Proof. It is easy to see that $\phi $ extends to a well-defined additive map after we quotient the image of $1+w$ in the target. To see that it is an isomorphism, choose a basis $k\cong \oplus _XS$ of k as an S-vector space. This induces an isomorphism of $C_2$ -equivariant abelian groups

$$\begin{align*}k\otimes_Sk\cong \bigoplus_{X\times X}S, \end{align*}$$

where the involution on the right-hand side sends a basis element $(x, y)$ of $X\times X$ to $(y, x)$ . Under this isomorphism, the map $\phi $ corresponds to the map

$$\begin{align*}\bigoplus_{X\times X}S\cong \bigoplus_{X}(\bigoplus_XS)\cong \bigoplus_{X}k\cong(\bigoplus_{X\times X}S)^{C_2}/Im(1+w), \end{align*}$$

where the second isomorphism is the sum over X of the isomorphism $k\cong \oplus _XS$ , and the last isomorphism sends the summand x to the summand $(x, x)$ via the square map $(-)^2\colon k\xrightarrow {\cong }S$ .

Proof. Since k is a field, the Frobenius module structure on $\underline {k}^{{\phi \mathbb {Z}/2}}$ provides an equivalence of k-modules

$$\begin{align*}{\mathrm{H}}\underline{k}^{{\phi \mathbb{Z}/2}} \simeq \bigoplus_{n \geq 0} \Sigma^{n} {\mathrm{H}}(\pi_n({\mathrm{H}}\underline{k}^{{\phi \mathbb{Z}/2}})).\end{align*}$$

Since the Frobenius module structure on ${\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ comes from a k-algebra ${\mathrm {H}} k\to {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ , the action of k on $\pi _n({\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}})$ is obtained by restricting, along the ring map $k=\pi _0{\mathrm {H}} k\to \pi _0({\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}})$ , the action of $\pi _0({\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}})$ on $\pi _n({\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}})$ induced by the ring structure of ${\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ . The $\pi _0({\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}})$ -module $\pi _n({\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}})$ can be computed from the isotropy separation sequence as follows. The canonical ring homomorphism ${\mathrm {H}} k={\mathrm {H}}\underline {k}^{\mathbb {Z}/2}\to {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ induces a long exact sequence of k-modules

$$\begin{align*}\dots \xrightarrow{\partial} \pi_{1}{\mathrm{H}} k_{h \mathbb{Z}/2} \to\pi_1 {\mathrm{H}} k\to \pi_1 {\mathrm{H}}\underline{k}^{{\phi \mathbb{Z}/2}} \xrightarrow{\partial} \pi_{0}{\mathrm{H}} k_{h \mathbb{Z}/2}\to \pi_{0}{\mathrm{H}} k\to \pi_{0} {\mathrm{H}}\underline{k}^{{\phi \mathbb{Z}/2}} \to 0. \end{align*}$$

Since $\pi _n{\mathrm {H}} k=0$ for $n>0$ and since the transfer map $k=k_{\mathbb {Z}/2}\cong \pi _{0}{\mathrm {H}} k_{h \mathbb {Z}/2}\to \pi _{0}{\mathrm {H}} k=k$ is multiplication by $2$ and hence also zero, all the connecting homomorphisms are isomorphisms of k-modules $\pi _n {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}\cong \pi _{n-1}{\mathrm {H}} k_{h \mathbb {Z}/2}$ for $n>0$ . The homotopy groups of the homotopy-orbit spectra are equivalent to group-cohomology, and since k is of characteristic $2$ , the standard resolution

$$\begin{align*}\dots \xrightarrow{0}k \xrightarrow{2}k \xrightarrow{0}k \xrightarrow{2}k\to 0 \end{align*}$$

gives an isomorphism of k-modules $\pi _{n-1}{\mathrm {H}} k_{h \mathbb {Z}/2}\cong H^{n-1}(\mathbb {Z}/2;k)\cong k$ for every $n>0$ . Moreover, again because the transfer map is zero, the canonical map $k=\pi _0 {\mathrm {H}} k\to \pi _0 {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ is an isomorphism of rings. Thus, we have completely identified the $\pi _0 {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ -module structure of $\pi _n {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ .

It finally remains to show that under the isomorphism $k\cong \pi _0 {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ above, the ring map ${{\mathrm {H}} k\to {\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}}$ defining the Frobenius module structure induces the Frobenius $\varphi $ in $\pi _0$ . This follows either from identifying this map with the Tate-valued Frobenius (see [Reference Nikolaus and ScholzeNS18, Example IV.1.2. (i)]) or by the following direct calculation. The counit $\epsilon \colon N^{\mathbb {Z}/2}_e({\mathrm {H}} k)\to {\mathrm {H}}\underline {k}$ induces a map on isotropy separation sequences

where the map $\tau $ is the external norm map. We need to identify the composite $\epsilon ^{\mathbb {Z}/2}\tau $ of the two vertical maps in the middle column. This is the norm of the constant Tambara functor $\underline {k}$ associated to the commutative ring k, and it is therefore the Frobenius $\varphi $ (see also [Reference Dotto, Krause, Nikolaus and PatchkoriaDKNP23, Example 2.18] for an explicit identification of the target of $\tau $ ).

Proof. Let ${\mathrm {H}}\underline {k}$ be the Eilenberg MacLane $C_2$ -spectrum of the ring with trivial involution k. Using the splitting of Lemma 2.2, we obtain from [Reference Dotto, Moi and PatchkoriaDMP24, Lemma 4.3] an equivalence of $C_2$ -spectra

$$\begin{align*}{{\mathrm{THR}}(k)}^{{\phi \mathbb{Z}/2}}\simeq \bigoplus_{n \geq 0} \Sigma^{n\rho} {\mathrm{H}}\underline{k}\otimes_{N_{e}^{C_2}{\mathrm{H}} k}N_{e}^{C_2}{\mathrm{H}}(\varphi^\ast k) \oplus \bigoplus_{ \begin{smallmatrix} (n, m) \\ 0 \leq n < m \end{smallmatrix}} \Sigma^{n+m} {C_2}_{+} \otimes {\mathrm{H}}(\varphi^\ast k\otimes_k\varphi^\ast k).\end{align*}$$

This equivalence is moreover natural in k since the decomposition of ${\mathrm {H}}\underline {k}^{{\phi \mathbb {Z}/2}}$ of Lemma 2.2 is natural. Clearly, $\varphi ^\ast k\otimes _k\varphi ^\ast k=k\otimes _S k$ , and therefore, to obtain the first decomposition of the proposition, it is sufficient to show that the canonical map

$$\begin{align*}{\mathrm{H}}\underline{k}\otimes_{N_{e}^{C_2}{\mathrm{H}} k}N_{e}^{C_2}{\mathrm{H}}(\varphi^\ast k)\longrightarrow {\mathrm{H}}(k\otimes_{k\otimes k}(\varphi^\ast k\otimes \varphi^\ast k))\cong {\mathrm{H}}(k\otimes_{S}k, w) \end{align*}$$

is an equivalence, where the middle term is $\pi _0$ of the underlying spectrum of the left term, with the induced involution.

Let us choose a basis of the k-vector space $\varphi ^\ast k$ ; that is, we write $\varphi ^\ast k$ as a direct sum

$$\begin{align*}\varphi^\ast k\cong \bigoplus_{X}k \end{align*}$$

over some set X. Since the norm commutes with direct sums, we obtain an equivalence of $C_2$ -spectra

$$ \begin{align*} {\mathrm{H}}\underline{k}\otimes_{N_{e}^{C_2}{\mathrm{H}} k}N_{e}^{C_2}{\mathrm{H}}(\varphi^\ast k)&\simeq {\mathrm{H}}\underline{k}\otimes_{N_{e}^{C_2}{\mathrm{H}} k}N_{e}^{C_2}{\mathrm{H}}(\bigoplus_{X}k) \\&\simeq \bigoplus_{X\times X} {\mathrm{H}}\underline{k}\otimes_{N_{e}^{C_2}{\mathrm{H}} k}N_{e}^{C_2}{\mathrm{H}} k\simeq \bigoplus_{X\times X} {\mathrm{H}}\underline{k}, \end{align*} $$

where the last term is the indexed sum of ${\mathrm {H}}\underline {k}$ with the involution on $X\times X$ that swaps the product factors. Under this equivalence, the canonical map above corresponds to the equivalence

$$\begin{align*}\bigoplus_{X\times X} {\mathrm{H}}\underline{k}\simeq {\mathrm{H}}((\bigoplus_{X}k)\otimes_k(\bigoplus_{X}k), w)\simeq {\mathrm{H}}((\varphi^\ast k)\otimes_k(\varphi^\ast k), w)={\mathrm{H}}(k\otimes_Sk, w), \end{align*}$$

where the middle equivalence is the tensor product of two copies of the choice of basis above.

