Rainy Day Lemmas # 12 and 35reznick/eccad42917.pdf · 2017. 4. 30. · Rainy Day Lemmas # 12 and 35 Bruce Reznick University of Illinois at Urbana-Champaign East Coast Computer Algebra

Rainy Day Lemmas # 12 and 35

Bruce ReznickUniversity of Illinois at Urbana-Champaign

East Coast Computer Algebra DayWolfram World HQ April 29, 2017

Bruce Reznick University of Illinois at Urbana-Champaign Rainy Day Lemmas # 12 and 35

Thank you for coming to the talk. I speak not as a producer ofcomputer algebra, but as a consumer. Richard Hamming, who gothis PhD from UIUC in 1942, liked to say that “The purpose ofcomputing is insight, not numbers.” In part, the justification ofthis talk is to present a few examples where insight allows you todo computations which are otherwise unwieldy.

My main interest here is homogeneous polynomials (forms) over Cand their representation as a sum of powers of forms of lowerdegree. Most of the work I’m talking about is on papers which canbe found on my website, or in the other talks from which this onehas been Frankenstein’d.

The organization of this talk after the introduction is (basically):canonical forms, sums of powers of linear forms, sums of powers ofquadratic forms.










I want to focus on a beautiful conjecture due to Boris Shapiro. LetHm(C2) denote the vector space of binary forms (homeogeneouspolynomials) p(x , y) of degree m with complex coefficients, andsuppose m = de, d , e ∈ N.

Conjecture

Every p ∈ Hde(C2) can be written as a sum of at most d d-thpowers of forms in He(C2).

The first point to make is that this assertion is a universalstatement, not a generic one.

If e = 1, this conjecture is a familiar statement to those who workwith Waring rank, and the binary forms of degree d which required d-th powers of linear forms are precisely those of the shape`d−1`′, where ` and `′ are non-proportional linear forms. (More onthis later.)



Conjecture






Conjecture





If d = 1, there is nothing to prove.

If d = 2, then m = 2e is even, and p can be factored into linearfactors in

(2e−1e

)ways, so that p = fg for f , g ∈ He(C2) and

p = fg =

(f + g

2

)2

−(

f − g

2

)2

=

(f + g

2

)2

+

(f − g

2i

)2

.

The conjecture is true generically. Using the classical Lasker-Wakeford approach, if de + 1 = k(e + 1) + s, 0 ≤ s ≤ e, then

k∑j=1

(αj0xe + . . . )d + (β0xe + · · ·+ βs−1xe−(s−1)y s−1)d

is a “canonical form” for binary forms of degree de, and

k + 1 =

⌈de + 1

e + 1

⌉≤⌈

de + d

e + 1

⌉= d .

I will say a bit more about Lasker-Wakeford later.




(2e−1e


p = fg =

(f + g

2

)2

−(

f − g

2

)2

=

(f + g

2

)2

+

(f − g

2i

)2

.


k∑j=1



k + 1 =

⌈de + 1

e + 1

⌉≤⌈

de + d

e + 1

⌉= d .





(2e−1e


p = fg =

(f + g

2

)2

−(

f − g

2

)2

=

(f + g

2

)2

+

(f − g

2i

)2

.


k∑j=1



k + 1 =

⌈de + 1

e + 1

⌉≤⌈

de + d

e + 1

⌉= d .





(2e−1e


p = fg =

(f + g

2

)2

−(

f − g

2

)2

=

(f + g

2

)2

+

(f − g

2i

)2

.


k∑j=1



k + 1 =

⌈de + 1

e + 1

⌉≤⌈

de + d

e + 1

⌉= d .

I will say a bit more about Lasker-Wakeford later.Bruce Reznick University of Illinois at Urbana-Champaign Rainy Day Lemmas # 12 and 35

I will also give you a proof of the first “new” case: (d , e) = (3, 2):every binary sextic is a sum of three cubes of quadratic forms. Aswe’ll see, this proof would be impossible without some kind ofcomputer algebra.

The conjecture is easily true if you remove the restriction to forms(but lose the information about degrees). By a 1979 result ofNewman-Slater, every complex polynomial in any number ofvariables is an explicit sum of d d-th powers of polynomials.Let ζd denote a primitive d-th root of unity and p be a polynomialin any number of variables. Then the usual orthogonalityproperties of roots of unity imply that

d2p =d−1∑k=0

ζ−kd (1 + ζkd p)d .

This equation fails to be helpful if you want both sides to behomogeneous.


I will also give you a proof of the first “new” case: (d , e) = (3, 2):every binary sextic is a sum of three cubes of quadratic forms. Aswe’ll see, this proof would be impossible without some kind ofcomputer algebra.

The conjecture is easily true if you remove the restriction to forms(but lose the information about degrees). By a 1979 result ofNewman-Slater, every complex polynomial in any number ofvariables is an explicit sum of d d-th powers of polynomials.Let ζd denote a primitive d-th root of unity and p be a polynomialin any number of variables. Then the usual orthogonalityproperties of roots of unity imply that

d2p =d−1∑k=0

ζ−kd (1 + ζkd p)d .

This equation fails to be helpful if you want both sides to behomogeneous.


Some notation.Let Hd(Cn) denote the set of forms p(x1, . . . , xn) of degree d withcoefficients in C. The dimension of the vector space Hd(Cn) isN(n, d) :=

(n+d−1d

). Let I(n, d) be the index set:

I(n, d) =

(i1, . . . , in) : 0 ≤ ik ∈ Z,

∑k

ik = d

.

Let x i = x i11 · · · x in

n and c(i) = d!∏ik !

denote the multinomial

coefficient. If p ∈ Hd(Cn), then we can write

p(x1, . . . , xn) =∑

i∈I(n,d)

c(i)a(p; i)x i .


Some notation.Let Hd(Cn) denote the set of forms p(x1, . . . , xn) of degree d withcoefficients in C. The dimension of the vector space Hd(Cn) isN(n, d) :=

(n+d−1d

). Let I(n, d) be the index set:

I(n, d) =

(i1, . . . , in) : 0 ≤ ik ∈ Z,

∑k

ik = d

.

Let x i = x i11 · · · x in

n and c(i) = d!∏ik !

denote the multinomial

coefficient. If p ∈ Hd(Cn), then we can write

p(x1, . . . , xn) =∑

i∈I(n,d)

c(i)a(p; i)x i .


The following amazing theorem is not as well known as it shouldbe. The only accessible proof of (ii) I know is in Ehrenborg-Rota,attributed to Artin and Mattuck. Please let me know of others!

Theorem

Suppose F : Cm → Cr is a polynomial map; that is,

F (t1, . . . , tm) = (f1(t1, . . . , tm), . . . , fr (t1, . . . , tm))

where each fj ∈ C[t1, . . . , tm]. Then either (i) or (ii) holds:(i) The polynomials fj : 1 ≤ j ≤ r are algebraically dependentand F (Cm) lies in some non-trivial P = 0 in Cr .(ii) The polynomials fj : 1 ≤ j ≤ r are algebraically independentand (F (Cm))c lies in some non-trivial P = 0 in Cr .Furthermore, the second case occurs if and only there is a point

u ∈ Cm at which the Jacobian matrix[∂fj∂tk

(u)]

has rank r .



Theorem


F (t1, . . . , tm) = (f1(t1, . . . , tm), . . . , fr (t1, . . . , tm))

where each fj ∈ C[t1, . . . , tm]. Then either (i) or (ii) holds:

(i) The polynomials fj : 1 ≤ j ≤ r are algebraically dependentand F (Cm) lies in some non-trivial P = 0 in Cr .(ii) The polynomials fj : 1 ≤ j ≤ r are algebraically independentand (F (Cm))c lies in some non-trivial P = 0 in Cr .Furthermore, the second case occurs if and only there is a point


(u)]

has rank r .



Theorem


F (t1, . . . , tm) = (f1(t1, . . . , tm), . . . , fr (t1, . . . , tm))

where each fj ∈ C[t1, . . . , tm]. Then either (i) or (ii) holds:

(i) The polynomials fj : 1 ≤ j ≤ r are algebraically dependentand F (Cm) lies in some non-trivial P = 0 in Cr .(ii) The polynomials fj : 1 ≤ j ≤ r are algebraically independentand (F (Cm))c lies in some non-trivial P = 0 in Cr .Furthermore, the second case occurs if and only there is a point


(u)]

has rank r .



Theorem


F (t1, . . . , tm) = (f1(t1, . . . , tm), . . . , fr (t1, . . . , tm))

where each fj ∈ C[t1, . . . , tm]. Then either (i) or (ii) holds:(i) The polynomials fj : 1 ≤ j ≤ r are algebraically dependentand F (Cm) lies in some non-trivial P = 0 in Cr .

(ii) The polynomials fj : 1 ≤ j ≤ r are algebraically independentand (F (Cm))c lies in some non-trivial P = 0 in Cr .Furthermore, the second case occurs if and only there is a point


(u)]

has rank r .



Theorem


F (t1, . . . , tm) = (f1(t1, . . . , tm), . . . , fr (t1, . . . , tm))

where each fj ∈ C[t1, . . . , tm]. Then either (i) or (ii) holds:(i) The polynomials fj : 1 ≤ j ≤ r are algebraically dependentand F (Cm) lies in some non-trivial P = 0 in Cr .(ii) The polynomials fj : 1 ≤ j ≤ r are algebraically independentand (F (Cm))c lies in some non-trivial P = 0 in Cr .

Furthermore, the second case occurs if and only there is a point


(u)]

has rank r .



Theorem


F (t1, . . . , tm) = (f1(t1, . . . , tm), . . . , fr (t1, . . . , tm))

where each fj ∈ C[t1, . . . , tm]. Then either (i) or (ii) holds:(i) The polynomials fj : 1 ≤ j ≤ r are algebraically dependentand F (Cm) lies in some non-trivial P = 0 in Cr .(ii) The polynomials fj : 1 ≤ j ≤ r are algebraically independentand (F (Cm))c lies in some non-trivial P = 0 in Cr .Furthermore, the second case occurs if and only there is a point


(u)]

has rank r .


When m = r = N(n, d), we may interpret such an F as a mapfrom CN to Hd(Cn) by indexing I(n, d) as ij : 1 ≤ j ≤ N andmaking the interpretation in an abuse of notation that

F (t, x) =N∑j=1

fj(t1 . . . , tN)x ij .

Definition

A canonical form for Hd(Cn) is any polynomial map F from CN

to Hd(Cn) so that “almost every” p ∈ Hd(Cn) is in the range of F .

If φj is a basis of Hd(Cn), e.g. φj(x) = c(ij)x ij , then

F (t, x) =∑N

j=1 tjφj(x) is technically, though not traditionally, acanonical form.



F (t, x) =N∑j=1

fj(t1 . . . , tN)x ij .

Definition




F (t, x) =∑N




F (t, x) =N∑j=1

fj(t1 . . . , tN)x ij .

Definition




F (t, x) =∑N




F (t, x) =N∑j=1

fj(t1 . . . , tN)x ij .

Definition




F (t, x) =∑N



Here’s a really simple example: Take a form in H2(C2):

p(x , y) = ax2 + 2bxy + cy2.

As long as a 6= 0, we can find tj ∈ C by completing the square sothat

p(x , y) = (t1x + t2y)2 + (t3y)2.

Using our machinery with F (t1, t2, t3) = (t21 , t1t2, t22 + t23 ), then at

“most” choices of t, the Jacobian has rank 3 and, indeed,

F (C3) = (z1, z2, z3) : z1 6= 0 ∪ (0, 0, z3).

which is case (ii).


Here’s a really simple example: Take a form in H2(C2):

p(x , y) = ax2 + 2bxy + cy2.

As long as a 6= 0, we can find tj ∈ C by completing the square sothat

p(x , y) = (t1x + t2y)2 + (t3y)2.

Using our machinery with F (t1, t2, t3) = (t21 , t1t2, t22 + t23 ), then at

“most” choices of t, the Jacobian has rank 3 and, indeed,

F (C3) = (z1, z2, z3) : z1 6= 0 ∪ (0, 0, z3).

which is case (ii).


Why can’t we just constant-count in the non-obvious cases?Historically, the simplest example occurs in H4(C3). SinceN(3, 4) =

(62

)= 15, can a general ternary quartic can 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 theαkj ’s at some chosen value span H4(C3). Using “apolarity”, whichI don’t have time to explain, this is equivalent to there being nonon-zero quartic which is singular at the αk = (αk1, αk2, αk3)’s.

However, as Clebsch argued in the 1860’s, since N(3, 2) = 6, anychoice of five αk ’s pass through a non-zero quadratic h(x1, x2, x3),and h2 is such a singular quartic. Thus, a sum of five 4th powers isnot a canonical form for ternary quartics.


Why can’t we just constant-count in the non-obvious cases?Historically, the simplest example occurs in H4(C3). SinceN(3, 4) =

(62

)= 15, can a general ternary quartic can 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 theαkj ’s at some chosen value span H4(C3). Using “apolarity”, whichI don’t have time to explain, this is equivalent to there being nonon-zero quartic which is singular at the αk = (αk1, αk2, αk3)’s.

However, as Clebsch argued in the 1860’s, since N(3, 2) = 6, anychoice of five αk ’s pass through a non-zero quadratic h(x1, x2, x3),and h2 is such a singular quartic. Thus, a sum of five 4th powers isnot a canonical form for ternary quartics.


