Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have a question on the "natural metric" on the space of Bridgeland stability condition.

A stability condition $\sigma=(Z,\mathcal{P})$ on a triangulated category $\mathcal{D}$ consists of a linear map $Z:K(\mathcal{D})\rightarrow \mathbb{C}$ called the central charge, and full additive subcacegories $\mathcal{P}(\phi) \subset \mathcal{D}$ for each $\phi \in \mathbb{R}$, satisfying the following four conditions:

  1. if non-zero $E \in \mathcal{P}(\phi)$, then we have $Z(E) = m_\sigma(E)exp(i\pi \phi)$ for some $m(E)\in \mathbb{R}_{+}$
  2. forall $\phi \in \mathbb{R}$, we have $\mathcal{P}(\phi+1)=\mathcal{P}(\phi)[1]$
  3. if $\phi_1 > \phi_2$ and $E_j \in \mathcal{P}(\phi_i)$ then $Hom_{\mathcal{D}}(E_1,E_2) = 0$
  4. for non-zero $E \in \mathcal{D}$ there exists a finite sequence of real numbers $\phi_1 >\phi_2 >\dots>\phi_n$ and $E$ obtained as an "iterated extension" of objects $A_i \in \mathcal{P}(\phi_i)$.

There is a "natural metric" on $Stab(\mathcal{D})$ defined on page 7 of this paper.

A celebrated result by Bridgeland says the forgetful map $$ \mathcal{Z}:Stab(\mathcal{D})\longrightarrow Hom_{\mathbb{Z}}(K(\mathcal{D}),\mathbb{C}) $$ induces a local homeomorphism on each connected component of $Stab(\mathcal{D})$. This seems a really nice theorem.

This generalized metric is at this point beyond my intuition and I cannot really follow the proof of the theorem above, so let me now ask

Why is the generalized metric above is "natural"?

Of course some people may say it is the right one because the theorem holds. But I guess it is not the only reason. My problem is that I cannot really see why "distance" of two stability condition is measured by the only three quantities in $\sup_{0\ne E\in \mathcal{D}}$.

Edit
Are there any good toy example with which one can appreciate the metric or topology above? e

share|improve this question
    
I cannot write down the natural metric on this post. THe equation does not appear well. So I put a link to a paper by Bridgeland. –  Koopa Jan 2 '13 at 23:13
    
@Angelo It seems you can type the explicit equation. I would appreciate your edit. –  Koopa Jan 2 '13 at 23:15
    
Read "How to write math" on the side. –  Angelo Jan 3 '13 at 19:44
add comment

1 Answer

The metric may be important for physics (and as far a I understand it comes from some physical considerations), but mathematically the important thing is the topology. From this point of view the definition of the metric just expresses the fact that two stability conditions are close to each other if all objects have similar phases ranges in those (in particular an object stable in one stability condition is almost stable in the other), and have close masses. The particular form of the metric is chosen to satisfy the triangle inequality (I think this is an explanation of the logarithm there).

share|improve this answer
    
Thank you for the answer, Sasha. Logarithm part now makes sense to me. But I am not totally convinced by the phase part yet. I will think about it more. –  Koopa Jan 3 '13 at 21:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.