border ranks of monomials - inria · the basic construction, and a certi cate for high border rank...

Border Ranks of Monomials

Luke Oeding, Auburn University

Different expressions for monomials:

Some minimal rank expressions of monomials as sums of powers of linear forms:

4 · xy = (x + y)2 − (x − y)2 (Rank 2)

24 · xyz = (x + y + z)3 + (x − y − z)3 + (−x − y + z)3 + (−x + y − z)3 (Rank 4)

192 · xyzw = (x + y + z + w)4 − (−x + y + z + w)4 − (x − y + z + w)4 − (x + y − z + w)4

−(x + y + z − w)4 + (−x − y + z + w)4 + (−x + y − z + w)4 + (−x + y + z − w)4

(Rank 8)

6x2y = (x + y)3 + (−x + y)3 − 2y 3 (Rank 3)

Rank(x2y) = 3 but you can use only 2 cubes with a limit (so it has border rank 2): x2y = limε→0

x3 − (x − εy)3

3ε

How many powers of linear forms do you need in a limiting expression for xyzw?

Special cases of previous results: [explicit expression ] ≤ 8. [Landsberg–Teitler’09]: ≥ 6. [Guan’15] ≥ 7.

Find upper bounds via explicit expressions, lower bounds via (determinantal) polynomial certificates.

How about other monomials?

Semi-continuity, and the Geometry of Border Rank

Let f ∈ k[x0, . . . , xn]d = SdCn+1 (homogeneous polynomials of total degree d).

The Waring rank of f is the minimal r in an expression f =∑r

i=1 ci`di , with ci scalars and `i linear forms.

The fact that we have minimal expressions like

6x2y = (x + y)3 + (−x + y)3 − 2y 3 = limε→0

x3 − (x − εy)3

ε

says that Waring rank is generally not upper semi-continuous (unlike matrix rank).

The border rank of f is the minimal r such that f is in the closure of forms of rank r .

Geometrically, σ0r is the set of forms of rank ≤ r , its closure σr = σ0

r is the r -secant variety of the Veronese.

The border rank of f is the smallest r such that f ∈ σr \ σr−1 , and border rank is upper semi-continuous.

Non-trivial equations in the ideal of σr , allow us to falsify membership, and provide lower bounds for border rank.

Previous results for x0 · · · xn

[Landsberg-Teitler 2009]

[Guan 2015]

Find upper bounds via explicit expressions, lower bounds via (determinantal) polynomial certificates.How about other monomials?

Let α = (α0 ≥ · · · ≥ αn > 0) ∈ Nn+1 be a partition of d .

Conjecture

The border rank of xα = xα00 · · · x

αnn is

∏ni=1(1 + αi ).

Landsberg and Teitler [LT10] gave the upper bound and proved the conjecture in the case α0 ≥ α1 + · · ·+ αn.About the Waring rank of monomials, the following appeared in 2012.

Theorem (Carlini–Catalisano–Geramita, Buczynska–Buczynski–Teitler)

The rank of xα = xα00 · · · x

αnn is

∏n−1i=0 (1 + αi ).

Combining the conjecture and theorem with [Ranestad–Schreyer’11, Bernardi–Brachat–Mourrain ’14]:

Corollary

For monomials the border rank, smoothable rank, and cactus rank all agree.In addition, for a monomial xα the rank exceeds the border rank by a factor of α0+1

αn+1. In particular,

Brank xα = Rank xα if and only if α = (d , . . . , d).

Theorem (Oeding)

Suppose xα = xα00 · · · x

αnn with α0 ≥ α1 ≥ · · · ≥ αn. If (α1, . . . , αn) � (α0, αn, αn−1, . . . , α2) (dominance order),

then the border rank of xα = xα00 · · · x

αnn is evidenced by a Young flattening and is equal to

∏ni=1(1 + αi ).

Some examples:

xyzw had border rank (1 + 1)(1 + 1)(1 + 1) = 8 since α = (1, 1, 1, 1): (1, 1, 1) � (1, 1, 1).

Similarly x0x1 · · · xn has border rank 2n since α = (1, . . . , 1): (1, . . . , 1) � (1, . . . , 1).

And (x0x1 · · · xn)e has border rank (e + 1)n since α = (e, . . . , e): (e, . . . , e) � (e, . . . , e).

Other examples include monomials like x4y 3z2w , which has border rank (3 + 1) · (2 + 1) · (1 + 1) = 24,because α = (4, 3, 2, 1): (3, 2, 1) � (4, 1, 2).

note: 3 ≤ 4, 3 + 2 ≤ 4 + 1, 3 + 2 + 1 ≤ 4 + 1 + 2

Flattenings, Catalecticants, Hankel matrices, Moments, . . .