A few years later, Sylvester gave another proof. Given

p(x1, x2, x3) =∑

r+s+t=4

4!

r !s!t!arstx

r1x s

2x t3,

define the catalecticant Hp as a quadratic form in 6 variables (or a6× 6 symmetric matrix defined linearly in terms of p).

Hp =

a400 a220 a202 a310 a301 a211a220 a040 a022 a130 a121 a031a202 a022 a004 a112 a103 a013a310 a130 a112 a220 a211 a121a301 a121 a103 a211 a202 a112a211 a031 a013 a121 a112 a022

(This also has an apolar interpretation as q2(D)p.)


Under this definition, H(αk ·)4 is a perfect square. Thus if p is asum of five fourth powers, then rank(Hp) ≤ 5, so Hp is singular.This can’t happen for a general ternary quartic, where thedeterminant is non-zero. The vanishing of the determinant isprecisely the algebraic relation of the 15 coefficients in a sum offive fourth powers. Clebsch’s proof and Sylvester’s proof are reallythe same, but that would take another talk.

In his original paper, Sylvester apologized for introducing this term:“Meicatalecticizant would more completely express the meaning ofthat which, for the sake of brevity, I denominate the catalecticant.”Sylvester was very interested in the technical aspects of poetry anda “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.










Our 19th century ancestors saw that funny things happen in sumsof powers of linear forms. when (n, d) = (3, 4), (4, 4), (5, 4), (5, 3).In the early 1990s, Alexander and Hirschowitz proved that theseare the only cases in which this can happen. This is the basis ofanother talk, and probably one that’s better than this one,especially if someone else gives it.

The attribution “Lasker-Wakeford” comes from The theory ofdeterminants, matrices and invariants by H. W. Turnbull, who wasone of the last “pre-Hilbert” invariant theorists. This 1960 book isa Rosetta Stone for understanding 19th century algebra. Turnbulldescribed this theorem as “paradoxical and very curious”. It is the“apolar” version of the earlier result.

Theorem (Lasker-Wakeford)

If F : CN → Hd(Cn), then F is a canonical form if and only ifthere is a point u ∈ CN so that there is no non-zero form q whichis apolar to all N forms ∂F∂t1 (u), . . . , ∂F∂tN (u).


Our 19th century ancestors saw that funny things happen in sumsof powers of linear forms. when (n, d) = (3, 4), (4, 4), (5, 4), (5, 3).In the early 1990s, Alexander and Hirschowitz proved that theseare the only cases in which this can happen. This is the basis ofanother talk, and probably one that’s better than this one,especially if someone else gives it.The attribution “Lasker-Wakeford” comes from The theory ofdeterminants, matrices and invariants by H. W. Turnbull, who wasone of the last “pre-Hilbert” invariant theorists. This 1960 book isa Rosetta Stone for understanding 19th century algebra. Turnbulldescribed this theorem as “paradoxical and very curious”. It is the“apolar” version of the earlier result.

Theorem (Lasker-Wakeford)

If F : CN → Hd(Cn), then F is a canonical form if and only ifthere is a point u ∈ CN so that there is no non-zero form q whichis apolar to all N forms ∂F∂t1 (u), . . . , ∂F∂tN (u).


And now, a biographical interlude.

Emanuel Lasker (1868-1941) received his Ph.D. under MaxNoether at Gottingen in 1902. He first developed the concept of aprimary ideal and proved the primary decomposition theorem for anideal of a polynomial ring in terms of primary ideals.

He is probably better known for being the world chess championfor 27 years (1894-1921), which spanned the life of ...

Edward Kingsley Wakeford (1894-1916).

If you remember European history, those dates will give you pause.














The memorial article by J. H. Grace about Wakeford in the1917/1918 volume of Proceedings of the London MathematicalSociety may be the angriest obituary I’ve ever read in a scholarlyjournal, and it can be found in its entirety on my webpage:

http://www.math.uiuc.edu/∼reznick/wakeford.pdf

“He [EKW] was slightly wounded early in 1916, and soon aftercoming home was busy again with Canonical Forms.... [H]ediscovered a paper of Hilbert’s which contained the very theoremhe had long been in want of – first vaguely, and later quitedefinitely. This was in March; April found him, full of the mostjoyous 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 onlyneeded a chance, and he never got it.”












The question of canonical forms for binary forms as a sum ofpowers of linear forms was completely settled by Sylvester in 1851.

Theorem (Sylvester’s Canonical Forms)

(i) A general binary form of odd degree d = 2k − 1 can be writtenuniquely (i.e., up to indexing and roots of unity) as

k∑j=1

(αjx + βjy)2k−1.

(ii) A general binary form of even degree d = 2k can be writtenuniquely as

λ · x2k +k∑

j=1

(αjx + βjy)2k .

for some λ ∈ C.


In 1869, James Joseph Sylvester (1814-1897) reflected on thediscovery of his canonical forms, done while he was working as anactuary in the office next to Arthur Cayley’s.

“I discovered and developed the whole theory of canonical binaryforms for odd degrees, and, as far as yet made out, for evendegrees too, at one evening sitting, with a decanter of port wine tosustain nature’s flagging energies, in a back office in Lincoln’s InnFields. The work was done, and well done, but at the usual cost ofracking thought — a brain on fire, and feet feeling, or feelingless,as if plunged in an ice-pail. That night we slept no more.”

The ”λx2k” must be what Sylvester meant by “as far as yet madeout”. As if anticipating modern mathematics, Sylvester proved histheory with one brilliant algorithm.










Theorem (Sylvester)

Suppose p(x , y) =∑d

j=0

(dj

)ajx

d−jy j and αkx + βky is a set of

non-proportional linear factors. Let h(x , y) =∑r

t=0 ctxr−ty t

=∏r

j=1(βjx − αjy). Then there exist λk ∈ C so that

p(x , y) =r∑

k=1

λk(αkx + βky)d

if and only ifa0 a1 · · · ara1 a2 · · · ar+1...

.... . .

...ad−r ad−r+1 · · · ad

·

c0c1...

cr

=

00...0

.

(This is an algorithm, because you start with r = 1 and incrementuntil you find a null-vector which gives a square-free h.)


Theorem (Sylvester)


j=0

(dj

)ajx



t=0 ctxr−ty t

=∏r


p(x , y) =r∑

k=1

λk(αkx + βky)d


.... . .

...ad−r ad−r+1 · · · ad

·

c0c1...

cr

=

00...0

.



Theorem (Sylvester)


j=0

(dj

)ajx



t=0 ctxr−ty t

=∏r


p(x , y) =r∑

k=1

λk(αkx + βky)d


.... . .

