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Let $\mathfrak t$ be a $t$-structure on a triangulated category $\cal T$. Let $\cal S$ be a thick (or even non-thick) triangulated subcategory, and ${\cal T}/\cal S$ the Verdier quotient.

Is there a canonical way to induce a $t$-structure on such a quotient, provided some assumptions on $\mathfrak t$ and $\cal S$?

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Often, the answer is that there is a unique way to induce a $t$-structure on $T/S$ such that the quotient functor $T\rightarrow T/S$ is right $t$-exact. This is explained well in the sections on $t$-structures in Lurie's Higher algebra.

Here is one case where this works. Suppose that in fact $S\rightarrow T$ is a fully faithful functor of stable presentable $\infty$-categories. All examples of coproduct-closed triangulated categories I know are homotopy categories of such objects. Suppose additionally that $T$ is equipped with a $t$-structure $(T_{\geq 0},T_{\leq 0})$ such that $T_{\geq 0}\subseteq T$ is closed under colimits and $T_{\geq 0}$ is itself presentable (but not, of course, stable in general). As you'll see, I'm indexing homologically, mainly because that's the way I remember all the definitions. In the quotient stable $\infty$-category $T/S$ we have the image of $T_{\geq 0}$. This generates a unique $t$-structure by Proposition 1.4.4.11 of Lurie's book. The functor $T\rightarrow T/S$ is right $t$-exact by definition (it preserves the $\geq 0$ objects).

One case where it is much less clear is for small triangulated categories. Suppose now that $S$ and $T$ are small triangulated categories with $t$-structures and that the fully faithful inclusion $S\rightarrow T$ is exact. Hence, there is an exact functor $S^\heartsuit\rightarrow T^\heartsuit$ on hearts, and $S^\heartsuit$ is a so-called weak Serre subcategory of $T^\heartsuit$ (it is closed under extensions, kernels, and cokernels). In order for there to be a $t$-structure on $T/S$ such that $T\rightarrow T/S$ is exact, it is clear that $S^\heartsuit$ must be a Serre subcategory of $T^\heartsuit$, which means that it is closed under subobjects as well.

A concrete case where this doesn't happen is as follows. Suppose that $R$ is a coherent commutative ring. This means that every finitely generated ideal of $R$ is finitely presented, and it has the consequence that the category of finitely presented $R$-modules is abelian. Let's call this category $Coh(R)$ and view it as a full subcategory of $Mod(R)$. We can consider $D^b_{Coh(R)}(Mod(R))\subseteq D^b(Mod(R))$, where the category consists of bounded complexes of $R$-modules with homology modules in $Coh(R)$. There are clearly bounded $t$-structures on these. However, if $R$ is not noetherian, then the heart of the first, namely $Coh(R)$, will not be Serre inside the heart of the second, namely $Mod(R)$.

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