Proof. Say $S$ is a free ring on the set $X$ of generators, and $Q$ is the free $S$-bimodule on the set $Y$ of generators, i.e. $\bigoplus _{Y} S\otimes _{\operatorname {\mathbb {Z}}} S$. Then it is easily seen that $Q^{\circledcirc _{S}n}$ is a direct sum $\bigoplus _{Y^{\times n}} (S\otimes _{\operatorname {\mathbb {Z}}} S)^{\circledcirc _{S} n}$, where $C_n$ acts on the index set $Y^{\times n}$ by permuting the factors cyclically, and on the summands by the $C_n$ action on the cyclic tensor product. The cyclic tensor product $(S\otimes _{\operatorname {\mathbb {Z}}} S)^{\circledcirc _{S} n}$ is equivalent to $S^{\otimes _{\operatorname {\mathbb {Z}}} n}$ with $C_n$ acting by cyclic permutation.

As an abelian group $S$ is free on a set $T$, and $S^{\otimes _{\operatorname {\mathbb {Z}}} n}$ is free abelian on the set $T^{\times n}$, with $C_n$ acting by permutation. Thus, the whole $Q^{\circledcirc _{S}n}$ is a free abelian group on the set $Y^{\times n} \times T^{\times n}$, with $C_n$ acting by cyclic permutation on both factors. Thus, the $C_n$-invariants are torsion free, because they are a subgroup, and the coinvariants are the free abelian group on the set $(Y^{\times n} \times T^{\times n})/ C_n$, thus also torsion free.

The transfer map is injective because the composite $(Q^{\circledcirc _{S}n})_{C_n}\to (Q^{\circledcirc _{S} n})^{C_n}\to (Q^{\circledcirc _{S}n})_{C_n}$ of the transfer and the quotient map is multiplication by $n$, which is injective if $(Q^{\circledcirc _{S}n})_{C_n}$ is torsion free.

Proof. A monoid structure on a bimodule $(R;M)$ is a morphism

\[ \mu=(\mu_R;\mu_M) : (R\otimes_{\operatorname{\mathbb{Z}}} R;M\otimes_{\operatorname{\mathbb{Z}}} M)\longrightarrow (R;M), \]

and a unit map $\eta =(\eta _R;\eta _M) : (\operatorname {\mathbb {Z}};\operatorname {\mathbb {Z}})\to (R;M)$, subject to the associativity and unitality axioms. The map $\mu _R$ and the unit $\eta _R$ then endow the ring $R$ with the structure of a monoid with respect to the tensor product of rings. By the Eckmann–Hilton argument [Reference Eckmann and HiltonEH61, Theorem 4.17], $\mu _R$ is the multiplication of $R$ and $R$ must be a commutative ring. The map $\mu _M$ is a map of $R\otimes _{\operatorname {\mathbb {Z}}} R$-bimodules

\[ \mu_{M} : M\otimes_{\operatorname{\mathbb{Z}}} M\longrightarrow \mu^{*}_RM, \]

which endows $M$ with a ring structure $m\star n:=\mu _M(m\otimes n)$. The bimodule structure determines and is determined by the ring homomorphisms $\eta _l(a)=a\cdot 1$ and $\eta _r(b)=1\cdot b$ so that we have $a \cdot m = \eta _l(a) \star m$ and $m \cdot b = m \star \eta _r(b)$. As $\mu _M$ is a map of left $R\otimes _{\operatorname {\mathbb {Z}}} R$-modules we also have

\[ \eta_l(a) \star m=a \cdot m = (1 \cdot a)\cdot (m \star 1) = (1\cdot m) \star (a\cdot 1) = m \star \eta_l(a), \]

which shows that $\eta _l$ is central. Similarly, we see that $\eta _r$ is central.

Conversely, for arbitrary central ring morphisms

\[ \eta_l, \eta_r: R \to M \]

we equip $M$ with the bimodule structure $rms := \eta _l(r) \star m \star \eta _r(s)$ and one directly checks that then $\star$ is a map in $\operatorname {biMod}$.

Proof. Assume we have a special unit in filtration $\geq n$, that is, one of the form

\[ 1 + a_n t^{n} + a_{n+1} t^{n+1} + \cdots. \]

Then the coefficient $a_n$ can be written as a finite sum of elements in $G_n$, and we can split off corresponding factors of the form $(1 + g_n t^{n})$. This allows us to write any such special unit as a product of ones of the form $(1 + g_n t^{n})$ and a remainder term of higher filtration. Inductively, this proves that, up to a remainder term of arbitrarily high filtration, any element of $\widehat {S}(R;M)$ can be written as a product of terms of the form $(1 + g_k t^{k})$. This proves the claim.

Proof. The kernel of $\widehat {S}(R;M)\to W(R;M)$ is, by definition, closed and the group $\widehat {S}(R;M)$ is first countable, hence metrizable by the Birkhoff–Kakutani theorem. Moreover, the induced filtration generates the quotient topology. Thus, the claim follows from [Reference BourbakiBou07, Chapitre IX, § 3, Proposition 4].

Proof. We first check that $\widehat {S}(-;-)$ commutes with reflexive coequalizers. To see this, we need to check that if

is a reflexive coequalizer of bimodules, then $\widehat {S}(R;M)$ is obtained from $\widehat {S}(R_0;M_0)$ by quotienting by the closed normal subgroup $N$ generated by all $f(y)g(y)^{-1}$ for $y\in \widehat {S}(R_1;M_1)$. Surjectivity is clear, so we have to check that the kernel of $\widehat {S}(R_0;M_0)\to \widehat {S}(R;M)$ agrees with $N$. The subgroup $N$ is clearly contained in the kernel. Given an element $x$ in the kernel, it is of the form $(1+a_n t^{n} + \cdots )$, with $a_n$ in the kernel of the right map in the following diagram.

