Lemma 57.5.1. With $k$, $n$, and $R$ as above, for an object $K$ of $D(R)$ the following are equivalent

$\sum _{i \in \mathbf{Z}} \dim _ k H^ i(K) < \infty $, and

$K$ is a compact object.

This section is in some sense a continuation of the discussion in Derived Categories of Schemes, Section 36.34 and More on Morphisms, Section 37.66.

Before we can state the result we need some notation. Let $k$ be a field. Let $n \geq 0$ be an integer. Let $S = k[X_0, \ldots , X_ n]$. For an integer $e$ denote $S_ e \subset S$ the homogeneous polynomials of degree $e$. Consider the (noncommutative) $k$-algebra

\[ R = \left( \begin{matrix} S_0
& S_1
& S_2
& \ldots
& \ldots
\\ 0
& S_0
& S_1
& \ldots
& \ldots
\\ 0
& 0
& S_0
& \ldots
& \ldots
\\ \ldots
& \ldots
& \ldots
& \ldots
& \ldots
\\ 0
& \ldots
& \ldots
& \ldots
& S_0
\end{matrix} \right) \]

(with $n + 1$ rows and columns) with obvious multiplication and addition.

Lemma 57.5.1. With $k$, $n$, and $R$ as above, for an object $K$ of $D(R)$ the following are equivalent

$\sum _{i \in \mathbf{Z}} \dim _ k H^ i(K) < \infty $, and

$K$ is a compact object.

**Proof.**
If $K$ is a compact object, then $K$ can be represented by a complex $M^\bullet $ which is finite projective as a graded $R$-module, see Differential Graded Algebra, Lemma 22.36.6. Since $\dim _ k R < \infty $ we conclude $\sum \dim _ k M^ i < \infty $ and a fortiori $\sum \dim _ k H^ i(M^\bullet ) < \infty $. (One can also easily deduce this implication from the easier Differential Graded Algebra, Proposition 22.36.4.)

Assume $K$ satisfies (1). Consider the distinguished triangle of trunctions $\tau _{\leq m}K \to K \to \tau _{\geq m + 1}K$, see Derived Categories, Remark 13.12.4. It is clear that both $\tau _{\leq m}K$ and $\tau _{\geq m + 1} K$ satisfy (1). If we can show both are compact, then so is $K$, see Derived Categories, Lemma 13.37.2. Hence, arguing on the number of nonzero cohomology modules of $K$ we may assume $H^ i(K)$ is nonzero only for one $i$. Shifting, we may assume $K$ is given by the complex consisting of a single finite dimensional $R$-module $M$ sitting in degree $0$.

Since $\dim _ k(M) < \infty $ we see that $M$ is Artinian as an $R$-module. Thus it suffices to show that every simple $R$-module represents a compact object of $D(R)$. Observe that

\[ I = \left( \begin{matrix} 0
& S_1
& S_2
& \ldots
& \ldots
\\ 0
& 0
& S_1
& \ldots
& \ldots
\\ 0
& 0
& 0
& \ldots
& \ldots
\\ \ldots
& \ldots
& \ldots
& \ldots
& \ldots
\\ 0
& \ldots
& \ldots
& \ldots
& 0
\end{matrix} \right) \]

is a nilpotent two sided ideal of $R$ and that $R/I$ is a commutative $k$-algebra isomorphic to a product of $n + 1$ copies of $k$ (placed along the diagonal in the matrix, i.e., $R/I$ can be lifted to a $k$-subalgebra of $R$). It follows that $R$ has exactly $n + 1$ isomorphism classes of simple modules $M_0, \ldots , M_ n$ (sitting along the diagonal). Consider the right $R$-module $P_ i$ of row vectors

\[ P_ i = \left( \begin{matrix} 0
& \ldots
& 0
& S_0
& \ldots
& S_{i - 1}
& S_ i
\end{matrix} \right) \]

with obvious multiplication $P_ i \times R \to P_ i$. Then we see that $R \cong P_0 \oplus \ldots \oplus P_ n$ as a right $R$-module. Since clearly $R$ is a compact object of $D(R)$, we conclude each $P_ i$ is a compact object of $D(R)$. (We of course also conclude each $P_ i$ is projective as an $R$-module, but this isn't what we have to show in this proof.) Clearly, $P_0 = M_0$ is the first of our simple $R$-modules. For $P_1$ we have a short exact sequence

\[ 0 \to P_0^{\oplus n + 1} \to P_1 \to M_1 \to 0 \]

which proves that $M_1$ fits into a distinguished triangle whose other members are compact objects and hence $M_1$ is a compact object of $D(R)$. More generally, there exists a short exact sequence

\[ 0 \to C_ i \to P_ i \to M_ i \to 0 \]

where $C_ i$ is a finite dimensional $R$-module whose simple constituents are isomorphic to $M_ j$ for $j < i$. By induction, we first conclude that $C_ i$ determines a compact object of $D(R)$ whereupon we conclude that $M_ i$ does too as desired. $\square$

Lemma 57.5.2. Let $k$ be a field. Let $n \geq 0$. Let $K \in D_\mathit{QCoh}(\mathcal{O}_{\mathbf{P}^ n_ k})$. The following are equivalent

$K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_{\mathbf{P}^ n_ k})$,

$\sum _{i \in \mathbf{Z}} \dim _ k H^ i(\mathbf{P}^ n_ k, E \otimes ^\mathbf {L} K) < \infty $ for each perfect object $E$ of $D(\mathcal{O}_{\mathbf{P}^ n_ k})$,

$\sum _{i \in \mathbf{Z}} \dim _ k \mathop{\mathrm{Ext}}\nolimits ^ i_{\mathbf{P}^ n_ k}(E, K) < \infty $ for each perfect object $E$ of $D(\mathcal{O}_{\mathbf{P}^ n_ k})$,

$\sum _{i \in \mathbf{Z}} \dim _ k H^ i(\mathbf{P}^ n_ k, K \otimes ^\mathbf {L} \mathcal{O}_{\mathbf{P}^ n_ k}(d)) < \infty $ for $d = 0, 1, \ldots , n$.

