Supports and Support Varieties
Billy Sanders
University of Kansas
[email protected]
13 January 2015
Billy Sanders (KU)
Support
13 January 2015
1 / 23
Thank You!
Thank You For The Invitation
Billy Sanders (KU)
Support
13 January 2015
2 / 23
Collaborator
Joint work with Hailong Dao
University of Kansas
[email protected]
Billy Sanders (KU)
Support
13 January 2015
3 / 23
Set up
Let (Q, n, k) be a regular local ring, assume k is algebraically closed
Billy Sanders (KU)
Support
13 January 2015
4 / 23
Set up
Let (Q, n, k) be a regular local ring, assume k is algebraically closed
Let f1 , . . . , fc ∈ n2 be a regular sequence
Billy Sanders (KU)
Support
13 January 2015
4 / 23
Set up
Let (Q, n, k) be a regular local ring, assume k is algebraically closed
Let f1 , . . . , fc ∈ n2 be a regular sequence
Set R = Q/(f1 , . . . , fc )Q and m = n/(f1 , . . . , fc )Q
(R, m, k) is a local complete intersection of codimension c
Billy Sanders (KU)
Support
13 January 2015
4 / 23
Support Varieties
Definition
For M ∈ mod(R)
o
n
V ∗ (M) = (a1 , . . . , ac ) ∈ Pc−1
|
pd
M
=
∞
Q/(ã1 f1 +···+ãc fc )Q
k
where ãi is a lift of ai
Billy Sanders (KU)
Support
13 January 2015
5 / 23
Support Varieties
Definition
For M ∈ mod(R)
o
n
V ∗ (M) = (a1 , . . . , ac ) ∈ Pc−1
|
pd
M
=
∞
Q/(ã1 f1 +···+ãc fc )Q
k
where ãi is a lift of ai
Note: We may also define V ∗ (M) as the support of Ext• (M, k) in the ring
of cohomological operators.
Billy Sanders (KU)
Support
13 January 2015
5 / 23
Support Varieties
Definition
For M ∈ mod(R)
o
n
V ∗ (M) = (a1 , . . . , ac ) ∈ Pc−1
|
pd
M
=
∞
Q/(ã1 f1 +···+ãc fc )Q
k
where ãi is a lift of ai
Note: We may also define V ∗ (M) as the support of Ext• (M, k) in the ring
of cohomological operators.
Note: We are considering support varieties as subsets of projective space
instead of cones in affine space
Billy Sanders (KU)
Support
13 January 2015
5 / 23
Example
V ∗ (R) = ∅
Billy Sanders (KU)
Support
13 January 2015
6 / 23
Example
V ∗ (R) = ∅
Why?
Billy Sanders (KU)
Support
13 January 2015
6 / 23
Example
V ∗ (R) = ∅
Why? Every g := a1 f + · · · ac fc ∈
/ n2 can be extended to a regular
sequence g , g2 , . . . , gc which generates (f1 , . . . , fc ). Hence
pdQ/gQ R = pdQ/gQ Q/(g , g2 , . . . , gc )Q < ∞.
Billy Sanders (KU)
Support
13 January 2015
6 / 23
Example
V ∗ (R) = ∅
Why? Every g := a1 f + · · · ac fc ∈
/ n2 can be extended to a regular
sequence g , g2 , . . . , gc which generates (f1 , . . . , fc ). Hence
pdQ/gQ R = pdQ/gQ Q/(g , g2 , . . . , gc )Q < ∞.
Example
V ∗ (M) = ∅ if and only if pdR M < ∞
Billy Sanders (KU)
Support
13 January 2015
6 / 23
Example
V ∗ (R) = ∅
Why? Every g := a1 f + · · · ac fc ∈
/ n2 can be extended to a regular
sequence g , g2 , . . . , gc which generates (f1 , . . . , fc ). Hence
pdQ/gQ R = pdQ/gQ Q/(g , g2 , . . . , gc )Q < ∞.
