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Notes on condensed mathematics

Kiran S. Kedlaya

Section 10 Analytic rings

In this section, we introduce the concept of an analytic ring, which in some sense gives the precise category of base rings over which one can do analytic geometry. This will correspond roughly to specifying an underlying condensed commutative ring and a category of “complete modules” over this ring. We have seen a few examples already coming from solid abelian groups, but using an analogue of solidification we will generate many more, including a key example associated to the real numbers.
Note: a previous version of these notes used the term liquid, but this created a terminological conflict with [6], in which “liquid” refers to a class of analytic rings that is not covered in these notes (see [7], or for a brief taste [27]). To resolve this, we are now using the term coalescent. See Remark 10.7.1 for some discussion of the metaphors underlying the terminology.

Reference.

This section is based on [8], Lectures 13 and 14. Remark 10.3.5 is taken from [8], Lecture 11.

Subsection 10.1 Analytic rings

Definition 10.1.1.

By a condensed (commutative) ring, we will mean a commutative ring object in the symmetric monoidal category \(\CAb\text{.}\) Similarly, for \(R\) a condensed commutative ring (meaning a commutative ring object in \(\CAb\)), by a condensed \(R\)-module we will mean an \(R\)-module object in \(\CAb\) Write \(\Mod_R\) for the category of condensed \(R\)-modules.
This notation must be used carefully to avoid creating confusion. For instance, for \(R\) a discrete commutative ring, there is a natural full embedding \(\Mod_R \to \Mod_{\underline{R}}\) which is not an equivalence: its essential image consists solely of the discrete \(R\)-modules.
To define an analytic ring, we specify a base ring and a category of “complete” modules over this base ring.

Definition 10.1.2.

A pre-analytic ring is a pair \(R = (R^{\triangleright}, \Mod_R)\) where \(R^\triangleright\) is a condensed commutative ring and \(\Mod_R\) is a full subcategory of \(\Mod_{R^{\triangleright}}\) closed under limits, colimits, and extensions, and satisfying \(\iExt^i_{R^\triangleright}(M, N) \in \Mod_R\) for all \(i \geq 0\) whenever \(M \in \Mod_{R^{\triangleright}}, N \in \Mod_R\text{.}\) A morphism of pre-analytic rings \(R \to S\) is a morphism \(R^{\triangleright} \to S^{\triangleright}\) of condensed commutative rings such that for any \(M \in \Mod_S\text{,}\) the pullback of \(M\) from \(\Mod_{S^{\triangleright}}\) to \(\Mod_{R^{\triangleright}}\) belongs to \(\Mod_R\text{.}\)
An analytic ring is a pre-analytic ring in which in addition \(R^{\triangleright} \in \Mod_R\text{.}\) We then have a symmetric monoidal structure on \(\Mod_R\) with unit \(R^{\triangleright}\) (see Remark 10.1.4). A morphism of analytic rings is a morphism of the underlying pre-analytic rings. Let \(\AnRing\) denote the category of analytic rings.
When all underlying objects of \(\Mod_R\) belong to \(\CAb_\solid\text{,}\) we refer to the result as a solid analytic ring; note that it is not enough to simply have \(R^{\triangleright} \in \CAb_\solid\) (Remark 10.2.2). Let \(\AnRing_\solid\) denote the full subcategory of \(\AnRing\) consisting of solid analytic rings; we are free to describe solid analytic rings using the interpretation of \(\CAb_\solid\) in terms of solid abelian groups (Theorem 7.4.1).

Definition 10.1.3.

Given a pre-analytic ring, the embedding functor \(\Mod_R \to \Mod_{R^{\triangleright}}\) admits a left adjoint which we will call analytic completion.
For example, forming the analytic completion of \(R^{\triangleright} \otimes \ZZ[\underline{T}]\) for \(T \in \Prof\) yields an object we call \(R[\underline{T}]\text{;}\) these objects form a set of compact (but not necessarily projective) generators of the category \(\Mod_R\text{.}\) A morphism of pre-analytic rings can then be interpreted as a map \(R^{\triangleright} \to S^{\triangleright}\) such that for each \(T \in \Prof\text{,}\) the image of \(S[\underline{T}]\) in \(\Mod_{R^{\triangleright}}\) via restriction of scalars belongs to \(\Mod_R\text{.}\)
We define a tensor product \(\otimes\) on \(\Mod_R\) by taking the tensor product in \(\Mod_{R^{\triangleright}}\) and then applying analytic completion to get back to \(\Mod_R\text{.}\)

Remark 10.1.4.