...ad−r ad−r+1 · · · ad

·

c0c1...

cr

=

00...0

.



If d = 2s − 1 and r = s, then the matrix is s × (s + 1) and has anon-trivial null-vector. The corresponding h (given in terms of thecoefficients of p) has distinct factors unless its discriminantvanishes, giving the canonical form in odd degree.

If d = 2s and r = s, then the matrix is square, and in general,there exists λ so that p(x , y)− λx2s has a matrix with a non-trivialnull-vector as above, giving the canonical form in even degree.It can be shown without too much difficulty that if r = d , so thereis a single equation, then there is a null-vector whose resultingform is square-free, and thus, every binary form of degree d is asum of at most d d-th powers of linear forms.It is not hard to see that xd−1y is not a sum of d − 1 powers,because

(0 1 · · · 01 0 · · · 0

)·

c0c1...

cd−1

=

00...0

=⇒ c0 = c1 = 0.


If d = 2s − 1 and r = s, then the matrix is s × (s + 1) and has anon-trivial null-vector. The corresponding h (given in terms of thecoefficients of p) has distinct factors unless its discriminantvanishes, giving the canonical form in odd degree.If d = 2s and r = s, then the matrix is square, and in general,there exists λ so that p(x , y)− λx2s has a matrix with a non-trivialnull-vector as above, giving the canonical form in even degree.It can be shown without too much difficulty that if r = d , so thereis a single equation, then there is a null-vector whose resultingform is square-free, and thus, every binary form of degree d is asum of at most d d-th powers of linear forms.

It is not hard to see that xd−1y is not a sum of d − 1 powers,because

(0 1 · · · 01 0 · · · 0

)·

c0c1...

cd−1

=

00...0

=⇒ c0 = c1 = 0.


If d = 2s − 1 and r = s, then the matrix is s × (s + 1) and has anon-trivial null-vector. The corresponding h (given in terms of thecoefficients of p) has distinct factors unless its discriminantvanishes, giving the canonical form in odd degree.If d = 2s and r = s, then the matrix is square, and in general,there exists λ so that p(x , y)− λx2s has a matrix with a non-trivialnull-vector as above, giving the canonical form in even degree.It can be shown without too much difficulty that if r = d , so thereis a single equation, then there is a null-vector whose resultingform is square-free, and thus, every binary form of degree d is asum of at most d d-th powers of linear forms.It is not hard to see that xd−1y is not a sum of d − 1 powers,because

(0 1 · · · 01 0 · · · 0

)·

c0c1...

cd−1

=

00...0

=⇒ c0 = c1 = 0.


Here is an example of Sylvester’s algorithm in action. Let

p(x , y) = 3x5 − 20x3y2 + 10xy4 =(5

0

)· 3 x5 +

(5

1

)· 0 x4y +

(5

2

)· (−2) x3y2

+

(5

3

)· 0 x2y3 +

(5

4

)· 2 xy4 +

(5

5

)· 0 y5;

3 0 −2 00 −2 0 2−2 0 2 0

·

0101

=

000

.

We have h(x , y) = y(x2 + y2) = y(y − ix)(y + ix).Accordingly, there exist λk ∈ C so that

p(x , y) = λ1x5 + λ2(x + iy)5 + λ3(x − iy)5.

Indeed, λ1 = λ2 = λ3 = 1, as may be checked.


Here is an example of Sylvester’s algorithm in action. Let

p(x , y) = 3x5 − 20x3y2 + 10xy4 =(5

0

)· 3 x5 +

(5

1

)· 0 x4y +

(5

2

)· (−2) x3y2

+

(5

3

)· 0 x2y3 +

(5

4

)· 2 xy4 +

(5

5

)· 0 y5;

3 0 −2 00 −2 0 2−2 0 2 0

·

0101

=

000

.

We have h(x , y) = y(x2 + y2) = y(y − ix)(y + ix).Accordingly, there exist λk ∈ C so that

p(x , y) = λ1x5 + λ2(x + iy)5 + λ3(x − iy)5.

Indeed, λ1 = λ2 = λ3 = 1, as may be checked.


It would take us too far afield to spend much time on this, but thenumber of summands required depends on the field in which oneallows coefficients. A few years ago, I proved that the quintic fromthe last page is a sum of three fifth powers over a field F iff i ∈ F ,but is a sum of four fifth powers over Q(

√−2) and at best a sum

of at least five fifth powers over Q. I also showed that any binaryform of degree d in F [x , y ] can be written as a linear combinationof d d-th powers of linear forms over F .

My PhD student Neriman Tokcan and I have shown in aforthcoming paper that this phenomenon of three different“lengths” occurs in all degrees ≥ 5. Our examples are xkyk(x − y)for k ≥ 2 and xkyk for k ≥ 3.Another theorem of Sylvester (this time from 1863) implies that ifp(x , y) is a hyperbolic real form of degree d (that is, p splits overR), then p cannot be written as a sum of fewer than d real d-thpowers.




of at least five fifth powers over Q. I also showed that any binaryform of degree d in F [x , y ] can be written as a linear combinationof d d-th powers of linear forms over F .My PhD student Neriman Tokcan and I have shown in aforthcoming paper that this phenomenon of three different“lengths” occurs in all degrees ≥ 5. Our examples are xkyk(x − y)for k ≥ 2 and xkyk for k ≥ 3.

Another theorem of Sylvester (this time from 1863) implies that ifp(x , y) is a hyperbolic real form of degree d (that is, p splits overR), then p cannot be written as a sum of fewer than d real d-thpowers.




of at least five fifth powers over Q. I also showed that any binaryform of degree d in F [x , y ] can be written as a linear combinationof d d-th powers of linear forms over F .My PhD student Neriman Tokcan and I have shown in aforthcoming paper that this phenomenon of three different“lengths” occurs in all degrees ≥ 5. Our examples are xkyk(x − y)for k ≥ 2 and xkyk for k ≥ 3.Another theorem of Sylvester (this time from 1863) implies that ifp(x , y) is a hyperbolic real form of degree d (that is, p splits overR), then p cannot be written as a sum of fewer than d real d-thpowers.


Getting back to constant-counting, it is easy to see that if m < r ,then any polynomial map F : Cm → Cr must have the propertythat fj(t1, . . . tm) must be algebraically dependent. In fact,Hilbert proved this, and it was considered a highly abstracttheorem at the time. Folks in the 19th century really didn’t knowthe dimension of a vector space!

To simplify matters, Suppose d = max(deg fj) and k is a positive

integer. There are(r+k

r

)“monomials” f k1

1 . . . f krr with

∑j kj ≤ k.

