Section 5. (0MVG) · ScalingStacks

5. The coheart of the co- $t$ -structure of Theorem 4.1

In Theorem 2.3, Hoshino, Kato and Miyachi not only obtain a non-degenerate $t$ -structure on a triangulated category $\mathcal{T}$ , but they also show that its heart, which is admissible abelian, is equivalent to the module category $\textnormal{Mod}(\textnormal{End}(S)^{\textnormal{op}})$ . We shall show that the coheart of the co- $t$ -structure obtained in Theorem 4.1 is equivalent to $\textnormal{Mod}(\textnormal{End}(S)^{\textnormal{op}})$ , where $S$ is the object from Theorem 4.1 and where $\textnormal{End}(S)^{\textnormal{op}}$ indicates that this is the category of right $\textnormal{End}(S)$ -modules. This is the second half of Theorem 2.4, and will complete the proof of the cochain analogue of Theorem 2.3.

0MUV

Setup 5.1. Throughout this section, we shall consider the co- $t$ -structure of Theorem 4.1; that is, let $\mathcal{T}$ be a triangulated category with set indexed coproducts, suppose $S$ is a compact simply connected corigid object of $\mathcal{T}$ . Furthermore, assume that $\{\Sigma^{i}S\ |\ i\in\mathbb{Z}\}$ is a generating set for $\mathcal{T}$ . Then, by Theorem 4.1, the following is a non-degenerate co- $t$ -structure on $\mathcal{T}$ :

	$\displaystyle\mathcal{A}$	$\displaystyle=$	$\displaystyle\{X\in\mathcal{T}\ \|\ \textnormal{Hom}(S,\Sigma^{i}X)=0% \textnormal{ for }i<0\},$
	$\displaystyle\mathcal{B}$	$\displaystyle=$	$\displaystyle\{X\in\mathcal{T}\ \|\ \textnormal{Hom}(S,\Sigma^{i}X)=0% \textnormal{ for }i>0\}.$

Let $\mathcal{C}=\mathcal{A}\cap\mathcal{B}$ be the coheart of this co- $t$ -structure.

0MUW

Lemma 5.2. Under the hypotheses of Setup 5.1 the functor

\textnormal{Hom}(S,-):\mathcal{C}\rightarrow\textnormal{Mod}(k^{\textnormal{op% }}),

where $k=\textnormal{End}(S)$ , is dense.

0MUX

Proof: Consider the object $S$ of $\mathcal{T}$ . Since $(\mathcal{A},\mathcal{B})$ forms a co- $t$ -structure on $\mathcal{T}$ , there is a distinguished triangle

(5.1)

\Sigma^{-1}A\stackrel{{\scriptstyle\alpha}}{{\longrightarrow}}S\longrightarrow B% \longrightarrow A{}{}

with $A\in\mathcal{A}$ and $B\in\mathcal{B}$ . Applying the functor $\textnormal{Hom}(S,-)$ to (5.1) gives the following long exact sequence:

(5.2)

\displaystyle\textnormal{Hom}(S,\Sigma^{i}S)\rightarrow\textnormal{Hom}(S,% \Sigma^{i}B)\rightarrow\textnormal{Hom}(S,\Sigma^{i}A).

In (5.2) we have $\textnormal{Hom}(S,\Sigma^{i}S)=\textnormal{Hom}(S,\Sigma^{i}A)=0$ for all $i<0$ since $A\in\mathcal{A}$ , so that $\textnormal{Hom}(S,\Sigma^{i}B)=0$ for all $i<0$ . We know that $\textnormal{Hom}(S,\Sigma^{i}B)=0$ for all $i>0$ since $B\in\mathcal{B}$ . Therefore, $\textnormal{Hom}(S,\Sigma^{i}B)=0$ for all $i\neq 0$ . By Lemma 4.3, $\textnormal{Hom}(S,S)\rightarrow\textnormal{Hom}(S,B)$ is an isomorphism. Hence we have:

\textnormal{Hom}(S,\Sigma^{i}B^{(m)})=\left\{\begin{array}[]{ll}0&\textrm{if $% i\neq 0$}\\ k^{(m)}&\textrm{if $i=0$.}\end{array}\right.

where $k=\textnormal{End}(S)$ . Hence $\textnormal{Hom}(S,-):\mathcal{C}\rightarrow\textnormal{Mod}(k^{\textnormal{op% }})$ is dense. $\Box$

Note that the fact that $\textnormal{Hom}(S,S)\rightarrow\textnormal{Hom}(S,B)$ is an isomorphism forces $\textnormal{Hom}(S,A)=0$ . Therefore, $A$ is an object of $\Sigma^{-1}\mathcal{A}$ rather than just an object of $\mathcal{A}$ .

The object $B$ introduced in triangle (5.1) above has the useful property that any object in the coheart $\mathcal{C}$ can be described in terms of it; we show this in the next lemma.

