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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

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

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This is an addition to Angelo's comment. You Given a vector bundle $E$, you can consider the its Harder-Narasimhan filtration of your vector bundle 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.Shatz S.S.Shatz discussed the behaviour of the HN-polygon under specialisation in [The decomposition and specialisation of algebraic families of vector bundles][1]. bundles I beleive this is where the term orginated, see also [Atiyah-Bott][2], 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
$$\deg \underline{Hom}(F,E)= \textrm{rk}F\deg E-\textrm{rk}E \deg F= \left| \begin{array}{cc} \deg E & \deg F\ rk E & rk F\ \end{array} \right|.$$ Also, $F$ destabilises $E$ exactly when the above determinant has negative sign.