The Stacks project

13.41 Postnikov systems

A reference for this section is [Orlov-K3]. Let $\mathcal{D}$ be a triangulated category. Let

\[ X_ n \to X_{n - 1} \to \ldots \to X_0 \]

be a complex in $\mathcal{D}$. In this section we consider the problem of constructing a “totalization” of this complex.

Definition 13.41.1. Let $\mathcal{D}$ be a triangulated category. Let

\[ X_ n \to X_{n - 1} \to \ldots \to X_0 \]

be a complex in $\mathcal{D}$. A Postnikov system is defined inductively as follows.

  1. If $n = 0$, then it is an isomorphism $Y_0 \to X_0$.

  2. If $n = 1$, then it is a choice of an isomorphism $Y_0 \to X_0$ and a choice of a distinguished triangle

    \[ Y_1 \to X_1 \to Y_0 \to Y_1[1] \]

    where $X_1 \to Y_0$ composed with $Y_0 \to X_0$ is the given morphism $X_1 \to X_0$.

  3. If $n > 1$, then it is a choice of a Postnikov system for $X_{n - 1} \to \ldots \to X_0$ and a choice of a distinguished triangle

    \[ Y_ n \to X_ n \to Y_{n - 1} \to Y_ n[1] \]

    where the morphism $X_ n \to Y_{n - 1}$ composed with $Y_{n - 1} \to X_{n - 1}$ is the given morphism $X_ n \to X_{n - 1}$.

Given a morphism

13.41.1.1
\begin{equation} \label{derived-equation-map-complexes} \vcenter { \xymatrix{ X_ n \ar[r] \ar[d] & X_{n - 1} \ar[r] \ar[d] & \ldots \ar[r] & X_0 \ar[d] \\ X'_ n \ar[r] & X'_{n - 1} \ar[r] & \ldots \ar[r] & X'_0 } } \end{equation}

between complexes of the same length in $\mathcal{D}$ there is an obvious notion of a morphism of Postnikov systems.

Here is a key example.

Example 13.41.2. Let $\mathcal{A}$ be an abelian category. Let $\ldots \to A_2 \to A_1 \to A_0$ be a chain complex in $\mathcal{A}$. Then we can consider the objects

\[ X_ n = A_ n \quad \text{and}\quad Y_ n = (A_ n \to A_{n - 1} \to \ldots \to A_0)[-n] \]

of $D(\mathcal{A})$. With the evident canonical maps $Y_ n \to X_ n$ and $Y_0 \to Y_1[1] \to Y_2[2] \to \ldots $ the distinguished triangles $Y_ n \to X_ n \to Y_{n - 1} \to Y_ n[1]$ define a Postnikov system as in Definition 13.41.1 for $\ldots \to X_2 \to X_1 \to X_0$. Here we are using the obvious extension of Postnikov systems for an infinite complex of $D(\mathcal{A})$. Finally, if colimits over $\mathbf{N}$ exist and are exact in $\mathcal{A}$ then

\[ \text{hocolim} Y_ n[n] = (\ldots \to A_2 \to A_1 \to A_0 \to 0 \to \ldots ) \]

in $D(\mathcal{A})$. This follows immediately from Lemma 13.33.7.

Given a complex $X_ n \to X_{n - 1} \to \ldots \to X_0$ and a Postnikov system as in Definition 13.41.1 we can consider the maps

\[ Y_0 \to Y_1[1] \to \ldots \to Y_ n[n] \]

These maps fit together in certain distinguished triangles and fit with the given maps between the $X_ i$. Here is a picture for $n = 3$:

\[ \xymatrix{ Y_0 \ar[rr] & & Y_1[1] \ar[dl] \ar[rr] & & Y_2[2] \ar[dl] \ar[rr] & & Y_3[3] \ar[dl] \\ & X_1[1] \ar[lu]_{+1} & & X_2[2] \ar[ll]_{+1} \ar[lu]_{+1} & & X_3[3] \ar[ll]_{+1} \ar[lu]_{+1} } \]

We encourage the reader to think of $Y_ n[n]$ as obtained from $X_0, X_1[1], \ldots , X_ n[n]$; for example if the maps $X_ i \to X_{i - 1}$ are zero, then we can take $Y_ n[n] = \bigoplus _{i = 0, \ldots , n} X_ i[i]$. Postnikov systems do not always exist. Here is a simple lemma for low $n$.