Input: f = x2y ∈ S3(C2)

Flattening: S2(C2)∗C(f )−→ C2 ∗

∗ ∗

Catalecticant (Hankel) C(f ) =

∂2x ∂x∂y ∂2

y( )∂x 0 2 0∂y 2 0 0

Rank(C(f )) = 2 implies that Brank(f ) ≥ 2.

For low rank (border rank) Sylvester’s algorithm (1850’s) decomposes f into a sum of powers of linear forms.See [Iarrobino–Kanev 1999], [Brachat–Comon–Mourrain–Tsigaridas 2010], [Oeding–Ottaviani 2013].

Easily implemented in Macaulay2:

R = QQ[x,y]

C = transpose basis(1,R)* basis(2,R)

f = x^2*y

Cf = diff(C,f)

Catalecticants in general [Iarrobino-Kanev 1999]Input: f ∈ Sd(V ) V = Cn+1.

1 Construct Cm(f ), (typically m = d d2e)

Cm(f ) : SmV ∗ −→ Sd−mV n + 1

d︷︸︸︷∗ ··· ∗

∗ ∗ ··· ∗

xi1 · · · xim 7−→ ∂mf

∂xi1 · · · ∂xim2 Macaulay (1916) showed: Brank(f ) ≥ RankCm(f ) for all 1 ≤ m ≤ d .

I For each m, Cm(f ) is a linear map depending linearly on f .I RankCm(xd ) = 1. Sub-additivity of matrix rank implies Macaulay’s result.

Example: xyz flattened via S2C3 → C3:

C(f ) =

∂2x ∂x∂y ∂x∂z ∂2

y ∂y∂z ∂2z( )

∂x 0 0 0 0 1 0∂y 0 0 1 0 0 0∂z 0 1 0 0 0 0

∗

∗ ∗

So Rank(xyz) ≥ 3 and hence Brank(xyz) ≥ 3. But this is all usual flattenings can tell us in this case.

Koszul flattenings get a bit further and show that Brank(xyz) ≥ 4 (Ottaviani’09), but don’t tell the whole story.

Landsberg and Ottaviani’s “Young flattenings”

The basic construction, and a certificate for high border rank

Let A, B be vector spaces over C.

Let F(p) : A→ B be linear, depending linearly on a (polynomial) input p.

If F(xd) has rank m (the multiplier), then F (`d1 + `d2 + · · ·+ `dr ) has rank at most m · r .

Thus, if F(p) has rank R, then the border rank of p is at least R/m.

Use Representation Theory for clever choices of F , A, and B for large R/m.

SπV is the Schur module associated with a partition π; a representation of SL(V ).The Pieri rule tells how to multiply some representations:

SπV⊗SdV =⊕ν`d+e

ν−π∈HS(d)

SνV

Restrict to one SνV , get a linear map F : SπV⊗SdV → SνV .Re-interpretation: Fπ,ν( ) : SπV → SνV is a linear map depending linearly on SdV .

Our tasks: Find RankFπ,ν(xd), and compute RankFπ,ν(p) for interesting polynomials p.

Some Combinatorics (see Fulton-Harris)

SSYT= Semi-Standard Young Tableaux of shape π and content {0, . . . , n}.

If V ∼= Cn+1, dim SπV = #SSYTπ{0, . . . , n} =∏

(i,j)∈Yπ

n + 1 + j − i

hi,j,

where hi,j is the hook-length of the hook with corner at (i , j).

It is convenient to express this as a ratio of tableaux, taking the product of the contents of each of the tableaux.For instance,

dimS4,3,2,1C4 =

4 5 6 73 4 52 31

/7 5 3 15 3 13 11

= 64.

Proposition

Suppose π = (n, n − 1, n − 2, . . . , 1, 0), and V ∼= Cn+1. Then dim SπV = 2(n+12 ).

Young Flattenings in practice

The labelled Pieri product: (Think of a b c as a ⊗ b ⊗ c + permutations.)

Example

The labeled Pieri product of 0 01 and 2 2 2 is

0 01⊗ 2 2 2 = 6 ·

(0 0 2 2 21

+ 0 0 2 21 2

+0 0 2 212

+0 0 21 22

).

The labeled Pieri product of 0 01 and 0 1 2 is

0 01 ⊗ 0 1 2 = 6 · 0 0 0 1 2

1 + 2 ·

0 0 0 11 2

+0 0 0 21 1

+ 2 ·0 0 0 112

+0 0 01 12

,

where we have deleted those tableaux with repeats in the columns and we have collected redundancies due topermutations of columns of the same size.

Example (A matrix representing F(2;1;0);(2;1;0)(x32 ))

All Pieri products of the form2

22

, produce the following after deleting012

F(x32 )

0 01

0 02

0 11

0 12

0 21

0 22

1 12

1 22

0 01

0 02

0 11

0 12

0 21

0 22

1

1 12

1 22

1

where we only record the non-zero entries.