Now let us identify the $C_2$ -fixed points of ${{\mathrm {THR}}(k)}^{{\phi \mathbb {Z}/2}}$ . Notice that ${\mathrm {H}}(k\otimes _Sk, w) $ is a module over ${\mathrm {H}}\underline {k}$ (via the ring map $k\to k\otimes _Sk$ that sends a to $a^2\otimes 1$ ), and therefore, its $C_2$ -fixed-points spectrum is an ${\mathrm {H}} k$ -module. Therefore, it decomposes canonically as a wedge of Eilenberg-MacLane spectra. Its homotopy groups are isomorphic to the Bredon homology groups

$$\begin{align*}\pi_i^{C_2}(\Sigma^{n\rho} {\mathrm{H}}(k\otimes_Sk, w)) =H^{C_2}_i(S^{n\rho}; (k\otimes_Sk, w)), \end{align*}$$

which in turn are the homology groups of the chain complex

$$\begin{align*}0\xleftarrow{}(k\otimes_Sk)^{C_2}\xleftarrow{1+w}k\otimes_Sk\xleftarrow{1+w}k\otimes_Sk\xleftarrow{1+w}\dots \xleftarrow{1+w}k\otimes_Sk\xleftarrow{}0, \end{align*}$$

where the first nonzero group on the left is in degree n and the last nonzero group on the right is in degree $2n$ (notice that all the signs on the arrows are $+$ since k has characteristic $2$ ). It follows that all the groups below n and above $2n$ vanish, that

$$\begin{align*}\pi_{2n}^{C_2}(\Sigma^{n\rho} {\mathrm{H}}(k\otimes_Sk, w))\cong (k\otimes_Sk)^{C_2} \end{align*}$$

and that

$$\begin{align*}\pi_{i}^{C_2}(\Sigma^{n\rho}{\mathrm{H}}(k\otimes_Sk, w))\cong (k\otimes_Sk)^{C_2}/Im(1+w)\xleftarrow{\cong}k\otimes_Sk \end{align*}$$

for every $n\leq i<2n$ , where the left-pointing isomorphism is the map $\phi $ from Lemma 2.1.

Proof. By [Reference Dotto, Moi and PatchkoriaDMP24, Lemma 4.3], the map r vanishes on the summands $(n, m)$ with $n<m$ . By the same lemma, under the identification of Proposition 2.3, it is given on the other summands, for a fixed $n\geq 0$ , by the outer composite of the maps in the diagram

Here, the left map on the top row is the canonical map, and the right map on the top row is the equivalence of the proof of Proposition 2.3. In the right column, the top vertical map is the monoidality of the geometric fixed points, the second map is the diagonal equivalence, and the third one is the splitting induced by the Frobenius module structure. The two bottom horizontal maps are the canonical equivalences.

Let us now consider the top left square. Its right vertical equivalence is given by splitting ${\mathrm {H}}(k\otimes _Sk, w)^{\phi C_2}$ as the sum of its homotopy groups using the ${\mathrm {H}} k$ -module induced by the map $k\to k\otimes _Sk$ as we did in Proposition 2.3 for ${\mathrm {H}}(k\otimes _Sk, w)^{C_2}$ , and then by identifying these homotopy groups with the homology of the chain complex

$$\begin{align*}0\xleftarrow{}(k\otimes_Sk)^{C_2}\xleftarrow{1+w}k\otimes_Sk\xleftarrow{1+w}k\otimes_Sk\xleftarrow{1+w}\cdots\ , \end{align*}$$

where the first nonzero group on the left is in degree zero. The horizontal map on the second row sends the summand $j<n$ to the summand j via $\phi $ , and it maps the last summand to the summand $j=2n$ via the projection map (here, $\phi $ appears because we used it to identify the homotopy groups of the source of the map in the proof of Proposition 2.3). The square commutes by the naturality of the canonical map from fixed points to geometric fixed points.

Thus, the identification of the map r follows once we prove that the equivalence from the bottom left corner of the diagram to the second entry of the second row is the map $\phi $ on homotopy groups. Let $a, b\colon \mathbb {S}\to {\mathrm {H}} k$ , so that the suspension of $a\otimes b\colon \mathbb {S}\to {\mathrm {H}}(k\otimes _S k)$ is a generator of a homotopy group of the bottom left entry of the diagram. The composite of the equivalences up to the top right corner of the diagram sends $a\otimes b$ to the element of the homotopy group represented by $a\otimes N_e^{C_2}(b)$ . The remaining two equivalences send this to the element represented by $b\cdot a\otimes b$ , where the multiplication is with respect to the k-module action on $\varphi ^\ast k$ , and this is precisely $ba^2\otimes b=\phi (a\otimes b)$ .

The identification of f is simpler: by [Reference Dotto, Moi and PatchkoriaDMP24, Lemma 4.3], it is the diagonal on the summands $(n, m)$ with $n<m$ . The identification on the other summands follows from the fact that the restriction map

$$\begin{align*}{\mathrm{res}}^{C_2}_e \colon H^{C_2}_*(S^{n\rho}; (k\otimes_S k, w)) \to H_*(S^{2n};k\otimes_S k) \end{align*}$$

is the inclusion of fixed points in degree $*=2n$ , and zero otherwise.

Proof. By Proposition 2.4, the map $r-f$ is an isomorphism in $\pi _\ast $ when restricted and corestricted to the summands $(n, m)$ with $n\neq m$ . It is therefore an isomorphism in odd degrees, and its long exact sequence decomposes into exact sequences

$$\begin{align*}0\to\pi_{2l} {\mathrm{TCR}}(k;2)^{\phi\mathbb{Z}/2}\to (k\otimes_S k)^{C_2}\oplus\bigoplus_{\begin{smallmatrix}(n, m)\\ n+m=2l \\ n, m \geq 0\\ n\neq m\end{smallmatrix}} k\otimes_S k\xrightarrow{r-f} \bigoplus_{\begin{smallmatrix}(n, m)\\ n+m=2l\\ n, m \geq 0\end{smallmatrix}} k\otimes_S k\to \pi_{2l-1} {\mathrm{TCR}}(k;2)^{\phi\mathbb{Z}/2}\to 0 \end{align*}$$

for every $l\geq 0$ . Again by Proposition 2.4, the kernel and cokernel of $r-f$ are the same as those of

$$\begin{align*}\iota-\phi^{-1}\pi\colon (k\otimes_S k)^{C_2}\longrightarrow k\otimes_S k,\end{align*}$$

where $\iota $ is the fixed-points inclusion. These are respectively isomorphic to the kernel and cokernel of $\pi -\phi $ , by applying the isomorphism $\phi $ of Lemma 2.1 to the target.

In [Reference KatoKat82], Kato exhibits an exact sequence

$$\begin{align*}0\to {\mathrm{W}}^s(k)\to k\otimes_S k\xrightarrow{\pi-\phi}(k\otimes_S k)/Im(1+w)\to {\mathrm{W}}^q(k)\to 0, \end{align*}$$

where ${\mathrm {W}}^s(k)$ and ${\mathrm {W}}^q(k)$ are respectively the symmetric and quadratic Witt groups of k. It is easy to see that the kernel and cokernel of $\pi -\phi $ agree with those above, by restricting and corestricting the maps to the fixed points.

Proof. By [Reference Dotto, Moi and PatchkoriaDMP24, Theorem 2.7] and §1, for every $l\geq 1$ , there is a pullback square of ring spectra

where $\sigma _l$ denotes the action of the generator of the Weyl group of $\mathbb {Z}/2$ in $D_{2^{l}}/C_{2^{l-1}}$ , which is of order $2$ . We prove, by induction on l, that the connecting homomorphism in the Mayer-Vietoris long exact sequence of this pullback square vanish, and therefore that the square gives a pullback square of homotopy groups. One can see, again by induction, that these pullbacks of homotopy groups indeed match the description of the homotopy groups of the Theorem. However, we will need to prove the vanishing of the connecting maps and the explicit description of the pullback in the same induction step.