Example
V ∗ (M) = ∅ if and only if pdR M < ∞
Example
V ∗ (k) = Pkc−1
Billy Sanders (KU)
Support
13 January 2015
6 / 23
Support Variety Facts
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
cx M = dim V ∗ (M) + 1 where cx M is the degree of polynomial
growth of the Betti numbers
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
cx M = dim V ∗ (M) + 1 where cx M is the degree of polynomial
growth of the Betti numbers
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if Ext0
R (M, N) = 0
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
cx M = dim V ∗ (M) + 1 where cx M is the degree of polynomial
growth of the Betti numbers
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if Ext0
R (M, N) = 0
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if TorR
0 (M, N) = 0
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
cx M = dim V ∗ (M) + 1 where cx M is the degree of polynomial
growth of the Betti numbers
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if Ext0
R (M, N) = 0
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if TorR
0 (M, N) = 0
If 0 → M1 → M2 → M3 → 0 is exact,
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
cx M = dim V ∗ (M) + 1 where cx M is the degree of polynomial
growth of the Betti numbers
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if Ext0
R (M, N) = 0
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if TorR
0 (M, N) = 0
If 0 → M1 → M2 → M3 → 0 is exact, then for {i, j, l} = {1, 2, 3}
V ∗ (Mi ) ⊆ V ∗ (Mj ) ∪ V ∗ (Ml )
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Support Variety Facts
See Avramov Buchweitz 2000
V ∗ (M) is a Zariski closed set of Pc−1
k
For every Zariski closed set U ⊆ Pc−1
there exists an M ∈ mod(R)
k
∗
such that V (M) = U [Bergh07] [Avramov,Iyengar07]
cx M = dim V ∗ (M) + 1 where cx M is the degree of polynomial
growth of the Betti numbers
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if Ext0
R (M, N) = 0
V ∗ (M) ∩ V ∗ (N) = ∅ if and only if TorR
0 (M, N) = 0
If 0 → M1 → M2 → M3 → 0 is exact, then for {i, j, l} = {1, 2, 3}
V ∗ (Mi ) ⊆ V ∗ (Mj ) ∪ V ∗ (Ml )
V ∗ (M) = V ∗ (Ω M)
Billy Sanders (KU)
Support
13 January 2015
7 / 23
Understanding V ∗ (M) is hard
Billy Sanders (KU)
Support
13 January 2015
8 / 23
Understanding V ∗ (M) is hard
Why?
Billy Sanders (KU)
Support
13 January 2015
8 / 23
Understanding V ∗ (M) is hard
Why?
Because understanding V ∗ (M) requires an understanding the action of the
cohomological operators on Ext• (M, k) which requires a detailed
understanding of a free resolution of M.
Billy Sanders (KU)
Support
13 January 2015
8 / 23
Main Result
Theorem (Dao, -)
If Tor>0 (M, N) = 0 and M, N 6= 0, then
V ∗ (M ⊗ N) = Join(V ∗ (M), V ∗ (N))
Billy Sanders (KU)
Support
13 January 2015
9 / 23
Main Result
Theorem (Dao, -)
If Tor>0 (M, N) = 0 and M, N 6= 0, then
V ∗ (M ⊗ N) = Join(V ∗ (M), V ∗ (N))
Definition
For U, V ⊆ Pn with U, V 6= ∅ and U ∩ V = ∅
[
Join(U, V ) =
l(x, y )
x∈U y ∈V
where l(x, y ) is the projective line containing x and y . Set Join(∅, U) = U.
Billy Sanders (KU)
Support
13 January 2015
9 / 23
Basics about joins
Billy Sanders (KU)
Support
13 January 2015
10 / 23
Basics about joins
The join of two closed sets is again closed
Billy Sanders (KU)
Support
13 January 2015
10 / 23
Basics about joins
The join of two closed sets is again closed
The join of two linear spaces is the smallest linear space containing
both of them
Billy Sanders (KU)
Support
13 January 2015
10 / 23
Basics about joins
The join of two closed sets is again closed
The join of two linear spaces is the smallest linear space containing
both of them
Join is associative
Billy Sanders (KU)
Support
13 January 2015
10 / 23
Basics about joins
The join of two closed sets is again closed
The join of two linear spaces is the smallest linear space containing
both of them
Join is associative
U, V ⊆ Pnk closed implies
dim V + dim U + 1 = dim Join(U, V )
Billy Sanders (KU)
Support
13 January 2015
10 / 23
Basics about joins
The join of two closed sets is again closed
The join of two linear spaces is the smallest linear space containing
both of them
Join is associative
U, V ⊆ Pnk closed implies
dim V + dim U + 1 = dim Join(U, V )
Understanding the defining equations of the join is difficult. In fact,
our theorem shows that this is the same hard problem of computing
support varieties.
