Several sources I see speak of a "left t-structure", but lack a precise definition. Where can I find a reference for this?
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1$\begingroup$ I don't think there is some general structure on a triangulated category called a "left/right t-structure". I think that the author of your source defined two different t-structures on a category and wanted to distinguish between the two. The motivation for the definition appears to be that one is defined in terms of what the "coconnective" objects are and the other in terms $\endgroup$– Dylan WilsonCommented Apr 16, 2017 at 17:18
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$\begingroup$ ... of the "connective" objects. $\endgroup$– Dylan WilsonCommented Apr 16, 2017 at 17:18
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$\begingroup$ I was thinking about something like a $t$-structure, but with only half of the closure properties with respect to de/suspension. Is there anything like this? $\endgroup$– foscoCommented Apr 16, 2017 at 17:30
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$\begingroup$ Well one thing that comes up is filtrations $\{\mathcal{C}_{\ge n}\}$ of a category by coreflective subcategories with the further property that $\Sigma \mathcal{C}_{\ge n} \subset \mathcal{C}_{\ge n+1}$. The slice filtration in equivariant homotopy theory is an example of such a thing. If I ever get off mathoverflow and finish my thesis, you can read more about these things :) $\endgroup$– Dylan WilsonCommented Apr 16, 2017 at 17:56
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When you try to put a t-structure, say, on the homotopy category of complexes $H(\mathcal{A})$, where $\mathcal{A}$ is "almost an abelian category" (more than additive, but the coimage must not be isomorphic to the image; I do not remember all the assumptions, but they are easy to find), there are two ways to do that. The first way to consider the subcategory $H^{\leq0}(\mathcal{A})$ of complexes split in positive degrees (make sure to define correctly what split is) and proceed with the usual construction. This is how you get the left one. Now, you can repeat this construction for the opposite category $\mathcal{A}^{op}$ and look at the left t-structure on $H(\mathcal{A}^{op})$. Under the equivalence $H(\mathcal{A})^{op}\simeq H(\mathcal{A}^{op})$ you get a second t-structure on $H(\mathcal{A})$, the right one. If you start with an abelian category, these will be the same. However, if you category is "almost abelian", these might be different (one can describe the hearts of both explicitly).
As for references, the key words are Schneiders and quasi-abelian categories.
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$\begingroup$ This was the answer I was looking for! $\endgroup$– foscoCommented Apr 17, 2017 at 22:36