For $l=1$ , the pullback above describing ${{\mathrm {TRR}}^2(k;2)}^{{\phi \mathbb {Z}/2}}$ is equivalent to the pullback

$$\begin{align*}({\mathrm{THR}}(k)^{\phi\mathbb{Z}/2})^{C_2}{\times_{{\mathrm{THR}}(k)^{\phi\mathbb{Z}/2}}} ({\mathrm{THR}}(k)^{\phi\mathbb{Z}/2})^{C_2}\end{align*}$$

along the maps r and $\sigma _1 r$ (since the right vertical map in the square above is the diagonal for $l=1$ ). By the characterisation of r of Proposition 2.4, the map $r-\sigma _1r$ in the corresponding Mayer-Vietoris sequence is surjective in every degree. Therefore, there is a pullback

$$\begin{align*}\pi_\ast {{\mathrm{TRR}}^2(k;2)}^{{\phi \mathbb{Z}/2}}\cong \big(\big(\bigoplus_{\begin{smallmatrix} (n, m), &\!\!\!\!n, m \geq 0 \\ n\neq m, &\!\!\!\! n+m=\ast \end{smallmatrix}} k\otimes_S k\big)\oplus (k\otimes_S k)^{C_2})\big) \ {}_{r}\times_{\sigma_1 r} \big(\big(\bigoplus_{\begin{smallmatrix} (n, m), &\!\!\!\!n, m \geq 0 \\ n\neq m, &\!\!\!\! n+m=\ast \end{smallmatrix}} k\otimes_S k\big)\oplus (k\otimes_S k)^{C_2})\big) \end{align*}$$

in even degrees and an analogous pullback without the summands $(k\otimes _S k)^{C_2}$ in odd degrees. Here, the subscripts of the product indicate which maps we are pulling back along. By the description of r from Proposition 2.4, this is isomorphic to the pullback of the statement of the Theorem. The characterisation of the maps $R, F$ and of the Weyl action follows by the description of the corresponding maps of [Reference Dotto, Moi and PatchkoriaDMP24, Theorem 2.7].

Now let $l \geq 2$ , and suppose that the decomposition above holds for $\pi _\ast {{\mathrm {TRR}}^h(k;2)}^{{\phi \mathbb {Z}/2}}$ for all $h \leq l$ , and that the maps $R, F\colon {{\mathrm {TRR}}^{h}(k;2)}^{{\phi \mathbb {Z}/2}}\to {{\mathrm {TRR}}^{h-1}(k;2)}^{{\phi \mathbb {Z}/2}}$ and $\sigma _h$ are given in homotopy groups by the formulas of the Theorem. We will show that the same holds for $l+1$ . The Mayer-Vietoris sequence of the pullback square above is then (we recall that $\sigma _1F =F$ )

for $\ast $ even, and a similar expression without the fixed points terms for $\ast $ odd. An argument completely analogous to that of the proof of [Reference Dotto, Moi and PatchkoriaDMP24, Theorem 4.7] shows that the bottom vertical map is surjective and identifies its kernel with the formula of the Theorem. The description of the maps R and F also follows by a similar argument.

Proof. Let us first show that the proposed generators belong to $\phi ^{n}\big ((k\otimes _S k)^{C_2}\big )$ . For the first, we see that for every $1\leq j\leq n$ , we have that

$$\begin{align*}(\phi^{-1}\pi)^{j}(\phi^{n+1}(a\otimes b))=\phi^{n+1-j}(a\otimes b), \end{align*}$$

which belongs to $(k\otimes _S k)^{C_2}$ since $n+1-j\geq 1$ . However, for all $0\leq i\leq n-1$ , we have that

$$\begin{align*}(\phi^{-1}\pi)^{j}(\phi^{i}(a\otimes b)+\phi^i(b\otimes a))=\phi^{i-j}(a\otimes b)+\phi^{i-j}(b\otimes a) \end{align*}$$

if $0\leq j\leq i$ , which is a fixed point, and for $i<j\leq n$ , this is

$$\begin{align*}(\phi^{-1}\pi)^{j}(\phi^{i}(a\otimes b)+\phi^i(b\otimes a))=(\phi^{-1}\pi)^{j-i}(a\otimes b+b\otimes a)=0 \end{align*}$$

since $\pi $ quotients off the image of $1+w$ .

The proof that these elements generate $\phi ^{n}\big ((k\otimes _S k)^{C_2}\big )$ is by induction on n. For $n=0$ , consider the exact sequence

$$\begin{align*}k\otimes_S k\xrightarrow{1+w}(k\otimes_S k)^{C_2}\longrightarrow (k\otimes_S k)^{C_2}/Im(1+w)\to 0. \end{align*}$$

By Lemma 2.1, the right term is generated by the equivalence classes of the elements of the form $\phi (a\otimes b)$ , and the image of $1+w$ is generated by the elements of the form $a\otimes b+b\otimes a$ , which proves the claim.

Now suppose that the claim holds for $n-1$ , and consider the exact sequence

$$\begin{align*}k\otimes_S k\xrightarrow{1+w} \phi^{n}((k\otimes_S k)^{C_2})\longrightarrow \phi^{n}((k\otimes_S k)^{C_2})/Im(1+w)\to 0. \end{align*}$$

By an argument analogous to the proof of Lemma 2.1, $\phi $ defines an isomorphism between $\phi ^{n-1}((k\otimes _S k)^{C_2})$ and $\phi ^{n}((k\otimes _S k)^{C_2})/Im(1+w)$ . Thus, by the inductive assumption, the classes of $\phi ^{n+1}(a\otimes b)$ and $\phi ^{i}(a\otimes b)+\phi ^i(b\otimes a)$ , for $1\leq i\leq n$ and $a\otimes b\in k\otimes _S k$ , generate the quotient. The image of $1+w$ is generated by the elements of the form $a\otimes b+b\otimes a$ , which concludes the induction.

The proof for $\pi _0\ {\mathrm {TRR}}^{n+1}(k;2)^{\phi \mathbb {Z}/2}$ is completely analogous, by induction on the exact sequences

Proof. Let us first suppose $h>0$ . We need to show that the unique map in the bottom row of the commutative square

agrees with $V^{l-h}$ , where the vertical maps are the isomorphisms of Theorem 2.7 and Remark 2.8. By the double coset formula of the $D_{2^l}$ -Mackey functor $\underline {\pi }_0{\mathrm {THR}}(k)$ , the upper composite has first component

$$ \begin{align*} F^l{\mathrm{tran}}^{D_{2^l}}_{D_{2^{h}}}&={\mathrm{res}}^{D_{2^l}}_{\mathbb{Z}/2}{\mathrm{tran}}^{D_{2^l}}_{D_{2^{h}}}=\sum_{g\in \mathbb{Z}/2/D_{2^l}/D_{2^h}}{\mathrm{tran}}^{\mathbb{Z}/2}_{^gD_{2^{h}}\cap \mathbb{Z}/2} c_g{\mathrm{res}}^{D_{2^{h}}}_{D_{2^{h}}\cap \mathbb{Z}/2^g}. \end{align*} $$

The double coset $\mathbb {Z}/2/D_{2^l}/D_{2^h}$ is the quotient of the cyclic group $C_{2^{l-h}}$ by the involution which acts by inversion. It therefore consists of two fixed points (the unit and the rotation $g_0$ of order $2$ in $D_l$ ) which conjugate $\mathbb {Z}/2$ to itself, and $(2^{l-h}-2)/2$ points whose corresponding intersection $D_{2^{h}}\cap \mathbb {Z}/2^g$ is trivial. Thus,

$$ \begin{align*} F^l{\mathrm{tran}}^{D_{2^l}}_{D_{2^{h}}}&={\mathrm{res}}^{D_{2^{h}}}_{\mathbb{Z}/2}+c_{g_0} {\mathrm{res}}^{D_{2^{h}}}_{\mathbb{Z}/2}+\sum_{1, g_0\neq g\in \mathbb{Z}/2/D_{2^l}/D_{2^h}}{\mathrm{tran}}^{\mathbb{Z}/2}_{e} c_g{\mathrm{res}}^{D_{2^{h}}}_{e} \\&={\mathrm{res}}^{D_{2^{h}}}_{\mathbb{Z}/2}+{\mathrm{res}}^{D_{2^{h}}}_{\mathbb{Z}/2}c_{g_0} =F^h+ F^h\sigma, \end{align*} $$

where the transfer ${\mathrm {tran}}^{\mathbb {Z}/2}_{e}$ is zero since k has characteristic $2$ (see [Reference Dotto, Moi, Patchkoria and ReehDMPR21, Theorem 5.1]), $\sigma $ is the action of the Weyl group of $D_h$ in $D_l$ , and the equality is regarded as elements of $k\otimes _S k$ . The map $F^h$ is determined in Theorem 2.7: it sends an element in the upper left corner of the square, corresponding to $(x, y)$ in the bottom left corner, to x. Thus, the unique bottom horizontal map in the square above sends $(x, y)$ to the pair with first component $x+y$ .

