research_codes

HSI

Text files:

\(\DeclareMathOperator{\HS}{HS}\DeclareMathOperator{\pdim}{pdim}\)

Instructions

The file n.txt contains information about the cochordal graph on nvertices.
For each such graph \(G\), there are three lines of information:

  1. A number denoting its position in the list.
  2. Its set of edges \(E(G)\). The vertex set is always {0, ..., n-1}.
  3. A dictionary d with keys as integers and True/False as values. This is to be interpreted as following:
    Consider the edge ideal \(I := I(G) \subset \mathbb{F}_{2}[x_{0}, \ldots, x_{n - 1}] =: R\).
    Let \(\HS_{n}(I)\) denote the \(n\)-th homological shift ideal of \(I\), as defined here.
    Then, the n-th entry of d is True of False depending on whether \(\HS_{n}(I)\) has a linear resolution or not.

Remarks

Example

Consider the following excerpt from 8.txt.

411
[(0, 4), (0, 5), (0, 7), (1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (3, 7), (4, 6), (5, 6), (5, 7), (6, 7)]
{2: True, 3: False, 4: True}

The above is the 411th graph in the file. It can be pictured as:

The above says that \(\HS_2\) and \(\HS_4\) have linear resolutions but \(\HS_3\) does not, and that the higher homological shift ideals are zero. In particular, \(\pdim = 4\). (All computations being done over \(\mathbb{F}_{2}\).)
Indeed, the Betti tables of the homological shift ideals are:

$$ \begin{align} \text{Betti}(\HS_{2}): &\qquad \begin{matrix} & 0 & 1 & 2 & 3 & 4 \\ 4: & 27 & 63 & 55 & 21 & 3 \end{matrix} \\~\\ \text{Betti}(\HS_{3}): &\qquad \begin{matrix} & 0 & 1 & 2 & 3\\ 5: & 14 & 22 & 11 & 2\\ 6: & . & 1 & 1 & . \end{matrix} \\~\\ \text{Betti}(\HS_{4}): &\qquad \begin{matrix} & 0 & 1\\ 6: & 3 & 2 \end{matrix} \end{align} $$

Dropping the differentials, the (necessarily linear) minimal resolution of \(I(G)\) is given as

$$ \underset{\vphantom{\Big|}0}{R^{13}} \leftarrow \underset{\vphantom{\Big|}1}{R^{32}} \leftarrow \underset{\vphantom{\Big|}2}{R^{33}} \leftarrow \underset{\vphantom{\Big|}3}{R^{16}} \leftarrow \underset{\vphantom{\Big|}4}{R^{3}} \leftarrow \underset{\vphantom{\Big|}5}{0}, $$

confirming that \(\pdim = 4\).