Macaulay2 / Macaulay2/M2

Consistency of BettiTally ZZ with nonstandard heft vectors

Open
#2,303 0 comments 0 reactions 0 assignees View on GitHub

Nobody has claimed this yet.

Core
Dominant language
Macaulay2
Stars
435
Forks
297
Avg merge
4d 20h
Merged PRs (30d)
11

Description

Here's a Betti table:
```m2
i1 : R = kk[x,y,z];

i2 : K = koszul vars R;

i3 : B = betti K

0 1 2 3
o3 = total: 1 3 3 1
0: 1 3 3 1

o3 : BettiTally

i4 : peek B

o4 = BettiTally{(0, {0}, 0) => 1}
(1, {1}, 1) => 3
(2, {2}, 2) => 3
(3, {3}, 3) => 1
```
The documentation says about the operations `BettiTally ZZ` and `BettiTally Array`:
> These operations mimic the corresponding operations on chain complexes.

However, compare the following:
```m2
i5 : B(10), peek B(10)

0 1 2 3
o5 = (total: 1 3 3 1, BettiTally{(0, {0}, -10) => 1})
-10: 1 3 3 1 (1, {1}, -9) => 3
(2, {2}, -8) => 3
(3, {3}, -7) => 1

o5 : Sequence

i6 : betti(K ** R^{10}), peek betti(K ** R^{10})

0 1 2 3
o6 = (total: 1 3 3 1, BettiTally{(0, {-10}, -10) => 1})
-10: 1 3 3 1 (1, {-9}, -9) => 3
(2, {-8}, -8) => 3
(3, {-7}, -7) => 1

o6 : Sequence
```
The reason is that `BettiTally ZZ` only shifts the weights, but not the multidegrees:
```m2
BettiTally ZZ := (C,n) -> applyKeys(C, (i,d,h) -> (i,d,h-n))
```
So I'm wondering whether the multidegrees should have been shifted as well.

In fact, I imagine I would have _only_ shifted the multidegrees, because if the ring has nonstandard heft vector, the effects of shifting the multidegrees of the complex by N is not going to simply shift the total weight by N, so even the Betti table would be incorrect if we only shift the total weights:
```m2
i1 : R = kk[x,y,Heft => {2}];

i2 : K = koszul vars R;

i3 : B = betti K

0 1 2
o3 = total: 1 2 1
0: 1 . .
1: . 2 .
2: . . 1

o3 : BettiTally

i4 : peek B

o4 = BettiTally{(0, {0}, 0) => 1}
(1, {1}, 2) => 2
(2, {2}, 4) => 1

i5 : B(10), peek B(10)

0 1 2
o5 = (total: 1 2 1, BettiTally{(0, {0}, -10) => 1})
-10: 1 . . (1, {1}, -8) => 2
-9: . 2 . (2, {2}, -6) => 1
-8: . . 1

o5 : Sequence

i6 : betti(K ** R^{10}), peek betti(K ** R^{10})

0 1 2
o6 = (total: 1 2 1, BettiTally{(0, {-10}, -20) => 1})
-20: 1 . . (1, {-9}, -18) => 2
-19: . 2 . (2, {-8}, -16) => 1
-18: . . 1

o6 : Sequence
```

Contributor guide

No contributing guide indexed for this repository

First steps

  1. Read the whole issue, then the project's contributing guide.
  2. Comment on the issue to say you are picking it up — it saves two people doing the same work.
  3. Fork the repository and make your change on a branch.
  4. Open a pull request that references the issue number.

Research direction

Reproduce the BettiTally ZZ and betti(K ** R^{10}) examples, including the nonstandard heft-vector case, and compare their weights and multidegrees. Then inspect the BettiTally ZZ operation and related betti behavior; done means the chosen shifting semantics are consistent and documented for both standard and nonstandard heft vectors.

Written by the indexing model from the issue text.

Assessment

Domain
tooling
Issue type
Bug
Difficulty
5/5
Estimated time
Over a week
Activity status
Stale
Clarity
Mostly clear
Newbie friendliness
25/100

Get new issues in your inbox

A short digest of beginner-friendly GitHub issues.