This number grows like k r/r ! in k . On the other hand, thesepolynomials live in the vector space of polynomials in (t1, . . . , tm)of degree ≤ dk, which has dimension

(m+dkm

), and grows like

(dk)m/m! in k . Since r > m,(r+k

r

)>(m+dk

m

)for sufficiently large

k, and so the monomials are linearly dependent. This gives thealgebraic dependence of the fj ’s.


Getting back to constant-counting, it is easy to see that if m < r ,then any polynomial map F : Cm → Cr must have the propertythat fj(t1, . . . tm) must be algebraically dependent. In fact,Hilbert proved this, and it was considered a highly abstracttheorem at the time. Folks in the 19th century really didn’t knowthe dimension of a vector space!

To simplify matters, Suppose d = max(deg fj) and k is a positive

integer. There are(r+k

r

)“monomials” f k1

1 . . . f krr with

∑j kj ≤ k.

This number grows like k r/r ! in k . On the other hand, thesepolynomials live in the vector space of polynomials in (t1, . . . , tm)of degree ≤ dk, which has dimension

(m+dkm

), and grows like

(dk)m/m! in k . Since r > m,(r+k

r

)>(m+dk

m

)for sufficiently large

k, and so the monomials are linearly dependent. This gives thealgebraic dependence of the fj ’s.


Here’s an example of how the proof overstates the degrees needed.Suppose you want to describe ternary quadratic forms which factoras

(x + ay + bz)(x + cy + dz) = x2 +∑i ,j ,k

cijkx iy jzk

Ignoring x2, the cijk ’s give five polynomials in the four variables(a, b, c , d): a + c , b + d , ac, bd , ad + bc. Using a more carefulversion of the previous argument, even accounting for two linearand three quadratic polynomials, the best we can say is that therehas to be an algebraic relation of degree 22. (Yes, I usedMathematica.)However any such quadratic form has rank two, so∣∣∣∣∣∣

1 (a + c)/2 (b + d)/2(a + c)/2 ac (ad + bc)/2(b + d)/2 (ad + bc)/2 bd

∣∣∣∣∣∣ = 0.



(x + ay + bz)(x + cy + dz) = x2 +∑i ,j ,k

cijkx iy jzk

Ignoring x2, the cijk ’s give five polynomials in the four variables(a, b, c , d): a + c , b + d , ac, bd , ad + bc. Using a more carefulversion of the previous argument, even accounting for two linearand three quadratic polynomials, the best we can say is that therehas to be an algebraic relation of degree 22. (Yes, I usedMathematica.)

However any such quadratic form has rank two, so∣∣∣∣∣∣1 (a + c)/2 (b + d)/2

(a + c)/2 ac (ad + bc)/2(b + d)/2 (ad + bc)/2 bd

∣∣∣∣∣∣ = 0.



(x + ay + bz)(x + cy + dz) = x2 +∑i ,j ,k

cijkx iy jzk

Ignoring x2, the cijk ’s give five polynomials in the four variables(a, b, c , d): a + c , b + d , ac, bd , ad + bc. Using a more carefulversion of the previous argument, even accounting for two linearand three quadratic polynomials, the best we can say is that therehas to be an algebraic relation of degree 22. (Yes, I usedMathematica.)However any such quadratic form has rank two, so∣∣∣∣∣∣

1 (a + c)/2 (b + d)/2(a + c)/2 ac (ad + bc)/2(b + d)/2 (ad + bc)/2 bd

∣∣∣∣∣∣ = 0.


Let’s talk about cubes.I’d like to begin with a simple question. Consider a sum of twocubes of quadratic forms:

2∑j=1

(αj0x2 + αj1xy + αj2y2)3 =6∑

k=0

ckx6−kyk ,

One can view the seven ck ’s as cubic polynomials in the six α′j`s,and since 7 > 6, we know that the ck ’s must be algebraicallydependent. There are

(n+66

)monomials in the cj ’s of degree n;

these are forms of degree 3n in the α′j`s, which comprise a vector

space of dimension(3n+5

5

). And, eventually,(

n + 6

6

)>

(3n + 5

5

)so there must be dependence at some degree n.


Let’s talk about cubes.I’d like to begin with a simple question. Consider a sum of twocubes of quadratic forms:

2∑j=1

(αj0x2 + αj1xy + αj2y2)3 =6∑

k=0

ckx6−kyk ,

One can view the seven ck ’s as cubic polynomials in the six α′j`s,and since 7 > 6, we know that the ck ’s must be algebraicallydependent. There are

(n+66

)monomials in the cj ’s of degree n;

these are forms of degree 3n in the α′j`s, which comprise a vector

space of dimension(3n+5

5

). And, eventually,(

n + 6

6

)>

(3n + 5

5

)so there must be dependence at some degree n.


Unfortunately, (1441 + 6

6

)>

(3× 1441 + 5

5

)is the first time this occurs. Various smart computational algebraicgeometers I’ve asked over the years have been unable to bruteforce this relation. It tends to kill the kernel.

Is there a better way to do this? The answer is yes, and it turnsout that there is a relation of degree n = 15 in the cj ’s, admittedlywith more than 1000 terms.

This quindecic form is derived from a simple characterization ofbinary sextics which are a sums of two cubes of binary quadratics.

Let’s go to the characterization. There are two equivalentstatements.



6

)>

(3× 1441 + 5

5







6

)>

(3× 1441 + 5

5






Theorem

Suppose p is a binary sextic. Then p is a sum of two cubes ofquadratics if and only if:(i) p is a perfect cube or p = f1f2f3, where the fi ’s are linearlydependent but non-proportional quadratic forms.(ii) There exists an invertible linear change of variables after whichp equals either g(x2, y2) or `3g for some linear form `, where g isa cubic which is a sum of two cubes (i.e., g 6= `21`2.)

The proof of (i) is part of a more general result about sums of twocubes of forms.The proof of (ii) relies on the ancient art of simultaneousdiagonalization: if q and r are two binary quadratic forms, theneither they share a common factor, or they can be simultaneouslydiagonalized.(Also note: an even form under (x , y) 7→ (x + y , x − y) becomes asymmetric form, and vice versa.)


Theorem

Suppose p is a binary sextic. Then p is a sum of two cubes ofquadratics if and only if:(i) p is a perfect cube or p = f1f2f3, where the fi ’s are linearlydependent but non-proportional quadratic forms.(ii) There exists an invertible linear change of variables after whichp equals either g(x2, y2) or `3g for some linear form `, where g isa cubic which is a sum of two cubes (i.e., g 6= `21`2.)

The proof of (i) is part of a more general result about sums of twocubes of forms.The proof of (ii) relies on the ancient art of simultaneousdiagonalization: if q and r are two binary quadratic forms, theneither they share a common factor, or they can be simultaneouslydiagonalized.(Also note: an even form under (x , y) 7→ (x + y , x − y) becomes asymmetric form, and vice versa.)


