for a general p, we can choose (u, v) so that the
coefficient of x r , say, disappears in the second sum, and
p(x, y) =
� r�
k=0
αkx r−k y k
� 2
+
� r�
k=1
βkx r−k y k
is another old canonical form for binary forms of even degree.
This also gives another proof of the Basic Example.
Suppose we don’t just look at maps C N → Hd(C n ), but also maps
from N-dimensional subspaces of C M for M > N. Here is the
simplest non-trivial case:
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
� 2
(9)

For which 0 �= c = (c1, . . . , c4) ∈ C 4 is it true that
(α1x + α2y) 2 + (α3x + α4y) 2 ,
is a canonical form for binary quadratic forms?
4�
cjαj = 0 (10)
j=1
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

For which 0 �= c = (c1, . . . , c4) ∈ C 4 is it true that
(α1x + α2y) 2 + (α3x + α4y) 2 ,
is a canonical form for binary quadratic forms?
Theorem
4�
cjαj = 0 (10)
A general binary quadratic form can be written as in (10), unless
c3 = ɛc1 and c4 = ɛc2, where ɛ ∈ {i, −i}.
j=1
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

For which 0 �= c = (c1, . . . , c4) ∈ C 4 is it true that
(α1x + α2y) 2 + (α3x + α4y) 2 ,
is a canonical form for binary quadratic forms?
Theorem
4�
cjαj = 0 (10)
A general binary quadratic form can be written as in (10), unless
c3 = ɛc1 and c4 = ɛc2, where ɛ ∈ {i, −i}.
In other words, (10) is canonically a canonical form. In the special
case (c1, c2, c3, c4) = (1, 0, i, 0), �4 j=1 cjαj = 0 ⇐⇒ α3 = iα1,
which is the “silly” example from the introduction.
j=1
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

We conjecture that for d = 2r and “most” c = (c1, . . . , c2r ) ∈ C 2r ,
r�
j=1
(α2j−1x + α2jy) 2r ,
�2r
j=1
cjαj = 0 (11)
is a canonical form for binary forms of even degree 2r. We’ve
proved this for 2r = 2, 4. In the special case where cj ≡ 1, it is
true for 2r = 6, 8.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Completing the d-th power, all props to Boris Reichstein
Let me give yet another ridiculous (but constructive) proof of
completing the square for a quadratic form. Suppose p ∈ H2(Cn ).
Then we have:
∂p
= α1x1 + · · · + αnxn.
∂xn
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Completing the d-th power, all props to Boris Reichstein
Let me give yet another ridiculous (but constructive) proof of
completing the square for a quadratic form. Suppose p ∈ H2(Cn ).
Then we have:
∂p
= α1x1 + · · · + αnxn.
∂xn
If we assume αn �= 0, and let
q(x1, . . . , xn) = p(x1, . . . , xn) − 1
2αn (α1x1 + · · · + αnxn) 2 ,
then it is easy to see that
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Completing the d-th power, all props to Boris Reichstein
Let me give yet another ridiculous (but constructive) proof of
completing the square for a quadratic form. Suppose p ∈ H2(Cn ).
Then we have:
∂p
= α1x1 + · · · + αnxn.
∂xn
If we assume αn �= 0, and let
q(x1, . . . , xn) = p(x1, . . . , xn) − 1
2αn (α1x1 + · · · + αnxn) 2 ,
then it is easy to see that
∂q
∂xn
= ∂p
−
∂xn
∂p
∂xn
= 0 =⇒ q = q(x1, . . . , xn−1)!
Now just iterate this, losing one variable at a time, to get the
traditional lower diagonal sum of squares.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

This argument can be repeated. Suppose p ∈ H3(Cn ) is cubic;
then ∂p
is a quadratic form, so we can complete the square in
∂xn
general, but now lets do it diagonally from left to right:
∂p
∂xn
=
n�
j=1
(αjjxj + · · · + αjnxn) 2 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

This argument can be repeated. Suppose p ∈ H3(Cn ) is cubic;
then ∂p
is a quadratic form, so we can complete the square in
∂xn
general, but now lets do it diagonally from left to right:
∂p
∂xn
=
n�
j=1
You can guess where this is going. Let
Then
q(x1, . . . , xn) = p(x1, . . . , xn) −
∂q
∂xn
= ∂p
−
∂xn
∂p
∂xn
(αjjxj + · · · + αjnxn) 2 .
n�
j=1
1
3αjn (αjjxj + · · · + αjnxn) 3 .
= 0 =⇒ q = q(x1, . . . , xn−1).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