Billy Sanders (KU)
Support
13 January 2015
10 / 23
Main Result Recap
Theorem (Dao, -)
If Tor>0 (M, N) = 0 and M, N 6= 0, then
V ∗ (M ⊗ N) = Join(V ∗ (M), V ∗ (N))
Billy Sanders (KU)
Support
13 January 2015
11 / 23
Why might this be true?
Suppose Tor>0 (M, N) = 0.
Billy Sanders (KU)
Support
13 January 2015
12 / 23
Why might this be true?
Suppose Tor>0 (M, N) = 0.
cx M ⊗ N = cx M + cx N
Billy Sanders (KU)
Support
13 January 2015
12 / 23
Why might this be true?
Suppose Tor>0 (M, N) = 0.
cx M ⊗ N = cx M + cx N
dim V ∗ (M ⊗ N) = dim V ∗ (M) + dim V ∗ (N) + 1
Billy Sanders (KU)
Support
13 January 2015
12 / 23
Why might this be true?
Suppose Tor>0 (M, N) = 0.
cx M ⊗ N = cx M + cx N
dim V ∗ (M ⊗N) = dim V ∗ (M)+dim V ∗ (N)+1 = dim Join(V ∗ (M), V ∗ (N))
Billy Sanders (KU)
Support
13 January 2015
13 / 23
Simple examples
Example
Suppose c = 2
Billy Sanders (KU)
Support
13 January 2015
14 / 23
Simple examples
Example
Suppose c = 2
=⇒ support varieties live in P1k
Billy Sanders (KU)
Support
13 January 2015
14 / 23
Simple examples
Example
Suppose c = 2
=⇒ support varieties live in P1k
Suppose Tor>0 (M, N) = 0 and pd M = pd N = ∞.
Billy Sanders (KU)
Support
13 January 2015
14 / 23
Simple examples
Example
Suppose c = 2
=⇒ support varieties live in P1k
Suppose Tor>0 (M, N) = 0 and pd M = pd N = ∞.
=⇒ dim V ∗ (M) = dim V ∗ (N) = 0
Billy Sanders (KU)
Support
13 January 2015
14 / 23
Simple examples
Example
Suppose c = 2
=⇒ support varieties live in P1k
Suppose Tor>0 (M, N) = 0 and pd M = pd N = ∞.
=⇒ dim V ∗ (M) = dim V ∗ (N) = 0
=⇒ dim V ∗ (M ⊗ N) = 1
Billy Sanders (KU)
Support
13 January 2015
14 / 23
Simple examples
Example
Suppose c = 2
=⇒ support varieties live in P1k
Suppose Tor>0 (M, N) = 0 and pd M = pd N = ∞.
=⇒ dim V ∗ (M) = dim V ∗ (N) = 0
=⇒ dim V ∗ (M ⊗ N) = 1
=⇒ V ∗ (M ⊗ N) = P1k = Join(V ∗ (M), V ∗ (N))
Billy Sanders (KU)
Support
13 January 2015
14 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
In this case
Join(V ∗ (M), V ∗ (N)) = Join(∅, V ∗ (N)) = V ∗ (N)
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
In this case
Join(V ∗ (M), V ∗ (N)) = Join(∅, V ∗ (N)) = V ∗ (N)
0 → R nt → · · · → R n0 → M → 0
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
In this case
Join(V ∗ (M), V ∗ (N)) = Join(∅, V ∗ (N)) = V ∗ (N)
0 → R nt → · · · → R n0 → M → 0
0 → N nt → · · · → N n0 → M ⊗ N → 0
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
In this case
Join(V ∗ (M), V ∗ (N)) = Join(∅, V ∗ (N)) = V ∗ (N)
0 → R nt → · · · → R n0 → M → 0
0 → N nt → · · · → N n0 → M ⊗ N → 0
=⇒ V ∗ (M ⊗ N) ⊆ V ∗ (N) = Join(V ∗ (M), V ∗ (N))
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
In this case
Join(V ∗ (M), V ∗ (N)) = Join(∅, V ∗ (N)) = V ∗ (N)
0 → R nt → · · · → R n0 → M → 0
0 → N nt → · · · → N n0 → M ⊗ N → 0
=⇒ V ∗ (M ⊗ N) ⊆ V ∗ (N) = Join(V ∗ (M), V ∗ (N))
If V ∗ (N) is irreducible
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Simple example
Example
Suppose pd M < ∞ and Tor>0 (M, N) = 0.