0MUY

Lemma 5.3. Under the hypotheses of Setup 5.1 we have that each $M\in\mathcal{C}$ is isomorphic to $B^{(m)}$ for some $m$ .

0MUZ

Proof: Consider the distinguished triangle (5.1) from Lemma 5.2:

\displaystyle\Sigma^{-1}A\rightarrow S\rightarrow B\rightarrow A

with $A\in\Sigma^{-1}\mathcal{A}$ and $B\in\mathcal{C}$ . Let $M\in\mathcal{C}$ ; since $k=\textnormal{Hom}(S,S)$ is a skew field, we can choose $m$ such that $S^{(m)}\rightarrow M$ becomes an isomorphism under $\textnormal{Hom}(S,-)$ . Again, by Lemma 4.3, the morphism $S\rightarrow B$ becomes an isomorphism under $\textnormal{Hom}(S,-)$ .

We may now apply the functor $\textnormal{Hom}(-,M)$ to (5.1) and from the long exact sequence notice that the morphism $\textnormal{Hom}(B,M)\rightarrow\textnormal{Hom}(S,M)$ is an isomorphism. So we obtain the commutative diagram:

where both $S^{(m)}\rightarrow M$ and $S^{(m)}\rightarrow B^{(m)}$ are isomorphisms under $\textnormal{Hom}(S,-)$ . Hence the unique map $B^{(m)}\rightarrow M$ making the diagram above commute becomes an isomorphism under $\textnormal{Hom}(S,-)$ .

Now extend this unique map $B^{(m)}\rightarrow M$ to a distinguished triangle

(5.3)

\displaystyle B^{(m)}\rightarrow M\rightarrow Z\rightarrow\Sigma B^{(m)}

and apply the functor $\textnormal{Hom}(S,-)$ to give a long exact sequence. One easily sees from this long exact sequence that $\textnormal{Hom}(S,\Sigma^{i}Z)=0$ for $i\neq 0$ . The fact that the morphism $B^{(m)}\rightarrow M$ becomes the isomorphism $\textnormal{Hom}(S,B^{(m)})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}% \textnormal{Hom}(S,M)$ forces $\textnormal{Hom}(S,Z)=0$ so that $\textnormal{Hom}(S,\Sigma^{i}Z)=0$ for all $i\in\mathbb{Z}$ . Since $\{\Sigma^{i}S\ |\ i\in\mathbb{Z}\}$ is a generating set for $\mathcal{T}$ , it follows that $Z=0$ . Hence $B^{(m)}\rightarrow M$ is an isomorphism. $\Box$

0MV0

Proposition 5.4. Under the hypotheses of Setup 5.1, the functor

\textnormal{Hom}(S,-):\mathcal{C}\rightarrow\textnormal{Mod}(k^{\textnormal{op% }})

is full and faithful.

0MV1

Proof: We first show that $\textnormal{Hom}(S,-)$ is faithful. Again, consider distinguished triangle (5.1):

\displaystyle\Sigma^{-1}A\rightarrow S\rightarrow B\rightarrow A.

By Lemma 5.3, any object $M\in\mathcal{C}$ is isomorphic to some coproduct $B^{(I)}$ , where $I$ is an indexing set and $B^{(I)}$ denotes the, possibly infinite, coproduct $\coprod_{i\in I}B$ . Hence, to show fidelity we can consider a morphism $B^{(I)}\rightarrow B^{(J)}$ which becomes zero under $\textnormal{Hom}(S,-)$ and show that it is itself necessarily zero. It is sufficient to show that the composite $B\hookrightarrow B^{(I)}\rightarrow B^{(J)}$ is zero, where the morphism $B\hookrightarrow B^{(I)}$ is just the coproduct inclusion into the $i^{\textnormal{th}}$ -summand for each $i\in I$ . This puts us in the following situation:

But, $A\in\Sigma^{-1}\mathcal{A}$ and $B^{(J)}\in\mathcal{C}=\mathcal{A}\cap\mathcal{B}$ , so that $\textnormal{Hom}(A,B^{(J)})=0$ . Therefore, the dotted arrow above is necessarily zero. Hence the composite $B\hookrightarrow B^{(I)}\rightarrow B^{(J)}$ is zero, showing that $\textnormal{Hom}(S,-)$ is faithful.

We must also show that $\textnormal{Hom}(S,-)$ is full. Suppose we have a morphism

\theta:\textnormal{Hom}(S,B^{(I)})\rightarrow\textnormal{Hom}(S,B^{(J)}),

where $I$ and $J$ are again indexing sets. We must construct a morphism $B^{(I)}\rightarrow B^{(J)}$ which induces $\theta$ under $\textnormal{Hom}(S,-)$ . We recall distinguished triangle (5.1) again:

\Sigma^{-1}A\longrightarrow S\stackrel{{\scriptstyle\sigma}}{{\longrightarrow}% }B\longrightarrow A.