As reflexive coequalizers of abelian groups commute with tensor products, this diagram is also a reflexive coequalizer of abelian groups, so $a_n$ is of the form $f(b_n)-g(b_n)$. Thus, the original $x$ can up to a term of higher filtration (which is also in the kernel) be written as $x = f(1+b_n t^{n}) g(1+b_nt^{n})^{-1}$. Inductively, we can write any element in the kernel as a convergent product of elements of the form $f(y)g(y)^{-1}$, so the kernel is contained in $N$ as desired. We now want to show that $W(-;-)$ also commutes with reflexive coequalizers. To that end, let $N(R;M)$ denote the closed normal subgroup of $\widehat {S}(R;M)$ generated by commutators and elements of the form $(1+rmt)(1+mrt)^{-1}$, so that $W(R;M) = \widehat {S}(R;M)/N(R;M)$. As $\widehat {S}(-)$ commutes with reflexive coequalizers, we see that the coequalizer of $W(R_1;M_1)\rightrightarrows W(R_0;M_0)$ can be described as the quotient of $\widehat {S}(R;M)$ by the closure of the image of $N(R_0;M_0)$. Thus, we have to check that this closure agrees with $N(R;M)$. However, this is clear: $N(R;M)$ is topologically generated by commutators and elements of the form $(1+rmt)(1+mrt)^{-1}$, all of which are in the image.

Proof. For the first claim, it suffices to show that $\log : \widehat {S}(R;M) \to \operatorname {\mathbb {Q}}\widehat {\otimes } \prod _{n\geq 1} (M^{\circledcirc _R n})_{C_n}$ is a homomorphism. We define an operator $\partial : \widehat {T}(R;M)\to \widehat {T}(R;M)$ that acts by multiplication with $n$ on the factor $M^{\otimes _R n}$. This satisfies $\partial (fg) = \partial f\cdot g + f\cdot \partial g$. In particular, we have

\[ \partial f^{n} = (\partial f) f^{n-1} + f (\partial f) f^{n-2} + \cdots + f^{n-1} (\partial f). \]

Now let us write $f \sim g$ when elements $f, g\in \operatorname {\mathbb {Q}}\widehat {\otimes } \widehat {T}(R;M)$ have the same image under the canonical map to $\operatorname {\mathbb {Q}}\widehat {\otimes } \prod _{n\geq 1} (M^{\circledcirc _R n})_{C_n}$. One easily sees by expanding that $fg \sim gf$ for any elements $f,g$. It follows that $\partial f^{n} \sim n (\partial f) f^{n-1}$, and for a special unit $u=1+f$:

\[ \partial \log u \sim (\partial f) - (\partial f)f + (\partial f)f^{2} - \cdots = (\partial f) \cdot (1 + f)^{-1} = (\partial u) \cdot u^{-1}. \]

Therefore, for any special units $u,v$, we see that

\begin{align*} \partial \log(uv) &\sim \partial(uv) \cdot (uv)^{-1} = ((\partial u) \cdot v + u\cdot (\partial v)) v^{-1} u^{-1} = (\partial u) \cdot u^{-1} + u \cdot (\partial v) \cdot v^{-1} \cdot u^{-1}\\ &= \partial \log u + u \cdot (\partial \log v) \cdot u^{-1}\sim \partial \log u + \partial \log v. \end{align*}

This shows that in the diagram

the lower composite from the leftmost node to the lower-right node is a homomorphism. However, because the rightmost vertical map is an isomorphism, the top horizontal composite is a homomorphism as well.

For the second claim, we calculate explicitly

\begin{align*} \operatorname{tlog}(1-a_nt^{n}) &= \operatorname{tr}_{e}^{C_{n}} a_n t^{n} + \operatorname{tr}_{e}^{C_{2n}} \frac{a_n^{2}}{2} t^{2n} + \operatorname{tr}^{C_{3n}}_{e}\frac{a_n^{3}}{3} t^{3n} + \cdots\\ &= \operatorname{tr}_{e}^{C_{n}} a_n t^{n} + \operatorname{tr}_{C_2}^{C_{2n}} \operatorname{tr}_{e}^{C_{2}} \frac{a_n^{2}}{2} t^{2n} + \operatorname{tr}^{C_{3n}}_{C_3} \operatorname{tr}^{C_{3}}_{e}\frac{a_n^{3}}{3} t^{3n} + \cdots\\ &= \operatorname{tr}^{C_n}_e a_n t^{n} + \operatorname{tr}^{C_{2n}}_{C_2} a_n^{2} t^{2n} + \operatorname{tr}^{C_{3n}}_{C_3}a_n^{3} t^{3n} + \cdots, \end{align*}

where the last equality comes from the fact that $a_n^{k}$ is already invariant under the action of the subgroup $C_k \subseteq C_{nk}$, so $\operatorname {tr}_{e}^{C_k}$ just acts by multiplication with $k$.

For the third claim, it suffices again to check this for the map $\log$. We have

\[ \log(1-fg) = -fg - \frac{fgfg}{2} - \cdots \sim -gf - \frac{gfgf}{2} - \cdots = \log(1-gf), \]

and, thus, they agree in $\operatorname {\mathbb {Q}}\widehat {\otimes }\prod _{n\geq 1} (M^{\circledcirc _R n})_{C_n}$.

For the last claim, we first observe that the image of $\operatorname {tlog}$ is integral, that is, contained in the image of the rationalisation $\prod _{n\geq 1} (M^{\circledcirc _R n})^{C_n}\to \operatorname {\mathbb {Q}}\widehat {\otimes }\prod _{n\geq 1} (M^{\circledcirc _R n})^{C_n}$. As $\widehat {S}(R;M)$ is topologically generated by elements of the form $(1+a_nt^{n})$, this follows immediately from the first two claims. For a pair $(R;M)$ where $(M^{\circledcirc _R n})^{C_n}$ is torsion free, the rationalisation is injective. Thus, on the full subcategory of those $(R;M)$ with torsion-free $(M^{\circledcirc _R n})^{C_n}$, $\operatorname {tlog}$ factors to a unique natural transformation as desired. As we are mapping to a Hausdorff topological group, $\widehat {S}(R;M)$ commutes with reflexive coequalizers in Hausdorff topological groups, and we can resolve every bimodule $(R;M)$ as a reflexive coequalizer of $(R_1;M_1)$ and $(R_0;M_0)$ with torsion-free $({M_i}^{\circledcirc _{R_i} n})^{C_n}$ (see Lemma 1.4), this natural transformation extends uniquely to all $(R;M)$.

