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Whenever one studies moduli spaces of vector bundles on curves, one of the first things to be introduced is the "slope" of a vector bundle, i.e., its degree:rank ratio. Is there a nice (preferably geometric) intuition behind the use of the word "slope" for this?

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Upvote this comment if you think this should be made CW. –  Charles Staats Jun 29 '11 at 6:00
The reason is rather trivial. It is called "slope" because one usually pictures rank and degree on a cartesian plane, and then the slope is the slope of a bundle is the slope of the corresponding point. –  Angelo Jun 29 '11 at 6:45
Well, that begs the question: What insight is gained by picturing the rank and degree on a cartesian plane? (dalakov's answer may provide some help here.) –  Charles Staats Jun 29 '11 at 18:20
Side comment: after the works of Dawei Chen (see arxiv.org/abs/1003.0731), it is known that the slope of divisors are intimately related to the Lyapunov exponents of the Kontsevich-Zorich cocycle and, in the language of Lyapunov exponents, the name slope'' is justified by the fact that it measures the relative growth of polyvectors (along orbits of the Kontsevich-Zorich cocycle) inside flat subbundles of the Hodge bundle over the moduli space of Abelian differentials with respect to their counterparts'' inside the tautological subbundle. –  Matheus Oct 3 '12 at 17:11

1 Answer 1

This is an addition to Angelo's comment. Given a vector bundle $E$, you can consider its Harder-Narasimhan filtration and assign to each element of the filtration a point in the degree-rank plane. The HN-polygon is the polygon obtained by connecting the dots. S.S.Shatz discussed the behaviour of the HN-polygon under specialisation in The decomposition and specialisation of algebraic families of vector bundles I beleive this is where the term orginated, see also Atiyah-Bott, section 7 (p.565).


You can gain minor visual gratification from looking at the degree-rank plane as follows. If $F\subset E$ is a subbundle, then

$$ \deg \underline{Hom}(F,E)= \textrm{rk}F\deg E-\textrm{rk}E \deg F= \left| \begin{array}{cc} \deg E & \deg F\\\ \textrm{rk} E & \textrm{rk} F\\\ \end{array} \right|. $$ Also, $F$ destabilises $E$ exactly when the above determinant has negative sign.

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