The see-saw inequality for the μ-slope on Coh X #
For a short exact sequence 0 ⟶ B ⟶ C ⟶ Q ⟶ 0 of coherent sheaves, the slope of the middle
term lies between the slopes of the ends. This file proves the half of that statement the
Harder–Narasimhan recursion needs: if the sub has slope at most the quotient, then it has slope
at most the whole.
No such inequality exists for the abstract weak slope in Abelian/Stability/**, and
this one is not abstract: it is the mediant inequality for the ratio of two additive integer
invariants, and its ⊤ cases are decided by the geometric input carried by MuPositivityData.
It is stated here, for the recursion that consumes it.
The three cases #
Multiplicity and degree coefficient are both additive on short exact sequences, so with
m = multiplicity and d = hilbertDegreeCoefficient the middle term has m C = m B + m Q and
d C = d B + d Q.
m B = 0. The sub has slope⊤, so the hypothesis forces the quotient to slope⊤as well, hencem Q = 0, hencem C = 0and the whole has slope⊤.m B > 0andm Q = 0. The whole hasm C = m B, and the inequality reduces to0 ≤ d Q, which is exactlyMuPositivityData.degree_nonneg_of_multiplicity_zero. This is the case that fails without that geometric input.- Both positive. The mediant inequality:
d B / m B ≤ d Q / m Qgivesd B / m B ≤ (d B + d Q) / (m B + m Q).
Multiplicity is additive on a short exact sequence.
The degree coefficient is additive on a short exact sequence.
The see-saw inequality. In a short exact sequence, if the sub has slope at most the quotient then it has slope at most the whole. See the module docstring for the three cases; the multiplicity-zero quotient is the one where the geometric input is used, and it is the only case needing the quotient to be nonzero — the sub may be zero.