Proof. We have to show that every element in the kernel of $\operatorname {tlog}$ can be written as a convergent product of elements in $G$. Suppose we have an element of the form $f_n=(1+a_nt^{n} + \cdots )$ in the kernel of $\operatorname {tlog}$, with $a_n\in M^{\otimes _R n}$. Then, because

\[ \operatorname{tlog}(1 + a_n t^{n} + \cdots) = - \operatorname{tr}_e^{C_n} a_n t^{n} + \cdots, \]

we have that $a_n$ is in the kernel of the composite $M^{\otimes _R n} \to (M^{\circledcirc _R n})_{C_n} \to (M^{\circledcirc _R n})^{C_n}$. As we assumed the latter map to be injective, $a_n$ is in the kernel of the quotient map $M^{\otimes _R n}\to (M^{\circledcirc _R n})_{C_n}$. This kernel is generated by differences of the form $x_i \otimes y_j - y_j \otimes x_i$ for $i+j=n$, with $x_i \in M^{\otimes _R i}$ and $y_j\in M^{\otimes _R j}$, so $a_n$ can be written as a sum of such elements. Now, by property 3, this implies that we can write $f_n$ as a product of elements in $G$ of filtration $\geq n$, and a remainder term of filtration $\geq n+1$, which by property 2 is also in the kernel of $\operatorname {tlog}$. Iterating this argument, property 1 implies that every element in the kernel of $\operatorname {tlog}$ is in $G$.

Proof. We first compute the leading term for a commutator of $(1+a_n t^{n})$ and $(1+b_m t^{m})$. We have

\begin{align*} (1+a_n t^{n})(1+b_m t^{m}) &= (1 + a_n t^{n} + b_m t^{m} + a_n b_m t^{n+m})\\ &= (1 + a_n b_m t^{n+m} + \cdots) (1 + a_n t^{n} + b_m t^{m}). \end{align*}

Multiplying this with the inverse of $(1+b_m t^{m})(1+a_n t^{n})$, we obtain

\begin{align*} [(1+a_n t^{n}), (1+b_m t^{m})] &= (1 + a_n b_m t^{n+m} + \cdots) \cdot (1 + b_ma_n t^{n+m} + \cdots)^{-1}\\ &= 1 + (a_nb_m - b_ma_n)\cdot t^{n+m}+\cdots. \end{align*}

In particular, this shows that elements $(1+a_k t^{k})$ and $(1+b_l t^{l})$ commute up to terms of filtration $\geq k+l$. By continuity, we also obtain that arbitrary elements of filtration $\geq k$ and $\geq l$ commute up to terms of filtration $\geq k+l$. Thus, if we have

\begin{gather*} x = (1 + a_n t^{n} + \cdots) = (1+ a_n t^{n}) \cdot x',\\ y = (1+ b_m t^{m} + \cdots) = (1+b_m t^{m}) \cdot y', \end{gather*}

with $x'$ of filtration $>n$, and $y'$ of filtration $>m$, we see that, up to terms of filtration $>n+m$, $x'$ commutes with $(1+b_mt^{m})$, $y'$ commutes with $(1+a_nt^{n})$ and $x'$ commutes with $y'$. We thus obtain that $[x,y]$ and $[(1+a_n t^{n}), (1+b_mt^{m})]$ agree up to order $n+m$, from which the result follows.

Proof. The map clearly factors through the abelianisation, and by Proposition 1.15(3), we have $\operatorname {tlog}(1- rm \cdot t) = \operatorname {tlog}(1-mr\cdot t)$, so it factors through $W(R;M)$.

For injectivity, note that by Lemmas 1.16 and 1.17, the closed subgroup generated by commutators and elements of the form $(1-rm\cdot t)(1-mr\cdot t)^{-1}$ actually agrees with the kernel of $\operatorname {tlog}$ if the transfers are injective.

For the last part, it suffices to check the following stronger version of injectivity: if an element $f\in W(R;M)$ has the property that $\operatorname {tlog} f\in \prod _{n\geq 1} (M^{\circledcirc _R n})^{C_n}$ has filtration ${\geq}k$, then $f$ has filtration ${\geq}k$ as well. However, observe that this is exactly what the argument in the proof of Lemma 1.16 gives us.

Proof. We have to show that all these subgroups of $\widehat {S}(R;M)$ agree. By resolving via reflexive coequalizers, we can reduce to the case where the transfers $\operatorname {tr}: (M^{\circledcirc _R n})_{C_n} \to (M^{\circledcirc _R n})^{C_n}$ are injective. In that case, we claim they all agree with the kernel of $\operatorname {tlog}: \widehat {S}(R;M) \to \prod _{n\geq 1} (M^{\circledcirc _R n})^{C_n}$. By Proposition 1.15, they are all contained in the kernel of $\operatorname {tlog}$, and using Lemma 1.17, we see that they also satisfy condition 3 of Lemma 1.16, which then implies the claim.

Proof. The map $\widehat {S}(R\times S;M\times N) \to \widehat {S}(R;M)\times \widehat {S}(S;N)$ is an isomorphism of topological groups. By Lemma 1.19, it suffices to check that it sends the closed subgroups generated by elements of the form $(1 - x_i y_j t^{i+j}) \cdot (1 - y_j x_i t^{i+j})^{-1}$ to each other. This follows from

\begin{align*} & (1 - (a_i,b_i) (a_j,b_j) t^{i+j}) \cdot (1 - (a_j,b_j) (a_i,b_i) t^{i+j})^{-1}\\ &\quad =(1- (a_i,0) (a_j,0) t^{i+j}) \cdot (1 - (a_j,0)(a_i,0) t^{i+j})^{-1}\\ &\qquad \cdot (1- (0,b_i) (0,b_j) t^{i+j}) \cdot (1 - (0,b_j)(0,b_i) t^{i+j})^{-1}. \end{align*}

Proof. This follows immediately from Lemmas 1.12 and 1.19.

