Some numerical results on decompositions of
Betti diagrams
Chris Francisco
Oklahoma State University
Joint work with
Jeff Mermin and Jay Schweig
Halifax
October 2014
Borel ideals
Let S = k [x1 , . . . , xn ], k a field.
Borel ideals
Let S = k [x1 , . . . , xn ], k a field.
Definition: A monomial ideal M ⊂ S is a Borel ideal if
• given any monomial m ∈ M,
• a variable xj dividing m, and
• an index i < j,
Borel ideals
Let S = k [x1 , . . . , xn ], k a field.
Definition: A monomial ideal M ⊂ S is a Borel ideal if
• given any monomial m ∈ M,
• a variable xj dividing m, and
• an index i < j,
then m ·
xi
∈ M.
xj
Also known as strongly stable or 0-Borel ideals.
Example: (a3 , a2 b, a2 c, ab2 , b3 ) is Borel. (Borel generated by
a2 c and b3 .)
Eliahou-Kervaire resolution of Borel ideals
The Eliahou-Kervaire resolution provides a minimal free
resolution of any Borel ideal B.
Eliahou-Kervaire resolution of Borel ideals
The Eliahou-Kervaire resolution provides a minimal free
resolution of any Borel ideal B.
Let max(m) be the largest index of a variable dividing m.
Eliahou-Kervaire resolution of Borel ideals
The Eliahou-Kervaire resolution provides a minimal free
resolution of any Borel ideal B.
Let max(m) be the largest index of a variable dividing m.
Basis elements: (m, α)
Eliahou-Kervaire resolution of Borel ideals
The Eliahou-Kervaire resolution provides a minimal free
resolution of any Borel ideal B.
Let max(m) be the largest index of a variable dividing m.
Basis elements: (m, α)
• m is a minimal monomial generator of B
Eliahou-Kervaire resolution of Borel ideals
The Eliahou-Kervaire resolution provides a minimal free
resolution of any Borel ideal B.
Let max(m) be the largest index of a variable dividing m.
Basis elements: (m, α)
• m is a minimal monomial generator of B
• α is a squarefree monomial with max(α) < max(m).
Eliahou-Kervaire resolution of Borel ideals
The Eliahou-Kervaire resolution provides a minimal free
resolution of any Borel ideal B.
Let max(m) be the largest index of a variable dividing m.
Basis elements: (m, α)
• m is a minimal monomial generator of B
• α is a squarefree monomial with max(α) < max(m).
Thus if gens(B) represents the minimal monomial generators of
B, then
X max(m) − 1
βi (B) =
.
i
m∈gens(B)
Example of resolution
If I is Borel, then to compute βi (I), it is enough to answer:
For each j, how many minimal generators m have max(m) = j?
Let wj (I) be this quantity.
Example of resolution
If I is Borel, then to compute βi (I), it is enough to answer:
For each j, how many minimal generators m have max(m) = j?
Let wj (I) be this quantity.
Example: Let I = (a3 , a2 b, a2 c, ab2 , b3 ).
Example of resolution
If I is Borel, then to compute βi (I), it is enough to answer:
For each j, how many minimal generators m have max(m) = j?
Let wj (I) be this quantity.
Example: Let I = (a3 , a2 b, a2 c, ab2 , b3 ).
Then w1 (I) = 1, w2 (I) = 3, and w3 (I) = 1.
Example of resolution
If I is Borel, then to compute βi (I), it is enough to answer:
For each j, how many minimal generators m have max(m) = j?
Let wj (I) be this quantity.
Example: Let I = (a3 , a2 b, a2 c, ab2 , b3 ).
Then w1 (I) = 1, w2 (I) = 3, and w3 (I) = 1.
Thus the vector of total Betti numbers of I is
1(1, 0, 0) + 3(1, 1, 0) + 1(1, 2, 1) = (5, 5, 1).
Basic question
Question: Suppose I is an equigenerated Borel ideal. How can
we decompose the Betti diagram of S/I as a positive rational
linear combination of Betti diagrams of the form
S/Vi,r := S/(x1 , . . . , xi )r ?
Basic question
Question: Suppose I is an equigenerated Borel ideal. How can
we decompose the Betti diagram of S/I as a positive rational
linear combination of Betti diagrams of the form
S/Vi,r := S/(x1 , . . . , xi )r ?
Example: I = (a3 , a2 b, a2 c, ab2 , b3 ).
total:
0:
1:
2:
1
1
.
.
5
.
.
5
5
.
.
5
1
.
.
1
Basic question
Question: Suppose I is an equigenerated Borel ideal. How can
we decompose the Betti diagram of S/I as a positive rational
linear combination of Betti diagrams of the form
S/Vi,r := S/(x1 , . . . , xi )r ?
Example: I = (a3 , a2 b, a2 c, ab2 , b3 ).
total:
0:
1:
2:
1
1
.
.
5
.
.
5
5
.
.
5
1
.
.
1
Start with the Veronese with the largest projective dimension:
S/(a, b, c)3 has Betti numbers (1, 10, 15, 6), so to get a 1 in the
last spot, use 61 βV3,3 .
Basic question
total:
0:
1:
2:
1
1
.
.
5
.
.
5
5
.
.
5
1
.
.
1
S/(a, b, c)3 has Betti numbers (1, 10, 15, 6). Get rightmost 1:
1
1 10 15
βS/V3,3 =
, , ,1 .
6
6 6 6
Basic question
total:
0:
1:
2:
1
1
.
.
5
.
.
5
5
.
.
5
1
.
.
1
S/(a, b, c)3 has Betti numbers (1, 10, 15, 6). Get rightmost 1:
1
1 10 15
βS/V3,3 =
, , ,1 .
6
6 6 6
Last Betti number matches, so go to the previous one.
Basic question
total:
0:
1:
2:
1
1
.
.
5
.
.
5
5
.
.
5
1
.
.
1
S/(a, b, c)3 has Betti numbers (1, 10, 15, 6). Get rightmost 1:
1
1 10 15
βS/V3,3 =
, , ,1 .
6
6 6 6
Last Betti number matches, so go to the previous one.
S/(a, b)3 has Betti numbers (1, 4, 3), so take
5
5 20 15
β
=
, ,
.
6 S/V2,3
6 6 6
Basic question
total:
0:
1:
2:
1
1
.
.
5
.
.
5
5
.
.
5
1
.
.
1
S/(a, b, c)3 has Betti numbers (1, 10, 15, 6). Get rightmost 1:
1
1 10 15
βS/V3,3 =
, , ,1 .
6
6 6 6
Last Betti number matches, so go to the previous one.
S/(a, b)3 has Betti numbers (1, 4, 3), so take
5
5 20 15
β
=
, ,
.
6 S/V2,3
6 6 6
Now note
βS/I =
5
1
β
+ β
.
6 S/V2,3 6 S/V3,3
Generating function relationship
For an equigenerated Borel ideal I, let
WI (t) = w1 (I) + w2 (I)t + · · · + wn (I)t n−1 .
Generating function relationship
For an equigenerated Borel ideal I, let
WI (t) = w1 (I) + w2 (I)t + · · · + wn (I)t n−1 .
Let
BI (t) = β0 (I) + β1 (I) + · · · + βn−1 t n−1 .
Generating function relationship
For an equigenerated Borel ideal I, let
WI (t) = w1 (I) + w2 (I)t + · · · + wn (I)t n−1 .
Let
BI (t) = β0 (I) + β1 (I) + · · · + βn−1 t n−1 .
Proposition: WI (t + 1) = BI (t).
Generating function relationship
For an equigenerated Borel ideal I, let
WI (t) = w1 (I) + w2 (I)t + · · · + wn (I)t n−1 .
Let
BI (t) = β0 (I) + β1 (I) + · · · + βn−1 t n−1 .
Proposition: WI (t + 1) = BI (t).
Example: For I = (a3 , a2 b, a2 c, ab2 , b3 ), we have
WI (t) = 1 + 3t + t 2 , so WI (t + 1) = 5 + 5t + t 2 .
Generating function relationship
For an equigenerated Borel ideal I, let
WI (t) = w1 (I) + w2 (I)t + · · · + wn (I)t n−1 .
Let
BI (t) = β0 (I) + β1 (I) + · · · + βn−1 t n−1 .
Proposition: WI (t + 1) = BI (t).
Example: For I = (a3 , a2 b, a2 c, ab2 , b3 ), we have
WI (t) = 1 + 3t + t 2 , so WI (t + 1) = 5 + 5t + t 2 .
Note: We can define WI (t) for any ideal with a linear resolution
by substituting the reverse-lex gin J and using its
W -polynomial.
Coefficients
Write BS/I (t) = λ1 BS/V1,d (t) + λ2 BS/V2,d (t) + · · · + λn BS/Vn,d (t),
where the λi are the coefficients from above in the
decomposition of the Betti diagram of S/I.