Now let $\mathbb {Z}/2^{\prime }$ be the subgroup of $D_{l}$ generated by a reflection non-conjugate to $\mathbb {Z}/2$ . Similarly to the calculation above, the second component of the top composite is

$$ \begin{align*} F^l\sigma{\mathrm{tran}}^{D_{2^l}}_{D_{2^{h}}}&={\mathrm{res}}^{D_{2^l}}_{\mathbb{Z}/2^{\prime}}{\mathrm{tran}}^{D_{2^l}}_{D_{2^{h}}}=\sum_{g\in \mathbb{Z}/2^{\prime}/D_{2^l}/D_{2^h}}{\mathrm{tran}}^{\mathbb{Z}/2}_{gD_{2^{h}}\cap \mathbb{Z}/2} c_g{\mathrm{res}}^{D_{2^{h}}}_{D_{2^{h}}\cap \mathbb{Z}/2^g}. \end{align*} $$

Now the double coset $\mathbb {Z}/2^{\prime }/D_{2^l}/D_{2^h}$ is the quotient of the cyclic group $C_{2^{l-h}}$ by the free involution, and none of the conjugates of $\mathbb {Z}/2$ is contained in $D_{2^h}$ . Thus,

$$ \begin{align*} F^l\sigma{\mathrm{tran}}^{D_{2^l}}_{D_{2^{h}}}&=\sum_{g\in \mathbb{Z}/2^{\prime}/D_{2^l}/D_{2^h}}{\mathrm{tran}}^{\mathbb{Z}/2}_{e} c_g{\mathrm{res}}^{D_{2^{h}}}_{e}=0, \end{align*} $$

again since ${\mathrm {tran}}^{\mathbb {Z}/2}_{e}=0$ . Thus, the second component of the bottom horizontal map is null as claimed.

The proof of the case $h=0$ is similar, by calculating the upper composite of the diagram

where the left vertical map is the isomorphism of [Reference Dotto, Moi, Patchkoria and ReehDMPR21, Theorem 5.1].

Proof. The identification of $N^n(a\otimes b)$ is similar to the proof of Proposition 2.10. It is sufficient to show that the unique map in the bottom row of the commutative square

agrees with $N^n$ on the elementary tensors $a\otimes b$ . Indeed, the value on a general element in the tensor product is determined by iterations of the relation

$$\begin{align*}N(x+y)=N(x)+N(y)+V(xy). \end{align*}$$

By the multiplicative double coset formula of the $D_{2^l}$ -Tambara functor $\underline {\pi }_0{\mathrm {THR}}(k)$ , the upper composite has first component

$$ \begin{align*} F^nN^{D_{2^n}}_{\mathbb{Z}/2}&={\mathrm{res}}^{D_{2^n}}_{\mathbb{Z}/2}N^{D_{2^n}}_{\mathbb{Z}/2}=\prod_{g\in \mathbb{Z}/2/D_{2^n}/\mathbb{Z}/2}N^{\mathbb{Z}/2}_{^g\mathbb{Z}/2\cap \mathbb{Z}/2} c_g{\mathrm{res}}^{\mathbb{Z}/2}_{\mathbb{Z}/2\cap \mathbb{Z}/2^g}. \end{align*} $$

The double coset $\mathbb {Z}/2/D_{2^n}/\mathbb {Z}/2$ is the quotient of the cyclic group $C_{2^{n}}$ by the involution which acts by inversion, and consists of two fixed points (the unit and the rotation $g_0$ of order $2$ in $D_n$ ) which conjugate $\mathbb {Z}/2$ to itself, and $(2^{n}-2)/2$ points whose corresponding intersection $\mathbb {Z}/2\cap \mathbb {Z}/2^g$ is trivial. Moreover, since the cyclic group acts trivially on $\pi _0{\mathrm {THH}}(k)=k$ , the conjugation $c_g$ is trivial except for $g=g_0$ . Thus,

$$ \begin{align*} F^nN^{D_{2^n}}_{\mathbb{Z}/2}&=({\mathrm{id}})\cdot (c_{g_0})\cdot(N^{\mathbb{Z}/2}_{e}{\mathrm{res}}^{\mathbb{Z}/2}_e)^{2^{n-1}-1}. \end{align*} $$

Since the restriction map ${\mathrm {res}}^{\mathbb {Z}/2}_e$ corresponds to the multiplication map $\mu \colon k\otimes _S k\to k$ and $N^{\mathbb {Z}/2}_{e}$ to the map $k\to k\otimes _Sk$ which sends a to $a^2\otimes 1$ (see [Reference Dotto, Moi, Patchkoria and ReehDMPR21, Corollary 5.2]), this sends $a\otimes b$ to

$$\begin{align*}a\otimes b\cdot b\otimes a\cdot ((ab)^2\otimes 1)^{2^{n-1}-1}=(ab)^{2^{n}-1}\otimes ab=\phi_n(1\otimes ab). \end{align*}$$

Similarly, by letting $\mathbb {Z}/2^{\prime }$ be the subgroup of $D_{n}$ generated by a reflection non-conjugate to $\mathbb {Z}/2$ , the second component of the upper composite in the square above is

Now the double coset $\mathbb {Z}/2^{\prime }/D_{2^n}/\mathbb {Z}/2$ is the quotient of the cyclic group $C_{2^{n}}$ by the free involution, and since $\mathbb {Z}/2$ and $\mathbb {Z}/2^{\prime }$ are not conjugate,

$$ \begin{align*} F^n\sigma N^{D_{2^n}}_{\mathbb{Z}/2}&=(N^{\mathbb{Z}/2^{\prime}}_{e}{\mathrm{res}}^{\mathbb{Z}/2}_{e})^{2^{n-1}}. \end{align*} $$

Thus, the second component of the bottom horizontal map of the square sends $a\otimes b$ to

$$\begin{align*}((ab)^2)^{2^{n-1}}\otimes 1=\phi_n(ab\otimes 1). \end{align*}$$

This identifies the map $N^n$ as claimed. Finally, observe that

$$ \begin{align*} N^n(a\otimes 1)\cdot \sigma N^n(b\otimes 1)&=(a^{2^{n}}\otimes 1, a^{2^{n}-1}\otimes a)\cdot (b^{2^{n}-1}\otimes b, b^{2^{n}}\otimes 1) \\&=(a^{2^{n}}b^{2^{n}-1}\otimes b, a^{2^{n}-1}b^{2^{n}}\otimes a)=\tau_n(a\otimes b).\\[-42pt] \end{align*} $$

Proof. Since $\underline {\pi _0}\ {\mathrm {THR}}(k)$ is a $D_{2^n}$ -Tambara functor, the restriction ${\mathrm {res}}^{D_{2^n}}_{C_{2^n}}$ is a ring homomorphism, and by the double-coset formulas, it commutes with norms and transfers, and with the Weyl action. The operators $V^{n-i}$ and $\tau _i$ are described in terms of norms and transfers by Propositions 2.10 and 2.12, and by [Reference Hesselholt and MadsenHM97, Theorem 3.3] for the usual Witt vectors. It therefore follows that

$$\begin{align*}{\mathrm{res}}^{D_{2^n}}_{C_{2^n}}\sigma V^{n-i}\tau_i(a\otimes b)=[w V^{n-i}\tau_i{\mathrm{res}}^{\mathbb{Z}/2}_{e}(a\otimes b)]=[V^{n-i}\tau_{i}(ab)], \end{align*}$$

where ${\mathrm {res}}^{\mathbb {Z}/2}_{e}$ is the multiplication map of $k\otimes _S k$ by [Reference Dotto, Moi, Patchkoria and ReehDMPR21, Theorem 5.1]. The proof for the other generators is similar.

Proof. The proof is by induction on n. For $n=0$ , this is the claim that the kernel of the multiplication map

$$\begin{align*}\mu\colon k\otimes_Sk\longrightarrow k \end{align*}$$

is generated by $a\otimes b+ab\otimes 1$ for $a\otimes b\in k\otimes _S k$ , which is clear.

Now suppose the claim holds for n, and consider the commutative diagram with exact rows

where the vertical maps from the top row to the middle row are kernel inclusions. The middle row is exact by [Reference Dotto, Moi and PatchkoriaDMP24, Proof of Theorem 4.9], or by the explicit calculation of Theorem 2.7 and Proposition 2.10. Thus, if we find a set of generators for K and show that the map R in the first row is surjective, then $J_{\langle 2^{n+1}\rangle }$ is generated by the image by $V^{n+1}+\sigma V^{n+1}$ of the generators of K, and by a choice of lifts of the generators of $J_{\langle 2^{n}\rangle }$ given by the inductive assumption. The kernel K consists of those elements $(x, y)$ such that $\mu (x)+\mu (y)$ is a square in k. Since every square $c^2$ in k is hit by $c\otimes c$ under $\mu $ , K is the subgroup of elements of the form $(x, y)$ , where

$$\begin{align*}x=y+c\otimes c+z \end{align*}$$

for some $c\in k$ and $z\in \ker (\mu )$ . Since the kernel of $\mu $ is generated by elements of the form $a\otimes b+ab\otimes 1$ , we conclude that K is generated by elements of the form $(a\otimes b, a\otimes b)$ , and elements of the form $(a\otimes b+ab\otimes 1+c\otimes c, 0)$ . The images of these generators by $V^{n+1}+\sigma V^{n+1}$ are respectively of the form

$$\begin{align*}V^{n+1}(a\otimes b)+\sigma V^{n+1}(a\otimes b) \end{align*}$$

and

$$\begin{align*}V^{n+1}(a\otimes b+ab\otimes 1+c\otimes c)+\sigma V^{n+1}(0)=V^{n+1}(a\otimes b)+V^{n+1}(ab\otimes 1)+V^{n+1}(c\otimes c), \end{align*}$$

where $V^{n+1}(c\otimes c)=0$ by Proposition 2.10. It therefore remains to show that by applying R to the elements $\tau _{n+1}(a\otimes b)+\tau _{n+1}(ab\otimes 1)$ , $V^{n+1-i}\tau _i(a\otimes b)+\sigma V^{n+1-i}\tau _i(a\otimes b)$ and $V^{n+1-i}\tau _i(a\otimes b)+V^{n+1-i}\tau _i(ab\otimes 1)$ , for $1\leq i\leq n$ , we hit all the generators of $J_{\langle 2^{n}\rangle }$ given by the inductive assumption. This is the case since $R\tau _{i+1}=\tau _i$ , $RV^{n+1-i}=V^{n-i}R$ , and $R\sigma =\sigma R$ , by Theorem 2.7 and Proposition 2.10.

Proof. By Proposition 2.16, the corollary follows from the identities

$$ \begin{align*} \tau_n(a\otimes b)+\tau_n(ab\otimes 1)&=\tau_{n}(a\otimes 1)\cdot(\tau_n(1\otimes b)+\tau_n(b\otimes 1)), \\ V^{n-i}\tau_i(a\otimes b)+V^{n-i}\tau_i(ab\otimes 1)&=(V^{n-i}\tau_i(a\otimes 1)+\sigma V^{n-i}\tau_i(b\otimes 1))\cdot (V^{n-i}\tau_i(b\otimes 1)+\sigma V^{n-i}\tau_i(b\otimes 1)), \end{align*} $$

which can be easily verified from the definitions.

Proof. The claim about $R, F$ and $\sigma $ are clear since these maps are multiplicative. For the map V, we employ Corollary 2.17. First suppose that $n\geq 1$ , so that the power $J_{\langle 2^{n}\rangle }^q$ is additively generated by elements of the form

$$\begin{align*}(x, y)\cdot (x_1, x_1)\cdot\dots\cdot (x_q, x_q)=(xx_1\dots x_q, yx_1\dots x_q), \end{align*}$$

where $(x, y)$ is a generator of $\pi _0\ {\mathrm {TRR}}^{n+1}(k;2)^{\phi \mathbb {Z}/2}$ from Proposition 2.9, and $(x_l, x_l)$ is a generator of $J_{\langle 2^{n}\rangle }$ from Corollary 2.17, which is diagonal since they are invariant by the Weyl action $\sigma $ . Since V is additive, it is sufficient to show that V sends these elements to $J_{\langle 2^{n+1}\rangle }^q $ . Now by Proposition 2.10, we have that

$$ \begin{align*} V(xx_1\dots x_q, yx_1\dots x_q)&=((x+y)x_1\dots x_q, 0) \\&=(x+y, x+y)\cdot (x_1, x_1)\cdot\dots\cdot (x_{q-2}, x_{q-2})\cdot (x_{q-1}, x_{q-1})\cdot(x_q, 0). \end{align*} $$

The first factor is

$$\begin{align*}(x+y, x+y)=V(x, y)+\sigma V(x, y), \ \ \ \ \ \end{align*}$$

which belongs to $J_{\langle 2^{n+1}\rangle }$ since it is sent to zero by the restriction map by Proposition 2.14. By Corollary 2.17, each of the factors $(x_1, x_1), \dots , (x_{q-1}, x_{q-1})$ is of the form

$$\begin{align*}V^{n-i}\tau_i(a\otimes b)+\sigma V^{n-i}\tau_i(a\otimes b) \end{align*}$$

for some $0\leq i\leq n$ , which as an element of $\pi _0\ {\mathrm {TRR}}^{n+2}(k;2)^{\phi \mathbb {Z}/2}$ is of the form

$$\begin{align*}V^{n+1-i}\tau_i(a\otimes b)+\sigma V^{n+1-i}\tau_i(a\otimes b) \end{align*}$$

for $0\leq i\leq n$ , and therefore belongs to $J_{\langle 2^{n+1}\rangle }$ . Thus, it suffices to show that $(x_{q-1}, x_{q-1})\cdot (x_q, 0)$ is also in $J_{\langle 2^{n+1}\rangle }$ . But since $(x_q, x_q)$ is of the form $V^{n-i}\tau _i(a\otimes b)+\sigma V^{n-i}\tau _i(a\otimes b)$ , we have that $(x_q, 0)$ is a well-defined element of $\pi _0\ {\mathrm {TRR}}^{n+2}(k;2)^{\phi \mathbb {Z}/2}$ , and since $J_{\langle 2^{n+1}\rangle }$ is an ideal, $ (x_{q-1}, x_{q-1})\cdot (x_q, 0)$ indeed belongs to $J_{\langle 2^{n+1}\rangle }$ .

If $n=0$ , the ideal $J_{\langle 1\rangle }^q$ of $k\otimes _Sk$ is additively generated by elements of the form $xx_1\dots x_q$ with $x\in k\otimes _S k$ and $x_1, \dots , x_q$ fixed by the involution w. Then by Proposition 2.10,

$$\begin{align*}V(xx_1\dots x_q)=(xx_1\dots x_q+w(xx_1\dots x_q), 0)=((x+w(x))x_1\dots x_q, 0), \end{align*}$$

and one can repeat the argument used in the case $n\geq 1$ .

Finally, let us show that $1+\sigma $ sends $J_{\langle 2^{n}\rangle }^q$ to $J_{\langle 2^{n}\rangle }^{q+1}$ . By Corollary 2.17, every element of $J_{\langle 2^{n}\rangle }^q $ is a sum of elements of the form $z\cdot g_1\cdot \dots \cdot g_q$ with $z\in \pi _0\ {\mathrm {TRR}}^{n+1}(k;2)^{\phi \mathbb {Z}/2}$ and each $g_i\in J_{\langle 2^{n}\rangle }$ fixed by $\sigma $ . Since $1+\sigma $ is additive, we only need to show that these elements are sent to $J_{\langle 2^{n}\rangle }^{q+1}$ . Since $\sigma $ is multiplicative, we have that

$$\begin{align*}(1+\sigma)(z\cdot g_1\cdot\dots\cdot g_q)=z\cdot g_1\cdot\dots\cdot g_q+\sigma(z)\cdot \sigma(g_1)\cdot\dots\cdot \sigma(g_q)=(z+\sigma(z))\cdot g_1\cdot\dots\cdot g_q. \end{align*}$$

It therefore suffices to show that $z+\sigma (z)$ belongs to $J_{\langle 2^{n}\rangle }$ , which is the case since the restriction map to the Witt vectors modulo $2$ is invariant under the action of $\sigma $ . Clearly, since $\sigma ^2$ is the identity, we have that $(1+\sigma )^2=0$ .

Proof. First of all, the maps $R, F, V$ and $ d:=(1+\sigma )$ are well defined on the quotients of the powers of the ideals by Proposition 3.1. Axioms i)–iv) of Definition 3.2 follow immediately from either the fact that $F, V$ and ${\mathrm {res}}^{D_{2^n}}_{C_{2^n}}$ are induced from the maps of a Mackey functor, or from their explicit formulas from Theorem 2.7 and Propositions 2.10 and 2.14. This is except from the identity $FV=2$ (which in our case is zero), since by these arguments, we only know that $FV=1+\sigma $ . However, for every $x\in J_{\langle 2^{n}\rangle }^q$ , we have that

$$\begin{align*}FV(x)=x+\sigma(x) \end{align*}$$

belongs to $J_{\langle 2^{n}\rangle }^{q+1}$ by Proposition 3.1, and it is therefore indeed zero in $J_{\langle 2^{n}\rangle }^q/J_{\langle 2^{n}\rangle }^{q+1}$ .