Proof. As the images of the maps $M^{\times n k}\to W(R; M^{\otimes _R n})$ topologically generate $W(R; M^{\otimes _R n})$, there is at most one $V_n$ with the desired properties.

For the existence, consider that the homomorphism $\widehat {S}(R;M^{\otimes _R n}) \to \widehat {S}(R;M)$ given by sending

\[ 1 + \sum_i a_i t^{i} \mapsto 1 + \sum_i a_i t^{ni} \]

preserves the relations given in Lemma 1.19, which were of the form

\[ (1 - x_i y_j t^{i+j}) \sim (1 - y_j x_i t^{i+j}). \]

Thus, this homomorphism factors to a homomorphism $V_n: W(R; M^{\otimes _R n}) \to W(R; M)$ as desired. Next, we compute that this $V_n$ is compatible with the given description on ghosts. However, it suffices to check this on generators. The ghost map sends

\begin{gather*} \operatorname{tlog}(1 - a_k t^{k}) = \operatorname{tr}_e^{C_k} a_k t^{k} + \operatorname{tr}_{C_2}^{C_{2k}} a_k^{2} t^{2k} + \cdots,\\ \operatorname{tlog} (V_n(1 - a_k t^{k})) = \operatorname{tlog}(1 - a_k t^{nk}) = \operatorname{tr}_e^{C_{nk}} a_k t^{nk} + \operatorname{tr}_{C_2}^{C_{2nk}} a_k^{2} t^{2nk} + \cdots. \end{gather*}

As $\operatorname {tr}_{C_{ik}}^{C_{nik}} \operatorname {tr}_{C_i}^{C_{ik}} = \operatorname {tr}_{C_i}^{C_{nik}}$, the described map on ghosts sends $\operatorname {tlog}(1 - a_k t^{k})$ to $\operatorname {tlog}(V_n(1-a_k t^{k}))$.

Finally, to check that $V_n V_m = V_{nm}$, it suffices that they agree on the image of $\tau _k$, which follows from the defining properties of $V_i$.

Proof. Again, the images of the upper horizontal maps (jointly for all $k$) generate $W(R; M^{\otimes _R n})$ topologically, and so there is at most one homomorphism $\sigma$. To see one exists, it is sufficient to do so for $(R;M)$ with torsion-free $(M^{\circledcirc nk})_{C_k}$, because the target is Hausdorff and we can resolve any $(R;M)$ as a reflexive coequalizer of $(R_0;M_0)$ and $(R_1; M_1)$ with torsion-free $(M_i^{\circledcirc nk})_{C_k}$.

In the torsion-free case, we know by Proposition 1.18 that $\operatorname {tlog}$ is a homeomorphism onto its image. It is, therefore, sufficient to check that the described $C_n$-action on ghost components restricts to an action on the image of $\operatorname {tlog}$ or, more precisely, sends $\operatorname {tlog}(1 - m_1 \otimes \cdots \otimes m_{nk} t^{k})$ to $\operatorname {tlog}(1 - m_{nk} \otimes m_1\otimes \cdots \otimes m_{nk-1}t^{k})$.

The $ik$th coefficient of $\operatorname {tlog}(1 - m_1 \otimes \cdots \otimes m_{nk} t^{k})$ is given (Proposition 1.15) by

\[ \operatorname{tr}_{C_i}^{C_{ik}} (m_1 \otimes \cdots \otimes m_{nk})^{\otimes i}, \]

which is shifted by a generator of $C_{nik}$ (representing the residual action of a generator of $C_n = C_{nik}/C_{ik}$) to the element

\[ \operatorname{tr}_{C_i}^{C_{ik}} (m_{nk} \otimes m_1\otimes \cdots \otimes m_{nk-1})^{\otimes i}, \]

which is the $ik$th coefficient of $\operatorname {tlog}(1 -m_{nk} \otimes m_1\otimes \cdots \otimes m_{nk-1} t^{k})$.

The $n$th power of $\sigma$ acts as identity on ghost components, and because of naturality, this implies that $\sigma$ always has order $n$.

Proof. Because of the first property, it suffices to define $F_p$ for each prime $p$. Uniqueness then follows from the fact that $W(R;M)$ is generated by elements of the form $\tau _k(a_k) = V_k(\tau (a_k))$: if $k$ is prime to $p$, then the value of $F_pV_k(\tau (a_k))$ is determined by properties 2 and 4, as

\[ F_pV_k(\tau(a_k)) = V_kF_p(\tau(a_k)) = V_k(\tau(a_k^{p})). \]

If $k$ is divisible by $p$, say $k=p\cdot l$, then

\[ F_pV_k(\tau(a_k)) = F_pV_pV_l(\tau(a_k)) = \sum_{\sigma\in C_p} \sigma V_l(\tau(a_k)). \]

To show existence of $F_n$, it suffices to do this again in the case of injective $\operatorname {tlog}$, by resolving $(R,M)$ via reflexive coequalizers (Propositions 1.14 and 1.18). In this case, we define the map on ghost components by

\begin{align*} \prod_{i\geq 1} (M^{\circledcirc_R i})^{C_i} &\to \prod_{i\geq 1} (M^{\circledcirc_R in})^{C_i} \\ \sum_i w_i t^{i} &\mapsto \sum_j w_{jn} t^{j}. \end{align*}

This map is clearly continuous. By Proposition 1.18, we know that $\operatorname {tlog}$ exhibits $W(R; M)$ and $W(R; M^{\otimes _R n})$ as subspaces with the subspace topology. Thus, it suffices to show that this map sends the image of $\operatorname {tlog}$ to the image of $\operatorname {tlog}$. To see this it suffices to show (similarly to the argument for uniqueness of $F_p$ above) that this map on ghosts makes the ghost component analogues of the diagrams in properties 2, 3 and 4 commute.