Example (A matrix representing F(2;1;0);(2;1;0)(x1x22 ))

All possible Pieri products of the form2

21

+2

12

+1

22

produce:

F(x1x22 )

0 01

0 02

0 11

0 12

0 21

0 22

1 12

1 22

0 01

0 02

0 11

0 12

1

0 21

1

0 22

-1

1 12

1

1 22

-1 -1

RankF(x1x22 ) = 4⇒ Brank(x1x

22 ) ≥ 4

2= 2

Example (A matrix representing F(2;1;0);(2;1;0)(x0x1x2))

All Pieri products of the form0

12

+0

21

+1

02

+1

20

+2

01

+2

10

F(x0x1x2)0 01

0 02

0 11

0 12

0 21

0 22

1 12

1 22

0 01

1

0 02

1

0 11

-1

0 12

-1 -2

0 21

2 1

0 22

1

1 12

1

1 22

-1

RankF(x0x1x2) = 8⇒ Brank(x0x1x2) ≥ 82

= 4. Block-diagonal structure?

The only non-obvious entries are the columns labeled 0 12 and 0 2

1 , which we now explain. We have the product

0 12 ⊗ 0 1 2 =

0 1 02 21

+0 1 22 01

≡ −(

1 02 + 1 2

0

)≡ −

(−2 · 0 2

1 + 0 12

),

where we have used the shuffling rules to put the tableau in semi-standard format. Similarly

0 21 ⊗ 0 1 2 =

0 2 11 02

+0 2 01 12

≡ 2 10 + 2 0

1

≡ −2 · 0 12 + 0 2

1 .

F4321(x40 ) has rank 8. F4321(x3

0 x1) has rank 16.

F4321(x20 x1x2) has rank 32. F4321(x0x1)x2x3 has rank 64.

We implemented Young flattenings in Macaulay2 using PieriMaps developed by Steven Sam. Here is F4,3,2,1

evaluated at quintic monomials:

loadPackage"PieriMaps"

A = QQ[x_0..x_4]

MX = pieri({5,4,3,2,1},{5,4,3,2,1},A);

diff(x_0^5, MX)

diff(x_0*x_1*x_2*x_3*x_4, MX)

monomial phi rank F_{4,3,2,1}(phi)

x_0^5 64

x_0^4*x_1 128

x_0^3*x_1^2 176

x_0^3*x_1*x_2 256

x_0^2*x_1^2*x_2 324

x_0^2*x_1*x_2*x_3 512

x_0*x_1*x_2*x_3*x_4 1024

A single flattening gives the best lower bound for all quintic monomials. This is not expected to hold in general.

Brank x40 x1 ≥ 2 = 128

64

Brank x30 x

21 ≥ 3 = d2.75e = d 176

64e

Brank x30 x1x2 ≥ 4 = 256

64

Brank x20 x

21 x2 ≥ 6 = d5.0625e = d 324

64e

Brank x20 x1x2x3 ≥ 8 = 512

64

Brank x0x1x2x3x4 ≥ 16 = 102464

The multiplier

Let π be a partition with #π ≤ n, and π1 ≥ d . Let V ∼= Cn+1 and V0 := V /〈x0〉.

Proposition

The flattening Fπ(xd0 ) : SπV → Sd,πV has image SπV0⊗〈xd

0 〉 ⊂ Sd,πV .

Proof.

Image is spanned by elements of SSYT d,π{0, . . . , n} with all 0 ’s in the first row.This set is equivalent to SSYTπ{1, . . . , n}.

The analogue of Macaulay’s result for Young flattenings:

If ϕ has rank r then Fπ(ϕ) has rank at most r · dim SπV0.

So lower bounds on border rank can be obtained using Young flattenings.

Sextic monomials and examples of optimal shapes

’ flattening size RankF(x60 ) RankF(’)

x50 x1

∗∗∗∗∗∗ 2× 6 1 2

x40 x

21

∗∗∗∗∗∗ 3× 5 1 3

x30 x

31

∗∗∗∗∗∗ 4× 4 1 4

x40 x1x2

∗∗∗∗∗∗ 8× 35 2 8

x30 x

21 x2

∗∗∗∗∗∗ 15× 24 2 12

x20 x

21 x

22

∗∗∗∗∗∗ 27× 27 3 27

x30 x1x2x3

∗∗∗∗∗∗64× 256 8 64

x20 x

21 x2x3

∗∗∗∗∗∗140× 140 8 96

x20 x1x2x3x4

∗∗∗∗∗∗1024× 2560 64 1024

x0x1x2x3x4x5

∗∗∗∗∗∗215 × 215 210 215

Optimal shapes

Suppose α0 ≥ α1 ≥ · · · ≥ αn. An optimal shape for a flattening of xα is

λ =

(n∑

i=1

αi ,

n−1∑i=1

αi ,

n−2∑i=1

αi , . . . , α2 + α1, α1

)(αi columns of height n + 1− i)

with d =∑n

i=0 αi and µ = (d , λ). Then Fλ,µ is a Young flattening.

