I am reading Bridgeland's paper on stability condition, where he defined a slicing of a triangulated category. See


After the definition he remarked that the decomposition in axiom (c) is unique and I am having trouble showing this.

I tried to follow the proof of uniqueness of Harder Narasimha filtration, but the property of semi-stability played an important role there and I could not find an analogue.

Moreover, would there be a problem if we let some of the $A_i$ to be 0 and artificially make the decomposition 'longer'?

Thanks in advance for any help!


The first step is the following. Assume $E'$ and $E''$ are two objects with filtrations as in axiom (d). Then if $\phi_n(E') > \phi_1(E'')$ then $Hom(E',E'') = 0$. This follows from (c) by induction.

The second step. Assume $E$ and $F$ are two objects, $\phi \in \mathbb{R}$, and $E' \to E \to E''$, $F' \to F \to F''$ are two triangles, such that $E'$, $F'$, $E''$, $F''$ have filtrations as in axiom (d) with $$ \phi_n(E') > \phi \ge \phi_1(E''),\qquad \phi_m(F') > \phi \ge \phi_1(F''). $$ Then for any morphism $f:E \to F$ there are unique morphisms $f':E' \to F'$ and $f'':E'' \to F''$ such that $(f',f,f'')$ is a morphism of triangles. Existence follows from $Hom(E',F'') = 0$ and axioms of triangulated category. Uniqueness follows from long exact sequence of $Ext$'s and $Hom(E',F''[-1]) = 0$ (note that the phases of $F''[-1]$ are strictly smaller than those of $F''$ by (b)).

Note, that from the uniqueness above it follows that if $f$ is an isomorphism, then so are $f'$ and $f''$. Indeed, if $g = f^{-1}$, then there are unique $g'$ and $g''$, and since $(f'\circ g',f\circ g,f'',circ g'')$ and $(1_{F'},1_F,1_{F''})$ are morphisms of triangles with the same middle map, uniqueness implies that $f'\circ g' = 1_{F'}$ and $f'' \circ g'' = 1_{F''}$.

Now one can prove a uniqueness of the filtration. Assume there are two. Choose some $\phi$ and split the filtrations at $\phi$. Take $f = 1$, and extend it to $(f',f,f'')$. By the argument above $f'$ and $f''$ are isomorphisms. Continue by induction.

Finally, you cannot artificially add $A_i = 0$, as $P(\phi)$ never contains $0$ by axiom (a), as $m(E)$ is assumed to be strictly positive!

  • $\begingroup$ Thanks for the great answer. One follow up: it seems you are using a different reference for the def of slicing than me( I noticed the difference in the indices, but of course the main contents of the def are the same), would you be able to share a link of it with me so that I can compare them? $\endgroup$ – Xuqiang QIN Feb 29 '16 at 17:55
  • $\begingroup$ I used your link (bottom of page 2). $\endgroup$ – Sasha Feb 29 '16 at 18:07
  • $\begingroup$ I think there is a problem with the 0 object. Bridgeland says that the $\mathcal P (\phi)$'s are full additive subcategories. In particular they have a zero object. He really means additive, and not preadditive, as can be observed in the proof of Lemma 5.2 in the article. Nevertheless this seems easy to fix, just asking objects to be nonzero in the relevant axioms. $\endgroup$ – Andrés Ibáñez Núñez Mar 25 at 23:05

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