Checking that the diagrams commute on ghosts is straightforward: for property 2, we have

where the condition that $n,m$ are coprime was used for evaluating the right vertical map: $im$ is divisible by $n$ if and only if $im$ is of the form $jnm$.

For properties 3 and 4, we similarly have the following.

Proof. Uniqueness follows from the fact that elements of the form $\tau _k(a_k)=V_k\tau (a_k)$ form a set of topological generators. Indeed, the first and second conditions together show that

\[ V_k(\tau(a_k)) * \tau(b) = V_k(s(\tau(a_k) * F_k\tau(b))) = V_k(s(\tau(a_k) *\tau(b^{\otimes k}))) = V_k(\tau(s(a_k\otimes b^{\otimes k}))) \]

(where we have slightly overloaded the notation $s$ for the various shuffle isomorphisms), so by bilinearity and continuity, multiplication of arbitrary elements with elements of the form $\tau (b)$ is completely determined. The flipped version of the first condition now shows that

\[ f * V_k\tau(b_k) = V_k(s(F_k(f) * \tau(b_k))), \]

which is completely determined by the properties, following the first step.

As a reflexive coequalizer diagram in Hausdorff abelian groups is also an underlying reflexive coequalizer diagram in Hausdorff spaces, and reflexive coequalizers in Hausdorff abelian groups commute with finite products, if we choose resolutions of $(R;M)$ and $(S;N)$ by reflexive coequalizers, the diagram

is a reflexive coequalizer diagram in Hausdorff spaces. Thus, a continuous bilinear map $*$ as desired can be extended from the case of free rings and bimodules to all (and is then easily seen to be bilinear in general).

To construct $*$ in the free case, we proceed similarly to the proof of Proposition 1.26. We define the map on ghost components by the shuffle map as desired. This map is continuous if we restrict it to the product, analogously to the continuity of the map $*$ in question. Using Proposition 1.18 it therefore suffices to show that this map in ghost components sends the image of $\operatorname {tlog}$ to the image of $\operatorname {tlog}$, so that is gives rise to a well-defined map on Witt vectors which is continuous (in the appropriate sense). For this, it suffices to check that the described map on ghost components satisfies the ghost analogues of properties 1 and 2, because using both properties we can express $*$ again in terms of Verschiebung and Teichmüller maps.

On ghosts, we simply check

\begin{align*} &V_k\bigg(\sum_i a_{ik} t^{i}\bigg) * \bigg(\sum_i b_i t^{i}\bigg)\nonumber\\ &\quad =\bigg(\sum_i \operatorname{tr}_{C_i}^{C_{ik}} a_{ik} t^{ik}\bigg) * \bigg(\sum_i b_i t^{i}\bigg) =\sum_i s((\operatorname{tr}_{C_i}^{C_{ik}} a_{ik}) \otimes b_{ik}) t^{ik}\\ &\quad =\sum_i \operatorname{tr}_{C_i}^{C_{ik}} s(a_{ik} \otimes b_{ik}) t^{ik} =V_k\bigg(\sum_i s(a_{ik}\otimes b_{ik}) t^{i}\bigg)\\ &\quad=V_k\bigg(\bigg(\sum_i a_{ik} t^{i}\bigg) * \bigg(\sum_i b_{ik}t^{i}\bigg)\bigg) =V_k\bigg(\bigg(\sum_i a_{ik} t^{i}\bigg) * F_k \bigg(\sum_i b_i t^{i}\bigg)\bigg), \end{align*}

where the crucial equality is the third one, and uses that the transfer is linear with respect to multiplication with invariant elements (and $b_{ik}$ is, by assumption, $C_{ik}$-invariant).

Proof. One checks that the map $(\mu _R,l_M) : (R \otimes R;R \otimes M) \to (R;M)$ is a map in $\operatorname {biMod}$ which is straightforward.Footnote ⁴ Then it follows that $(R;M)$ is a module in $\operatorname {biMod}$ over the commutative monoid $(R; R)$. Thus, the claim follows because $W(-;-)$ is lax symmetric monoidal.

Proof. Most of these formulas, namely properties 1, 2, 5 and 7, were proved in the construction of the Verschiebung, Frobenius and symmetric monoidal structure. The rest is easily checked explicitly on ghost components, by reducing to the case of injective $\operatorname {tlog}$. For example, properties 3 and 4 correspond to invariance of transfer and restriction maps.

Proof. Just as in the proof of Proposition 1.25, uniqueness follows immediately because the images of the $\tau _k$ form a system of generators. Existence is checked in the case of suitably free $R_i$, $M_{i,i+1}$, such that the $\operatorname {tlog}$ is injective, by computing that the claimed action on ghost components acts correctly on elements of the form $\operatorname {tlog}(\tau _k(a_k))$. Finally, the statement about the $n$-fold iterate of this isomorphism also follows by observing that the corresponding map on ghosts is the identity.

Proof. Any additive functor $A : \operatorname {Proj}_R \to \operatorname {Proj}_S$ is of the form $A \cong (-) \otimes _R M$, where $M$ is the $R$-$S$-bimodule $M := A(R)$. Let $N$ be the $S$-$R$-bimodule $N := \operatorname {Hom}_S(M, S)$. There are bimodule maps

\[ \eta : R \to M \otimes_S N \quad \text{and} \quad \operatorname{ev} : N \otimes_R M \to S, \]

where the second map is the evaluation, whereas the first map corresponds under the isomorphism

(5)\begin{equation} M \otimes_S N \cong M \otimes_S \operatorname{Hom}_S(M, S) \cong \operatorname{Hom}_S(M,M) \end{equation}

to the map which sends $1 \in R$ to the identity. The isomorphism (5) uses the fact that $M$ is finitely generated projective over $S$. The desired map is defined as the composite

\[ W(R) = W(R;R) \stackrel{\eta_*}{\longrightarrow} W(R; M \otimes_S N) \cong W(S; N \otimes_R M) \stackrel{\operatorname{ev}_*}{\longrightarrow} W(S; S) = W(S), \]

where the middle isomorphism is from Proposition 1.34.