Note that $\sigma:S\rightarrow B$ becomes an isomorphism under $\textnormal{Hom}(-,B^{(I)})$ because $B^{(I)}\in\mathcal{C}$ . Hence we get the following commutative diagram:

(5.4)

where $\phi=\textnormal{Hom}(\sigma,B^{(J)})^{-1}\circ\theta\circ\textnormal{Hom}(% \sigma,B^{(I)})$ . Let $q_{i}:B\hookrightarrow B^{(I)}$ be the $i^{\textrm{th}}$ -inclusion of the coproduct and consider its image $\phi(q_{i}):B\rightarrow B^{(J)}$ . By the universal property of the coproduct there exists a unique map $\langle\phi(q_{i})\rangle:B^{(I)}\rightarrow B^{(J)}$ such that the following diagram commutes for each $i\in I$ :

Let us show that $\langle\phi(q_{i})\rangle$ induces $\theta$ under $\textnormal{Hom}(S,-)$ . The map $\textnormal{Hom}(S,\sigma):\textnormal{Hom}(S,S)\rightarrow\textnormal{Hom}(S,B)$ is an isomorphism, therefore, it takes a set of generators for $\textnormal{Hom}(S,S)$ to a set of generators for $\textnormal{Hom}(S,B)$ . The vector space $\textnormal{Hom}(S,S)$ is one-dimensional and generated by the identity map on $S$ , $1_{S}$ , whose image under $\textnormal{Hom}(S,\sigma)$ is $\sigma:S\rightarrow B$ . Hence $\textnormal{Hom}(S,B)$ is generated by $\sigma$ . By the compactness of $S$ , we have

\textnormal{Hom}(S,B^{(I)})\cong\coprod_{I}\textnormal{Hom}(S,B)

and $B^{(I)}$ is generated by $|I|$ copies of $\sigma$ . It follows that $\textnormal{Hom}(S,B^{(I)})$ is generated by the family $\{\sigma\circ q_{i}\}_{i\in I}$ . Therefore, we now only need to check that $\theta$ and the map, $\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)$ , induced by $\langle\phi(q_{i})\rangle$ coincide on this set of generators.

By the commutativity of diagram (5.4) we have:

$\displaystyle\theta(q_{i}\circ\sigma)$	$\displaystyle=$	$\displaystyle\textnormal{Hom}(\sigma,B^{(J)})\circ\phi(q_{i})$
	$\displaystyle=$	$\displaystyle\phi(q_{i})\circ\sigma$
	$\displaystyle=$	$\displaystyle(\langle\phi(q_{i})\rangle\circ q_{i})\circ\sigma$
	$\displaystyle=$	$\displaystyle\langle\phi(q_{i})\rangle\circ(q_{i}\circ\sigma)$
	$\displaystyle=$	$\displaystyle\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)(q_{i}\circ\sigma)$

Hence, $\theta$ and $\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)$ coincide on a basis of $\textnormal{Hom}(S,B^{(I)})$ , thus

\theta=\textnormal{Hom}(S,\langle\phi(q_{i})\rangle)

with $\langle\phi(q_{i})\rangle\in\textnormal{Hom}(B^{(I)},B^{(J)})$ . Therefore, the functor $\textnormal{Hom}(S,-)$ is full and faithful. $\Box$

0MV2

Theorem 5.5. Under the hypotheses of Setup 5.1, the functor

\textnormal{Hom}(S,-):\mathcal{C}\rightarrow\textnormal{Mod}(k^{\textnormal{op% }}),

is an equivalence of categories, and hence, the coheart $\mathcal{C}$ of the non-degenerate co- $t$ -structure obtained in Theorem 4.1 is an abelian category.

0MV3

Proof: By Lemma 5.2 and Proposition 5.4, $\textnormal{Hom}(S,-)$ is dense and fully faithful. Hence, by [14, Theorem IV.4.1], $\textnormal{Hom}(S,-)$ is an equivalence of categories. $\Box$

Theorems 4.1 and 5.5 now combine to give Theorem 2.4.

Although it is known that the coheart of a co- $t$ -structure is not always an abelian subcategory of $\mathcal{T}$ , see [7], Theorem 5.5 leads us to pose the following question.

0MV4

Question 5.6. Under what circumstances is the coheart of a co- $t$ -structure on a triangulated category $\mathcal{T}$ an abelian subcategory of $\mathcal{T}$ ?

Acknowledgment. The author would like to thank his supervisor, Peter Jørgensen, for all the help and advice he has given during the preparation of this paper, and also to thank the University of Leeds and EPSRC of the United Kingdom for financial support. In addition, the author is particularly grateful for the useful comments made by the referees.

David Pauksztello

Original source: arXiv:0705.0102v2

5. The coheart of the co-t-structure of Theorem 4.1

5. The coheart of the co- $t$ -structure of Theorem 4.1