Lemma 13.41.3. Let $\mathcal{D}$ be a triangulated category. Consider Postnikov systems for complexes of length $n$.

  1. For $n = 0$ Postnikov systems always exist and any morphism (13.41.1.1) of complexes extends to a unique morphism of Postnikov systems.

  2. For $n = 1$ Postnikov systems always exist and any morphism (13.41.1.1) of complexes extends to a (nonunique) morphism of Postnikov systems.

  3. For $n = 2$ Postnikov systems always exist but morphisms (13.41.1.1) of complexes in general do not extend to morphisms of Postnikov systems.

  4. For $n > 2$ Postnikov systems do not always exist.

Proof. The case $n = 0$ is immediate as isomorphisms are invertible. The case $n = 1$ follows immediately from TR1 (existence of triangles) and TR3 (extending morphisms to triangles). For the case $n = 2$ we argue as follows. Set $Y_0 = X_0$. By the case $n = 1$ we can choose a Postnikov system

\[ Y_1 \to X_1 \to Y_0 \to Y_1[1] \]

Since the composition $X_2 \to X_1 \to X_0$ is zero, we can factor $X_2 \to X_1$ (nonuniquely) as $X_2 \to Y_1 \to X_1$ by Lemma 13.4.2. Then we simply fit the morphism $X_2 \to Y_1$ into a distinguished triangle

\[ Y_2 \to X_2 \to Y_1 \to Y_2[1] \]

to get the Postnikov system for $n = 2$. For $n > 2$ we cannot argue similarly, as we do not know whether the composition $X_ n \to X_{n - 1} \to Y_{n - 1}$ is zero in $\mathcal{D}$. $\square$

Lemma 13.41.4. Let $\mathcal{D}$ be a triangulated category. Given a map (13.41.1.1) consider the condition

13.41.4.1
\begin{equation} \label{derived-equation-P} \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], X'_ j) = 0 \text{ for }i > j + 1 \end{equation}

Then

  1. If we have a Postnikov system for $X'_ n \to X'_{n - 1} \to \ldots \to X'_0$ then property (13.41.4.1) implies that

    \[ \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], Y'_ j) = 0 \text{ for }i > j + 1 \]

  2. If we are given Postnikov systems for both complexes and we have (13.41.4.1), then the map extends to a (nonunique) map of Postnikov systems.

Proof. We first prove (1) by induction on $j$. For the base case $j = 0$ there is nothing to prove as $Y'_0 \to X'_0$ is an isomorphism. Say the result holds for $j - 1$. We consider the distinguished triangle

\[ Y'_ j \to X'_ j \to Y'_{j - 1} \to Y'_ j[1] \]

The long exact sequence of Lemma 13.4.2 gives an exact sequence

\[ \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], Y'_{j - 1}[-1]) \to \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], Y'_ j) \to \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], X'_ j) \]

From the induction hypothesis and (13.41.4.1) we conclude the outer groups are zero and we win.