In this case
Join(V ∗ (M), V ∗ (N)) = Join(∅, V ∗ (N)) = V ∗ (N)
0 → R nt → · · · → R n0 → M → 0
0 → N nt → · · · → N n0 → M ⊗ N → 0
=⇒ V ∗ (M ⊗ N) ⊆ V ∗ (N) = Join(V ∗ (M), V ∗ (N))
If V ∗ (N) is irreducible =⇒ V ∗ (M ⊗ N) = V ∗ (N) = Join(V ∗ (M), V ∗ (N))
since the dimensions are the same.
Billy Sanders (KU)
Support
13 January 2015
15 / 23
Interesting consequences
Corollary
Suppose TorR
>0 (M, N) = 0 and N, M 6= 0
Billy Sanders (KU)
Support
13 January 2015
16 / 23
Interesting consequences
Corollary
Suppose TorR
>0 (M, N) = 0 and N, M 6= 0
1
V ∗ (M) ⊆ V ∗ (M ⊗ N)
Billy Sanders (KU)
Support
13 January 2015
16 / 23
Interesting consequences
Corollary
Suppose TorR
>0 (M, N) = 0 and N, M 6= 0
1
V ∗ (M) ⊆ V ∗ (M ⊗ N)
2
If R is an isolated singularity, M ∈ Thick M ⊗ N [Stevenson12].
Billy Sanders (KU)
Support
13 January 2015
16 / 23
Interesting consequences
Corollary
Suppose TorR
>0 (M, N) = 0 and N, M 6= 0
1
2
V ∗ (M) ⊆ V ∗ (M ⊗ N)
If R is an isolated singularity, M ∈ Thick M ⊗ N [Stevenson12]. This
means we can build M from M ⊗ N from extensions, syzygies, and
other similar operations
Billy Sanders (KU)
Support
13 January 2015
16 / 23
Interesting consequences
Corollary
Suppose TorR
>0 (M, N) = 0 and N, M 6= 0
1
2
3
V ∗ (M) ⊆ V ∗ (M ⊗ N)
If R is an isolated singularity, M ∈ Thick M ⊗ N [Stevenson12]. This
means we can build M from M ⊗ N from extensions, syzygies, and
other similar operations
R
if TorR
0 (M ⊗ N, L) = 0, then Tor0 (M, L) = 0
Billy Sanders (KU)
Support
13 January 2015
16 / 23
Interesting consequences
Corollary
Suppose TorR
>0 (M, N) = 0 and N, M 6= 0
1
2
V ∗ (M) ⊆ V ∗ (M ⊗ N)
If R is an isolated singularity, M ∈ Thick M ⊗ N [Stevenson12]. This
means we can build M from M ⊗ N from extensions, syzygies, and
other similar operations
3
R
if TorR
0 (M ⊗ N, L) = 0, then Tor0 (M, L) = 0
4
0
if ExtR
(M ⊗ N, L) = 0, then Ext0
R (M, L) = 0
Billy Sanders (KU)
Support
13 January 2015
16 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
First assume each V ∗ (Mi ) = xi , a point.
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
First assume each V ∗ (Mi ) = xi , a point. Suppose not.
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
First assume each V ∗ (Mi ) = xi , a point. Suppose not.
Tor>0 (M1 , M2 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ) = Join(x1 , x2 ) a line
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
First assume each V ∗ (Mi ) = xi , a point. Suppose not.
Tor>0 (M1 , M2 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ) = Join(x1 , x2 ) a line
Tor>0 (M1 ⊗ M2 , M3 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ⊗ M3 ) = Join(x1 , x2 , x3 ) a plane
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
First assume each V ∗ (Mi ) = xi , a point. Suppose not.