Let \(R\) be a pre-analytic ring and let \(\hat{R}^{\triangleright}\) be the analytic completion of \(R^{\triangleright}\text{.}\) Since \(R^{\triangleright} \otimes_{R^{\triangleright}} R^{\triangleright}\) and \(\hat{R}^{\triangleright} \otimes_{R^{\triangleright}} \hat{R}^{\triangleright}\) have the same analytic completion, the multiplication map \(R^{\triangleright} \otimes_{R^{\triangleright}} R^{\triangleright} \to R^{\triangleright}\) induces a map \(\hat{R}^{\triangleright} \otimes_{R^{\triangleright}} \hat{R}^{\triangleright} \to \hat{R}^{\triangleright}\text{.}\) That is, \(\hat{R}^{\triangleright}\) canonically promotes to an \(R^{\triangleright}\)-algebra.
Similarly, for \(M \in \Mod_R\text{,}\) we have \(M = R^{\triangleright} \otimes_{R^{\triangleright}} M\text{;}\) since \(R^{\triangleright} \otimes_{R^{\triangleright}} M\) and \(\hat{R}^{\triangleright} \otimes_{R^{\triangleright}} M\) have the same analytic completion, we also have \(M = \hat{R}^{\triangleright} \otimes_{R^{\triangleright}} M\text{.}\) Consequently, the \(R^{\triangleright}\)-module structure on \(M\) canonically promotes to an \(\hat{R}^{\triangleright}\)-module structure.
Continuing in this vein, for \(M \in \Mod_{\hat{R}^{\triangleright}}, N \in \Mod_R\) we have
\begin{align*} \Hom_{R^{\triangleright}}(M,N) &= \Hom_{\hat{R}^{\triangleright}}(\hat{R}^{\triangleright} \otimes_{R^{\triangleright}} M,N)\\ &= \Hom_{\hat{R}^{\triangleright}}(\hat{R}^{\triangleright} \otimes_{R^{\triangleright}} \hat{R}^{\triangleright} \otimes_{\hat{R}^{\triangleright}} M,N)\\ &= \Hom_{\hat{R}^{\triangleright}}(\hat{R}^{\triangleright} \hat{\otimes}_{R^{\triangleright}} \hat{R}^{\triangleright} \otimes_{\hat{R}^{\triangleright}} M,N)\\ &= \Hom_{\hat{R}^{\triangleright}}(\hat{R}^{\triangleright} \otimes_{\hat{R}^{\triangleright}} M,N)\\ & = \Hom_{\hat{R}^{\triangleright}}(M,N)\text{.} \end{align*}
(The introduction of \(\hat{\otimes}_{R^{\triangleright}}\) in place of \(\otimes_{R^{\triangleright}}\) is where we are using the fact that the target \(N\) belongs to \(\Mod_R\text{.}\)) Since this applies to \(M \in \Mod_R\) in particular, it confirms that \(\Mod_R \to \Mod_{R^{\triangleright}}\) factors through \(\Mod_{\hat{R}^{\triangleright}}\text{.}\)
By the same logic with \(M\) replaced by \(M \otimes \ZZ[\underline{S}]\) for \(S \in \Prof\text{,}\) we see that also \(\iHom_{R^{\triangleright}}(M,N) \cong \iHom_{\hat{R}^{\triangleright}}(M,N)\text{.}\) However, note that it is not automatic that
\begin{equation} \iExt^i_{R^{\triangleright}}(M,N) \cong \iExt^i_{\hat{R}^{\triangleright}}(M,N)\tag{10.1} \end{equation}
for \(i \gt 0\) (or even the corresponding statement at the level of global sections). When this holds, \(\hat{R} := (\hat{R}^{\triangleright}, \Mod_{\hat{R}} := \Mod_R)\) is an analytic ring.
The condition (10.1) for all \(i \geq 0\) and all \(M,N \in \Mod_R\) formally reduces to the case \(M = \hat{R}^{\triangleright}\) (by writing \(M\) as a suitable colimit). It can also be interpreted as saying that \(\hat{R}^\triangleright\) is also the derived analytic completion of \(R^{\triangleright}\text{,}\) or that
\begin{equation*} \hat{R}^{\triangleright} \hat{\otimes}^L_{R^{\triangleright}} \hat{R}^{\triangleright} = \hat{R}^{\triangleright}\text{.} \end{equation*}
This condition will automatically satisfied in the examples we consider; see Remark 10.3.5.

Remark 10.1.5.

Building on Remark 10.1.4, we verify (10.1) for \(M = \hat{R}^{\triangleright}, i=1\text{;}\) in this case the claim is that \(\iExt^1_{R^{\triangleright}}(\hat{R}^{\triangleright},N) = 0\text{.}\) For any exact sequence in \(\Mod_{R}\) of the form
\begin{equation*} 0 \to N \to N' \to \hat{R}^{\triangleright} \to 0 \end{equation*}
we have \(N' \in \Mod_{R}\text{.}\) The pullback sequence
\begin{equation*} 0 \to N \to N' \times_{\hat{R}^{\triangleright}} R^{\triangleright} \to R^{\triangleright} \to 0 \end{equation*}
(where the term in the middle is the fiber product, not the tensor product!) is split because \(R^{\triangleright}\) is projective as a module over itself; any splitting then extends by adjunction to a splitting of the original sequence. This proves that \(\Ext^1_{R^{\triangleright}}(\hat{R}^{\triangleright},N) = 0\text{;}\) we deduce the claimed statement by replacing \(N\) with \(\iHom_{R^{\triangleright}}(R^{\triangleright} \otimes \ZZ[\underline{S}], N)\text{.}\)

Remark 10.1.6.

The notation \(R = (R^{\triangleright}, \Mod_R)\) is meant to simulate the notation used by Huber in his development of the theory of adic spaces. We will integrate this theory into our work later, by showing that Huber pairs can be viewed as a full subcategory of solid analytic rings (Proposition 11.3.13).

Remark 10.1.7.

To take a colimit of a system of analytic rings \((R_i^{\triangleright}, \Mod_{R_i})\text{,}\) one first takes the colimit in pre-analytic rings: the base ring is \(\colim_i R_i^{\triangleright}\) and the module category is the limit of \(\Mod_{R_i}\text{.}\) One then replaces the base ring with its analytic completion as usual.
For filtered colimits, the second step is not necessary: if \(i \leq j\) then \(R_j^{\triangleright} \in \Mod_{R_i}\text{,}\) so the colimit automatically belongs to \(\Mod_{R_i}\) for every \(i\text{.}\)

Remark 10.1.8.

For some applications, it is useful to extend the notion of an analytic ring to the case where the base is no longer a condensed commutative ring, but a more homotopical object such as an animated ring, a simplicial ring, or an \(E_\infty\)-commutative ring. We will not touch on these topics in these notes.

Subsection 10.2 Examples of analytic rings

Example 10.2.1.

For every commutative ring \(A\text{,}\) the pair \((\underline{A}, \Mod_{\underline{A}})\) is an analytic ring. An analytic ring of this form will be said to be discrete.
More generally, if \(A\) is a condensed commutative ring, we may treat \((A, \Mod_A)\) as an analytic ring. It is natural to label this also as \(A\text{,}\) so that the two resulting interpretations of \(\Mod_A\) coincide.

Remark 10.2.2.

Beware that while discrete abelian groups are solid, discrete analytic rings are not solid analytic rings because the module category includes non-solid objects.
Note also that for \(A\) a commutative ring, the pair \((\underline{A}, \Mod_A)\) (i.e., taking the discrete modules) is not an analytic ring: it fails the condition that \(\iHom_{\underline{A}}(M, N) \in \Mod_A\) for \(M \in \Mod_{\underline{A}}, N \in \Mod_A\text{.}\) For example, take \(R = \ZZ, M = \bigoplus_\NN \underline{\ZZ}, N = \underline{\ZZ}\text{;}\) we then have \(\iHom_{\underline{\ZZ}}(M, N) = \prod_\NN \underline{\ZZ}\) which is not a discrete \(\ZZ\)-module. This is corrected for using solid modules instead (Example 10.2.3).

Example 10.2.3.

For \(R\) a condensed commutative ring, define the category of solid \(R\)-modules by
\begin{equation*} \Mod_{R\solid} := \Mod_R(\CAb_\solid) = \Mod_R \times_{\CAb} \CAb_\solid\text{.} \end{equation*}
By Proposition 6.2.4, \((R, \Mod_{R \solid})\) is a pre-analytic ring.
If \(R\) is itself solid as a condensed abelian group, then \((R, \Mod_{R \solid})\) is an analytic ring. By Lemma 6.2.2, this applies if \(R\) is discrete; comparing with Example 10.2.1 gives us examples of analytic rings where the underlying ring is the same but the module category is different.

Example 10.2.4.

For \(R\) a commutative ring and \(f \in R\) not a zero-divisor, let \(\hat{R}_{\solid}\) denote the classical \(f\)-completion of \(R\) in \(\Ab_\solid\) (i.e., the usual \(f\)-adic completion carrying the \(f\)-adic topology). The pair consisting of \(\hat{R}_\solid\) and the category of derived \(f\)-complete objects in \(\Mod_{R \solid}\) is a solid analytic ring.
Similarly, the pair consisting of \(\hat{R}_\solid[1/f]\) and the category of derived \(f\)-complete objects in \(\Mod_{R \solid}\) on which \(f\) is invertible is a solid analytic ring.

Subsection 10.3 Coalescence

We define a new process for constructing pre-analytic rings. We start with a bit of motivation.

Remark 10.3.1.

For \(R\) a commutative topological ring, \(f \in R\) an element, \(M\) a Hausdorff topological \(R\)-module, and \((a_n)\) a sequence in \(M\text{,}\) we say that \((a_n)\) is summable against \(f\) if the series \(\sum_n a_n f^n\) converges in \(M\text{.}\) In this language, \(M\) is classically \(f\)-complete if and only every series is summable against \(f\text{.}\)
By contrast, \(M\) is solid if and only if every null sequence is summable against \(f=1\text{.}\) This suggests that we might find it productive to also enforce summability of null sequences against other \(f\text{.}\) For example, in \(\RR\) every null sequence is summable against any \(f\) with \(|f| \lt 1\text{,}\) but not in general against \(f=1\text{.}\)
We now turn to a formal definition.

Definition 10.3.2.

For any condensed commutative ring \(R\text{,}\) using the tensor product in \(\CAb\) we obtain a base extension functor \(\bullet \otimes R\colon \CAb \to \Mod_R\) which is left adjoint to the forgetful functor \(\Mod_R \to \CAb\text{.}\) By the adjunction plus Proposition 5.4.8, the object \(P \otimes R \in \Mod_R\) (the relative sequence space over \(R\)) is internally projective.
For \(f \in R(*)\text{,}\) let \(\Delta_f\colon P \otimes R \to P \otimes R\) be the map \(1 - f \sigma\text{.}\) An object \(M \in \Mod_R\) is \(f\)-coalescent or coalescent around \(f\), if \(\Delta^*\) induces an isomorphism on \(\iHom_{\CAb}(P, M) = \iHom_R(P \otimes R, M)\text{.}\) It is again equivalent to consider \(R\iHom_R(P \otimes R, M)\) as by Proposition 5.4.8, we have \(\iExt_R^i(P \otimes R, M) = 0\) for \(i \gt 0\text{.}\) Let \(\Mod_{R \liquid f}\) denote the full subcategory of \(f\)-coalescent objects of \(\Mod_R\text{;}\) we write \(*_{R \liquid f}\) instead of \(*_{\Mod_{R \liquid f}}\) for \(* \in \{\Hom, \Ext^i, \iHom, \iExt^i\}\text{.}\)
We say that an object of the derived category \(D(\Mod_R)\) is \(f\)-coalescent if its cohomology groups are \(f\)-coalescent. Let \(D(\Mod_R)_{\liquid f}\) be the subcategory of \(f\)-coalescent objects. (The corresponding term in [8] is \(f\)-gaseous.)

Proof.

Proof.

Left to the reader.

Remark 10.3.5.

Picking up a thread from Remark 6.1.4, we compare the split exact sequence
\begin{equation} 0 \to R \to P \otimes R \to (P/\underline{\ZZ}) \otimes R \to 0\tag{10.2} \end{equation}
with the nonsplit exact sequences
\begin{equation} 0 \to P \otimes R \stackrel{\Delta_f}{\to} P \otimes R \to \coker(\Delta_f) \to 0\tag{10.3} \end{equation}
and
\begin{equation} 0 \to P \otimes R \stackrel{\Delta_f}{\to} (P/\underline{\ZZ}) \otimes R \to \coker(\Delta_f)/R \to 0\text{.}\tag{10.4} \end{equation}
The latter gives an internally projective resolution of \(\coker(\Delta_f)/R\) of length \(1\text{.}\)
Let us see what tensoring with \(M \in \Mod_{R \liquid f}\) does to these sequences. For (10.2), since the sequence is split we get for every \(i \geq 0\) an exact sequence
\begin{equation*} 0 \to \iExt^i_R((P/\underline{\ZZ}) \otimes R, M) \to \iExt^i_R(P \otimes R, M) \to \iExt^i_R(R, M) \to 0\text{;} \end{equation*}
that is, the map \(\iHom_R((P/\underline{\ZZ}) \otimes R, M) \to \iHom_R(P \otimes R, M)\) is injective with cokernel \(M\) and the map \(\iExt^i_R((P/\underline{\ZZ}) \otimes R, M) \to \iExt^i_R(P \otimes R, M)\) is an isomorphism for all \(i \gt 0\text{.}\)
For (10.3), \(\Delta_f^*\colon \iExt^i_R(P \otimes R, M) \to \iExt^i_R(P \otimes R, M)\) is an isomorphism for all \(i \geq 0\text{,}\) so \(\iExt^i_R(\coker(\Delta_f), M) = 0\) for all \(i \geq 0\text{.}\) In other words, the derived \(f\)-coalescence of \(\coker(\Delta_f)\) is zero in all degrees.
For (10.4), we refactor \(\Delta_f^*\colon \iExt^i_R((P/\underline{\ZZ}) \otimes R, M) \to \iExt^i_R(P \otimes R, M)\) as the isomorphism \(\iExt^i_R((P/\underline{\ZZ}) \otimes R, M) \to \iExt^i_R((P/\underline{\ZZ}) \otimes R, M)\) followed by the pullback \(\iExt^i_R((P/\underline{\ZZ}) \otimes R, M) \to \iExt^i_R(P \otimes R, M)\) arising from (10.2). We deduce that \(\iExt^1_R(\coker(\Delta_f)/R, M) = M\) and \(\iExt^i_R(\coker(\Delta_f)/R, M) = 0\) for \(i \neq 1\text{;}\) in particular, the derived \(f\)-coalescence of \(\coker(\Delta_f)/R\) is concentrated in degree \(-1\text{.}\)
Finally, using the exact sequence
\begin{equation*} 0 \to R \to \coker(\Delta_f) \to \coker(\Delta_f)/R \to 0 \end{equation*}
we deduce that the derived \(f\)-coalescence of \(R\) is concentrated in degree \(0\text{.}\) We are thus in the favorable situation of Remark 10.1.4, that is, \((R_{\liquid f}, \Mod_{R \liquid f})\) is an analytic ring.
The operation of \(f\)-coalescence bears a strong formal resemblance to derived \(f\)-completion. This drives the next few remarks.

Remark 10.3.6.

If \(S \to R\) is a homomorphism of commutative ring objects in \(\CAb\text{,}\) then for any \(f \in S\) we have
\begin{equation*} \Mod_{R \liquid f} = \Mod_{S \liquid f} \times_{\Mod_S} \Mod_R\text{.} \end{equation*}
This applies in particular when \(f = 1\text{,}\) in which case we may take \(S = \ZZ\) to obtain
\begin{equation*} \Mod_{R \liquid 1} = \Mod_{R\solid}\text{.} \end{equation*}
It follows that for \(M \in \Mod_R\text{,}\) forming the \(f\)-coalescence \(M_{\liquid f}\) commutes with restriction of scalars from \(S\) to \(R\text{,}\) so there is no ambiguity in the notation. However, this is not necessarily true of the derived \(f\)-coalescence unless \(S\) is projective in \(\Mod_R\text{.}\)

Remark 10.3.7.

For \(f,g \in R\text{,}\) (derived) \(f\)-coalescence commutes with (derived) \(g\)-coalescence because the composition in either order can be written in a symmetric form using the Koszul complex associated to \(\Delta_f\) and \(\Delta_g\text{.}\) In particular, (derived) \(f\)-coalescence acts on the (derived) category of \(g\)-coalescent modules.
Consequently, for any subset \(S \subset R\text{,}\) taking \(R\) together with the full subcategory \(\Mod_{R \liquid S} := \bigcap_{f \in S} \Mod_{R \liquid f}\) forms a pre-analytic ring. An important special case is when \(1 \in S\text{;}\) thanks to Theorem 7.4.1, we can treat this case in the language of solid abelian groups.

Subsection 10.4 The core of an analytic ring

We study the effect of coalescing with respect to different elements. This involves picking up the thread from Remark 6.1.3, now algebraizing various statements about summability of null sequences against \(f\text{.}\)

Proof.

For \(S \in \Prof\text{,}\) we must check that we get a map on \(\Hom_{\CAb}(\ZZ[\underline{S}], \iHom_{\CAb}(P, M))\text{;}\) using Hom-tensor adjunction, we may rewrite this as \(\Hom_{\CAb}(P, \Hom_{\CAb}(\ZZ[\underline{S}], M))\text{.}\) Since again \(\Hom_{\CAb}(\ZZ[\underline{S}], M) \in \Mod_{R \liquid f}\text{,}\) we formally reduce to the case where \(S\) is a singleton. The claimed statement now becomes the observation that for any null sequence \((m_0, m_1, \dots)\) in \(M\text{,}\) the inclusion \(M \in \Mod_{R \liquid f}\) means that the sequence \((m_0, f m_1, \dots)\) is summable against 1, and so must itself be a null sequence.

Proof.

We algebraize the following statement: given a null sequence \((m_0, m_1, \dots)\text{,}\) if we know that \((m_0, f_1 m_1, \dots)\) is again a null sequence and that every null sequence is summable against \(g\text{,}\) we may conclude that the original sequence is summable against \(fg\text{.}\)
Let \(h_g^* \colon \iHom_R(P \otimes R, M) \to \iHom_R(P \otimes R, M)\) be the map induced by \(R[T] \stackrel{T \mapsto gT}{\to} R[T]\text{,}\) whose existence has been assumed. Note that the notation is mnemonic in nature: we are not assuming the existence of a map \(h_g\) on \(P \times R\text{,}\) but \(h_g^*\) functions as the “virtual pullback” by such a nonexistent map.
We have an equality of maps on \(\iHom_{\CAb}(P, M) \cong \iHom_R(P \otimes R, M)\text{:}\)
\begin{equation*} \Delta_f^* \circ h_g^* = h_g^* \circ \Delta_{fg}^*\text{:} \end{equation*}
both compositions have the effect
\begin{equation*} (m_0, m_1, \dots) \mapsto (m_0-fgm_1, gm_1-fg^2m_2, \dots).\text{.} \end{equation*}
Next, define the map
\begin{equation*} \lambda\colon P \to P, \qquad [0] \to 0, [n+1] \mapsto [n] \end{equation*}
and form the composition
\begin{equation*} \lambda^* \oplus h_g^*\colon \iHom_R(P \otimes R, M) \to \iHom_R((P \oplus P) \otimes R, M) \end{equation*}
where \(\lambda^*\colon \iHom_{\CAb}(P, M) \to \iHom_{\CAb}(P, M)\) is a genuine pullback. The map \(\lambda^* \oplus h_g^*\) is a split injection: we can first project from \(\iHom_R((P \oplus P) \otimes R, M)\) to \(\iHom_R((P \oplus \ZZ) \otimes R, M)\) by restricting along \(\ZZ \to \ZZ[0] \subset P\text{.}\) The resulting composition \(\iHom_R(P \otimes R, M) \to \iHom_R((P \oplus \ZZ) \otimes R, M)\) is now the pullback along a genuine map \(P \oplus \ZZ \to P\) which is a split surjection, so the pullback is a split injection.
We now have an equality of maps on \(\iHom_{\CAb}(P \oplus P, M) \cong \iHom_R((P \oplus P) \otimes R, M)\text{:}\)
\begin{equation*} (\Delta_f^* \oplus \Delta_f^*) \circ (h_g^* \oplus \lambda^*) = (h_g^*\oplus \lambda^*) \circ \Delta_{fg}^*\text{.} \end{equation*}
Since \((\Delta_f \oplus \Delta_f)^*\) is an isomorphism and \(h_g^* \oplus \lambda^*\) is a split injection, tracing around the square as in Lemma 7.2.8 (but with the sense of arrows reversed) shows that \(\Delta_{fg}^*\) is also an isomorphism.

Proof.

Proof.

The fact that \(\Mod_{R \liquid f} \subseteq \Mod_{R \liquid f^n}\) follows from Lemma 10.4.3. To see that \(\Mod_{R \liquid f^n} \subseteq \Mod_{R \liquid f}\text{,}\) we algebraize the statement that a null sequence \((m_0, m_1, \dots)\) is summable against \(f\) if and only if the decimated sequences \((m_i, m_{i+n}, \dots)\) for \(i=0,\dots,n-1\) are summable against \(f^n\text{,}\) in which case the original sum coincides with the sum of \(f^i\) times the sum of the \(i\)-th decimated sequence for \(i=1,\dots,n-1\text{.}\) This corresponds to the fact that \(\tilde{\Delta}_{f^n} = \Delta_{f} \circ g = g \circ \Delta_{f}\) where
\begin{gather*} \tilde{\Delta}_{f^n} \colon P \otimes R \to P \otimes R, \qquad [m] \mapsto [m] - f^n [m+n]\\ g \colon P \otimes R \to P \otimes R, \qquad [m] \mapsto [m] + f [m+1] + \cdots + f^{n-1} [m+n-1]\text{;} \end{gather*}
via the identification \(P \cong \underline{\ZZ[x]}\) (for the \(x\)-adic topology on the latter) this corresponds to the factorization
\begin{align*} 1-f^n x^n &= (1-fx)(1+fx + \cdots + f^{n-1}x^{n-1})\\ &= (1+fx + \cdots + f^{n-1}x^{n-1})(1-fx)\text{.} \end{align*}
For any \(M \in \Mod_{R \liquid f^n}\text{,}\) \(\Delta_{f^n}^*\) is an isomorphism on \(\iHom_{\CAb}(P, M)^{\oplus n}\) and so \(\tilde{\Delta}_{f^n}^*\) is an isomorphism on \(\iHom_{\CAb}(P, M)\text{.}\) Since \(\tilde{\Delta}_{f^n}^*\) factors through \(\Delta_f^*\) both ways, this implies that \(\Delta_f^*\) is both injective and surjective.

Definition 10.4.5.

Let \(R\) be a condensed commutative ring. We say that \(g \in R(*)\) is topologically nilpotent if there is an element of \(\Hom_{\CAb}(P, R)\) taking \([n]\) to \(g^n\) for all \(n \in \NN\text{.}\) If \(R\) is a topological ring, this says that the powers of \(g\) form a null sequence.
We say that \(g \in R(*)\) is universally power-bounded if there is an element of \(\Hom_{\CAb}(P, P \otimes R)\) taking \([n]\) to \(g^n[n]\) for all \(n \in \NN\text{;}\) the corresponding element of \(\Hom_R(P \otimes R, P \otimes R)\) is then a ring homomorphism. If \(R\) is a topological ring, this implies that for every topological \(R\)-module \(M\) and every null sequence \((m_n)_n\) in \(M\text{,}\) \((g^n m_n)_n\) is also a null sequence; this is in general strictly stronger than making the same assertion only for \(M=R\text{.}\)
If \(g\) is topologically nilpotent, then it is also universally power-bounded: we may tensor the map \(\Hom_{\CAb}(P, R)\) taking \([n]\) to \(g^n\) with the identity map on \(P\) to obtain a map \(P \to P \otimes R\) taking \([n]\) to \(g^n [n]\text{.}\) Moreover, both properties are “integrally closed” in \(R(*)\text{:}\) if \(g \in R(*)\) and \(g^n\) is universally power-bounded (resp. topologically nilpotent), then so is \(g\text{.}\)

Proof.

The universally power-bounded property implies that the hypothesis of Lemma 10.4.2 is satisfied by every object of \(\Mod_R\text{,}\) so in particular for objects of \(\Mod_{R \liquid f}\text{.}\)

Proof.

Since \(g\) is topologically nilpotent, so is \(g^n f^{-1}\text{.}\) By Lemma 10.4.6 and Lemma 10.4.4,
\begin{equation*} \Mod_{R \liquid f} \subseteq \Mod_{R \liquid f(g^n f^{-1}} = \Mod_{R \liquid g^n} = \Mod_{R \liquid g}\text{.} \end{equation*}
We may summarize the previous discussion as follows.

Definition 10.4.8.

For \(R\) an analytic ring, an element \(f \in R^{\triangleright}(*)\) is gaseous if \(\Mod_R = \Mod_{R \liquid f}\text{.}\) Define the gaseous core (or more commonly just the core) of \(R\text{,}\) denoted \(R^{\gas}\text{,}\) to be the set of gaseous elements of \(R^{\triangleright}(*)\text{.}\)

Proof.

Remark 10.4.10.

In general, there is no reason to expect the core of an analytic ring to have any meaningful additive structure. This changes if we restrict attention to solid analytic rings; see Proposition 11.3.10.

Subsection 10.5 The gaseous real numbers

We next introduce a key example that does not factor through solidification (although we will not deploy it in earnest until Section 16).

Definition 10.5.1.

By Corollary 10.4.7, the category \(\Mod_{\underline{\RR} \liquid f}\) is the same for any two choices of \(f \in (0,1)\text{.}\) We denote it also as \(\Mod_{\underline{\RR} \liquid}\text{.}\)

Proof.

For \(S \in \Prof\text{,}\) we must check that \(\Delta^*\) acts isomorphically on
\begin{equation*} \Hom_{\CAb}(P \otimes \ZZ[\underline{S}], \underline{\RR})= \Hom_{\CAb}(P, \iHom_{\CAb}(\ZZ[\underline{S}], \underline{\RR}))\text{.} \end{equation*}
Using (5.3) this becomes
\begin{equation*} \Hom_{\CAb}(P, \underline{\Cts(S, \RR)})\text{.} \end{equation*}
Since the compact-open topology on \(\Cts(S, \RR)\) is induced by the supremum norm, we are reduced to checking that for any null sequence of functions \(g_n\colon S \to \RR\) for the supremum norm and any \(f \in (0,1)\text{,}\) the sequence \((f^n g_n)_n\) is summable in \(\Cts(S, \RR)\text{.}\) This follows by observing that the supremum norms \(c_n\) of \(g_n\) form a null sequence in \(\RR\) and \(f^n c_n\) is summable, so \(\sum_n f^n g_n\) is absolutely convergent.

Remark 10.5.3.

The same logic as in Proposition 10.5.2 shows that for any real Banach space \(V\text{,}\) \(\underline{V} \in \Mod_{\underline{\RR} \liquid}\text{.}\) Since we can take limits in \(\Mod_{\underline{\RR} \liquid}\text{,}\) we also obtain Frechet spaces and various other classical examples from functional analysis, but now with a specific tensor product that does not preserve these subcategories.
However, in order to work with the category \(\Mod_{\underline{\RR}\liquid}\text{,}\) we really need to compute the coalescence of \(\RR[\underline{S}]\) for \(S \in \Prof\text{.}\) Possibly the easiest way to do this is to compute the analogue in a certain universal setting; see Proposition 10.6.6 for the analogue and Proposition 10.6.8 for the result over \(\RR\text{.}\)

Subsection 10.6 The initial analytic ring

For some calculations, it is helpful to describe a “universal” version of coalescence.

Definition 10.6.1.

Let \(\ZZ[\hat{q}]\) be a copy of \(P := \ZZ[\underline{\NN_\infty}]/\ZZ[\underline{\infty}]\) viewed as an algebra over \(\ZZ[q]\) via \(q^n \mapsto [n]\text{.}\) Then set \(A_0^{\triangleright} := \ZZ[\hat{q}][q^{-1}]\text{;}\) this condensed ring is represented by the topological ring with underlying ring \(\ZZ[q^{\pm}]\text{,}\) which is initial among condensed rings \(R\) equipped with a choice of a topologically nilpotent unit \(q \in R(*)^\times\text{.}\) By Proposition 10.3.3, the pair \((A_0^{\triangleright}, \Mod_{A_0^{\triangleright} \liquid q})\) is a pre-analytic ring.
The ring \(A_0^{\triangleright}\) is not itself \(q\)-coalescent because we cannot take any infinite sums over \(q\text{.}\) To remedy this, we allow certain controlled infinite sums.

Definition 10.6.2.

For \(k=1,2,\dots\text{,}\) define the topological ring
\begin{equation} A_k^{\triangleright} := \bigcup_{c, n \gt 0} \prod_{m \geq -n} \left(\ZZ \cap [-c(m+n)^{k-1}, c(m+n)^{k-1}] \right) q^m\tag{10.5} \end{equation}
and set \(A^{\triangleright} = \bigcup_k A_k^{\triangleright}\text{.}\) That is, \(A^{\triangleright}\) consists of Laurent series whose coefficients grow polynomially in the exponent.
For \(S = \varprojlim_i S_i \in \Prof\text{,}\) we similarly have
\begin{equation} A_k^{\triangleright}\otimes \ZZ[\underline{S}] := \bigcup_{c, n \gt 0} \varprojlim_i \prod_{m \geq -n} \ZZ[S_i]_{\leq c(m+n)^{k-1}} q^m\tag{10.6} \end{equation}
where the subscript \(\leq c(m+n)^{k-1}\) means that we bound the sum of the absolute values of the coefficients accordingly.

Proof.

We treat the case \(S = \{*\}\) in detail, the general case being similar but with more notation. Given a null sequence \((s_i)_i\) in \(A_{k-1}^{\triangleright}\text{,}\) for some fixed \(n,c\) we have \(s_i = \sum_{m \geq -n} s_{i,m} q^m\) where we have \(|s_{i,m}| \leq c(m+n)^{k-2}\) for all \(i,m\) if \(k \gt 1\text{,}\) and for \(k =1\) we instead have that only finitely many of the \(s_{i,m}\) are nonzero. Now write
\begin{align*} \sum_{i=0}^\infty s_i q^i &= \sum_{i=0}^\infty \sum_{m \geq -n} s_{i,m} q^{m+i}\\ &= \sum_{j \geq -n} \left( \sum_{i=0}^{j+n} s_{i,j-i} \right) q^j\text{.} \end{align*}
If \(k=1\) it is obvious that the resulting series has bounded coefficients and so belongs to \(A_0^{\triangleright}\text{.}\) Otherwise, since \(|s_{i,j-i}| \leq c(j+n-i)^k\) for all \(i,j\text{,}\) we have
\begin{equation*} \left| \sum_{i=0}^{j+n} s_{i,j-i} \right| \leq c(j+n+1)^{k-1} \end{equation*}
from which we infer that \(\sum_{i=0}^\infty s_i q^i \in A_{k}^{\triangleright}\text{.}\)
To prove the converse, we employ a trick introduced in Remark 9.2.13. Without loss of generality we may consider an element \(t = \sum_{m=0}^\infty t_m q^m \in A_k^{\triangleright}\) such that for some \(c\gt 0\text{,}\) we have \(|t_m| \leq c(m+1)^{k-1}\) for all \(m\text{.}\) If \(k=1\text{,}\) we write
\begin{equation*} t = \sum_{m=0}^\infty (t_{2m} q^m + t_{2m+1}q^{m+1}) q^m \end{equation*}
and the quantities \(t_{2m} q^m + t_{2m+1}q^{m+1}\) form a null sequence in \(A_0^{\triangleright}\text{.}\) If \(k \gt 1\text{,}\) set
\begin{equation*} s_{i,j} := \left \lfloor \frac{t_{i+j} + j}{i+j+1} \right\rfloor\text{,} \end{equation*}
so that \(|s_{i,j}| \leq c(i+j+1)^{k-2} +1 \leq (c+1)(i+j+1)^{k-2}\) and \(t_m = \sum_{i=0}^m s_{i,m-i}\text{.}\) Set \(s_i := \sum_{j=0}^\infty s_{i,j} q^j\text{,}\) so that
\begin{equation*} t = \sum_{i=0}^\infty s_i q^i = \sum_{i=0}^\infty (s_{2i} q^i + s_{2i+1}q^{i+1}) q^i\text{.} \end{equation*}
It now suffices to check that as \(i\) varies, the quantities \(s_{2i} q^i + s_{2i+1}q^{i+1}\) form a null sequence in \(A_{k-1}^{\triangleright}\text{:}\) write
\begin{equation*} s_{2i+*} q^{i+*} = \sum_{j=i+*}^\infty s_{2i+*,j-i-*} q^j \end{equation*}
and then note that because \(j \geq i\)
\begin{align*} |s_{2i+*,j-i-*}| &\leq (c+1)(i+j+1)^{k-1}\\ &\leq (c+1) (2j+1)^{k-1} \leq 2^{k-1} (c+1) (j+1)^{k-1}\text{.} \end{align*}

Proof.

For \(S = \varprojlim_i S_i \in \Prof\text{,}\) we may apply Lemma 10.6.3 to see that the action of \(\Delta^*\) on
\begin{equation*} \Hom_{\CAb}(P \otimes \ZZ[\underline{S}], A^{\triangleright}) = \Hom_{\CAb}(P, \iHom_{\CAb}(\ZZ[\underline{S}], A^{\triangleright})) \end{equation*}
is an isomorphism. This implies that the action on \(\iHom_{\CAb}(P, A^{\triangleright})\) is an isomorphism, and hence the desired result.

Proof.

By replacing \(M\) with \(\iHom_{\CAb}(\ZZ[\underline{S}], M)\) and applying Hom-tensor adjunction, we reduce to the case \(S = \{*\}\text{.}\) From Lemma 10.6.3, we see that for \(k \geq 0\text{,}\)
\begin{equation} \Hom_{A_0^{\triangleright}}(A_{k+1}^{\triangleright}, M) \cong \Hom_{A_0^{\triangleright}}(A_k^{\triangleright}, M)\text{.}\tag{10.8} \end{equation}
In particular, for \(k \geq 0\text{,}\)
\begin{equation*} \Hom_{A_0^{\triangleright}}(A_k^{\triangleright}, M) \cong \Hom_{A_0^{\triangleright}}(A_0^{\triangleright}, M) \end{equation*}
and taking a colimit over \(k\) yields the claim.

Proof.

For the underived coalescence, this follows directly from Lemma 10.6.5; for the derived coalescence, we must also apply Remark 10.3.5. Alternatively, one can give a more careful argument to directly ensure that
\begin{equation*} A^{\triangleright} \widehat{\otimes}_{A_0^{\triangleright}}^L A^{\triangleright} \cong A^{\triangleright}\text{;} \end{equation*}
see [8], Lecture 14, timestamp 17:00.

Remark 10.6.7.

In addition to its interest for studying specific analytic rings like \(\RR_{\liquid}\text{,}\) one can also make certain constructions directly over the analytic ring of Proposition 10.6.6, notably the Tate elliptic curve. We will consider this example later.
As promised, we may apply this to compute the coalescence of \(\RR[\underline{S}] := \RR \otimes \ZZ[\underline{S}]\text{.}\)

Proof.

This follows by applying Proposition 10.6.6 and tensoring along the map \(A_0^{\triangleright} \to \RR\) sending \(q\) to any value in \((0,1)\text{.}\) For more details, see [8], Lecture 14, timestamp 1:15:00.

Remark 10.6.9.

Just as with solid modules, we can take this further and coalesce around other parameters. For instance, we can start with \(\RR[T]\) and coalesce around some \(q \in (0,1)\) and \(T\text{.}\)

Subsection 10.7 A remark on terminology

Remark 10.7.1.

We insert a note on the metaphor underlying terminology here. We want to think of the module category of a discrete analytic ring as consisting of an inchoate elemental plasma, as found in the empty areas of outer space. By contrast, we think of the module category of a solid analytic ring as consisting of more concrete objects, such as rocky planets. The process of adding elements to the core of an analytic ring amounts to coalescing loose matter into something more structured; that is, the core is the source of “gravity” that pulls the matter together.
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