# stability conditions in the sense of Kontsevich-Soibelman

What are the stability conditions in the sense of Kontsevich-Soibelman.

I am reading Bridgeland's stability conditions and I've heard people talking about the Kontsevich-Soibelman Stability. I would appreciate a brief introduction on this, in particular my questions are :

1. What are the Kontsevich-Soibelman Stability conditions ?

2. How is it related to Bridgeland's Stability (or Douglas' $\pi$ - stability on D-branes) ?

3. Why do we need to consider Kontsevich-Soibelman stability.

I've to admit my ignorance of the field. Please suggest some references. Thanks.

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Kontsevich-Soibelman's version is a version of Bridgeland's stability given for triangulated $A_\infty$ categories (with a few additional properties; Kontsevich and Soibelman call "non-commutative proper algebraic variety" such an $A_\infty$-category) rather than for triangulated categories as in the original Bridgeland's version.
The main point in Kontsevich-Soibelman definition is that once one thinks of the relevant $A_\infty$ categories as non-commutative analogues of algebraic varieties, one sees that a Bridgeland stability condition can be seen as the datum of a polarization on these non-commutative varieties. The reason for considering $A_\infty$ rather than ordinary categories is that categorical structures arising from branes are naturally $A_\infty$ categories (the most classical example to be done here is probably Fukaya's $A_\infty$ category of a symplectic manifold $X$).
I haven't looked closely at the KS paper, but is the issue really $A_\infty$ vs triangulated? i.e. I would imagine a stability condition (like a t-structure) is really a structure on the underlying (triangulated) homotopy category in any case. KS do talk about stability structures for Lie algebras (abstracting Hall algebras attached to categories) and add a support condition.. –  David Ben-Zvi Dec 27 '10 at 22:01
@ Domenico - thanks for your response. What I understood from this is that : KS stability is the Bridgeland type stability defined on $A_{\infty}$ categories rather than triangulated categories. Is KS version a generalization of Bridgeland's in a sense that for triangulated categories KS = Bridgeland. –  J Verma Dec 27 '10 at 22:16
also KS categories are triangulated: they are $A_\infty$ triangulated categories. for the particular case of triangulated categories KS and Bridgeland definition basically coincide (there could be a few minor differences; e.g., as David points out, KS introduce a support condition) –  domenico fiorenza Dec 27 '10 at 22:45