**Proof.**
Parts (2) and (3) are equivalent by Cohomology, Lemma 20.47.5. If (1) is true, then for $E$ perfect the derived tensor product $E \otimes ^\mathbf {L} K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_{\mathbf{P}^ n_ k})$ and we see that (2) holds by Derived Categories of Schemes, Lemma 36.11.3. It is clear that (2) implies (4) as $\mathcal{O}_{\mathbf{P}^ n_ k}(d)$ can be viewed as a perfect object of the derived category of $\mathbf{P}^ n_ k$. Thus it suffices to prove that (4) implies (1).

Assume (4). Let $R$ be as in Lemma 57.5.1. Let $P = \bigoplus _{d = 0, \ldots , n} \mathcal{O}_{\mathbf{P}^ n_ k}(-d)$. Recall that $R = \text{End}_{\mathbf{P}^ n_ k}(P)$ whereas all other self-Exts of $P$ are zero and that $P$ determines an equivalence $- \otimes ^\mathbf {L} P : D(R) \to D_\mathit{QCoh}(\mathcal{O}_{\mathbf{P}^ n_ k})$ by Derived Categories of Schemes, Lemma 36.20.1. Say $K$ corresponds to $L$ in $D(R)$. Then

\begin{align*} H^ i(L) & = \mathop{\mathrm{Ext}}\nolimits ^ i_{D(R)}(R, L) \\ & = \mathop{\mathrm{Ext}}\nolimits ^ i_{\mathbf{P}^ n_ k}(P, K) \\ & = H^ i(\mathbf{P}^ n_ k, K \otimes P^\vee ) \\ & = \bigoplus \nolimits _{d = 0, \ldots , n} H^ i(\mathbf{P}^ n_ k, K \otimes \mathcal{O}(d)) \end{align*}

by Differential Graded Algebra, Lemma 22.35.4 (and the fact that $- \otimes ^\mathbf {L} P$ is an equivalence) and Cohomology, Lemma 20.47.5. Thus our assumption (4) implies that $L$ satisfies condition (2) of Lemma 57.5.1 and hence is a compact object of $D(R)$. Therefore $K$ is a compact object of $D_\mathit{QCoh}(\mathcal{O}_{\mathbf{P}^ n_ k})$. Thus $K$ is perfect by Derived Categories of Schemes, Proposition 36.17.1. Since $D_{perf}(\mathcal{O}_{\mathbf{P}^ n_ k}) = D^ b_{\textit{Coh}}(\mathcal{O}_{\mathbf{P}^ n_ k})$ by Derived Categories of Schemes, Lemma 36.11.8 we conclude (1) holds. $\square$

Lemma 57.5.3. Let $X$ be a scheme proper over a field $k$. Let $K \in D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ and let $E$ in $D(\mathcal{O}_ X)$ be perfect. Then $\sum _{i \in \mathbf{Z}} \dim _ k \mathop{\mathrm{Ext}}\nolimits ^ i_ X(E, K) < \infty $.

**Proof.**
This follows for example by combining Derived Categories of Schemes, Lemmas 36.11.7 and 36.18.2. Alternative proof: combine Derived Categories of Schemes, Lemmas 36.11.6 and 36.11.3.
$\square$

Lemma 57.5.4. Let $X$ be a projective scheme over a field $k$. Let $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. The following are equivalent

$K \in D^ b_{\textit{Coh}}(\mathcal{O}_ X)$, and

$\sum _{i \in \mathbf{Z}} \dim _ k \mathop{\mathrm{Ext}}\nolimits ^ i_ X(E, K) < \infty $ for all perfect $E$ in $D(\mathcal{O}_ X)$.

**Proof.**
The implication (1) $\Rightarrow $ (2) follows from Lemma 57.5.3.

Assume (2). Choose a closed immersion $i : X \to \mathbf{P}^ n_ k$. It suffices to show that $Ri_*K$ is in $D^ b_{\textit{Coh}}(\mathbf{P}^ n_ k)$ since a quasi-coherent module $\mathcal{F}$ on $X$ is coherent, resp. zero if and only if $i_*\mathcal{F}$ is coherent, resp. zero. For a perfect object $E$ of $D(\mathcal{O}_{\mathbf{P}^ n_ k})$, $Li^*E$ is a perfect object of $D(\mathcal{O}_ X)$ and

\[ \mathop{\mathrm{Ext}}\nolimits ^ q_{\mathbf{P}^ n_ k}(E, Ri_*K) = \mathop{\mathrm{Ext}}\nolimits ^ q_ X(Li^*E, K) \]

Hence by our assumption we see that $\sum _{q \in \mathbf{Z}} \dim _ k \mathop{\mathrm{Ext}}\nolimits ^ q_{\mathbf{P}^ n_ k}(E, Ri_*K) < \infty $. We conclude by Lemma 57.5.2. $\square$

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