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

Question

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.

Answer 1

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.