Both compactness and dualizability express a finiteness property of an object, but they are defined differently: compactness concerns mapping out of the object, whereas dualizability concerns tensoring with it. We will now give conditions under which these notions agree, and then apply them to modules over ring spectra.
Lemma 11.2.1. Let \(C\) be a symmetric monoidal \(\infty \)-category with filtered colimits, and assume that the tensor product in \(C\) preserves filtered colimits in both variables. If the monoidal unit \(\unit \in C\) is compact, then every dualizable object is compact.
Proof. Let \(X\) be a dualizable object in \(C\), and let \(X^{\vee }\) denote its dual. Then the functor \(X \otimes -\colon C \to C\) admits a right adjoint of the form \(X^{\vee } \otimes -\colon C \to C\). In particular, we have \[ \Hom _C(X,-) \simeq \Hom _C(X \otimes \unit ,-) \simeq \Hom _C(\unit ,X^{\vee } \otimes -), \] and the right-hand side preserves filtered colimits by the assumptions on \(C\). □
In practice, the converse frequently holds as well: all compact objects are dualizable.
Lemma 11.2.2 (cf. Hovey et al. (1997)). Let \(C\) be a presentably symmetric monoidal stable \(\infty \)-category in the sense of Definition 22.5.1, and let \(\Gg \) be a small collection of compact generators of \(C\). If every object of \(\Gg \) is dualizable, then every compact object of \(C\) is dualizable.
Proof. The tensor product admits internal homs by Theorem 22.2.5. The result on compact generation, Proposition 22.3.4 from Chapter 22, which we use as a black box, identifies the compact objects of \(C\) with the thick subcategory generated by \(\Gg \). The claim now follows from Lemma 11.1.16. □
Corollary 11.2.3. Let \(C\) be a presentably symmetric monoidal stable \(\infty \)-category in the sense of Definition 22.5.1, and let \(\Gg \) be a small collection of compact generators. Then the compact objects of \(C\) are precisely its dualizable objects if and only if the unit \(\unit \) is compact and every object of \(\Gg \) is dualizable.
Proof. If the unit is compact and every object of \(\Gg \) is dualizable, the claim follows from Lemma 11.2.1, Lemma 11.2.2. Conversely, if the compact and dualizable objects coincide, then the unit is compact because it is always dualizable, while every object of \(\Gg \) is dualizable because it is compact. □
We now apply this discussion to the perfect modules introduced in Definition 8.1.7 as the thick subcategory of left \(R\)-modules generated by \(R\). When \(R\) is commutative, Proposition 8.1.5 identifies left modules with the symmetric monoidal category \(\Mod _R\), and perfectness admits a third characterization in addition to its descriptions by finite construction and compactness from Lemma 8.1.8.
Proposition 11.2.4. Let \(R\) be a commutative ring spectrum. An \(R\)-module is perfect if and only if it is dualizable in the symmetric monoidal \(\infty \)-category \((\Mod _R,\otimes _R)\).
Proof. The category \(\Mod _R\) is presentable and its tensor product preserves colimits, so it admits internal homs by Proposition 8.1.5, Theorem 22.2.5. By Lemma 11.1.16, the dualizable \(R\)-modules therefore form a thick subcategory of \(\Mod _R\). Since this subcategory contains the monoidal unit \(R\), every perfect \(R\)-module is dualizable. Conversely, the unit \(R\) is compact and the tensor product of \(R\)-modules preserves colimits in both variables. Hence every dualizable \(R\)-module is compact by Lemma 11.2.1, and therefore perfect by Lemma 8.1.8. □
Generated from the authoritative LaTeX source.