3
$\begingroup$

We know that an additive full subcategory $\mathcal{S}$ of a triangulated category $\mathcal{T}$ is called a triangulated subcategory if it is closed under isomorphism, shift and if any two objects in a distinguished triangle in $\mathcal{T}$ are in $\mathcal{S}$, then so is the third. See Neeman's book section 1.5. Why is it equivalent to that the inclusion functor is exact in the sense that it preserves distinguished triangles? And what if assuming any additive subcategory which may not be full?

$\endgroup$
3
  • 1
    $\begingroup$ Crossposted: mathoverflow.net/questions/206829/… $\endgroup$ May 17, 2015 at 3:38
  • $\begingroup$ Neeman's definition (1.5.1 here) does assume that $S$ is full. $\endgroup$ Mar 13 at 9:38
  • $\begingroup$ I would like to note that if one uses 05QM as definition for triangulated subcategory (more general than Neeman's definition), then a triangulated subcategory (not necessarily full) $\mathcal{D'}$ of $\mathcal{D}$ is replete in $\mathcal{D}$ if and only if for all distinguished triangles $X\xrightarrow{f}Y\xrightarrow{g}Z\xrightarrow{h}X[1]$ of $\mathcal{D}$, if at least one of $f,g,h$ is in $\mathcal{D}'$, then so are the other two. $\endgroup$ Mar 13 at 9:53

2 Answers 2

1
$\begingroup$

Let's contemplate the two possible definitions of triangulated subcategory that OP is hinting at. For simplicity, the translation functor of a triangulated category (that I shall denote as $[1]$) is always assumed to be an automorphism. The class of distinguished triangles of a (pre)triangulated category $\mathcal{D}$ is called the (pre)triangulated structure of $\mathcal{D}$.

Definition 1. Let $\mathcal{D}$ be a (pre)triangulated category. Let $\mathcal{D}'\subset\mathcal{D}$ be an additive subcategory such that $\mathcal{D}'[1]=\mathcal{D}'$.

  • $\mathcal{D}'$ is a Neeman (pre)triangulated subcategory of $\mathcal{D}$ if it is a full replete subcategory satisfying that for every distinguished triangle $X\to Y\to Z\to X[1]$ in $\mathcal{D}$, if $X,Y$ are in $\mathcal{D}'$, then $Z$ too (by TR2, this is the same as saying that if two out of three among $X,Y,Z$ are in $\mathcal{D}'$, then so is the third).
  • $\mathcal{D}'$ is an SP (pre)triangulated subcategory of $\mathcal{D}$ if $\mathcal{D}'$ is a (pre)triangulated category whose (pre)triangulated structure is contained in the (pre)triangulated structure of $\mathcal{D}$.

(The second definition is the Stacks Project one, 05QM; it is also the same as Definition 10.1.9.iii in Kashiwara, Schapira, Categories and Sheaves.) OP is asking why does the first definition imply the second one. That is:

(Q) If $\mathcal{D}'\subset\mathcal{D}$ is a Neeman (pre)triangulated subcategory, what is the (pre)triangulated structure on $\mathcal{D}'$ turning it into an SP (pre)triangulated subcategory of $\mathcal{D}$?

The answer to (Q) is: it is the distinguished triangles in $\mathcal{D}$ whose objects all lie in $\mathcal{D}'$. Let's see why.

Remark 2. Let $\mathcal{D}$ be a pretriangulated category. If $\mathcal{D}'\subset\mathcal{D}$ is a full SP pretriangulated subcategory, then the triangulated structure of $\mathcal{D}'$ is as big as it can get, and it is fully determined by the objects of $\mathcal{D}'$. More precisely, the distinguished triangles of $\mathcal{D}'$ turn out to be exactly the distinguished triangles $\Delta=(X,Y,Z,f,g,h)$ in $\mathcal{D}$ such that $X,Y,Z\in\operatorname{ob}\mathcal{D}'$. Indeed, by (TR1)$_{\mathcal{D}'}$ there is is a d.t. $\Delta'=(X,Y,Z',f,g',h')$ in $\mathcal{D}'$. By Lemma 014B, $\Delta\cong\Delta'$. Since $\mathcal{D}'$ is full, this is an isomorphism of triangles in $\mathcal{D}'$. By (TR1)$_{\mathcal{D}'}$, the triangle $\Delta$ is distinguished in $\mathcal{D}'$.