Conjecture

Suppose λ is the optimal shape for xα. Then the flattening Fλ(xα) has rank

Rank(Fλ(xα)) =n∏

i=1

(αi + 1) · dimSλV0.

Remark:Suppose ωi are the fundamental weights GL(V ). Consider the weight α = α0ω0 + . . . ωnαn.The optimal shape λ is the partition indexing the Schur module dual to Γα: i.e. Γα = (SλV )∗.

We found that the conjecture is only partially true.

A surprise in degree 7

ϕ flattening size F(x70 ) F(ϕ)

x60 x1

∗∗∗∗∗∗∗ 7× 2 1 2

x50 x

21

∗∗∗∗∗∗∗ 6× 3 1 3

x50 x1x2

∗∗∗∗∗∗∗ 48× 8 2 8

x40 x

31

∗∗∗∗∗∗∗ 5× 4 1 4

x40 x

21 x2

∗∗∗∗∗∗∗ 35× 15 2 12

x40 x1x2x3

∗∗∗∗∗∗∗420× 64 8 64

x30 x

31 x2

∗∗∗∗∗∗∗ 24× 24 2 16

x30 x

21 x

22

∗∗∗∗∗∗∗ 42× 27 3 27

ϕ flattening size F(x70 ) F(ϕ)

x30 x

21 x2x3

∗∗∗∗∗∗∗256× 140 8 96

x30 x1x2x3x4

∗∗∗∗∗∗∗5120× 1024 64 1024

x20 x

21 x

22 x3

∗∗∗∗∗∗∗360× 300 15 255

x20 x

21 x2x3x4

∗∗∗∗∗∗∗2520× 2520 64 1536

x20 x1x2x3x4x5x6

∗∗∗∗∗∗∗88704× 215 210 215

x0x1x2x3x4x5x6x7

∗∗∗∗∗∗∗

221 × 221 215 221

ϕ shape λ shape λ∗ pairing

x50 x1

∗∗∗∗∗∗ ∗∗∗∗∗∗x4

0 x21

∗∗∗∗∗∗ ∗∗∗∗∗∗x3

0 x31

∗∗∗∗∗∗ ∗∗∗∗∗∗

x40 x1x2

∗∗∗∗∗∗∗∗∗∗∗∗

x30 x

21 x2

∗∗∗∗∗∗∗∗∗∗∗∗

x20 x

21 x

22

∗∗∗∗∗∗∗∗∗∗∗∗

ϕ shape λ shape λ∗ pairing

x30 x1x2x3

∗∗∗∗∗∗

∗∗∗∗∗∗

x20 x

21 x2x3

∗∗∗∗∗∗

∗∗∗∗∗∗

x20 x1x2x3x4

∗∗∗∗∗∗

∗∗∗∗∗∗

x0x1x2x3x4x5

∗∗∗∗∗∗

∗∗∗∗∗∗

Note: all contragradients of optimal shapes have optimal shape in the degree 6 case.

A surprise in degree 7

∗∗∗∗∗∗∗

∗∗∗∗∗∗∗flattening contragradient pairing

For x20 x

21 x

22 x3 the optimal partition (5, 4, 2) has contragradient (5, 3, 2), which is not an optimal shape. This is

the “cause” of the rank deficiency in the flattening.

This feature only shows up when d > 6. There is only one failure d = 7.

When d ≥ 27 more than half of the optimal partitions fail to have optimal contragradients.

So Young flattenings of monomials rarely have maximal rank for large degree.

However, many classes of monomials, like (x0 · · · xn)k , do have optimal Young flattenings of maximal rank.

Now we ask how to compute the rank of a Young flattening in general?

An action of gl(V ) on the dominance poset

X 10

++Brank x6

0 = 1

X 10

++X 2

0��

Brank x50 x1 = 2

X 10

**X 2

1

ttX 2

0��

Brank x40 x

21 = 3

Brank x40 x1x2 = 4

X 10

**X 3

0��

X 21

tt

Brank x30 x

31 = 4

X 20

))

X 31

uuX 3

0

��

Brank x30 x

21 x2 = 6

Brank x30 x1x2x3 = 8

X 10

))X4,0

��

X 32

uu

Brank x20 x

21 x

22 = 9

X 41

vv

Brank x20 x

21 x2x3 = 12

X 50

xx

Brank x20 x1x2x3x4 = 16

Brank x0x1x2x3x4x5 = 32

An action of gl(V ) on the dominance poset

X 10

++Brank x6

0 = 1

X 10

++X 2

0��

Brank x50 x1 = 2

X 10

**X 2

1

ttX 2

0��

Brank x40 x

21 = 3

Brank x40 x1x2 = 4

X 10

**X 3

0��

X 21

tt

Brank x30 x

31 = 4

X 20

))

X 31

uuX 3

0

��

Brank x30 x

21 x2 = 6

Brank x30 x1x2x3 = 8

X 10

))X4,0

��

X 32

uu

Brank x20 x

21 x

22 = 9

X 41

vv

Brank x20 x

21 x2x3 = 12

X 50

xx

Brank x20 x1x2x3x4 = 16

Brank x0x1x2x3x4x5 = 32 X ij = xi · ∂∂xj

A gl(V ) action on flattenings

Proposition

Fix ϕ ∈ SdV and let F = Fλ,µ. For X ∈ gl(V ) and T ∈ SλV we have

F(X .ϕ)(T ) = F(ϕ)(X .T )− X .(F(ϕ)(T )),

and in general

Fλ(xα)(T ) =∑

µ+ν=α′

(−1)|ν|X ν(F(xd0 )(Xµ.T )), (1)

where α′ = (α1, . . . , αn).

Proof.

The result follows from basic algebra and an application of X ∈ gl(V ):

X .(F(ϕ)(T )) = X .(T⊗ϕ)

= (X .T )⊗ϕ+ T⊗(X .ϕ)

= F(ϕ)(X .T ) + F(X .ϕ)(T ).

Question: When does the sum in (1) split the map SλV → Sd,λV as a direct sum?If λ is optimal, the source contains “the right” direct sum.If λ∗ is optimal, the target contains “the right” direct sum.

Consider the flattening F(x40 x1x2) : S2,1V → S4,2,1V , Lie algebra elements X 1

0 and X 20 , and vector spaces

V = C3, and V0 := V /〈x0〉.We want to use the expression

F(x20 x1x2)(T ) =

F(X 20 X

10 .(x0)4)(T ) = X 2

0 X10 . F((x0)4)( T )

−X 10 . F((x0)4)( X 2

0 .T )

−X 20 . F((x0)4)( X 1

0 .T )

+ F((x0)4)( X 20 X

10 .T )

(2)

and obtain a direct sum decomposition

S2,1V = A(1, 2)⊕A(2)⊕A(1)⊕A(∅)→ B(1, 2)⊕ B(2)⊕ B(1)⊕ B(∅) = S4,2,1V .

The rank of each piece should be easy to compute. Let me describe the pieces.

A different kind of branching

A standard branching rule sl(3) ↓ sl(2) decomposes S2,1C3 into sl(2) modules as:

001

011 002

012 , or S2,1C3 = S2,1C2 ⊕ (S2C2 ⊕ C)⊕ S1C2,

112 022

122

where the connected components in the diagram indicate different sl(2) modules.

A different type of branching (The ghs algorithm tells when this works):

S2,1C3 = A(1, 2)⊕A(2)⊕A(1)⊕A(∅),

A(1, 2) =⟨

0 01

,0 02

⟩, A(1) =

⟨0 12

,0 22

⟩, A(2) =

⟨0 11

,0 21

⟩, A(∅) =

⟨1 12

,1 22

⟩.

Visualize the decomposition S2,1C3 = S2,1C2 ⊕ S2,1C2 ⊕ S2,1C2 ⊕ S2,1C2:

0 01 dd

$$0 11 dd

$$

0 02

0 21

0 12 dd$$

1 12 dd

$$

0 22

1 22

For each γ ⊂ {1, 2}, we obtain a direct sum decomposition

S2,1V = A(γ)⊕K(γ),

with K(γ) = kerH0Xγ0 : S2,1V → S2,1V ,

and A(γ) ∼= S2,1V /K(γ) ⊂ S2,1V .

Arrows oo // indicate modules for a different action of sl(2) = 〈X 21 ,X

12 ,H12〉

given by the action on the base⟨

1 12

, 1 22

⟩, after conjugating by different elements of U(gl(V )).

We can visualize the U(gl(V )) action moving the weight diagram. Weight spaces that end up outside the weightdiagram for S2,1C3 are in the kernel.

0 01 dd

$$0 11 dd

$$

0 02

0 21

0 12 dd$$

1 12 dd

$$

0 22

1 22

H0X10 X

20 moves the diagram down two steps.

The only remaining weights are 1 12 , 1 2

2 .

The only weights not in the kernel of H0X10 X

20 are in the top part

of the diagram,⟨

0 01 , 0 0

2

⟩= A(1, 2).

We can visualize the U(gl(V )) action moving the weight diagram. Weight spaces that end up outside the weightdiagram for S2,1C3 are in the kernel.

0 11 ff

&&− 0 2

1+ 2

0 12

1 12 ff

&&1 22

In the A(2) case, X 10 moves the diagram south-west one step, the

result is supported on the south-west part of the weight lattice.

The action of H0 annihilates⟨

0 11 ,− 0 2

1 + 2 0 12

⟩.

Note H0X10 maps

⟨0 12 , 0 2

2

⟩to⟨

1 12 , 1 2

2

⟩, and

⟨0 12 , 0 2

2

⟩is a complement of the kernel of H0X

10 .

A similar procedure is followed to construct each of A(∅),A(1),A(2),A(1, 2).

By construction only one term F(x40 )(X γ

0 T ), is non-zero on each A(γ), so the expression (2) splits over thedirect sum. By the fact that H0X

γ0 S2,1V = S2,1V0 and our knowledge of the flattening for (xd

0 ),

F(x40 )(X γ

0 ·) : A(γ)→ S4,2,1V

is an isomorphism onto its image, which is F2,1(x40 )(S2,1V0) ∼= S2,1V0.

Finally, one checks that the vector spaces

B(ν) := X ν0 F2,1(x4

0 )(S2,1V0)

for ν ⊂ {1, 2} are linearly independent subspaces of Sd,2,1V . In particular,S4,2,1C3 ⊃ B(∅) + B(1) + B(2) + B(1, 2), with

B(∅) =

⟨0 0 0 01 12

,0 0 0 01 22

⟩, B(1) = X 1

0

⟨0 0 0 01 12

,0 0 0 01 22

⟩,

B(2) = X 20

⟨0 0 0 01 12

,0 0 0 01 22

⟩, B(1, 2) = X 2

0 X10

⟨0 0 0 01 12

,0 0 0 01 22

⟩.

Moreover, the sum is direct.

So we have a direct sum

F2,1(x40 x1x2) : S2,1V =

A(1, 2) → B(∅)⊕

A(2) → B(1)⊕

A(1) → B(2)⊕

A(∅) → B(1, 2)

⊂ S4,2,1V .

Respectively choose ordered bases for the source (S2,1V ) and target (S4,2,1V ):{0 01 , 0 0

2 , 0 11 , 0 1

2 , 0 21 , 0 2

2 , 1 12 , 1 2

2

},

0 0 0 01 12

,0 0 0 01 22

,0 0 0 11 12

,0 0 0 11 22

,0 0 0 21 12

,0 0 0 21 22

,0 0 1 11 12

,

0 0 1 11 22

,0 0 1 21 12

,0 0 1 21 22

,0 0 2 21 12

,0 0 2 21 22

,0 1 1 11 22

,0 1 1 21 22

,0 1 2 21 22

.

One verifies that M12,∅ −M1,2 −M2,1 + M∅,12 = 12M :

−1 0 0 0 0 0 0 00 2 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0

-

0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 4 −2 0 0 00 0 0 0 0 4 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0

-

0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 −2 0 0 0 0 00 0 0 4 −8 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0

+

0 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 12 00 0 0 0 0 0 0 120 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0

=

−1 0 0 0 0 0 0 00 2 0 0 0 0 0 00 0 2 0 0 0 0 00 0 0 −4 8 0 0 00 0 0 −4 2 0 0 00 0 0 0 0 −4 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 12 00 0 0 0 0 0 0 120 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 00 0 0 0 0 0 0 0

.

The image is the vector subspace of S4,2,1V isomorphic to S2,1V0⊗〈x40 , x

30 x1, x

30 x2, x

20 x1x2〉.

So the border rank of x20 x1x2 is at least 4;

the ratio of the dimensions of S2,1V0⊗〈x40 , x

30 x1, x

30 x2, x

20 x1x2〉 and S2,1V0,

which is also the known upper bound, so the border rank is exactly 4.

Theorem (Oeding)

Let V ∼= Cn+1 and V0 = V /〈x0〉. Suppose α, β ∈ Nn+1 with α0 ≥ · · · ≥ αn and∑

i αi =∑

i βi = d . Let

λ = (∑n

i=1 αi ,∑n−1

i=1 αi , . . . , α1). � denotes dominance.

1 If (β1, β2, . . . , βn) � (α1, α2, . . . , αn) and (β1, . . . , βn) � (α0, αn, αn−1, . . . , α2), then the Young flattening

Fλ,(d,λ)(xβ00 · · · x

βnn ) : SλV → Sd,λV has rank dim SλV0 ·

n∏i=1

(βi + 1).

Proof idea.

Use the Lie algebra U(gl(V )) to express F(ϕ) as a sum of∏n

i=1(1 + αi ) simpler maps.Use combinatorics of Young tableaux (row addition and row subtraction) to show that if the hypotheses on αand β are satisfied, then the sum is direct and the rank of each summand is dim SλV0.