If $R$ and $S$ are Morita equivalent we can find an $R$-$S$-bimodule $M$ such that $\eta$ and $\operatorname {ev}$ are isomorphisms, and it follows that the map above is also an isomorphism.

Proof. Observe that the image filtration of the $\widehat {S}^{(n)}(R;M)$ on $W_S(R;M)$ is still Hausdorff and complete, by the same argument as in the proof of Lemma 1.13.

The map $W_S(R;M) \to W_{S_i}(R;M)$ is surjective, with kernel $K_i$ topologically generated by the elements of the form $\tau _n(x)$ with $n\not \in S_i$, and because elements of the form $\tau _n(x)$ with $n\not \in S$ are already zero in $W_S(R;M)$, $K_i$ is actually generated by those $\tau _n(x)$ with $n\in S\setminus S_i$. We let $d_i$ be the minimal element of $S\setminus S_i$. Thus, every element of $K_i$ has a representative of filtration $\geq d_i$. As $\bigcup S_i = S$, $d_i$ tends to $\infty$ with $i$ and, thus, the $K_i$ also form a Hausdorff and complete filtration of $W_S(R;M)$, which implies the claim.

Proof. To check that $\operatorname {tlog}$ factors as claimed, it suffices to show that $\operatorname {tlog}(1-a_kt^{k})$ for $k\not \in S$ is sent to $0$ under the projection map $\prod _{n\geq 1} (M^{\circledcirc _R n})^{C_n} \to \prod _{n\in S} (M^{\circledcirc _R n})^{C_n}$. As

\[ \operatorname{tlog}(1-a_kt^{k}) = \sum_i \operatorname{tr}_{C_i}^{C_{ki}} a_k^{i} t^{ki}, \]

and $S$ contains no multiples of $ki$, this is clear.

Now assume that for each $n\in S$, the transfer $(M^{\circledcirc _R n})_{C_n}\to (M^{\circledcirc _R n})^{C_n}$ is injective. We want to show that $\operatorname {tlog}_S$ is injective. Let $x$ be an element in the kernel, say with a representative of the form $(1-a_kt^{k} + \cdots )$. If $k\not \in S$, we can factor this in the form $(1-a_kt^{k}) \cdot (1 - a_{k+1} t^{k+1} + \cdots )$, with the second factor still in the kernel of $\operatorname {tlog}_S$. If, on the other hand, $k\in S$, then we have

\[ \operatorname{tlog}(1-a_kt^{k} + \cdots) = \operatorname{tr}_e^{C_k}a_kt^{k} + \cdots, \]

so $a_k$ lies in the kernel of the transfer $M^{\circledcirc _R k} \to (M^{\circledcirc _R k})^{C_k}$. By assumption, this means that $a_k$ vanishes in $(M^{\circledcirc _R k})_{C_k}$. As in the proof of Lemma 1.16, this shows that we can multiply $(1-a_kt^{k} + \cdots )$ by a series with filtration $\geq k$ that vanishes in $W(R;M)$, in order to obtain a representative of the form $(1-a_{k+1}t^{k+1}+\cdots )$. This shows that any element which gets mapped by $\operatorname {tlog}_S$ to something of filtration $\geq k$ admits a representative of filtration $\geq k$. In particular, $\operatorname {tlog}_S$ is an embedding, because any element in the kernel equals a convergent product of elements trivial in $W_S(R;M)$ and, thus, vanishes.

Proof. The formulas given on ghost components for the various structure maps are all seen to be compatible with the projections onto the respective index sets. Now note that if $\operatorname {tlog}_S$ is injective, the kernel of $W(R;M) \to W_S(R;M)$ is the same as the preimage of the kernel of the projection map $\prod _{n\geq 1} (M^{\circledcirc _R n})^{C_n} \to \prod _{n\in S} (M^{\circledcirc _R n})^{C_n}$ under $\operatorname {tlog}$. Thus, in the injective case, we see that the structure maps preserve these kernels and, thus, descend to structure maps on $W_S$. The statement for general pairs $(R;M)$ now follows by resolving by pairs where the relevant $\operatorname {tlog}$ are injective.

Proof. Recall that $W_{S'}(R;M)$ is, by definition, the quotient of $W(R;M)$ by the closed subgroup generated by all $\tau _d(a_d)$ for $a_d \in M^{\times d}$ and $d\not \in S'$. Equivalently, we can view this as the quotient of $W_S(R;M)$ by the image of that subgroup. As $\tau _d(a_d)$ for $d\in S$ is already zero in $W_S(R;M)$, we can obtain $W_{S'}(R;M)$ from $W_S(R;M)$ by quotienting just by all $\tau _d(a_d)$ for $d\in S\setminus S' = S\cap k\mathbb {N}$. We have to check that this coincides with the image of $V_k$. To see this, recall (Lemma 1.12) that $W_{S/k}(R;M^{\otimes _R k})$ is generated by elements of the form $\tau _l(a_{kl})$ with $a_{kl} \in M^{\times kl}$, and $l\in S/k$ or, equivalently, $kl\in S$. Now observe that

\[ V_k(\tau_l(a_{kl})) = \tau_{kl}(a_{kl}), \]

which proves the claim.

Proof. Assume $x\in W_{S/k}(R;M^{\otimes _R k})_{C_k}$ is in the kernel. Assume $x$ is not $0$, so there exists a maximal $l$ such that $x$ has filtration $\geq l$. We write $x = \tau _{l}(a_{kl}) + x'$ with $x'$ of filtration $\geq l+1$. If $kl\not \in S$, then $\tau _{l}(a_{kl}) = 0$ in $W_{S/k}$, and so $x=x'$ and $x$ has filtration $\geq l+1$, contradicting the maximality of $l$. Thus, $kl \in S$.

