# A question on the “natural metric” on the space of Bridgeland stability condition

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 $\mathcal{P}(\phi+1)=\mathcal{P}(\phi)$
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

• 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