In other words, if $(\mathcal{D},[1])$ is a triangulated category and $\mathcal{D}'\subset\mathcal{D}$ is a full additive subcategory such that $\mathcal{D}'[1]\subset\mathcal{D}'$ and $[1]:\mathcal{D}'\to\mathcal{D}'$ is an automorphism, then either $\mathcal{D}'$ can be endowed with a unique triangulated structure turning it into a triangulated subcategory of $\mathcal{D}$ or there is no such triangulated structure at all. Here is a criterion to detect the positive situation.

(A preadditive subcategory of a preadditive category is a subcategory with a preadditive structure such that the inclusion functor is additive.)

Lemma 3. Let $(\mathcal{D},[1],\mathcal{T})$ be a (pre)triangulated category. Let $\mathcal{D}'$ be a non-empty full preadditive subcategory of $\mathcal{D}$ preserved under $[1]$ and $[-1]$. Then $\mathcal{D}'$ is additive and an SP (pre)triangulated subcategory of $\mathcal{D}$ if and only $\mathcal{D}'$ is compatible with the (pre)triangulated structure of $\mathcal{D}$ in the following sense:

$(*)$ Given any morphism $f : X \to Y$ in $\mathcal{D}'$ there exists a distinguished triangle $(X, Y, Z, f, g, h)$ in $\mathcal{D}$ such that $Z$ is isomorphic to an object of $\mathcal{D}'$.

Note that in $(*)$ it doesn't matter which distinguished triangle we pick. If $\mathcal{D}'$ satisfies the compatibility condition and $f:X\to Y$ is in $\mathcal{D}'$, then by Lemma 014B, for any d.t. $(X,Y,Z,f,g,h)$ in $\mathcal{D}$, it holds that $Z$ is isomorphic to an object of $\mathcal{D}'$.

Proof. It is clear that if $\mathcal{D}'$ is additive and an SP (pre)triangulated subcategory of $\mathcal{D}$, the compatibility condition holds. Conversely, suppose the compatibility condition holds. Let's see first that $\mathcal{D}'$ is additive. Pick any object $X$ in $\mathcal{D}$. By (TR1), the triangle $X\xrightarrow{1_X}X\to 0\to X[1]$ is distinguished. Thus, $(*)$ implies that $\mathcal{D}'$ has a zero object. On the other hand, by Lemma 05QT(3) and (TR2), the triangle $Y[-1]\xrightarrow{0}X\to X\oplus Y\to Y$ is distinguished. Now $(*)$ implies that $\mathcal{D}'$ has the direct sum for $X$ and $Y$. This shows that $\mathcal{D}'$ is additive. If there is a (pre)triangulated structure $\mathcal{T}'$ on $\mathcal{D}'$ turning it into a a (pre)triangulated subcategory of $\mathcal{D}$, by Remark 2 it has to be $$ \mathcal{T}'=\{(X,Y,Z,f,g,h)\in\mathcal{T}\mid X,Y,Z\in\operatorname{ob}\mathcal{D}'\}. $$ Verification that $\mathcal{T}'$ constitutes a (pre)triangulated structure on $\mathcal{D}'$ is now an exercise on using the axioms. We will leave this to the reader. $\square$

It then follows that every Neeman (pre)triangulated subcategory is an SP (pre)triangulated subcategory in a canonical way, which answers (Q).


Bonus: We can weaken the definition of Neeman (pre)triangulated subcategory to a preadditive subcategory and drop the repleteness requirement, and still get the same (see Lemma 5 below).

