24
$\begingroup$

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?

$\endgroup$
3
  • 31
    $\begingroup$ 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. $\endgroup$
    – Angelo
    Commented Jun 29, 2011 at 6:45
  • 9
    $\begingroup$ 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.) $\endgroup$ Commented Jun 29, 2011 at 18:20
  • 1
    $\begingroup$ 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. $\endgroup$
    – Matheus
    Commented Oct 3, 2012 at 17:11

2 Answers 2

21
$\begingroup$

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).

Addendum:

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.

$\endgroup$
7
$\begingroup$

Mumford studied the quasi-projectivity of moduli space of vector bundle. But such big family is unbounded. So he needed to consider moduli space of vector bundles with some sort of GIT. He arrived to the notion of slope stability. One advantage to restricting to semistable bundles of fixed rank and degree is that the moduli problem is bounded. In fact Mumford extended Giesecker-stability notion. See section 2.5

Suppose $\mathcal F$ is a coherent sheaf on a projective $X$. Define the Euler characteristic $$\chi(\mathcal F)=\sum_{i=0}^{dim X}(-1)^i\dim H^i(X,\mathcal F)$$

Let $L$ be an ample line bundle with $c_1(L) =\beta$ on a smooth projective variety $X$ of dimension $g$. A torsion-free sheaf $E$ is Gieseker stable (respectively Gieseker semistable) with respect to $\beta$ if for each subsheaf $F$ we have $$P(F) < P(E) \;\; \;\; (P(F) \leq P(E))$$ where $P(E) = \frac{\chi(E \otimes L^n)}{r(E)} $, is the reduced Hilbert polynomial.

Using Riemann-Roch theorem we have

$$P(E\otimes \mathcal O_X(1)^{\otimes m})=rk E\frac{m^n}{n!}+(deg E-rk E\frac{deg K}{2})\frac{m^{n-1}}{(n-1)!}+\cdots$$

where $K$ is the canonical divisor.

We have nice interpretation between the notion of semi-stability for vector bundles and the notion of semistability coming from an associated GIT problem

We can interpret the slope of vector bundle over a curve of genus $g$ using Riemann-Roch formula. Let me explain it

From of differential geometric point of view the degree of a holomorphic vector bundle can be computed by Chern–Weil formula in terms of curvature, and the fact that curvature decreases in sub-bundles. We explain Chern-Weil formula which gives an effective way for degree of holomorphic vector bundle.

A reflexive sheaf (i.e its double dual is equal itself) is locally free (i.e., a holomorphic vector bundle) outside a subvariety of codimension greater than or equal to two. Let $\mathcal F$ be a coherent subsheaf of holomorphic vector bundle $E$, then there is an analytic subset $S \subset M$ of codimension bigger than two and a holomorphic vector bundle $F$ on $X \setminus S$ such that $$\mathcal F|_{X\setminus S}=\mathcal O(F)$$ and $F$ is a sub-bundle of $E|_{X\setminus S}$ and there is an orthogonal projection $\pi:E|_{X\setminus S}\to F$ which $\pi\in L_1^2(End(E))$ lying in the Sobolev space of $L^2$ sections of $End(E)$ with $L^2$ first-order weak derivatives and satisfying $\pi=\pi^*=\pi^2$ where $\pi^*$ denotes the adjoint of $\pi$. The Chern-Weil formula is $$deg_\omega \mathcal F=\frac{\sqrt[]{-1}}{2\pi n}\int_X tr(\pi\Lambda_\omega F_h)\omega^n-\frac{1}{2\pi n}\int_X|\nabla''\pi|^2\omega^n$$ where $\nabla''\pi$ is computed in the sense of currents using the $(0,1)$ part of the Chern connection of $E$.

We define the slope of $\mathcal E$, to be $$\mu(\mathcal E)=\frac{deg \mathcal E}{rk \mathcal E}$$

For any non-trivial vector bundle $E$ on curve $X$ by using the Riemann-Roch formula we can compute the slope of a vector bundle over a curve as follows,

$$\mu(E)=\frac{dim H^0(X,E)-dim H^1(X,E)}{rank E}+g_X-1$$ where $g_X$ is the genus of curve $X$.

Another interpretation about semi-stability via slope

It is known that a holomorphic line bundles on a compact connected Riemann surface $\Sigma_g$ do not admit non-zero global holomorphic sections if their degree is negative.

A non-zero homomorphism from line bundles $L_1$ and $L_2$(which is a section of the line bundle $L^∗_1 ⊗ L_2$) exist if $deg (L^∗_1 ⊗ L_2) ≥ 0$, which is equivalent to $deg L_1 ≤ deg L_2$

Note that, for higher rank vector bundles, the degree of $E^∗_1 ⊗ E_2$ is $$deg (E^∗_1 ⊗ E_2) = rk (E_1)deg (E_2) − deg (E_1)rk E_2$$

so the semi-positivity condition is equivalent to $$\frac{deg E_1}{rk E_1} ≤ \frac{deg E_2}{rk E_2} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .