Localization sequences for logarithmic topological Hochschild homology

Abstract

We study the logarithmic topological Hochschild homology of ring spectra with logarithmic structures and establish localization sequences for this theory. Our results apply, for example, to connective covers of periodic ring spectra like real and complex topological $K$-theory.

Appendix: Homotopy invariance of \(\mathbb {S}^{\mathcal {J}}\)

Being a left Quillen functor, $\mathbb {S}^{\mathcal {J}}$ takes $\mathcal {J}$-equivalences between cofibrant $\mathcal {J}$-spaces to stable equivalences. It is useful to know that $\mathbb {S}^{\mathcal {J}}$ is homotopically well-behaved on a larger class of $\mathcal {J}$-spaces that includes cofibrant $\mathcal {J}$-spaces and the underlying $\mathcal {J}$-spaces of cofibrant commutative $\mathcal {J}$-space monoids. For this purpose it is not sufficient to work with flat $\mathcal {J}$-spaces, because $\mathbb {S}^{\mathcal {J}}$ is not left Quillen with respect to the flat model structure [30, Remark 4.29].

Definition 8.1

A $\mathcal {J}$-space $X$ is $\mathbb S^{\mathcal {J}}\!$ -good if there exists a cofibrant $\mathcal {J}$-space $X'$ and a $\mathcal {J}$-equivalence $X'\rightarrow X$ such that $\mathbb S^{\mathcal {J}}[X']\rightarrow \mathbb S^{\mathcal {J}}[X]$ is a stable equivalence.

It is clear from the definition that cofibrant $\mathcal {J}$-spaces are $\mathbb S^{\mathcal {J}}\!$-good. The terminal $\mathcal {J}$-space $T$ is an example of a $\mathcal {J}$-space that is not $\mathbb S^{\mathcal {J}}\!$-good. Using that $\mathbb S^{\mathcal {J}}$ is a left Quillen functor we see that if $X$ is $\mathbb S^{\mathcal {J}}\!$-good and $Y\rightarrow X$ is any $\mathcal {J}$-equivalence with $Y$ cofibrant, then the induced map $\mathbb S^{\mathcal {J}}[Y]\rightarrow \mathbb S^{\mathcal {J}}[X]$ is a stable equivalence. This in turn has the following consequence.

Proposition 8.2

The functor $\mathbb S^{\mathcal {J}}$ takes $\mathcal {J}$-equivalences between $\mathbb S^{\mathcal {J}}\!$-good $\mathcal {J}$-spaces to stable equivalences. $\square$

The automorphism group of an object $({\mathbf {n}}_1,{\mathbf {n}}_2)$ in $\mathcal {J}$ may evidently be identified with $\Sigma _{n_1}\times \Sigma _{n_2}$.

Definition 8.3

A $\mathcal {J}$-space $X$ is $\Sigma$ -free in the second variable if $\Sigma _{n_2}$ acts freely on $X({\mathbf {n}}_1,{\mathbf {n}}_2)$ for each object $({\mathbf {n}}_1,{\mathbf {n}}_2)$ in $\mathcal {J}$.

Lemma 8.4

If a $\mathcal {J}$-space is $\Sigma$-free in the second variable, then it is $\mathbb {S}^{\mathcal {J}}\!$-good.

Proof

Let $X$ be $\Sigma$-free in the second variable and let $X' \rightarrow X$ be a cofibrant replacement, which we may assume is a level equivalence. Then $X'$ is also $\Sigma$-free in the second variable. The freeness assumptions imply that the quotients by the $\Sigma _k$-actions arising in the explicit description of $\mathbb {S}^{\mathcal {J}}$ given in (2.2) preserve weak equivalences. Hence $\mathbb {S}^{\mathcal {J}}[X'] \rightarrow \mathbb {S}^{\mathcal {J}}[X]$ is a level equivalence of symmetric spectra and therefore a stable equivalence. $\square$

The following condition for $\mathbb {S}^{\mathcal {J}}$-goodness can often be checked in practice.

Corollary 8.5

Let $X\rightarrow Y$ be a map of $\mathcal {J}$-spaces such that $Y$ is $\Sigma$-free in the second variable. Then $X$ is $\mathbb {S}^{\mathcal {J}}$-good.

Proof

If $Y$ is $\Sigma$-free in the second variable then it is automatic that also $X$ is $\Sigma$-free in the second variable. $\square$

Lemma 8.6

Let $M$ be a cofibrant commutative $\mathcal {J}$-space monoid. Then $M$ is $\Sigma$-free in the second variable and hence $\mathbb {S}^{\mathcal {J}}$-good.

Proof

Let us say that a $\mathcal {J}$-space $X$ is strongly free in the second variable if for every subgroup $G \subseteq \Sigma _{m_1}\times \Sigma _{m_2}$ such that the composite $G \rightarrow \Sigma _{m_1}\times \Sigma _{m_2} \rightarrow \Sigma _{m_1}$ with the projection is injective, the group $\Sigma _{n_2}$ acts freely on

$$\begin{aligned} \big (X \boxtimes \big (F^{\mathcal {J}}_{({\mathbf {m_1}},{\mathbf {m_2}})}(*)/G\big )\big )({\mathbf {n_1}},{\mathbf {n_2}}). \end{aligned}$$

(8.1)

We will prove the lemma by showing the stronger statement that the underlying $\mathcal {J}$-space of $M$ is strongly free in the second variable.

We first show that $U^{\mathcal {J}} = \mathcal {J}(({\mathbf {0}},{\mathbf {0}}),-)$ is strongly free in the second variable. Let $G \subseteq \Sigma _{m_1}\times \Sigma _{m_2}$ be a subgroup with $G \rightarrow \Sigma _{m_1}\times \Sigma _{m_2} \rightarrow \Sigma _{m_1}$ injective. If a morphism $(\alpha _1,\alpha _2,\rho ):({\mathbf {m_1}},{\mathbf {m_2}})\rightarrow ({\mathbf {n_1}},{\mathbf {n_2}})$ represents an element

$$\begin{aligned} {[(\alpha _1,\alpha _2,\rho )]} \in \mathcal {J}(({\mathbf {m_1}},{\mathbf {m_2}}),({\mathbf {n_1}},{\mathbf {n_2}}))/G \cong \big (U^{\mathcal {J}} \boxtimes \big (F^{\mathcal {J}}_{({\mathbf {m_1}},{\mathbf {m_2}})}(*)/G\big )\big )({\mathbf {n_1}},{\mathbf {n_2}}), \end{aligned}$$

then $\sigma [(\alpha _1,\alpha _2,\rho )] = [(\alpha _1,\alpha _2,\rho )]$ for a $\sigma \in \Sigma _{n_2}$ implies that there is a $(\gamma _1,\gamma _2)\in G$ with

$$\begin{aligned} ({\mathrm {id}}_{{\mathbf {n_1}}},\sigma ,{\mathrm {id}}_{\emptyset })(\alpha _1,\alpha _2,\rho )=(\alpha _1,\alpha _2,\rho )(\gamma _1,\gamma _2,{\mathrm {id}}_{\emptyset }). \end{aligned}$$

By definition of the composition in $\mathcal {J}$ (see [30, Definition 4.2]), this implies $\alpha _1 = \alpha _1 \gamma _1$. Since $\alpha _1$ is injective, $\gamma _1 = {\mathrm {id}}_{{\mathbf {m_1}}}$. Hence $\gamma _2 = {\mathrm {id}}_{{\mathbf {m_2}}}$ because $G \rightarrow \Sigma _{m_1}$ is injective. So we have $({\mathrm {id}}_{{\mathbf {n_1}}},\sigma ,{\mathrm {id}}_{\emptyset })(\alpha _1,\alpha _2,\rho )=(\alpha _1,\alpha _2,\rho )$. This implies $\sigma (i) = i$ for $i\in \alpha _2({\mathbf {m_2}})$. In the third variable, we have $\left( \sigma |_{{\mathbf {n_2}}{\setminus }\alpha _2}\right) \rho = \rho$ and hence $\sigma (i) = i$ for every $i\in {\mathbf {n_2}}{\setminus } \alpha _2$. Hence the $\Sigma _{n_2}$-action on $(U^{\mathcal {J}} \boxtimes (F^{\mathcal {J}}_{({\mathbf {m_1}},{\mathbf {m_2}})}(*)/G))({\mathbf {n_1}},{\mathbf {n_2}})$ is free.