Proof of (2). For $n = 1$ the existence of morphisms has been established in Lemma 13.41.3. For $n > 1$ by induction, we may assume given the map of Postnikov systems of length $n - 1$. The problem is that we do not know whether the diagram

\[ \xymatrix{ X_ n \ar[r] \ar[d] & Y_{n - 1} \ar[d] \\ X'_ n \ar[r] & Y'_{n - 1} } \]

is commutative. Denote $\alpha : X_ n \to Y'_{n - 1}$ the difference. Then we do know that the composition of $\alpha $ with $Y'_{n - 1} \to X'_{n - 1}$ is zero (because of what it means to be a map of Postnikov systems of length $n - 1$). By the distinguished triangle $Y'_{n - 1} \to X'_{n - 1} \to Y'_{n - 2} \to Y'_{n - 1}[1]$, this means that $\alpha $ is the composition of $Y'_{n - 2}[-1] \to Y'_{n - 1}$ with a map $\alpha ' : X_ n \to Y'_{n - 2}[-1]$. Then (13.41.4.1) guarantees $\alpha '$ is zero by part (1) of the lemma. Thus $\alpha $ is zero. To finish the proof of existence, the commutativity guarantees we can choose the dotted arrow fitting into the diagram

\[ \xymatrix{ Y_{n - 1}[-1] \ar[d] \ar[r] & Y_ n \ar[r] \ar@{..>}[d] & X_ n \ar[r] \ar[d] & Y_{n - 1} \ar[d] \\ Y'_{n - 1}[-1] \ar[r] & Y'_ n \ar[r] & X'_ n \ar[r] & Y'_{n - 1} } \]

by TR3. $\square$

Lemma 13.41.5. Let $\mathcal{D}$ be a triangulated category. Given a map (13.41.1.1) assume we are given Postnikov systems for both complexes. If

  1. $\mathop{\mathrm{Hom}}\nolimits (X_ i[i], Y'_ n[n]) = 0$ for $i = 1, \ldots , n$, or

  2. $\mathop{\mathrm{Hom}}\nolimits (Y_ n[n], X'_{n - i}[n - i]) = 0$ for $i = 1, \ldots , n$, or

  3. $\mathop{\mathrm{Hom}}\nolimits (X_{j - i}[-i + 1], X'_ j) = 0$ and $\mathop{\mathrm{Hom}}\nolimits (X_ j, X'_{j - i}[-i]) = 0$ for $j \geq i > 0$,

then there exists at most one morphism between these Postnikov systems.

Proof. Proof of (1). Look at the following diagram

\[ \xymatrix{ Y_0 \ar[r] \ar[d] & Y_1[1] \ar[r] \ar[ld] & Y_2[2] \ar[r] \ar[lld] & \ldots \ar[r] & Y_ n[n] \ar[lllld] \\ Y'_ n[n] } \]

The arrows are the composition of the morphism $Y_ n[n] \to Y'_ n[n]$ and the morphism $Y_ i[i] \to Y_ n[n]$. The arrow $Y_0 \to Y'_ n[n]$ is determined as it is the composition $Y_0 = X_0 \to X'_0 = Y'_0 \to Y'_ n[n]$. Since we have the distinguished triangle $Y_0 \to Y_1[1] \to X_1[1]$ we see that $\mathop{\mathrm{Hom}}\nolimits (X_1[1], Y'_ n[n]) = 0$ guarantees that the second vertical arrow is unique. Since we have the distinguished triangle $Y_1[1] \to Y_2[2] \to X_2[2]$ we see that $\mathop{\mathrm{Hom}}\nolimits (X_2[2], Y'_ n[n]) = 0$ guarantees that the third vertical arrow is unique. And so on.

Proof of (2). The composition $Y_ n[n] \to Y'_ n[n] \to X_ n[n]$ is the same as the composition $Y_ n[n] \to X_ n[n] \to X'_ n[n]$ and hence is unique. Then using the distinguished triangle $Y'_{n - 1}[n - 1] \to Y'_ n[n] \to X'_ n[n]$ we see that it suffices to show $\mathop{\mathrm{Hom}}\nolimits (Y_ n[n], Y'_{n - 1}[n - 1]) = 0$. Using the distinguished triangles

\[ Y'_{n - i - 1}[n - i - 1] \to Y'_{n - i}[n - i] \to X'_{n - i}[n - i] \]

we get this vanishing from our assumption. Small details omitted.

Proof of (3). Looking at the proof of Lemma 13.41.4 and arguing by induction on $n$ it suffices to show that the dotted arrow in the morphism of triangles

\[ \xymatrix{ Y_{n - 1}[-1] \ar[d] \ar[r] & Y_ n \ar[r] \ar@{..>}[d] & X_ n \ar[r] \ar[d] & Y_{n - 1} \ar[d] \\ Y'_{n - 1}[-1] \ar[r] & Y'_ n \ar[r] & X'_ n \ar[r] & Y'_{n - 1} } \]

is unique. By Lemma 13.4.8 part (5) it suffices to show that $\mathop{\mathrm{Hom}}\nolimits (Y_{n - 1}, X'_ n) = 0$ and $\mathop{\mathrm{Hom}}\nolimits (X_ n, Y'_{n - 1}[-1]) = 0$. To prove the first vanishing we use the distinguished triangles $Y_{n - i - 1}[-i] \to Y_{n - i}[-(i - 1)] \to X_{n - i}[-(i - 1)]$ for $i > 0$ and induction on $i$ to see that the assumed vanishing of $\mathop{\mathrm{Hom}}\nolimits (X_{n - i}[-i + 1], X'_ n)$ is enough. For the second we similarly use the distinguished triangles $Y'_{n - i - 1}[-i - 1] \to Y'_{n - i}[-i] \to X'_{n - i}[-i]$ to see that the assumed vanishing of $\mathop{\mathrm{Hom}}\nolimits (X_ n, X'_{n - i}[-i])$ is enough as well. $\square$

Lemma 13.41.6. Let $\mathcal{D}$ be a triangulated category. Let $X_ n \to X_{n - 1} \to \ldots \to X_0$ be a complex in $\mathcal{D}$. If

\[ \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 2], X_ j) = 0 \text{ for }i > j + 2 \]

then there exists a Postnikov system. If we have

\[ \mathop{\mathrm{Hom}}\nolimits (X_ i[i - j - 1], X_ j) = 0 \text{ for }i > j + 1 \]

then any two Postnikov systems are isomorphic.

Proof. We argue by induction on $n$. The cases $n = 0, 1, 2$ follow from Lemma 13.41.3. Assume $n > 2$. Suppose given a Postnikov system for the complex $X_{n - 1} \to X_{n - 2} \to \ldots \to X_0$. The only obstruction to extending this to a Postnikov system of length $n$ is that we have to find a morphism $X_ n \to Y_{n - 1}$ such that the composition $X_ n \to Y_{n - 1} \to X_{n - 1}$ is equal to the given map $X_ n \to X_{n - 1}$. Considering the distinguished triangle

\[ Y_{n - 1} \to X_{n - 1} \to Y_{n - 2} \to Y_{n - 1}[1] \]

and the associated long exact sequence coming from this and the functor $\mathop{\mathrm{Hom}}\nolimits (X_ n, -)$ (see Lemma 13.4.2) we find that it suffices to show that the composition $X_ n \to X_{n - 1} \to Y_{n - 2}$ is zero. Since we know that $X_ n \to X_{n - 1} \to X_{n - 2}$ is zero we can apply the distinguished triangle

\[ Y_{n - 2} \to X_{n - 2} \to Y_{n - 3} \to Y_{n - 2}[1] \]

to see that it suffices if $\mathop{\mathrm{Hom}}\nolimits (X_ n, Y_{n - 3}[-1]) = 0$. Arguing exactly as in the proof of Lemma 13.41.4 part (1) the reader easily sees this follows from the condition stated in the lemma.

The statement on isomorphisms follows from the existence of a map between the Postnikov systems extending the identity on the complex proven in Lemma 13.41.4 part (2) and Lemma 13.4.3 to show all the maps are isomorphisms. $\square$

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