Coefficients
Write BS/I (t) = λ1 BS/V1,d (t) + λ2 BS/V2,d (t) + · · · + λn BS/Vn,d (t),
where the λi are the coefficients from above in the
decomposition of the Betti diagram of S/I.
Theorem: Let I have a linear resolution, and let m be the
irrelevant ideal. Then the Boij-Söderberg coefficients of S/I are:
λn =
wn (I)
wi (I)
wi+1 (I)
and λi =
−
for i < n.
wn (md )
wi (md ) wi+1 (md )
Coefficients
Write BS/I (t) = λ1 BS/V1,d (t) + λ2 BS/V2,d (t) + · · · + λn BS/Vn,d (t),
where the λi are the coefficients from above in the
decomposition of the Betti diagram of S/I.
Theorem: Let I have a linear resolution, and let m be the
irrelevant ideal. Then the Boij-Söderberg coefficients of S/I are:
λn =
wn (I)
wi (I)
wi+1 (I)
and λi =
−
for i < n.
wn (md )
wi (md ) wi+1 (md )
Note: Cook has recent independent work with the same
formula written differently.
Coefficients
Write BS/I (t) = λ1 BS/V1,d (t) + λ2 BS/V2,d (t) + · · · + λn BS/Vn,d (t),
where the λi are the coefficients from above in the
decomposition of the Betti diagram of S/I.
Theorem: Let I have a linear resolution, and let m be the
irrelevant ideal. Then the Boij-Söderberg coefficients of S/I are:
λn =
wn (I)
wi (I)
wi+1 (I)
and λi =
−
for i < n.
wn (md )
wi (md ) wi+1 (md )
Note: Cook has recent independent work with the same
formula written differently.
Example: Let I = (a3 , a2 b, a2 c, ab2 , b3 ). Then w1 (m3 ) = 1,
w2 (m3 ) = 3, and w3 (m3 ) = 6. Thus
λ3 =
1
,
6
λ2 =
3 1
5
− = ,
3 6
6
and λ1 =
1 3
− = 0.
1 3
Multiplication by a form
Question: What happens to the Boij-Söderberg coefficients
when you multiply an ideal with linear resolution by a
homogeneous form?
Multiplication by a form
Question: What happens to the Boij-Söderberg coefficients
when you multiply an ideal with linear resolution by a
homogeneous form?
Example: Let I = (a, b, c, d)2 , and let µ be a linear form. The
Betti diagram of S/µI is:
total:
0:
1:
2:
3:
1
1
.
.
.
10
.
.
.
10
20
.
.
.
20
15
.
.
.
15
4
.
.
.
4
The Boij-Söderberg decomposition
of S/I is trivial, but for S/µI,
2 1 1 1
the coefficients are 5 , 10 , 6 , 3 .
Coefficient growth
Question: What are the coefficients for S/µk I?
Coefficient growth
Question: What are the coefficients for S/µk I? Let λ∗ be the
denominator that we’ll clear.
k
0
1
2
3
λ∗
24
60
120
210
λ1
0
20
60
126
λ2
0
10
24
42
λ3
0
6
12
18
λ4
24
24
24
24
Coefficient growth
Question: What are the coefficients for S/µk I? Let λ∗ be the
denominator that we’ll clear.
k
0
1
2
3
λ∗
24
60
120
210
λ1
0
20
60
126
λ2
0
10
24
42
λ3
0
6
12
18
λ4
24
24
24
24
For general k , λ∗ = (k + 2)(k + 3)(k + 4), λ1 = k (k + 3)(k + 4),
λ2 = 2k (k + 4), λ3 = 2 · 3 · k , and λ4 = 2 · 3 · 4.
Coefficient growth
Question: What are the coefficients for S/µk I? Let λ∗ be the
denominator that we’ll clear.
k
0
1
2
3
λ∗
24
60
120
210
λ1
0
20
60
126
λ2
0
10
24
42
λ3
0
6
12
18
λ4
24
24
24
24
For general k , λ∗ = (k + 2)(k + 3)(k + 4), λ1 = k (k + 3)(k + 4),
λ2 = 2k (k + 4), λ3 = 2 · 3 · k , and λ4 = 2 · 3 · 4.
More general result: Starting with an ideal with a pure
resolution, multiplying by a power yields polynomial growth in
the numerators of the Boij-Söderberg coefficients with similar
factorizations.