Theorem (Either mine, or very old and obscure, or both)

Suppose F ∈ C[x1, . . . , xn]. Then F = G 3 + H3 for forms G ,H ifand only if either F = K 3, or F = G1G2G3, where the Gj ’s arenon-proportional, but linearly dependent factors.

Proof.

First G 3 + H3 = (G + H)(G + ωH)(G + ω2H), where ω = e2πi3 ,

and if two of the factors G + ωjH are proportional, then so are Gand H, and hence F is a cube. In any event, please observe that(G + H) +ω(G + ωH) + ω2(G + ω2H) = 0.

Conversely, if F has such a factorization, there exist 0 6= α, β ∈ Cso that F = G1G2(αG1 + βG2). It is easily checked that

3αβ(ω − ω2)F = (ω2αG1 − ωβG2)3 − (ωαG1 − ω2βG2)3.

Note that 3αβ(ω − ω2) = 3√−3 αβ 6= 0.


Theorem (Either mine, or very old and obscure, or both)

Suppose F ∈ C[x1, . . . , xn]. Then F = G 3 + H3 for forms G ,H ifand only if either F = K 3, or F = G1G2G3, where the Gj ’s arenon-proportional, but linearly dependent factors.

Proof.

First G 3 + H3 = (G + H)(G + ωH)(G + ω2H), where ω = e2πi3 ,

and if two of the factors G + ωjH are proportional, then so are Gand H, and hence F is a cube. In any event, please observe that(G + H) +ω(G + ωH) + ω2(G + ω2H) = 0.Conversely, if F has such a factorization, there exist 0 6= α, β ∈ Cso that F = G1G2(αG1 + βG2). It is easily checked that

3αβ(ω − ω2)F = (ω2αG1 − ωβG2)3 − (ωαG1 − ω2βG2)3.

Note that 3αβ(ω − ω2) = 3√−3 αβ 6= 0.


In any particular case, if deg F = 3r , there are, up to multiple, only(3r)!3!(r !)3

ways to write F as a product of three factors of degree r , so

checking this condition is algorithmic.

In particular, if F (x , y) is a binary cubic form, then it has threelinear factors `j(x , y) = αjx + βjy , and these are alwaysdependent. Thus, as Sylvester and our 19th century predecessorsknew, a binary cubic F is a sum of two cubes unless it has a squarefactor (and isn’t a cube).


In any particular case, if deg F = 3r , there are, up to multiple, only(3r)!3!(r !)3

ways to write F as a product of three factors of degree r , so

checking this condition is algorithmic.

In particular, if F (x , y) is a binary cubic form, then it has threelinear factors `j(x , y) = αjx + βjy , and these are alwaysdependent. Thus, as Sylvester and our 19th century predecessorsknew, a binary cubic F is a sum of two cubes unless it has a squarefactor (and isn’t a cube).


The second case uses a simple old lemma whose proof is omitted.

Lemma

Two quadratic forms q1(x , y) and q2(x , y) either have a commonlinear factor, or can be simultaneously diagonalized; that is,qj(ax + by , cx + dy) = ρjx

2 + σjy2.

Thus, if p = qt1 + qt

2, where qj is quadratic, then either the qj ’shave a common linear factor (and p = `tg , where g is a sum oftwo linear t-th powers), or after a linear change of variables,

p(ax + by , cx + dy) =2∑

j=1

(ρjx2 + σjy

2)t ;

That is, p(ax + by , cx + dy) = g(x2, y2), where g again is a sumof two linear t-th powers (typical for t = 3, not for t > 3.)


The second case uses a simple old lemma whose proof is omitted.

Lemma

Two quadratic forms q1(x , y) and q2(x , y) either have a commonlinear factor, or can be simultaneously diagonalized; that is,qj(ax + by , cx + dy) = ρjx

2 + σjy2.

Thus, if p = qt1 + qt

2, where qj is quadratic, then either the qj ’shave a common linear factor (and p = `tg , where g is a sum oftwo linear t-th powers), or after a linear change of variables,

p(ax + by , cx + dy) =2∑

j=1

(ρjx2 + σjy

2)t ;

That is, p(ax + by , cx + dy) = g(x2, y2), where g again is a sumof two linear t-th powers (typical for t = 3, not for t > 3.)


Finding if p is even after a change of variables is also algorithmic.

p(x , y) =2d−1∏j=0

(x − λjy) =⇒

p(ax + by , cx + dy) = p(a,−c)2d−1∏j=0

(x −

(λjd − b

a− λjc

)y

)

:= p(a,−c)2d−1∏j=0

(x − µjy).

Thus, the roots of p (taking ∞ if y | p) are mapped by a Mobiustransformation. If p(x , y) = p(ax + by , cx + dy) is even, thenT (z) = −z is an involution on the roots, say T (µ2j) = µ2j+1. Itfollows that there is an involutory Mobius transformation Upermuting the d pairs of roots of p; to be specific:

λ2j+1 =2ad − (ad + bc)λ2j(ad + bc)− 2cdλ2j

.


The algorithm is this: Given p, find the roots λj , and for eachquadruple λi1 , λi2 , λi3 , λi4 , define the Mobius transformation U sothat U(λi1) = λi2 , U(λi2) = λi1 and U(λi3) = λi4 and see if itpermutes the others. There are instances in which more than oneU may work; for example, if p is both even and symmetric.

Don’t get me wrong. Complications abound. Here’s a simple one.Consider the even sextic

p(x , y) = x6 − x4y2 − x2y4 + y6 = (x2 − y2)2(x2 + y2).

Here, p(x , y) = g(x2, y2), where g(x , y) = (x − y)2(x + y)(having a square factor) is unfortunately not a sum of two cubes.On the other hand, if γ = 2√

3i , then

p(x , y) = (x2 + 2xy + y2)(x2 + y2)(x2 − 2xy + y2) =⇒2p(x , y) = (x2 + γxy + y2)3 + (x2 − γxy + y2)3.


Now let’s suppose our given cubic p is a sum of two cubes, factorit and expand it in the usual way. Write p as

6∑k=0

ckx6−kyk = c0

(x6 +

6∑k=1

ekx6−kyk

)= c0

6∏j=1

(x + rjy),

where the ek ’s are the elementary symmetric functions.

There are 15 ways to divide the 6 rj ’s into 3 pairs of roots, and thecondition that the quadratic factors be dependent for some choiceof factorization is equivalent to the vanishing of

H(r) :=15∏`=1

∣∣∣∣∣∣1 1 1

rσ`(1) + rσ`(2) rσ`(3) + rσ`(4) rσ`(5) + rσ`(6)rσ`(1)rσ`(2) rσ`(3)rσ`(4) rσ`(5)rσ`(6)

∣∣∣∣∣∣ .This is an I-really-hope-it’s-symmetric polynomial of degree 45 inthe rj ’s.