Lemma 4. Let $\mathcal{D}'$ be an SP (pre)triangulated subcategory of a (pre)triangulated category $\mathcal{D}$. Then $\mathcal{D}'$ is replete in $\mathcal{D}$ if and only if for all distinguished triangles $(X,Y,Z,f,g,h)$ in $\mathcal{D}$, if at least some morphism of $f,g,h$ is in $\mathcal{D}'$, then so are the other two.

Proof. ($\Rightarrow$). Suppose $(X,Y,Z,f,g,h)$ is a d.t. in $\mathcal{D}$ such that some of $f,g,h$ is in $\mathcal{D}$. By (TR2), without loss of generality $f\in\operatorname{Mor}(\mathcal{D})$. By (TR1) in $\mathcal{D}'$, there is a distinguished triangle $(X,Y,Z',f,g',h')$ in $\mathcal{D}'$. By Lemma 014B in $\mathcal{D}$, there is $c:Z\to Z'$ such that $(1,1,c):(X,Y,Z,f,g,h)\to (X,Y,Z',f,g',h')$ is an isomorphism of triangles. Thus both $Z$ and $c$ are in $\mathcal{D}'$. Hence $g=c^{-1}\circ g'$ and $h=h'\circ c$ are also in $\mathcal{D}'$.

($\Leftarrow$). Suppose $\alpha:X\to X'$ is an isomorphism in $\mathcal{D}$, and that $X'\in\operatorname{Ob}\mathcal{D}'$. By Lemma 05QR, the triangle $(X,X',0,1,1,0)$ is distinguished in $\mathcal{D}$. Since $0:0\to X$ is in $\mathcal{D}$, so is $\alpha$. $\square$

Lemma 5. Let $\mathcal{D}$ be a (pre)triangulated category. Suppose $\mathcal{D}'$ is a full preadditive subcategory of $\mathcal{D}$ satisfying that for every d.t. $X\to Y\to Z\to X[1]$ in $\mathcal{D}$, if $X,Y$ are in $\mathcal{D}'$, then $Z$ too. Then $\mathcal{D}'$ is a Neeman (pre)triangulated subcategory (i.e., $\mathcal{D}'$ is replete in $\mathcal{D}$ and $\mathcal{D}'$ is additive).

Proof. By Lemma 3, $\mathcal{D}'$ is additive and there is a unique (pre)triangulated structure on $\mathcal{D}'$ turning it into an SP (pre)triangulated subcategory of $\mathcal{D}$. By Lemma 4, $\mathcal{D}'$ is replete in $\mathcal{D}$. $\square$

$\endgroup$
1
0
$\begingroup$

A triangulated functor $F\colon \mathcal T'\to \mathcal T$ must preserve structures. So we have $F$ commutes with translations (aka shifts/suspensions) and takes triangles in $\mathcal T'$ to triangles in $\mathcal T$.

If $\mathcal T$ is a triangulated category and $\mathcal T'$ is a full additive subcategory of $\mathcal T'$, we ask when the inclusion functor $i\colon \mathcal T'\hookrightarrow \mathcal T$ gives us an induced triangulated subcategory as its essential image. By induced, we mean a definition of triangles and translation that comes from restrictions of translation and triangles in $\mathcal T$.

So let's look at why these three criteria are reasonable:

(1) Closed under isomorphisms: Since the substructure is given by the essential image, we should have $\mathcal T'$ closed under isomorphism.

(2) Closed under translations: Clearly, if the translation of $\mathcal T'$ is inherited from $\mathcal T$ the restriction to $\mathcal T'$ must factor through itself.

(3) 2 of 3: If $\mathcal T'$ has triangles consisting of all those triangles in $\mathcal T$ such that all objects of the triangle are in $\mathcal T'$, we need only consider two of three since cones of morphisms in $\mathcal T'$ must be in $\mathcal T'$ (it is triangulated) and translation allows us to consider any two.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .