\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

## 13.27 Ext groups

In this section we start describing the Ext groups of objects of an abelian category. First we have the following very general definition.

Definition 13.27.1. Let $\mathcal{A}$ be an abelian category. Let $i \in \mathbf{Z}$. Let $X, Y$ be objects of $D(\mathcal{A})$. The $i$th extension group of $X$ by $Y$ is the group

$\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X, Y) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(X, Y[i]) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(X[-i], Y).$

If $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ we set $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(A, B) = \text{Ext}^ i_\mathcal {A}(A, B)$.

Since $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(X, -)$, resp. $\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(-, Y)$ is a homological, resp. cohomological functor, see Lemma 13.4.2, we see that a distinguished triangle $(Y, Y', Y'')$, resp. $(X, X', X'')$ leads to a long exact sequence

$\ldots \to \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X, Y) \to \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X, Y') \to \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X, Y'') \to \mathop{\mathrm{Ext}}\nolimits ^{i + 1}_\mathcal {A}(X, Y) \to \ldots$

respectively

$\ldots \to \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X'', Y) \to \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X', Y) \to \mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X, Y) \to \mathop{\mathrm{Ext}}\nolimits ^{i + 1}_\mathcal {A}(X'', Y) \to \ldots$

Note that since $D^+(\mathcal{A})$, $D^-(\mathcal{A})$, $D^ b(\mathcal{A})$ are full subcategories we may compute the Ext groups by Hom groups in these categories provided $X$, $Y$ are contained in them.

In case the category $\mathcal{A}$ has enough injectives or enough projectives we can compute the Ext groups using injective or projective resolutions. To avoid confusion, recall that having an injective (resp. projective) resolution implies vanishing of homology in all low (resp. high) degrees, see Lemmas 13.18.2 and 13.19.2.

Lemma 13.27.2. Let $\mathcal{A}$ be an abelian category. Let $X^\bullet , Y^\bullet \in \mathop{\mathrm{Ob}}\nolimits (K(\mathcal{A}))$.

1. Let $Y^\bullet \to I^\bullet$ be an injective resolution (Definition 13.18.1). Then

$\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X^\bullet , Y^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(X^\bullet , I^\bullet [i]).$
2. Let $P^\bullet \to X^\bullet$ be a projective resolution (Definition 13.19.1). Then

$\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(X^\bullet , Y^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet [-i], Y^\bullet ).$

Proof. Follows immediately from Lemma 13.18.8 and Lemma 13.19.8. $\square$

In the rest of this section we discuss extensions of objects of the abelian category itself. First we observe the following.

Lemma 13.27.3. Let $\mathcal{A}$ be an abelian category.

1. Let $X$, $Y$ be objects of $D(\mathcal{A})$. Given $a, b \in \mathbf{Z}$ such that $H^ i(X) = 0$ for $i > a$ and $H^ j(Y) = 0$ for $j < b$, we have $\mathop{\mathrm{Ext}}\nolimits ^ n_\mathcal {A}(X, Y) = 0$ for $n < b - a$ and

$\mathop{\mathrm{Ext}}\nolimits ^{b - a}_\mathcal {A}(X, Y) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(H^ a(X), H^ b(Y))$
2. Let $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. For $i < 0$ we have $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A) = 0$. We have $\mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B, A) = \mathop{\mathrm{Hom}}\nolimits _\mathcal {A}(B, A)$.

Proof. Choose complexes $X^\bullet$ and $Y^\bullet$ representing $X$ and $Y$. Since $Y^\bullet \to \tau _{\geq b}Y^\bullet$ is a quasi-isomorphism, we may assume that $Y^ j = 0$ for $j < b$. Let $L^\bullet \to X^\bullet$ be any quasi-isomorphism. Then $\tau _{\leq a}L^\bullet \to X^\bullet$ is a quasi-isomorphism. Hence a morphism $X \to Y[n]$ in $D(\mathcal{A})$ can be represented as $fs^{-1}$ where $s : L^\bullet \to X^\bullet$ is a quasi-isomorphism, $f : L^\bullet \to Y^\bullet [n]$ a morphism, and $L^ i = 0$ for $i < a$. Note that $f$ maps $L^ i$ to $Y^{i + n}$. Thus $f = 0$ if $n < b - a$ because always either $L^ i$ or $Y^{i + n}$ is zero. If $n = b - a$, then $f$ corresponds exactly to a morphism $H^ a(X) \to H^ b(Y)$. Part (2) is a special case of (1). $\square$

Let $\mathcal{A}$ be an abelian category. Suppose that $0 \to A \to A' \to A'' \to 0$ is a short exact sequence of objects of $\mathcal{A}$. Then $0 \to A \to A' \to A'' \to 0$ leads to a distinguished triangle in $D(\mathcal{A})$ (see Lemma 13.12.1) hence a long exact sequence of Ext groups

$0 \to \mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B, A) \to \mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B, A') \to \mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B, A'') \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {A}(B, A) \to \ldots$

Similarly, given a short exact sequence $0 \to B \to B' \to B'' \to 0$ we obtain a long exact sequence of Ext groups

$0 \to \mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B'', A) \to \mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B', A) \to \mathop{\mathrm{Ext}}\nolimits ^0_\mathcal {A}(B, A) \to \mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {A}(B'', A) \to \ldots$

We may view these Ext groups as an application of the construction of the derived category. It shows one can define Ext groups and construct the long exact sequence of Ext groups without needing the existence of enough injectives or projectives. There is an alternative construction of the Ext groups due to Yoneda which avoids the use of the derived category, see [Yoneda].

Definition 13.27.4. Let $\mathcal{A}$ be an abelian category. Let $A, B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. A degree $i$ Yoneda extension of $B$ by $A$ is an exact sequence

$E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0$

in $\mathcal{A}$. We say two Yoneda extensions $E$ and $E'$ of the same degree are equivalent if there exists a commutative diagram

$\xymatrix{ 0 \ar[r] & A \ar[r] & Z_{i - 1} \ar[r] & \ldots \ar[r] & Z_0 \ar[r] & B \ar[r] & 0 \\ 0 \ar[r] & A \ar[r] \ar[u]^{\text{id}} \ar[d]_{\text{id}} & Z''_{i - 1} \ar[r] \ar[u] \ar[d] & \ldots \ar[r] & Z''_0 \ar[r] \ar[u] \ar[d] & B \ar[r] \ar[u]_{\text{id}} \ar[d]^{\text{id}} & 0 \\ 0 \ar[r] & A \ar[r] & Z'_{i - 1} \ar[r] & \ldots \ar[r] & Z'_0 \ar[r] & B \ar[r] & 0 }$

where the middle row is a Yoneda extension as well.

It is not immediately clear that the equivalence of the definition is an equivalence relation. Although it is instructive to prove this directly this will also follow from Lemma 13.27.5 below.

Let $\mathcal{A}$ be an abelian category with objects $A$, $B$. Given a Yoneda extension $E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0$ we define an associated element $\delta (E) \in \mathop{\mathrm{Ext}}\nolimits ^ i(B, A)$ as the morphism $\delta (E) = fs^{-1} : B \to A[i]$ where $s$ is the quasi-isomorphism

$(\ldots \to 0 \to A \to Z_{i - 1} \to \ldots \to Z_0 \to 0 \to \ldots ) \longrightarrow B$

and $f$ is the morphism of complexes

$(\ldots \to 0 \to A \to Z_{i - 1} \to \ldots \to Z_0 \to 0 \to \ldots ) \longrightarrow A[i]$

We call $\delta (E) = fs^{-1}$ the class of the Yoneda extension. It turns out that this class characterizes the equivalence class of the Yoneda extension.

Lemma 13.27.5. Let $\mathcal{A}$ be an abelian category with objects $A$, $B$. Any element in $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A)$ is $\delta (E)$ for some degree $i$ Yoneda extension of $B$ by $A$. Given two Yoneda extensions $E$, $E'$ of the same degree then $E$ is equivalent to $E'$ if and only if $\delta (E) = \delta (E')$.

Proof. Let $\xi : B \to A[i]$ be an element of $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A)$. We may write $\xi = f s^{-1}$ for some quasi-isomorphism $s : L^\bullet \to B$ and map $f : L^\bullet \to A[i]$. After replacing $L^\bullet$ by $\tau _{\leq 0}L^\bullet$ we may assume that $L^ i = 0$ for $i > 0$. Picture

$\xymatrix{ L^{- i - 1} \ar[r] & L^{-i} \ar[r] \ar[d] & \ldots \ar[r] & L^0 \ar[r] & B \ar[r] & 0 \\ & A }$

Then setting $Z_{i - 1} = (L^{- i + 1} \oplus A)/L^{-i}$ and $Z_ j = L^{-j}$ for $j = i - 2, \ldots , 0$ we see that we obtain a degree $i$ extension $E$ of $B$ by $A$ whose class $\delta (E)$ equals $\xi$.

It is immediate from the definitions that equivalent Yoneda extensions have the same class. Suppose that $E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0$ and $E' : 0 \to A \to Z'_{i - 1} \to Z'_{i - 2} \to \ldots \to Z'_0 \to B \to 0$ are Yoneda extensions with the same class. By construction of $D(\mathcal{A})$ as the localization of $K(\mathcal{A})$ at the set of quasi-isomorphisms, this means there exists a complex $L^\bullet$ and quasi-isomorphisms

$t : L^\bullet \to (\ldots \to 0 \to A \to Z_{i - 1} \to \ldots \to Z_0 \to 0 \to \ldots )$

and

$t' : L^\bullet \to (\ldots \to 0 \to A \to Z'_{i - 1} \to \ldots \to Z'_0 \to 0 \to \ldots )$

such that $s \circ t = s' \circ t'$ and $f \circ t = f' \circ t'$, see Categories, Section 4.26. Let $E''$ be the degree $i$ extension of $B$ by $A$ constructed from the pair $L^\bullet \to B$ and $L^\bullet \to A[i]$ in the first paragraph of the proof. Then the reader sees readily that there exists “morphisms” of degree $i$ Yoneda extensions $E'' \to E$ and $E'' \to E'$ as in the definition of equivalent Yoneda extensions (details omitted). This finishes the proof. $\square$

Lemma 13.27.6. Let $\mathcal{A}$ be an abelian category. Let $A$, $B$ be objects of $\mathcal{A}$. Then $\mathop{\mathrm{Ext}}\nolimits ^1_\mathcal {A}(B, A)$ is the group $\mathop{\mathrm{Ext}}\nolimits _\mathcal {A}(B, A)$ constructed in Homology, Definition 12.6.2.

Proof. This is the case $i = 1$ of Lemma 13.27.5. $\square$

Lemma 13.27.7. Let $\mathcal{A}$ be an abelian category and let $p \geq 0$. If $\mathop{\mathrm{Ext}}\nolimits ^ p_\mathcal {A}(B, A) = 0$ for any pair of objects $A$, $B$ of $\mathcal{A}$, then $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A) = 0$ for $i \geq p$ and any pair of objects $A$, $B$ of $\mathcal{A}$.

Proof. For $i > p$ write any class $\xi$ as $\delta (E)$ where $E$ is a Yoneda extension

$E : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_0 \to B \to 0$

This is possible by Lemma 13.27.5. Set $C = \mathop{\mathrm{Ker}}(Z_{p - 1} \to Z_ p) = \mathop{\mathrm{Im}}(Z_ p \to Z_{p - 1})$. Then $\delta (E)$ is the composition of $\delta (E')$ and $\delta (E'')$ where

$E' : 0 \to C \to Z_{p - 1} \to \ldots \to Z_0 \to B \to 0$

and

$E'' : 0 \to A \to Z_{i - 1} \to Z_{i - 2} \to \ldots \to Z_ p \to C \to 0$

Since $\delta (E') \in \mathop{\mathrm{Ext}}\nolimits ^ p_\mathcal {A}(B, C) = 0$ we conclude. $\square$

Lemma 13.27.8. Let $\mathcal{A}$ be an abelian category. Assume $\mathop{\mathrm{Ext}}\nolimits ^2_\mathcal {A}(B, A) = 0$ for any pair of objects $A$, $B$ of $\mathcal{A}$. Then any object $K$ of $D^ b(\mathcal{A})$ is isomorphic to the direct sum of its cohomologies: $K \cong \bigoplus H^ i(K)[-i]$.

Proof. Choose $a, b$ such that $H^ i(K) = 0$ for $i \not\in [a, b]$. We will prove the lemma by induction on $b - a$. If $b - a \leq 0$, then the result is clear. If $b - a > 0$, then we look at the distinguished triangle of truncations

$\tau _{\leq b - 1}K \to K \to H^ b(K)[-b] \to \tau _{\leq b - 1}K$

see Remark 13.12.4. By Lemma 13.4.10 if the last arrow is zero, then $K \cong \tau _{\leq b - 1}K \oplus H^ b(K)[-b]$ and we win by induction. Again using induction we see that

$\mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(H^ b(K)[-b], \tau _{\leq b - 1}K) = \bigoplus \nolimits _{i < b} \mathop{\mathrm{Ext}}\nolimits _\mathcal {A}^{b - i - 1}(H^ b(K), H^ i(K))$

Since $\mathop{\mathrm{Ext}}\nolimits ^ i_\mathcal {A}(B, A) = 0$ for $i \geq 2$ and any pair of objects $A, B$ of $\mathcal{A}$ by our assumption and Lemma 13.27.7 we are done. $\square$

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