Tor>0 (M1 , M2 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ) = Join(x1 , x2 ) a line
Tor>0 (M1 ⊗ M2 , M3 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ⊗ M3 ) = Join(x1 , x2 , x3 ) a plane
..
.
TorR
>0 (M1 ⊗ . . . ⊗ Mc , Mc+1 ) = 0
=⇒ V ∗ (M1 ⊗ · · · ⊗ Mc+1 ) = Join(x1 , . . . , xc+1 ) a c dim. linear space
Billy Sanders (KU)
Support
13 January 2015
17 / 23
More interesting consequences
Corollary
Suppose M1 , . . . , Mc+1 ∈ mod(R) are non free maximal Cohen-Macaulay.
Then there exists a t ∈ {1, . . . , c} such that
TorR
0 (M1 ⊗ . . . ⊗ Mt , Mt+1 ) 6= 0
Proof.
First assume each V ∗ (Mi ) = xi , a point. Suppose not.
Tor>0 (M1 , M2 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ) = Join(x1 , x2 ) a line
Tor>0 (M1 ⊗ M2 , M3 ) = 0
=⇒ V ∗ (M1 ⊗ M2 ⊗ M3 ) = Join(x1 , x2 , x3 ) a plane
..
.
TorR
>0 (M1 ⊗ . . . ⊗ Mc , Mc+1 ) = 0
=⇒ V ∗ (M1 ⊗ · · · ⊗ Mc+1 ) = Join(x1 , . . . , xc+1 ) a c dim. linear space
This is a contradiction.
Billy Sanders (KU)
Support
13 January 2015
17 / 23
Other Possible Applications
Billy Sanders (KU)
Support
13 January 2015
18 / 23
Other Possible Applications
Realizability:
Billy Sanders (KU)
Support
13 January 2015
18 / 23
Other Possible Applications
Realizability: In order to construct a module whose support variety is
a given linear space, it suffices to construct a module for a given
point.
Billy Sanders (KU)
Support
13 January 2015
18 / 23
Other Possible Applications
Realizability: In order to construct a module whose support variety is
a given linear space, it suffices to construct a module for a given
point.
Giving the defining equations for the join is poorly understood.
Billy Sanders (KU)
Support
13 January 2015
18 / 23
Other Possible Applications
Realizability: In order to construct a module whose support variety is
a given linear space, it suffices to construct a module for a given
point.
Giving the defining equations for the join is poorly understood. The
defining equations for support varieties are given by Avramov and
Buchweitz.
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Sketch of the proof
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
It suffices to show that for every hyper plane H ⊆ Pc−1
k :
Join(V ∗ (M)∩H, V ∗ (N)∩H) = Join(V ∗ (M), V ∗ (N))∩H ⊆ V ∗ (M ⊗N)∩H
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
It suffices to show that for every hyper plane H ⊆ Pc−1
k :
Join(V ∗ (M)∩H, V ∗ (N)∩H) = Join(V ∗ (M), V ∗ (N))∩H ⊆ V ∗ (M ⊗N)∩H
After changing coordinates, we may assume that H = V (xc ).
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Support
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
It suffices to show that for every hyper plane H ⊆ Pc−1
k :
Join(V ∗ (M)∩H, V ∗ (N)∩H) = Join(V ∗ (M), V ∗ (N))∩H ⊆ V ∗ (M ⊗N)∩H
After changing coordinates, we may assume that H = V (xc ).
Let T = Q/(f1 , . . . , fc−1 )Q.
Billy Sanders (KU)
Support
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
It suffices to show that for every hyper plane H ⊆ Pc−1
k :
Join(V ∗ (M)∩H, V ∗ (N)∩H) = Join(V ∗ (M), V ∗ (N))∩H ⊆ V ∗ (M ⊗N)∩H
After changing coordinates, we may assume that H = V (xc ).
Let T = Q/(f1 , . . . , fc−1 )Q. Note R = T /fc .
Billy Sanders (KU)
Support
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
It suffices to show that for every hyper plane H ⊆ Pc−1
k :
Join(V ∗ (M)∩H, V ∗ (N)∩H) = Join(V ∗ (M), V ∗ (N))∩H ⊆ V ∗ (M ⊗N)∩H
After changing coordinates, we may assume that H = V (xc ).