This argument can be repeated. Suppose p ∈ H3(Cn ) is cubic;
then ∂p
is a quadratic form, so we can complete the square in
∂xn
general, but now lets do it diagonally from left to right:
∂p
∂xn
=
n�
j=1
You can guess where this is going. Let
Then
q(x1, . . . , xn) = p(x1, . . . , xn) −
∂q
∂xn
= ∂p
−
∂xn
∂p
∂xn
(αjjxj + · · · + αjnxn) 2 .
n�
j=1
1
3αjn (αjjxj + · · · + αjnxn) 3 .
= 0 =⇒ q = q(x1, . . . , xn−1).
We repeat this running backwards inductively to get our second
theorem from the introduction, which is an algebra prelim question
in a 19th century graduate math program:
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Theorem
A canonical form for the general cubic is
p(x1, . . . , xn) = �
(α i,j
1≤i≤j≤n
i xi + · · · + α i,j
j
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
xj) 3

Theorem
A canonical form for the general cubic is
p(x1, . . . , xn) = �
(α i,j
1≤i≤j≤n
i xi + · · · + α i,j
j
xj) 3
Proof.
We can do this abstractly with Lasker-Wakeford, or directly by
using the previous construction. We leave as an exercise the
constant-counting via the combinatorial identity:
�
� �
n + 2
(j − i + 1) =
3
1≤i≤j≤n
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

There is a wonderful non-trivial way to complete the cube, but
almost nobody knows it. It appears in a paper by Boris Reichstein
from 1987 which (according to MathSciNet) has had no citations.
It is a truly beautiful theorem, though it was not transparently
presented and appears in the context of trilinear forms.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

There is a wonderful non-trivial way to complete the cube, but
almost nobody knows it. It appears in a paper by Boris Reichstein
from 1987 which (according to MathSciNet) has had no citations.
It is a truly beautiful theorem, though it was not transparently
presented and appears in the context of trilinear forms.
Theorem (Reichstein)
A general cubic p(x1, . . . , xn) can be written as
n�
k=1
(αk1x1 + · · · + αknxn) 3 + q(x1, . . . , xn−2). (12)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

There is a wonderful non-trivial way to complete the cube, but
almost nobody knows it. It appears in a paper by Boris Reichstein
from 1987 which (according to MathSciNet) has had no citations.
It is a truly beautiful theorem, though it was not transparently
presented and appears in the context of trilinear forms.
Theorem (Reichstein)
A general cubic p(x1, . . . , xn) can be written as
n�
k=1
(αk1x1 + · · · + αknxn) 3 + q(x1, . . . , xn−2). (12)
We have N(n, 3) − N(n − 2, 3) = n3 +3n2 +2n
6 − n3−3n2 +2n
6 = n2 , so
if (12) is generally possible, it is a canonical form. This can be
verified by Lasker-Wakeford, specializing at x1, x2, x1 + kx2 + xk
(for k ≥ 3), but Reichstein’s constructive proof is prettier.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
A general pair of quadratic forms can be simultaneously
diagonalized: for general p, there exist n linearly independent
forms Lj(x) = � n
k=1 αj,kxk and complex numbers cj so that
∂p
∂xn−1
=
n�
j=1
L 2 j ,
∂p
∂xn
=
n�
j=1
cjL 2 j ,
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Proof.
A general pair of quadratic forms can be simultaneously
diagonalized: for general p, there exist n linearly independent
forms Lj(x) = � n
k=1 αj,kxk and complex numbers cj so that
∂p
∂xn−1
=
n�
j=1
L 2 j ,
∂p
∂xn
=
n�
j=1
cjL 2 j ,
Mixed partials are equal, so 2 �n j=1 αj,nLj = 2 �n the Lj’s are linearly independent, so αj,n = cjαj,n−1. Now,
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms
j=1 cjαj,n−1Lj and