Theorem (Oeding)

Suppose xα = xα00 · · · x

αnn with α0 ≥ α1 ≥ · · · ≥ αn. If (α1, . . . , αn) � (α0, αn, αn−1, . . . , α2) (dominance order),

then the border rank of xα = xα00 · · · x

αnn is evidenced by a Young flattening and is equal to

∏ni=1(1 + αi ).

Proof.

The upper bound is by an explicit limit given by Landsberg-Teitler.The lower bound comes from Im(Fλ,(d,λ)(x

d0 )) ∼= SλV0 and the Macaulay-type bound on border rank.

Generalized Horizontal Strips (ghs)

Optimal shape tableau have Generalized Horizontal Strips with content µ ≺ α′.

Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1

i=1 αi , . . . , α1) and µ+ ν = α′. Every T inSSYTλ{1, . . . , n} has a ghs with content ν.

Proof by algorithm.

1 1 1 1 1 1 1 2 2 2 2 32 2 2 2 2 2 3 3 4 4 43 3 3 3 3 3 4 44 4 4 4 4

∅



Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1


1 1 1 1 1 1 1 2 2 2 2 32 2 2 2 2 2 3 3 4 4 43 3 3 3 3 3 4 44 4 4 4 4

2223

We choose the smallest content possible in the last block: 3 .



Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1


1 1 1 1 1 1 1 2 2 2 2 32 2 2 2 2 2 3 3 4 4 43 3 3 3 3 3 4 44 4 4 4 4

2223

We choose the smallest content possible in the second to last block: 2 2 2 .It remains to find a ghs with content (5, 0, 2, 1) in the first two blocks.



Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1


1 1 1 1 1 1 1 2 2 2 2 32 2 2 2 2 2 3 3 4 4 43 3 3 3 3 3 4 44 4 4 4 4

1132223

Choose the smallest content from the remaining in the 2nd block: 1 1 3 .It remains to find a ghs with content (3, 0, 1, 1) in the first block.



Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1


1 1 1 1 1 1 1 2 2 2 2 32 2 2 2 2 2 3 3 4 4 43 3 3 3 3 3 4 44 4 4 4 4

431111132223

We’ve found a ghs with the desired content. ♦



Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1


Let λ = (12, 11, 8, 5), α = (5, 5, 3, 3, 1).Let’s find a ghs with content α′ = (5, 3, 3, 1) in the following element of SSYTλ{1, . . . , 4}

For curiosity we can replace all selected content with 0’s and shuffle the 0’s upward to produce

1 1 0 0 0 0 0 2 0 0 0 02 2 2 2 2 2 3 0 4 4 43 0 3 3 3 3 4 40 4 4 4 4

and

0 0 0 0 0 0 0 0 0 0 0 01 1 2 2 2 2 3 2 4 4 42 2 3 3 3 3 4 43 4 4 4 4

,

the last of which happens to be almost semi-standard.



Lemma

Let α = (α0 ≥ · · · ≥ αn), α′ = (α1, . . . , αn), λ = (∑n

i=1 αi ,∑n−1


In general row additions (and row subtractions) tell when the terms X ν0 .F(xd

0 ).Xµ0 have full rank.

A special gl(V0) action is used to show that the summands are linearly independent.

The Koszul complex and Koszul matrices

The Koszul complex arises via the minimal free resolution of the maximal ideal 〈x0, . . . , xn〉. Let V be the spanof the xi .

0−→∧n+1V

kn+1−→∧nV−→· · · k3−→

∧2Vk2−→∧1V

k1−→C−→0

Some examples:

for n = 2, k1 =(w x y

), k2 =

−x −y 0w 0 −y0 w x

k3 =

y−xw

,

for n = 3, k1 =(w x y z

), k2 =

−x −y 0 −z 0 0w 0 −y 0 −z 00 w x 0 0 −z0 0 0 w x y

, ...

Koszul flattenings

Suppose T ∈ A⊗B⊗C . Have a natural inclusion A ⊂∧2A⊗A∗.

Construct a new linear map via the inclusionA⊗B⊗C ⊂

∧2A⊗A∗⊗B⊗C = (A∗⊗B)⊗(∧2A⊗C

).

Fix dim(A) = 3. With T =∑

i ui⊗Ti , we choose a good basis and write

ψ1,T : A⊗B∗

(0 T3 −T2−T3 0 T1T2 −T1 0

)−−−−−−−−−−−→

∧2A⊗C .

Basic idea (The analogue of Macaulay’s result for Koszul flattenings):

ψ1,T+T ′ = ψT + ψT ′ construction is linear in TRank(T ) = 1⇒ Rank(ψT ) = 2 base case (see next frame)

∴ Rank(T ) = r ⇒ Rank(ψT ) ≤ 2r subadditivity of matrix rank

The (2r + 1)× (2r + 1) minors of ψ1,T are necessary conditions for Brank(T ) ≤ r .