The leading term of $\operatorname {tlog} V_k(x)$ agrees with that of $\operatorname {tlog} V_k(\tau _{l}(a_{kl})) = \operatorname {tlog} \tau _{kl}(a_{kl})$, which is given by $\operatorname {tr}_{e}^{C_{kl}}(a_{kl})$. As $kl\in S$, the vanishing of $V_k(x)$ therefore implies that $\operatorname {tr}_{e}^{C_{kl}}(a_{kl})=0$. As we assumed that the transfers $\operatorname {tr}: (M^{\circledcirc _R kl})_{C_{kl}} \to (M^{\circledcirc _R kl})^{C_{kl}}$ are injective, this implies that $a_{kl} = 0$ in $(M^{\circledcirc _R kl})_{C_{kl}}$. Similarly to the proof of Lemma 1.16, one can then write $(1+a_{kl}t^{l})$ in $\widehat {S}(R; M^{\otimes _R k})$ as a product of elements of the form $(1+x_i y_j t^{l})\cdot (1+y_j x_i t^{l})^{-1}$ with $x_i\in M^{\otimes _R i}$, $y_j\in M^{\otimes _R j}$, $i+j=kl$, and a remainder term of higher filtration. Observe that, from the definition of the $C_k$ action on $W(R; M^{\otimes k})$, the element $(1+x_i y_j t^{l})(1+y_j x_i t^{l})^{-1}$ represents $\tau _l(x_i y_j) - \sigma ^{j} \tau _l(x_i y_j)$. It follows that $x\in W_{S/k}(R;M^{\otimes _R k})_{C_k}$ has filtration bigger than $l$, contradicting the maximality of $l$. Thus, $x=0$.

Proof. The functor $(-)\otimes _{\operatorname {End}_R(P)}P: \operatorname {Proj}_{\operatorname {End}_R(P)}\to \operatorname {Proj}_{R}$ has a right adjoint given by $(-)\otimes _R P^{\vee }$ and the counit of the adjunction is given by the map $M \otimes _R P^{\vee } \otimes _{\operatorname {End}_R(P)} P \to M$, which implies the criterion. For the second condition we note that we can check the first condition Zariski-locally, and for free modules it is equivalent to being non-trivial.

Finally, to see that $P$ is supported everywhere, we have to show that $P$ is a generator of $\operatorname {Proj}_{R}$, which is immediate if $P$ contains a free summand.

Proof. We abbreviate $E := \operatorname {End}_R(P)$. By definition and the proof of Corollary 1.36, the map (8) is the composite of the top row of the diagram

where the composite of the bottom row is, by definition, the product of the traces. The description in ghost components follows from the commutativity of this diagram. The three upper squares commute by the naturality of the ghost map and by Proposition 1.34. The lower vertical isomorphisms in the first and third columns are from Example 1.1, where the $C_n$-actions on the cyclic powers are trivial. The lower middle isomorphism which makes the lower left square commute is defined by iterating the multiplication maps

\[ (P\otimes_RP^{\vee})\otimes_{E}(P\otimes_RP^{\vee}) \xrightarrow{P\otimes_R\operatorname{ev}\otimes_RP^{\vee}}P\otimes_RR\otimes_R P^{\vee}\cong P\otimes_RP^{\vee} \]

which correspond to the multiplication of $E$ under $\eta$. It is then easy to see that both composites in the lower right rectangle are

\[ (p_1\otimes_R\lambda_1)\otimes_E\cdots \otimes_E(p_n\otimes_R\lambda_n) \longmapsto \lambda_n(p_1)\lambda_1(p_2)\cdots \lambda_{n-1}(p_n). \]

For the second property, we note that both maps $W(\operatorname {End}_R P \times \operatorname {End}_R Q) \to W(R)$ are induced by functors

\[ \operatorname{Proj}_{\operatorname{End}_R P \times \operatorname{End}_R Q} \to \operatorname{\operatorname{Proj}_R} \]

which are easily seen to agree with the functor that sends $(M,N)$ to $(M \oplus N) \otimes _{(\operatorname {End}_R P \times \operatorname {End}_R Q)} (P \oplus Q)$.

For the third property, consider $\varphi = x \circ \psi \in \operatorname {End}_R{P}$. Using property 2, we can assume that $P$ admits $R$ as a direct summand, by replacing it with $P\oplus R$ if necessary (and replacing $\varphi$ correspondingly by the map $\varphi \oplus 0: P\oplus R \to P\oplus R$). Thus, we can choose $f: P\to R$ and $e: R\to P$ with $f\circ e=\operatorname {id}$. Now the element

\[ \varphi\in \operatorname{End}_R P\cong P\otimes_R P^{\vee} \otimes_{\operatorname{End}_R P} \cdots \otimes_R P^{\vee} \]

can be represented by the elementary tensor

\[ x\otimes_R f\otimes_{\operatorname{End}_R P} e \otimes_R \cdots \otimes_{\operatorname{End}_R P} e \otimes_R \psi. \]

By Proposition 1.34, the map (8) therefore sends $(1-\varphi t^{n})$ to $(1-\psi (x)f(e)\cdots f(e)t^{n})=(1-\psi (x)t^{n})$ as claimed.