Proof.
A general pair of quadratic forms can be simultaneously
diagonalized: for general p, there exist n linearly independent
forms Lj(x) = � n
k=1 αj,kxk and complex numbers cj so that
∂p
∂xn−1
=
n�
j=1
L 2 j ,
∂p
∂xn
=
n�
j=1
cjL 2 j ,
Mixed partials are equal, so 2 �n j=1 αj,nLj = 2 �n the Lj’s are linearly independent, so αj,n = cjαj,n−1. Now,
q(x1, . . . , xn) = p(x1, . . . , xn) −
=⇒ ∂q
∂xn−1
= ∂q
∂xn
n�
j=1
j=1 cjαj,n−1Lj and
1
3αj,n (αj,1x1 + · · · + αj,nxn) 3
= 0 =⇒ q = q(x1, . . . , xn−2).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Reichstein’s construction gives a cubic as a sum of roughly n2
4
cubes, which is more than the actual minimum of roughly n2
6 .
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Reichstein’s construction gives a cubic as a sum of roughly n2
4
cubes, which is more than the actual minimum of roughly n2
6 .
We conclude the talk by veering off in another direction.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Reichstein’s construction gives a cubic as a sum of roughly n2
4
cubes, which is more than the actual minimum of roughly n2
6 .
We conclude the talk by veering off in another direction.
Imagine a general canonical form for quartics of “Reichstein-type”
p(x1, . . . , xn) =
r�
k=1
(αk1x1 + · · · + αknxn) 4 + q(x1, . . . , xm). (13)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Reichstein’s construction gives a cubic as a sum of roughly n2
4
cubes, which is more than the actual minimum of roughly n2
6 .
We conclude the talk by veering off in another direction.
Imagine a general canonical form for quartics of “Reichstein-type”
p(x1, . . . , xn) =
r�
k=1
(αk1x1 + · · · + αknxn) 4 + q(x1, . . . , xm). (13)
We would need N(n, 4) = r × n + N(m, 4).
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Reichstein’s construction gives a cubic as a sum of roughly n2
4
cubes, which is more than the actual minimum of roughly n2
6 .
We conclude the talk by veering off in another direction.
Imagine a general canonical form for quartics of “Reichstein-type”
p(x1, . . . , xn) =
r�
k=1
(αk1x1 + · · · + αknxn) 4 + q(x1, . . . , xm). (13)
We would need N(n, 4) = r × n + N(m, 4).
It turns out that if n = 12, there does not exist m < 12 so that
� � � � �
�
12 �
15 m + 3
� − ,
4 4
so number theory rules out universal Reichstein-type canonical
forms for quartics.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Reichstein’s construction gives a cubic as a sum of roughly n2
4
cubes, which is more than the actual minimum of roughly n2
6 .
We conclude the talk by veering off in another direction.
Imagine a general canonical form for quartics of “Reichstein-type”
p(x1, . . . , xn) =
r�
k=1
(αk1x1 + · · · + αknxn) 4 + q(x1, . . . , xm). (13)
We would need N(n, 4) = r × n + N(m, 4).
It turns out that if n = 12, there does not exist m < 12 so that
� � � � �
�
12 �
15 m + 3
� − ,
4 4
so number theory rules out universal Reichstein-type canonical
forms for quartics.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Now define
�
Ad = n : 0 ≤ m < n =⇒ n � | � � � � n+d−1 m+d−1
d − d
�
.
We have a few partial results.
If 3 � | k, then n = 2 2k · 3 ∈ A4.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Now define
�
Ad = n : 0 ≤ m < n =⇒ n � | � � � � n+d−1 m+d−1
d − d
�
.
We have a few partial results.
If 3 � | k, then n = 2 2k · 3 ∈ A4.
If p ≡ 1 (mod 144) is prime, then 12p ∈ A4.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Now define
�
Ad = n : 0 ≤ m < n =⇒ n � | � � � � n+d−1 m+d−1
d − d
�
.
We have a few partial results.
If 3 � | k, then n = 2 2k · 3 ∈ A4.
If p ≡ 1 (mod 144) is prime, then 12p ∈ A4.
If p is prime, then p | � � � � n+p−1 n
p − p , hence Ap is empty for
prime p (such as p = 2, 3.)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Now define
�
Ad = n : 0 ≤ m < n =⇒ n � | � � � � n+d−1 m+d−1
d − d
�
.
We have a few partial results.
If 3 � | k, then n = 2 2k · 3 ∈ A4.
If p ≡ 1 (mod 144) is prime, then 12p ∈ A4.
If p is prime, then p | � � � � n+p−1 n
p − p , hence Ap is empty for
prime p (such as p = 2, 3.)
The smallest elements of A6, A8, A10, A12, A14 and A15 are 10,
1792, 6, 242, 338 and 273 respectively. If A9 or A16 are
non-empty, then their smallest elements are at least 10 5 .
(Fortunately, steampunk allows Mathematica.)
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms

Now define
�
Ad = n : 0 ≤ m < n =⇒ n � | � � � � n+d−1 m+d−1
d − d
�
.
We have a few partial results.
If 3 � | k, then n = 2 2k · 3 ∈ A4.
If p ≡ 1 (mod 144) is prime, then 12p ∈ A4.
If p is prime, then p | � � � � n+p−1 n
p − p , hence Ap is empty for
prime p (such as p = 2, 3.)
The smallest elements of A6, A8, A10, A12, A14 and A15 are 10,
1792, 6, 242, 338 and 273 respectively. If A9 or A16 are
non-empty, then their smallest elements are at least 10 5 .
(Fortunately, steampunk allows Mathematica.)
Thus the way is clear for a Reichstein canonical form for quintics. I
hope one of you in the audience can find it.
Bruce Reznick University of Illinois at Urbana-Champaign Steampunk canonical forms