# Bridgeland stability to Fukaya stability on elliptic curve; geometric proof of no slope decreasing homs

For a bridgeland stability condition $$(P,Z)$$ on $$\mathcal{C}$$ and $$a > b$$ we know that $$Hom^0(A,B)=0$$ for $$A,B \in P(a), P(b)$$ respectively.

I would like to see the geometric incarnation of this under mirror symmetry, in the simplest case of an elliptic curve. Thus from now on we work with a torus\elliptic curve.

Namely, under many simplifications if I understand, we can think of a bridgeland stability condition as a choice of symplectic form $$\omega$$ on $$E$$ (compatible with a fixed complex structure $$J$$).

Now every curve in $$E$$ is a lagrangian, and it should be semistable (i.e from $$P(A)$$) sorta when it's a special lagrangian, which I found is correlative to being length minimizing wrt to the natural metric $$g$$ that $$\omega, J$$ induce in its homology class.

I am not sure how to geometrically see the slope, nor how to then see that if $$a>b$$, then the fukaya hom vanishes, and that is my question.

Until someone better comes along here is some info about mirror symmetry on elliptic curve.

The rational line of slope $$m/n$$ for coprime $$m,n$$ is the appropriate stable vector bundle.

Then if you have rational lines $$L_1 ,L_2$$ with $$L_1$$ having bigger slope than to show $$Hom(L_1, L_2)=0$$ we analyze $$L_1 \cap L_2$$. THere ie one intersection point $$(0,0)$$ and the bigger slope translates to an orientation statement about the intersection point.

Namely, recall there a $$\mathbb{Z}$$ grading on $$Hom(L_1,L_2)$$ coming from a grading on the intersection points but to get just the $$\mathbb{Z}/2$$ one you only care about orientation.

So we see that the orientation will be $$1 \pmod{2}$$ and so in particular there are no homs decreasing slope.

As a further fun point; rotating the elliptic curve by 90 degrees fukayawise is the auto equivalence of derived sheaf category of the poincare bundle.

What's curious about this argument is that it was a $$\mathbb{Z}/2$$ argument; the reason there are no homs is because the grading modulo 2 won't allow it. That surprises me and I wonder if in higher dimensions it's similiar or if it uses the whole $$\mathbb{Z}$$ grading.