Now let’s suppose our given cubic p is a sum of two cubes, factorit and expand it in the usual way. Write p as

6∑k=0

ckx6−kyk = c0

(x6 +

6∑k=1

ekx6−kyk

)= c0

6∏j=1

(x + rjy),

where the ek ’s are the elementary symmetric functions.There are 15 ways to divide the 6 rj ’s into 3 pairs of roots, and thecondition that the quadratic factors be dependent for some choiceof factorization is equivalent to the vanishing of

H(r) :=15∏`=1

∣∣∣∣∣∣1 1 1

rσ`(1) + rσ`(2) rσ`(3) + rσ`(4) rσ`(5) + rσ`(6)rσ`(1)rσ`(2) rσ`(3)rσ`(4) rσ`(5)rσ`(6)

∣∣∣∣∣∣ .This is an I-really-hope-it’s-symmetric polynomial of degree 45 inthe rj ’s.


It is symmetric.Mathematica can compute H(r) without too much difficulty, andin 11657.87 seconds transform it into a symmetric function in theek ’s of degree 15. Now write ek = ck/c0, make the substitutionand multiply by c15

0 to get the relation. It has 1360 terms, so Iwon’t write it here. I also need to express it in terms of thefundamental invariants of the binary sextic, and haven’t done soyet. It is isobaric in the old sense, each monomial

∏cmkk has∑

mk = 15,∑

kmk = 45.

You can use it to check that a generic pencil of sectic contains afinite number of sums of two cubes, but it is easy to get some veryugly equations. For example, (x2 + y2)3 + ax5y is a sum of twocubes if and only if

a ∈

0,±1

12√

3

125× (25±2

√−15)

.

The non-zero values have modulus 96√30

125 .


It is symmetric.Mathematica can compute H(r) without too much difficulty, andin 11657.87 seconds transform it into a symmetric function in theek ’s of degree 15. Now write ek = ck/c0, make the substitutionand multiply by c15

0 to get the relation. It has 1360 terms, so Iwon’t write it here. I also need to express it in terms of thefundamental invariants of the binary sextic, and haven’t done soyet. It is isobaric in the old sense, each monomial

∏cmkk has∑

mk = 15,∑

kmk = 45.You can use it to check that a generic pencil of sectic contains afinite number of sums of two cubes, but it is easy to get some veryugly equations. For example, (x2 + y2)3 + ax5y is a sum of twocubes if and only if

a ∈

0,±1

12√

3

125× (25±2

√−15)

.

The non-zero values have modulus 96√30

125 .


Part of an ongoing project with Samuel Lundqvist, AlessandroOneto and Boris Shapiro.

Theorem

There is an algorithm for writing every binary sextic in C[x , y ] as asum of three cubes of quadratic forms.

Write the binary sextic (warning: different notation) as

p(x , y) =6∑

k=0

(6

k

)akx6−kyk .

Given p 6= 0, we may always make an invertible change of variablesto ensure that p(0, 1)p(1, 0) 6= 0; hence, assume a0a6 6= 0.

By an observation of ad hoc,

q(x , y) = x2 + 2a1a0

x y +5a0a2−4a21

a20y2

=⇒ a0q3(x , y) = a0x6 + 6a1x5y + 15a2x4y2 + ...



Theorem



p(x , y) =6∑

k=0

(6

k

)akx6−kyk .



q(x , y) = x2 + 2a1a0

x y +5a0a2−4a21

a20y2

=⇒ a0q3(x , y) = a0x6 + 6a1x5y + 15a2x4y2 + ...



Theorem



p(x , y) =6∑

k=0

(6

k

)akx6−kyk .



q(x , y) = x2 + 2a1a0

x y +5a0a2−4a21

a20y2

=⇒ a0q3(x , y) = a0x6 + 6a1x5y + 15a2x4y2 + ...



Theorem



p(x , y) =6∑

k=0

(6

k

)akx6−kyk .



q(x , y) = x2 + 2a1a0

x y +5a0a2−4a21

a20y2

=⇒ a0q3(x , y) = a0x6 + 6a1x5y + 15a2x4y2 + ...


Thus there always exists a cubic c such that

p(x , y)− a0q(x , y)3 = y3c(x , y).

Usually, (p − a0q3)/y3 = c is a sum of 2 cubes of linear forms,from which it follows that p is a sum of 3 cubes. As we’ve seen,this only fails if c has a square factor. Trusting Mathematica, thediscriminant of c(x , y) is a non-zero polynomial in the ai ’s ofdegree 18, divided by a140 for general p.

We now consider the remaining cases in which this first approachfails. Such a failure will have the shape

p(x , y) = (ax2 + bxy + cy2)3 + y3(rx + sy)2(tx + uy)

where ru − st 6= 0, so that c(x , y) genuinely is not a sum of twocubes.


Let pT (x , y) = p(x ,Tx + y) and write

pT (x , y) =6∑

k=0

(6

k

)ak(T )x6−kyk .

Here, ak is a polynomial in T of degree 6− k and a6(T ) = a6 6= 0.There are at most 6 values of T which must be avoided to ensurethat a0(T ) 6= 0.

Repeating the same construction as above to pT , we find that thediscriminant is a polynomial of degree 72 in T with coefficients ina, b, c , r , s, t, u and tens of thousands of terms. It turns out,tediously, that for every non-trivial choice of (a, b, c , d , r , s, t, u),this discriminant gives a non-zero polynomial in T . (Cased out,not completely trusting in “Solve”. Sorry, Danny.)

Hence by avoiding finitely many values of T , the previousargument will work successfully on pT to give it as a sum of threecubes. We then reverse the invertible transformations and get anexpression for p itself.


For example, suppose p(x , y) = x6 + x5y + x4y2 + x3y3 + x2y4

+xy5 + y6. Then

p(x , y)−(x2 + 1

3xy + 29y2)3

=

7

729y3(54x3 + 81x2y + 99xy2 + 103y3).

An application of Sylvester’s algorithm shows that

54x3 + 81x2y + 99xy2 + 103y3 =

m1(78x + (173−√

20153)y)3 + m2(78x + (173 +√

20153)y)3,

m1 =20153 + 134

√20153

354209128, m2 =

20153− 134√

20153

354209128

This gives a simple sextic p as a sum of three cubes in an ugly wayand gives no hint about the existence of the formula

p(x , y) =∑±

(9±√−3

18

)(x2 + 1±

√−3

2 xy + y2)3.


For example, suppose p(x , y) = x6 + x5y + x4y2 + x3y3 + x2y4

+xy5 + y6. Then

p(x , y)−(x2 + 1

3xy + 29y2)3

=

7

729y3(54x3 + 81x2y + 99xy2 + 103y3).

An application of Sylvester’s algorithm shows that

54x3 + 81x2y + 99xy2 + 103y3 =

m1(78x + (173−√

20153)y)3 + m2(78x + (173 +√

20153)y)3,

m1 =20153 + 134

√20153

354209128, m2 =

20153− 134√

20153

354209128

This gives a simple sextic p as a sum of three cubes in an ugly wayand gives no hint about the existence of the formula

p(x , y) =∑±

(9±√−3

18

)(x2 + 1±

√−3

2 xy + y2)3.


An alternative approach is to observe that for a sextic p, there isusually a quadratic q so that p − q3 is even. (Look at thecoefficients of x5y , x3y3, xy5 and solve the equations for thecoefficients of q.) Then p − q3 is a cubic in x2, y2 and so isusually a sum of two cubes of even quadratic form. If this doesn’twork, apply it to pT .

We do not know how to completely characterize the symmetries ofthe sets of sums of three cubes for a given p. Or if a real binarysextic is a sum of three cubes of real quadratic forms. (Question ofLek-Heng Lim at a conference in China.)This heavy reliance on tools from Ecole de calcul ad hoc can onlytake you so far. There are two natural next steps; based on theobservation that 8 = 4× 2 and 9 = 3× 3. Is every binary octic asum of four 4th powers of quadratic forms? Is every binary nonic asum of three cubes of quadratic forms? One more fun fact:according to the Oxford English Dictionary (as well as wikipedia),an obsolete term for the 4th power is zenzizenzic.


An alternative approach is to observe that for a sextic p, there isusually a quadratic q so that p − q3 is even. (Look at thecoefficients of x5y , x3y3, xy5 and solve the equations for thecoefficients of q.) Then p − q3 is a cubic in x2, y2 and so isusually a sum of two cubes of even quadratic form. If this doesn’twork, apply it to pT .We do not know how to completely characterize the symmetries ofthe sets of sums of three cubes for a given p. Or if a real binarysextic is a sum of three cubes of real quadratic forms. (Question ofLek-Heng Lim at a conference in China.)

This heavy reliance on tools from Ecole de calcul ad hoc can onlytake you so far. There are two natural next steps; based on theobservation that 8 = 4× 2 and 9 = 3× 3. Is every binary octic asum of four 4th powers of quadratic forms? Is every binary nonic asum of three cubes of quadratic forms? One more fun fact:according to the Oxford English Dictionary (as well as wikipedia),an obsolete term for the 4th power is zenzizenzic.


An alternative approach is to observe that for a sextic p, there isusually a quadratic q so that p − q3 is even. (Look at thecoefficients of x5y , x3y3, xy5 and solve the equations for thecoefficients of q.) Then p − q3 is a cubic in x2, y2 and so isusually a sum of two cubes of even quadratic form. If this doesn’twork, apply it to pT .We do not know how to completely characterize the symmetries ofthe sets of sums of three cubes for a given p. Or if a real binarysextic is a sum of three cubes of real quadratic forms. (Question ofLek-Heng Lim at a conference in China.)This heavy reliance on tools from Ecole de calcul ad hoc can onlytake you so far. There are two natural next steps; based on theobservation that 8 = 4× 2 and 9 = 3× 3. Is every binary octic asum of four 4th powers of quadratic forms? Is every binary nonic asum of three cubes of quadratic forms? One more fun fact:according to the Oxford English Dictionary (as well as wikipedia),an obsolete term for the 4th power is zenzizenzic.


The octic (or zenzizenzizenzic) case is interesting because acanonical form for the octics gives them as a sum of three fourthpowers, and the Conjecture is still true even if some singular casesrequire one more.

In my earlier canonical form project, I noted that 7 = 3 + 4 and ageneral binary sextic is a sum of a cubic squared plus a quadraticcubed, but I don’t have an algorithm for it.

p(x , y) = f 2(x , y) + g3(x , y) =

(t1x3 + t2x2y + t3xy2 + t4y3)2 + (t5x2 + t6xy + t7y2)3.

Also, 15 = 3× 3 + 1× 6, and

p(x1, x2, x3) =3∑

k=1

(αk1x1 + αk2x2 + αk3x3)4 + (q(x1, x2, x3))2,

for quadratic q is a quadratic form for ternary quartics.


The octic (or zenzizenzizenzic) case is interesting because acanonical form for the octics gives them as a sum of three fourthpowers, and the Conjecture is still true even if some singular casesrequire one more.In my earlier canonical form project, I noted that 7 = 3 + 4 and ageneral binary sextic is a sum of a cubic squared plus a quadraticcubed, but I don’t have an algorithm for it.

p(x , y) = f 2(x , y) + g3(x , y) =

(t1x3 + t2x2y + t3xy2 + t4y3)2 + (t5x2 + t6xy + t7y2)3.

Also, 15 = 3× 3 + 1× 6, and

p(x1, x2, x3) =3∑

k=1

(αk1x1 + αk2x2 + αk3x3)4 + (q(x1, x2, x3))2,

for quadratic q is a quadratic form for ternary quartics.


Here are some references of papers referred to in the talk. Theseare all available on my website

Homogeneous polynomial solutions to constant coefficient PDE’sover fields, Adv. Math., 117 (1996), 179-192 (MR97a:12006).

On the length of binary forms, Quadratic and Higher DegreeForms, (K. Alladi, M. Bhargava, D. Savitt, P. Tiep, eds.),Developments in Math. 31 (2013), Springer, New York, pp.207-232, http://arxiv.org/pdf/1007.5485.pdf, MR3156559.

Some new canonical forms for polynomials, Pac. J. Math., 266(2013), 185220, http://arxiv.org/abs/1203.5722, MR3105781.

(with Neriman Tokcan) Binary forms with three different relativeranks, http://front.math.ucdavis.edu/1608.08560, to appear inProc. Amer. Math. Soc.

(with Samuel Lundqvist, Alessandro Oneto and Boris Shapiro), Ongeneric and maximal k-ranks of binary forms, in preparation.


Thank you for your attention. I want to thank Danny Lichtblau forthe invitation to speak.

I also want to thank him for arranging such comfortableaccommodations. I woke up this morning and I almost felt like Iwas home.


Thank you for your attention. I want to thank Danny Lichtblau forthe invitation to speak.

I also want to thank him for arranging such comfortableaccommodations. I woke up this morning and I almost felt like Iwas home.


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Documents

Rainy Day Lemmas # 12 and 35reznick/eccad42917.pdf · 2017. 4. 30. · Rainy Day Lemmas # 12 and 35 Bruce Reznick University of Illinois at Urbana-Champaign East Coast Computer Algebra