Timeline for What are some standard operations on the set of localizations of a triangulated category?
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11 events
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| Jun 8, 2016 at 16:37 | comment | added | David Treumann | Dragos, sorry for the long delay. You're right, the question's no good as there's no reason to think that $\rtimes$ should be closed under cones. Thanks! | |
| Mar 21, 2016 at 13:33 | comment | added | Dragos Fratila | David, do you want $D_1\rtimes D_2$ to be a triangulated subcategory by definition? (to me just the condition $x\to x_1\leftarrow x_2 ... \to x_r\leftarrow y$ with all the cones in $D_1$ and $y\in D_2$ doesn't seem to be closed under cones). If not, then I think, heuristically, $D_1\rtimes D_2$="objects in $C$ which are $D_1$-equivalent to an object of $D_2$" is formed of objects that have a filtration (iterations of triangles) with one graded piece in $D_2$ and all the others in $D_1$. And then $D_1\rtimes D_2$ seems different from $D_2\rtimes D_1$. Did I misunderstood the definition again? | |
| Mar 21, 2016 at 12:43 | comment | added | Dragos Fratila | Hi David. I agree with your first remark and what I had in mind was that the category $<D_1,D_2>$ has objects iterations of triangles as I wrote, but I didn't say it carefully. Anyway, I realise I misunderstood the "$D_1$-equivalent to $D_2$" so I take back the argument. Now I'm having doubts whether the two subcategories that you define are the same... | |
| Mar 21, 2016 at 2:51 | comment | added | David Treumann | Thanks Dragos. I'm not sure that $\langle D_1,D_2\rangle$ contains only those objects that are either extensions of $d_1$ by $d_2$ or that are extensions of $d_2$ by $d_1$. It could also contain an object that carries a filtration whose 1st, 3rd, 5th,... graded pieces belong to $D_1$ and whose 2nd, 4th, 6th,... graded pieces belong to $D_2$. I also don't think that $x$ and $y$ are $D$-equivalent only if there is either a $D$-equivalence $x \to y$ or a $D$-equivalence $y \to x$. But I still don't know whether $D_1 \rtimes D_2 = \langle D_1,D_2\rangle$, maybe my two objections cancel out. | |
| Mar 19, 2016 at 20:33 | comment | added | Dragos Fratila | So one can rewrite each of the two $D_1\rtimes D_2$ and $D_2\rtimes D_1$ as $x$ sitting in a triangle $d_1\to x\to d_2\to $ or in a triangle $d_2\to x\to d_1\to $. Therefore it looks like both categories are $<D_1,D_2>$ the triangulated subcat gen by $D_1$ and $D_2$. But again, maybe I'm missing something important... | |
| Mar 19, 2016 at 20:25 | comment | added | Dragos Fratila | Sorry for the late reply. So here's what I thought at first: $x$ is $D$-equivalent to $y$ if either there's a triangle $x\to y\to d\to $ or $y\to x\to d\to $ where $d\in D$. Now if $x$ is $D_1$-equivalent to an object of $D_2$ then there's a triangle $x\to d_2\to d_1\to $. But this implies also a triangle $d_1[-1]\to x\to d_2\to $ and so $x$ is $D_2$ equivalent to $d_1$. This seems to imply that the two equivalence relations "$D_1$ equivalent to an object of $D_2$" and "$D_2$ equivalent to an object of $D_1$" are the same... Or maybe I'm too naive and missing something? | |
| Mar 17, 2016 at 17:43 | comment | added | David Treumann | I think I misunderstood your first comment, and also maybe now I'm not sure what you mean by "symmetric closure." By "D_1-equivalent" in the second paragraph I mean, "is connected by a chain of D_1-equivalences, $x \to x' \leftarrow \cdots \to y$," which I hope is the same as "$x$ and $y$ are isomorphic in a quotient category $C/D$." If that's enough to prove $D_1 \rtimes D_2$ is subcategory generated by $D_1$ and $D_2$, can you spoon feed me the explanation? | |
| Mar 17, 2016 at 15:11 | comment | added | Dragos Fratila | Oh, so when you say $x$ is $D$-equivalent to $y$ you mean precisely for a map $x\to y$? (i.e. exactly in this direction) I thought by equivalence you meant the symmetric closure of this. So I guess $D_1\rtimes D_2$ are extension of an object in $D_1$ by an object in $D_2$, and the other way around for $D_2\rtimes D_1$. Makes me think of torsion pairs in triang cats but they are not translation invariant in general... | |
| Mar 17, 2016 at 13:31 | comment | added | David Treumann | Hi Dragos. I think a $D$-equivalence $x \to y$ does not guarantee a $D$-equivalence $y \to x$. For example when $C$ is the derived category of abelian groups and $D$ is the subcategory of torsion groups, then $\mathbf{Z} \to \mathbf{Q}$ is a $D$-equivalence. I don't know if this disproves your conclusion, though. | |
| Mar 17, 2016 at 8:08 | comment | added | Dragos Fratila | Isn't it in both cases just the subcategory formed by objects which are extensions of $D_1$ by $D_2$ or of $D_2$ by $D_1$? Provided that the equivalence is symmetric: $x$ is $D$-equivalent to $y$ if there is map $x\to y$ or $y\to x$ whose cone is in $D$. Then I think $D_1\rtimes D_2$ would be just the triangulated subcategory generated by $D_1$ and $D_2$. | |
| Mar 17, 2016 at 5:24 | history | asked | David Treumann | CC BY-SA 3.0 |