The matrix rank of a Koszul flattening certifies lower bounds on tensor border rank.

Koszul Flattenings: Rank(T ) = 1⇒ Rankψ1,T = 2

Let T1 =

(1 ··· 0

0. . . 0

), Ti =

(0 ... 0

0. . . 0

), i = 2..m.

ψ1,T =

(0 T3 −T2−T3 0 T1T2 −T1 0

)=

0

0 ... 0

0. . . 0

0 ... 0

0. . . 0

0 ... 0

0. . . 0

0

1 ··· 0

0. . . 0

, 0 ... 0

0. . . 0

−1 ··· 0

0. . . 0

, 0

So Rank(T ) = 1⇒ Rank(ψ1,T ) = 2.

Therefore Rank(T ) = r ⇒ Rank(ψ1,T ) ≤ 2r .

Ottaviani’s Example

Apply k2 ⊗ C to xyz , get a map k2⊗C(xyz) : C3⊗(C3)∗ →∧2C3⊗C3:

k2⊗C(xyz) =

0 0 −1 0 −1 0 0 0 00 0 0 −1 0 0 0 0 0−1 0 0 0 0 0 0 0 00 0 0 0 0 0 0 −1 00 0 1 0 0 0 −1 0 00 1 0 0 0 0 0 0 00 0 0 0 0 0 0 0 10 0 0 0 0 1 0 0 00 0 0 0 1 0 1 0 0

The matrix k2⊗C(xyz) has rank 8, implying that Brank(xyz) ≥ 4.

In appropriate bases, this map is skew-symmetric 9× 9 so always has at least a 1-dimensional kernel.The more efficient map is gotten by restricting

V⊗∧2V → V⊗

∧2V⊗∧3V or

(⊗

)⊗ ∗ ∗ ∗ → ∗⊗ ∗⊗ ∗

to irreducible representations

S2,1V → S3,2,1V or ⊗ ∗ ∗ ∗ →∗∗∗

Macauly2 implementations of (Koszul) flattenings

R = QQ[x,y,z]

b1 = basis(1,R)

r = res ideal b1

b2 = basis(2,R)

C = transpose b1**b2

diff(x*y*z, C)

K = r.dd_2**b1**transpose b1

diff(x*y*z,K)

Eventually Koszul flattenings don’t give good enough lower bounds for border rank of monomials.[Guan’15] and [Farnsworth’15] got the most possible out of Koszul Flattenings.

\begin{verbatim}

Macaulay2, version 1.9.2

i1 : R = QQ[x,y,z],;

i2 : b1 = basis(1,R)

o2 = | x y z |

1 3

o2 : Matrix R <--- R

i3 : r = res ideal b1

1 3 3 1

o3 = R <-- R <-- R <-- R <-- 0

0 1 2 3 4

o3 : ChainComplex

i4 : b2 = basis(2,R)

o4 = | x2 xy xz y2 yz z2 |

1 6


i5 : C = transpose b1**b2

o5 = {-1} | x3 x2y x2z xy2 xyz xz2 |

{-1} | x2y xy2 xyz y3 y2z yz2 |

{-1} | x2z xyz xz2 y2z yz2 z3 |

3 6


i6 : diff(x*y*z, C)

o6 = {-1} | 0 0 0 0 1 0 |

{-1} | 0 0 1 0 0 0 |

{-1} | 0 1 0 0 0 0 |

3 6


i7 : K = r.dd_2**b1**transpose b1

o7 = | -x2y -xy2 -xyz -x2z -xyz -xz2 0 0 0 |

| -xy2 -y3 -y2z -xyz -y2z -yz2 0 0 0 |

| -xyz -y2z -yz2 -xz2 -yz2 -z3 0 0 0 |

| x3 x2y x2z 0 0 0 -x2z -xyz -xz2 |

| x2y xy2 xyz 0 0 0 -xyz -y2z -yz2 |

| x2z xyz xz2 0 0 0 -xz2 -yz2 -z3 |

| 0 0 0 x3 x2y x2z x2y xy2 xyz |

| 0 0 0 x2y xy2 xyz xy2 y3 y2z |

| 0 0 0 x2z xyz xz2 xyz y2z yz2 |

9 9


i8 : diff(x*y*z,K)

o8 = | 0 0 -1 0 -1 0 0 0 0 |

| 0 0 0 -1 0 0 0 0 0 |

| -1 0 0 0 0 0 0 0 0 |

| 0 0 0 0 0 0 0 -1 0 |

| 0 0 1 0 0 0 -1 0 0 |

| 0 1 0 0 0 0 0 0 0 |

| 0 0 0 0 0 0 0 0 1 |

| 0 0 0 0 0 1 0 0 0 |

| 0 0 0 0 1 0 1 0 0 |

9 9


border ranks of monomials - inria · the basic construction, and a certi cate for high border rank...

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