Now we assume that $f :X \rightarrow Y$ is a generating cofibration for the positive $\mathcal {J}$-model structure on $\mathcal {S}^{\mathcal {J}}$ and that the square

is a pushout in $\mathcal {C}\mathcal {S}^{\mathcal {J}}$. We want to show that the underlying $\mathcal {J}$-space of $B$ is strongly free in the second variable if that of $A$ is. For this we use [30, Proposition 10.1]. It provides a filtration $A=F_0(B) \rightarrow F_1(B) \rightarrow \dots$ of $A \rightarrow B$ with ${{\mathrm{colim}}}_i F_i(B) = B$ such that there are pushout squares of $\mathcal {J}$-spaces

(8.2)

where $Q^i_{i-1}(f) \rightarrow Y^{\boxtimes i}$ is the iterated pushout product map associated with $f$. Here we use that since we only consider the commutative case, the functors $U_i^{\mathbb {C}}$ appearing in [30, Proposition 10.1] are the forgetful functors (see [30, Example 10.2]). Since the top horizontal map in (8.2) is a monomorphism of $\mathcal {J}$-spaces by [30, Proposition 7.1(vii)], it is enough to show that $A\boxtimes (Y^{\boxtimes i}/\Sigma _i)$ is strongly free in the second variable. Using that $Y$ is of the form $F^{\mathcal {J}}_{({\mathbf {k_1}},{\mathbf {k_2}})}(K)$ with $k_1 \ge 1$, it is therefore enough to show that $A\boxtimes ((F^{\mathcal {J}}_{({\mathbf {k_1}},{\mathbf {k_2}})}(*))^{\boxtimes i}/\Sigma _i)$ is strongly free in the second variable if $k_1 \ge 1$. This follows from the hypothesis on $A$ because

$$\begin{aligned} \big (F^{\mathcal {J}}_{({\mathbf {k_1}},{\mathbf {k_2}})}(*)^{\boxtimes i}/\Sigma _i\big ) \boxtimes \big (F^{\mathcal {J}}_{({\mathbf {m_1}},{\mathbf {m_2}})}(*)/G\big ) \cong F^{\mathcal {J}}_{({\mathbf {k_1}},{\mathbf {k_2}})^{\sqcup i}\sqcup ({\mathbf {m_1}},{\mathbf {m_2}})}(*)/(\Sigma _i \times G) \end{aligned}$$

and $\Sigma _i \times G \rightarrow \Sigma _{ik_1+m_1}$ is injective if $k_1\ge 1$ and $G \rightarrow \Sigma _{m_1}$ is injective.

For a general cofibrant commutative $\mathcal {J}$-space monoid $M$, we may without loss of generality assume that $M$ is a cell complex constructed from the generating cofibrations. This means that there is a $\lambda$-sequence $\{M_{\alpha } :\alpha < \lambda \}$ in $\mathcal {C}\mathcal {S}^{\mathcal {J}}$ for some ordinal $\lambda$ such that $M_0 = U^{\mathcal {J}}$ and $M_{\alpha } \rightarrow M_{\alpha +1}$ is the cobase change of a generating cofibration in $\mathcal {C}\mathcal {S}^{\mathcal {J}}$. In this situation, the above arguments imply that $M$ is strongly free in the second variable. $\square$

The following consequence of Lemma 8.6 can also easily be verified directly.

Corollary 8.7

Positive cofibrant $\mathcal {J}$-spaces are $\Sigma$-free in the second variable.

Proof

If $X$ is a positive cofibrant $\mathcal {J}$-space, then $\mathbb {C}(X)$ is a cofibrant commutative $\mathcal {J}$-space monoid. Since there is a canonical map of $\mathcal {J}$-spaces $X \rightarrow \mathbb {C}(X)$, the result follows by Lemma 8.6 and the proof of Corollary 8.5. $\square$

Combining Proposition 8.2, Corollary 8.5 and Lemma 8.6 provides the following result.

Corollary 8.8

If $M$ is cofibrant in $\mathcal {C}\mathcal {S}^{\mathcal {J}}$ and $X \rightarrow Y \rightarrow M$ is a sequence of maps of $\mathcal {J}$-spaces with $X \rightarrow Y$ a $\mathcal {J}$-equivalence, then $\mathbb {S}^{\mathcal {J}}[X]\rightarrow \mathbb {S}^{\mathcal {J}}[Y]$ is a stable equivalence. $\square$

Remark 8.9

When working with the topological version of $\mathcal {J}$-spaces, the $\Sigma$-freeness condition should be replaced by a suitable equivariant cofibrancy condition. The analogue of Lemma 8.6 then continues to hold, but we do not have a direct topological analogue of Corollary 8.5.