Let us show axiom v). To see that d satisfies the Leibniz rule, let $x\in J_{\langle 2^{n}\rangle }^q/J_{\langle 2^{n}\rangle }^{q+1}$ and $y\in J_{\langle 2^{n}\rangle }^{q^{\prime }}/J_{\langle 2^{n}\rangle }^{q^{\prime }+1}$ , and let us calculate

$$ \begin{align*} d(xy)+d(x) y+xd(y)&=xy+\sigma(x)\sigma(y)+(x+\sigma(x)) y+x(y+\sigma(y))\\ &=xy+\sigma(x)\sigma(y)+\sigma(x)y+x\sigma(y)=(x+\sigma(x)) (y+\sigma(y))\\ &=d(x)d(y). \end{align*} $$

Since $d(x)$ belongs to $J_{\langle 2^{n}\rangle }^{q+1}$ and $d(y)$ to $J_{\langle 2^{n}\rangle }^{q^{\prime }+1}$ by Proposition 3.1, we have that $d(x)d(y)$ belongs to $J_{\langle 2^{n}\rangle }^{q+q^{\prime }+2}$ , and therefore it vanishes in $J_{\langle 2^{n}\rangle }^{q+q^{\prime }+1}/J_{\langle 2^{n}\rangle }^{q+q^{\prime }+2}$ .

Let us now verify the last three identities involving d in axiom v). For the first one, let $x\in J_{\langle 2^{n}\rangle }^q/J_{\langle 2^{n}\rangle }^{q+1}$ . Then

$$\begin{align*}FdV(x)=FV(x)+F\sigma V(x)=FV(x)=(1+\sigma)(x)=d(x)\end{align*}$$

in $J_{\langle 2^{n}\rangle }^{q+1}/J_{\langle 2^{n}\rangle }^{q+2}$ , where $F\sigma V(x)=0$ by the double coset formula (or by direct calculation). For the second identity, we have that

$$\begin{align*}d^2=(1+\sigma)^2=1+2\sigma+\sigma^2=2+2\sigma =0 \end{align*}$$

since $\sigma $ has order $2$ . Finally, for the third one, let $a\in k={\mathrm {W}}_{\langle 1\rangle }(k)$ . On the one hand, by Proposition 2.14,

$$ \begin{align*} Fd\lambda\tau_n(a)&=Fd\tau_n(a\otimes 1)=F(\tau_n(a\otimes 1)+\sigma \tau_n(a\otimes 1)) \\&=\tau_n(a\otimes 1)+\sigma \tau_n(a\otimes 1)=(a^{2^{n}}\otimes 1+a^{2^{n}-1}\otimes a, a^{2^{n}-1}\otimes a+a^{2^{n}}\otimes 1), \end{align*} $$

where the third equality holds by the formula for F of Theorem 2.7. On the other hand,

$$ \begin{align*} &(\lambda\tau_{n-1}(a))\cdot (d\lambda\tau_{n-1}(a))=\tau_{n-1}(a\otimes 1)\cdot (\tau_{n-1}(a\otimes 1)+\sigma\tau_{n-1}(a\otimes 1)) \\&=(a^{2^{n-1}}\otimes 1, a^{2^{n-1}-1}\otimes a)\cdot (a^{2^{n-1}}\otimes 1+a^{2^{n-1}-1}\otimes a, a^{2^{n-1}-1}\otimes a+a^{2^{n-1}}\otimes 1) \\&=(a^{2^{n}}\otimes 1+a^{2^{n}-1}\otimes a, a^{2^{n}-2}\otimes a^2+a^{2^{n}-1}\otimes a), \end{align*} $$

and these are equal since we are tensoring over S.

Proof. For every $a\in k$ , let us denote $\Delta (a):=1\otimes a+a\otimes 1\in k\otimes _S k$ . We note that the map u is necessarily given by the formula

$$\begin{align*}u(ada_1\dots da_q):=a\Delta(a_1)\cdot \dots\cdot \Delta(a_q).\end{align*}$$

In order to show that this is an isomorphism, we choose suitable bases of the source and target as k-vector spaces. Let $\{x_{i}\}_{i\in I}$ be a $2$ -basis of k. We recall that this is a set of elements of k whose differentials $\{dx_i\}_{i\in I}$ form a basis of the k-vector space $\Omega ^1_k$ or, equivalently, such that the elements

$$\begin{align*}x^\xi:=\prod_{i\in \xi}x_i \end{align*}$$

form a basis of k as an S-vector space, where $\xi $ ranges through the finite subsets of I (see [Reference GrothendieckGro67, Chapter 0, §21.4]). Here, we use the convention that $x^\emptyset =1$ , and we will write $\xi \subset ^f \! I$ if $\xi $ is a finite subset of I. It is easy to see that the set $\{1\otimes x^\xi \}_{\xi \subset ^f I}$ is a basis of $k\otimes _S k$ as a k-vector space, where k acts by multiplication on the left tensor factor. Now let us denote

$$\begin{align*}\Delta(x)^{\xi}:=\prod_{i\in \xi}\Delta(x_i) \end{align*}$$

for every $\xi \subset ^f\! I$ (with the convention that $\Delta (x)^{\emptyset }=1\otimes 1$ ). These elements satisfy the identities

(3.1) $$ \begin{align} \Delta(x)^{\xi}&=\sum_{\nu\subset\xi}x^{\xi\setminus\nu}\cdot(1\otimes x^\nu) \end{align} $$

(3.2) $$ \begin{align} 1\otimes x^{\xi}&=\sum_{\nu\subset\xi}x^{\xi\setminus\nu}\cdot\Delta(x)^{\nu} \end{align} $$

for every finite subset $\xi $ of I. It follows that $\{\Delta (x)^{\xi }\}_{\xi \subset ^f I}$ is also a basis of $k\otimes _S k$ . Since the multiplication map sends $\Delta (x)^{\xi }$ to a nonzero element of k if and only if $\xi =\emptyset $ , it follows that $\{\Delta (x)^{\xi }\}_{\emptyset \neq \xi \subset ^f I}$ is a basis for $J_{\langle 1\rangle }$ as a k-vector space with respect to multiplication on the left factor. It then readily follows that the elements $\Delta (x)^{\xi }$ with $|\xi |\geq q$ form a basis of $J_{\langle 1\rangle }^q$ , and that the elements $\Delta (x)^{\xi }$ with $|\xi |= q$ form a basis of $J_{\langle 1\rangle }^q/J_{\langle 1\rangle }^{q+1}$ . Since for every $\xi \subset I$ with $|\xi |=q$ , the map u sends a basis element $(dx)^\xi :=\prod _{i\in \xi }dx_i$ of $\Omega ^q_k$ to $\Delta (x)^{\xi }$ , the claim follows.

Proof. Let us write $\mu _\nu \in k$ uniquely as a linear combination $\mu _\nu =\sum _{\delta \subset ^f I}s_{\nu , \delta }^2x^\delta $ with respect to the basis $\{x^\delta \}_{\delta \subset ^f I}$ of k as an S-vector space. Let us notice that, since we are tensoring over S, the map $\Delta \colon k\to k\otimes _Sk$ is S-linear. Then by applying formula (3.2),

$$ \begin{align*} \Delta(\mu_\nu)&=\sum_{\delta\subset^f I}s_{\nu, \delta}^2\Delta(x^\delta)=\sum_{\delta\subset^f I}s_{\nu, \delta}^2(1\otimes x^\delta+x^{\delta}\cdot (1\otimes 1)) \\ &=\sum_{\delta\subset^f I}s_{\nu, \delta}^2\sum_{\gamma\subset\delta}x^{\delta\setminus \gamma}\Delta(x)^\gamma+\sum_{\delta\subset^f I}s_{\nu, \delta}^2x^{\delta}\cdot \Delta(x)^{\emptyset} =\sum_{\emptyset\neq \gamma\subset\delta\subset^f I}s_{\nu, \delta}^2x^{\delta\setminus \gamma}\Delta(x)^\gamma, \end{align*} $$

where the last equality holds since the sum $\sum _{\delta \subset ^f I}s_{\nu , \delta }^2x^{\delta }\cdot \Delta (x)^{\emptyset }$ is equal to the term $\gamma =\emptyset $ in the previous sum. It follows that

$$ \begin{align*} \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} }\Delta(\mu_\nu)\Delta(x)^{\nu} &=\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } (\sum_{\emptyset\neq \gamma\subset\delta^f\subset I}s_{\nu, \delta}^2x^{\delta\setminus \gamma}\Delta(x)^\gamma)\Delta(x)^{\nu} =\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{ \begin{smallmatrix} \emptyset\neq \gamma\subset\delta\subset^f I \\ \gamma\cap\nu=\emptyset \end{smallmatrix} }s_{\nu, \delta}^2x^{\delta\setminus \gamma}\Delta(x)^{\gamma\amalg\nu}, \end{align*} $$

where in the last sum $\gamma $ and $\nu $ are disjoint, since $\Delta (a)\Delta (a)=0$ in $k\otimes _Sk$ for all $a\in k$ . Since, by assumption, this element belongs to $J_{\langle 1 \rangle }^{q+1}$ , and the $\Delta (x)^\xi $ with $|\xi |=q+1$ are a basis for $ J_{\langle 1 \rangle }^{q+1}$ , we must have that, for every $\nu \subset I$ with $|\nu |=q-1$ and every $j\in I\setminus \nu $ (corresponding to the terms $\gamma =\{j\}$ in the sum above),

$$\begin{align*}\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{\delta\subset^f I}\sum_{ j\in \delta\setminus\nu } s_{\nu, \delta}^2x^{\delta\setminus j}\Delta(x)^{j\amalg\nu} =0. \end{align*}$$

Since the $\Delta (x)^{\xi }$ are linearly independent, we have that, for every subset $\xi \subset I$ with $|\xi |=q$ (corresponding to the terms $\epsilon =j\amalg \nu $ ), the coefficient of $\Delta (x)^{\xi }$ must vanish – that is, that

$$\begin{align*}0=\sum_{j\in \xi} \sum_{ \begin{smallmatrix} \delta\subset^f I \\ j\in \delta \end{smallmatrix} } s_{\xi\setminus j, \delta}^2x^{\delta\setminus j} = \sum_{\alpha\subset^f I} \sum_{ j\in \xi\setminus \alpha } s_{\xi\setminus j, \alpha\amalg j}^2x^{\alpha}. \end{align*}$$

Since the $x^{\alpha }$ form a basis of k as an S-vector space, we find that for every finite $\alpha \subset I$ and $\xi \subset I$ with $|\xi |=q$ , the corresponding coefficient must vanish:

(3.3) $$ \begin{align} \sum_{ j\in \xi\setminus \alpha } s_{\xi\setminus j, \alpha\amalg j}^2=0. \end{align} $$

We now show that these relations among the coefficients $s_{\nu , \delta }$ imply that $V(\sum _{\nu \subset I, |\nu |=q-1}\mu _\nu (dx)^\nu )$ is divisible by $2$ in ${\mathrm {W}}_{\langle 2\rangle }\Omega ^{q-1}_k$ . We do this by showing that the sum $\sum _{\nu \subset I, |\nu |=q-1}\mu _\nu (dx)^\nu $ is in the image of the Frobenius map. The claim will then follow since, by the linearity of V of axiom iv) of Definition 3.2,

$$\begin{align*}V(F(z))=V(1)\cdot z \end{align*}$$

for every $z\in {\mathrm {W}}_{\langle 2\rangle }\Omega ^{q-1}_k$ , and $V(1)=2\in {\mathrm {W}}_{\langle 2\rangle }(k)$ since k has characteristic $2$ .

By rearranging the terms and grouping pairs $(\nu , \delta )$ with the same intersection $\beta $ , we can write

(3.4) $$ \begin{align} \begin{split} \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \mu_\nu (dx)^\nu &= \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{\delta\subset^f I}s^2_{\nu, \delta}x^{\delta} (dx)^\nu = \sum_{\beta\subset^f I} \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{ \begin{smallmatrix} \delta\subset^f I \\ \delta\cap\nu=\beta \end{smallmatrix} } s^2_{\nu, \delta}x^{\delta} (dx)^\nu \\&= \sum_{\beta\subset^f I} x^{\beta} (dx)^\beta \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{ \begin{smallmatrix} \delta\subset^f I \\ \delta\cap\nu=\beta \end{smallmatrix} } s^2_{\nu, \delta}x^{\delta\setminus\beta} (dx)^{\nu\setminus\beta}. \end{split} \end{align} $$

Each term $x^{\beta } (dx)^\beta $ is in the image of the Frobenius, since by axiom v) of Definition 3.2 (we recall that in the de Rham-Witt complex the map $\lambda $ is the identity),

$$ \begin{align*} x^\beta(dx)^\beta&=\prod_{i\in\beta}x_idx_i=\prod_{i\in\beta}\tau_0(x_i)d\tau_0(x_i)=\prod_{i\in\beta}F(d\tau_1(x_i))=F(\prod_{i\in\beta}d\tau_1(x_i)). \end{align*} $$

It is therefore sufficient to show that for every fixed $\beta \subset I$ , the double sum in equation (3.4) is in the image of the Frobenius. Let us now group those terms by the union $\lambda $ of $\nu $ and $\delta $ , and write

(3.5) $$ \begin{align} \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{ \begin{smallmatrix} \delta\subset^f I \\ \delta\cap\nu=\beta \end{smallmatrix} } s^2_{\nu, \delta}x^{\delta\setminus\beta} (dx)^{\nu\setminus\beta} &= \sum_{\lambda\subset^f I} \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } \sum_{ \begin{smallmatrix} \delta\subset^f I \\ \delta\cap\nu=\beta \\ \delta\cup \nu=\lambda \end{smallmatrix} } s^2_{\nu, \delta}x^{\delta\setminus\beta} (dx)^{\nu\setminus\beta}. \end{align} $$

We now show that for every fixed $\beta , \lambda \subset ^f I$ , the inner double sum in (3.5) is in the image of the Frobenius. Notice that, after fixing $\beta $ and $\lambda $ , the subset $\delta $ is determined by $\nu $ , and let us write $\delta _\nu :=(\lambda \setminus \nu )\amalg \beta $ . That is, we show that

(3.6) $$ \begin{align} \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \end{smallmatrix} } s^2_{\nu, \delta_\nu}x^{\delta_\nu\setminus\beta} (dx)^{\nu\setminus\beta}=\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \end{smallmatrix} } s^2_{\nu, \delta_\nu}x^{\lambda\setminus \nu} (dx)^{\nu\setminus\beta} \end{align} $$

is in the image of the Frobenius. Let us first treat the case where $\beta =\lambda $ (with $q-1$ elements, otherwise the sum is trivially zero). In this case, the sum is just $s^2_{\beta , \beta }$ , and every square of k is in the image of the Frobenius since $s^2=F(\tau _1(s))$ . Thus, suppose that $\beta $ is a proper subset of $\lambda $ , and choose an element $j_0\in \lambda \setminus \beta $ . We claim that (3.6) is equal to

$$\begin{align*}\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } s^2_{\nu, \delta_\nu}d(x^{(\lambda\setminus \nu)\amalg j_0})(dx)^{(\nu\setminus\beta)\setminus j_0}. \end{align*}$$

This will conclude the proof, since any square is in the image of the Frobenius by the argument above, and so is each differential by the relation $d=FdV$ of axiom v) (observe that $\lambda \setminus \nu $ and $\nu \setminus \beta $ are disjoint, with union $\lambda \setminus \beta $ , so that no summands contain the square of a differential). To see that the last claim holds, let us notice that by iterating the Leibniz rule,

$$\begin{align*}d(x^{\xi})=\sum_{j\in\xi}x^{\xi\setminus j}dx_j \end{align*}$$

for every finite subset $\xi \subset I$ , and therefore

$$ \begin{align*} &\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } s^2_{\nu, \delta_\nu}d(x^{(\lambda\setminus \nu)\amalg j_0})(dx)^{(\nu\setminus\beta)\setminus j_0} = \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } \sum_{j\in (\lambda\setminus \nu)\amalg j_0} s^2_{\nu, \delta_\nu}x^{((\lambda\setminus \nu)\amalg j_0)\setminus j}(dx_{j})(dx)^{(\nu\setminus\beta)\setminus j_0} \\ &= \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } \big( s^2_{\nu, \delta_\nu}x^{\lambda\setminus \nu}(dx)^{\nu\setminus\beta} + \sum_{j\in \lambda\setminus \nu} s^2_{\nu, \delta_\nu}x^{((\lambda\setminus \nu)\amalg j_0)\setminus j}(dx)^{(\nu\setminus\beta)\setminus j_0\amalg j} \big) \\ &= \big(\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } s^2_{\nu, \delta_\nu}x^{\lambda\setminus \nu}(dx)^{\nu\setminus\beta} \big) + \big(\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } \sum_{j\in \lambda\setminus \nu} s^2_{\nu, \delta_\nu}x^{((\lambda\setminus \nu)\amalg j_0)\setminus j}(dx)^{(\nu\setminus\beta)\setminus j_0\amalg j} \big). \end{align*} $$