Let T = Q/(f1 , . . . , fc−1 )Q. Note R = T /fc .
Note that V ∗ (M) ∩ H = VT∗ (M) etc
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Support
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Sketch of the proof
Lemma
∗
∗
∗
If TorR
>0 (M, N) = 0, then Join(V (M), V (N)) ⊆ V (M ⊗ N).
Proof.
We induct on c. When c = 1, 2 we are done.
It suffices to show that for every hyper plane H ⊆ Pc−1
k :
Join(V ∗ (M)∩H, V ∗ (N)∩H) = Join(V ∗ (M), V ∗ (N))∩H ⊆ V ∗ (M ⊗N)∩H
After changing coordinates, we may assume that H = V (xc ).
Let T = Q/(f1 , . . . , fc−1 )Q. Note R = T /fc .
Note that V ∗ (M) ∩ H = VT∗ (M) etc
Hence it suffices to show that
Join(VT∗ (N), VT∗ (M)) ⊆ VT∗ (M ⊗ N)
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Proof Continued.
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Proof Continued.
Now TorT
>1 (M, N) = 0
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M))
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
Tensoring 0 → ΩT N → T n → N → 0 yields:
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
Tensoring 0 → ΩT N → T n → N → 0 yields:
0 → M ⊗ N → M ⊗ ΩT N → M n → M ⊗ N → 0
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
Tensoring 0 → ΩT N → T n → N → 0 yields:
0 → M ⊗ N → M ⊗ ΩT N → M n → M ⊗ N → 0
V ∗ (M ⊗ ΩT N) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
Tensoring 0 → ΩT N → T n → N → 0 yields:
0 → M ⊗ N → M ⊗ ΩT N → M n → M ⊗ N → 0
V ∗ (M ⊗ ΩT N) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
Join(VT∗ (N), VT∗ (M)) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
Tensoring 0 → ΩT N → T n → N → 0 yields:
0 → M ⊗ N → M ⊗ ΩT N → M n → M ⊗ N → 0
V ∗ (M ⊗ ΩT N) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
Join(VT∗ (N), VT∗ (M)) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
Similarly Join(VT∗ (N), VT∗ (M)) ⊆ VT∗ (N) ∪ VT∗ (M ⊗ N)
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Proof Continued.
T
Now TorT
>1 (M, N) = 0 and Tor1 (M, N) = M ⊗ N
TorT
>0 (M, ΩT N) = 0
VT∗ (ΩT N) = VT∗ (N)
codim T = c − 1, so by induction
VT∗ (M ⊗ ΩT N) ⊇ Join(VT∗ (ΩT N), VT∗ (M)) = Join(VT∗ (N), VT∗ (M))
Tensoring 0 → ΩT N → T n → N → 0 yields:
0 → M ⊗ N → M ⊗ ΩT N → M n → M ⊗ N → 0
V ∗ (M ⊗ ΩT N) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
Join(VT∗ (N), VT∗ (M)) ⊆ VT∗ (M) ∪ VT∗ (M ⊗ N)
Similarly Join(VT∗ (N), VT∗ (M)) ⊆ VT∗ (N) ∪ VT∗ (M ⊗ N)
Since V ∗ (M) ∩ V ∗ (N) = ∅, then VT∗ (M ⊗ N) ⊆ Join(VT∗ (M), VT∗ (N))
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Rest of the proof
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Rest of the proof
First, reduce to the case where V ∗ (M) and V ∗ (N) are points.
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Rest of the proof
First, reduce to the case where V ∗ (M) and V ∗ (N) are points.
Then use induction.
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Thank You!
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References
Avramov, Luchezar L. and Buchweitz, Ragnar-Olaf. Support varieties
and cohomology over complete intersections. Invent. Math. 142
(2000), no. 2, 0020-9910.
Avramov, Luchezar L. and Iyengar, Srikanth B. Constructing modules
with prescribed cohomological support. Illinois J. Math. 51 (2007), no.
1, 0019-2082.
Bergh, Petter. On support varieties for modules over complete
intersections. Proc. Amer. Math. Soc. 135 (2007), no. 12, 3795-3803
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