Proof. The first statement is obvious from the definition. The second statement is an immediate consequence of the fact that $1 - fg t = 1 - gft$ in $W(\operatorname {End}_R(P))$ by definition. The third statement follows from Propositions 2.4(1) and 1.15. Indeed, the ghost components of $\tau (f)$ are given by

\[ \operatorname{tlog}(1-ft) = ft + f^{2} t^{2} + \cdots, \]

and so the ghost components of $\chi _f$ are given by

\[ \operatorname{tlog}(\chi_f) = \operatorname{tr}(f) t + \operatorname{tr}(f^{2}) t^{2} + \cdots. \]

For the fourth statement first consider the special case where the endomorphism splits, by which we mean that there exists a section $P_3 \to P_2$ such that, under the induced isomorphism $P_1\oplus P_3 \cong P_2$, $f_2$ corresponds to $f_1 \oplus f_3$. In other words, $1- f_2t$ is the image of $1- (f_1, f_3)t$ under the map $\oplus : W(\operatorname {End}_R(P_1) \times \operatorname {End}_R(P_3)) \to W(\operatorname {End}_R(P_2))$. Then the claim follows from property 2 of Proposition 2.4. For the general case, we choose a section $s: P_3\to P_2$, and under the isomorphism $P_2 \cong P_1\oplus P_3$ we write $f_2$ as a ‘block matrix’

\[ \bigg(\begin{array}{c|c} f_1 & \rho\\ \hline 0 & f_3 \end{array}\bigg), \]

where $f_i$ is an endomorphism of $P_i$, for $i=1,2$, and $\rho : P_3 \to P_1$ is $R$-linear. In $W(\operatorname {End}_R(P_1\oplus P_3))$ we see that

\[ \tau\bigg(\begin{array}{c|c} 0 & \rho\\ \hline 0 & 0 \end{array}\bigg) = \tau\bigg(\bigg(\begin{array}{c|c} 1 & 0\\ \hline 0 & 0 \end{array}\bigg)\bigg(\begin{array}{c|c} 0 & \rho\\ \hline 0 & 0 \end{array}\bigg)\bigg) = \tau \bigg(\bigg(\begin{array}{c|c} 0 & \rho\\ \hline 0 & 0 \end{array}\bigg)\bigg(\begin{array}{c|c} 1 & 0\\ \hline 0 & 0 \end{array}\bigg)\bigg) = \tau\bigg(\begin{array}{c|c} 0 & 0\\ \hline 0 & 0 \end{array}\bigg) = 1, \]

using the relation $\tau (ab) = \tau (ba)$ that holds in $W(-)$ by definition. Thus, the characteristic element of $\bigg (\begin {array}{c|c} 0 & \rho \\ \hline 0 & 0\end {array}\bigg )$ vanishes, and we further see that

\begin{gather*} \bigg(1 - \bigg(\begin{array}{c|c} f_1 & \rho\\ \hline 0 & f_3 \end{array}\bigg)t\bigg) = \bigg(1 - \bigg(\begin{array}{c|c} 0 & 0\\ \hline 0 & f_3 \end{array}\bigg)t\bigg)\bigg(1 - \bigg(\begin{array}{c|c} 0 & \rho\\ \hline 0 & 0 \end{array}\bigg)t\bigg)\bigg(1 - \bigg(\begin{array}{c|c} f_1 & 0\\ \hline 0 & 0 \end{array}\bigg)t\bigg),\\ \bigg(1 - \bigg(\begin{array}{c|c} 0 & 0\\ \hline 0 & f_3 \end{array}\bigg)t\bigg)\bigg(1 - \bigg(\begin{array}{c|c} f_1 & 0\\ \hline 0 & 0 \end{array}\bigg)t\bigg) = \bigg(1 - \bigg(\begin{array}{c|c} f_1 & 0\\ \hline 0 & f_3 \end{array}\bigg)t\bigg), \end{gather*}

which together imply that

\begin{align*} \tau\bigg(\begin{array}{c|c} f_1 & \rho\\ \hline 0 & f_3 \end{array}\bigg) &= \tau\bigg(\begin{array}{c|c} 0 & 0\\ \hline 0 & f_3 \end{array}\bigg)\tau\bigg(\begin{array}{c|c} 0 & \rho\\ \hline 0 & 0 \end{array}\bigg)\tau\bigg(\begin{array}{c|c} f_1 & 0\\ \hline 0 & 0 \end{array}\bigg)\\ &=\tau\bigg(\begin{array}{c|c} 0 & 0\\ \hline 0 & f_3 \end{array}\bigg)\tau\bigg(\begin{array}{c|c} f_1 & 0\\ \hline 0 & 0 \end{array}\bigg)=\tau\bigg(\begin{array}{c|c} f_1 & 0\\ \hline 0 & f_3 \end{array}\bigg)\end{align*}

Thus, the characteristic element of $f_2$ agrees with that of $f_1 \oplus f_3$, so by the previous case $\chi _{f_2} = \chi _{f_1} + \chi _{f_3}$.

Proof. By naturality, it is sufficient to check the claim in the universal case, $R=\operatorname {\mathbb {Z}}[a_{ij}\,|\, 1\leq i\leq n, 1\leq j\leq n]$, with $f$ the endomorphism given by the matrix $(a_{ij})$. As $R$ is a domain, it embeds into an algebraically closed field $K$ of characteristic zero. As the map $W(R) \to W(K)$ is injective, it suffices to check that $\chi _f(t)$ and $t^{n} \chi _f^{\mathrm {cl}}(t^{-1})$ agree in $W(K)$. Over the algebraically closed $K$, however, $f$ can be brought into triangular form by a base change. This does not affect $\chi _f^{\mathrm {cl}}$ nor $\chi _f$, the latter because of property 2 of Lemma 2.7. For triangular $f$, with diagonal entries $\lambda _1, \ldots, \lambda _n$, property 3 of Lemma 2.7 implies $\chi _f = \prod (1-\lambda _i t)$, which agrees with $t^{n} \prod (t^{-1} - \lambda _i) = t^{n} \chi _f^{\mathrm {cl}}$.

Proof. An isomorphism $P [\kern-0.15em[{t}]\kern-0.15em] \to Q [\kern-0.15em[{t}]\kern-0.15em] $ reduces to an isomorphism $P\to Q$. We can, thus, identify $P$ with $Q$ (note that the map (8) is clearly natural in isomorphisms of projective modules), and consider $\alpha$ as an automorphism of $P [\kern-0.15em[{t}]\kern-0.15em] $. Now the proof proceeds analogously to the proof of Lemma 2.7. Note that by Lemma 1.19, we have the relation