By setting $\zeta =(\nu \setminus j_0)\amalg j$ in the second sum, and observing that $\delta _\nu =(\delta _{\zeta }\setminus j_0)\amalg j$ , we find that this is equal to

(3.7) $$ \begin{align} &\big(\!\!\!\!\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } s^2_{\nu, \delta_\nu}x^{\lambda\setminus \nu}(dx)^{\nu\setminus\beta} \big) + \big(\sum_{ \begin{smallmatrix} \zeta\subset I \\ |\zeta|=q-1 \\ \beta\subset\zeta\subset\lambda \\ j_0\notin\zeta \end{smallmatrix} } \sum_{j\in \zeta\setminus\beta} s^2_{(\zeta\amalg j_0)\setminus j, (\delta_{\zeta}\setminus j_0)\amalg j}x^{((\lambda\setminus (\zeta\amalg j_0\setminus j))\amalg j_0)\setminus j}(dx)^{((\zeta\amalg j_0\setminus j)\setminus\beta)\setminus j_0\amalg j} \big) \end{align} $$

(3.8) $$ \begin{align} &= \big(\!\!\!\!\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\in\nu \end{smallmatrix} } s^2_{\nu, \delta_\nu}x^{\lambda\setminus \nu}(dx)^{\nu\setminus\beta} \big) + \big(\sum_{ \begin{smallmatrix} \zeta\subset I \\ |\zeta|=q-1 \\ \beta\subset\nu\subset\lambda \\ j_0\notin\zeta \end{smallmatrix} } \big( \sum_{j\in \zeta\setminus\beta} s^2_{(\zeta\amalg j_0)\setminus j, (\delta_{\zeta}\setminus j_0)\amalg j}\big)x^{\lambda\setminus \zeta}(dx)^{\zeta\setminus\beta} \big). \end{align} $$

Finally, by applying the relation (3.3) for $\xi =\zeta \amalg j_0$ and $\alpha =\delta _{\zeta }\setminus j_0$ , so that $\xi \setminus \alpha =(\zeta \setminus \beta )\amalg j_0$ , we find that

$$\begin{align*}\sum_{j\in \zeta\setminus\beta} s^2_{(\zeta\amalg j_0)\setminus j, (\delta_{\zeta}\setminus j_0)\amalg j}=s^2_{\zeta, \delta_\zeta}, \end{align*}$$

which identifies (3.7) and (3.6).

Proof. By the characterisation of $J_{\langle 2^{n} \rangle }$ of Proposition 2.16, we see that $(x, 0)$ can be expressed as a sum of elements $u_1\dots u_{q+1}$ , with $u_j\in J_{\langle 2^{n} \rangle }$ , of three types:

1. i) At least one of the $u_j$ is of the form $V^{n-i}\tau _i(a\otimes b)+V^{n-i}\tau _i(ab\otimes 1)$ for $a, b\in k$ and $0\leq i\leq n-1$ . We can then write this generator in components as

   $$\begin{align*}u_1\dots u_{q+1}=(w_1+w_1^{\prime}, 0)u_2\dots u_{q+1}, \end{align*}$$
   where $w_1$ is the first component of $V^{n-i}\tau _i(a\otimes b)$ and $w_1^{\prime }$ the first component of $V^{n-i}\tau _i(ab\otimes 1)$ .

2. ii) All of the $u_j$ are of the form $V^{n-i}\tau _i(a\otimes b)+\sigma V^{n-i}\tau _i(a\otimes b)$ for $a, b\in k$ and $0\leq i\leq n-1$ . Then each factor is diagonal; that is, $u_j=(v_j, v_j)$ where $v_j$ is in $J_{\langle 1\rangle }$ , and

   $$\begin{align*}u_1\dots u_{q+1}=(v_1\dots v_{q+1}, v_1\dots v_{q+1}). \end{align*}$$

3. iii) It is not of the first two types. In this case, at least one of the $u_j$ is of the form $\tau _n(a\otimes b)+\tau _n(ab\otimes 1)$ and the other factors are diagonal. We can then write such a generator as

   $$ \begin{align*} u_1\dots u_{q+1}&=(t_1, t_1^{\prime})\dots (t_l, t_l^{\prime})(s_{l+1}, s_{l+1})\dots (s_{q+1}, s_{q+1}) \\&=(t_1\dots t_ls_{l+1}\dots s_{q+1}, t_1^{\prime}\dots t_l^{\prime}s_{l+1}\dots s_{q+1}) \end{align*} $$
   for some $1\leq l\leq q+1$ , where $t_j=\phi _n(a_j\otimes b_j)+\phi _n(a_jb_j\otimes 1)$ , $t_j^{\prime }=\phi _n(b_j\otimes a_j)+\phi _n(1\otimes a_jb_j)$ , and $s_j$ is in $J_{\langle 1\rangle }$ .

Thus, let us write $(z, 0)$ as a sum of these types of generators, where we omit the indexing from the sums to make this expression more digestible:

$$\begin{align*}(z, 0)&=\sum (w_1+w_1^{\prime}, 0)u_2\dots u_{q+1} +\sum(v_1\dots v_{q+1}, v_1\dots v_{q+1}) \\&\quad+\sum (t_1\dots t_ls_{l+1}\dots s_{q+1}, t_1^{\prime}\dots t_l^{\prime}s_{l+1}\dots s_{q+1}). \end{align*}$$

Since the second component of $(z, 0)$ is null, we must have that

$$\begin{align*}\sum v_1\dots v_{q+1}=\sum t_1^{\prime}\dots t_l^{\prime}s_{l+1}\dots s_{q+1}. \end{align*}$$

By replacing the left-hand side in the first component above, we have that z is, by denoting $r_j$ the first component of $u_j$ , of the form

$$ \begin{align*} z&=\sum (w_1+w_1^{\prime})r_2\dots r_{q+1}+\sum t_1^{\prime}\dots t_l^{\prime}s_{l+1}\dots s_{q+1}+\sum t_1\dots t_ls_{l+1}\dots s_{q+1} \\&=\sum(w_1+w_1^{\prime})r_2\dots r_{q+1}+\sum (t_1\dots t_l+t_1^{\prime}\dots t_l^{\prime})s_{l+1}\dots s_{q+1}. \end{align*} $$

Thus, since the $r_j$ and $s_j$ belong to $J_{\langle 1 \rangle }$ , to conclude the proof, it is sufficient to show that $w_1+w_1^{\prime }$ belongs to $J_{\langle 1 \rangle }^2$ and that $(t_1\dots t_l+t_1^{\prime }\dots t_l^{\prime })$ belongs to $J_{\langle 1 \rangle }^{l+1}$ . For the first case, we have that

$$ \begin{align*} w_1+w_1^{\prime}&=\phi_i(a\otimes b)+\phi_i(b\otimes a)+\phi_i(ab\otimes 1)+\phi_i(1\otimes ab) \\&= a^{2^i-1}b^{2^i}\otimes a+b^{2^i-1}a^{2^i}\otimes b+(ab)^{2^i-1}\otimes ab+(ab)^{2^i}\otimes 1 \\&=(ab)^{2^i-1}(b\otimes a+a\otimes b+1\otimes ab+ab\otimes 1) \\&=(ab)^{2^i-1}(1\otimes a+a\otimes 1)(1\otimes b+b\otimes 1), \end{align*} $$

which indeed belongs to $J_{\langle 1 \rangle }^2$ . For the second case, we have that $(t_1\dots t_l+t_1^{\prime }\dots t_l^{\prime })$ is equal to

$$ \begin{align*} &\prod_{j=1}^l(\phi_n(a_j\otimes b_j)+\phi_n(a_jb_j\otimes 1))+\prod_{j=1}^l(\phi_n(b_j\otimes a_j)+\phi_n(1\otimes a_jb_j)) \\&=\prod_{j=1}^l(a_j^{2^n-1}b_j^{2^n}\otimes a_j+(a_jb_j)^{2^n-1}\otimes a_jb_j)+\prod_{j=1}^l(b_j^{2^n-1}a_j^{2^n}\otimes b_j+(a_jb_j)^{2^n}\otimes 1) \\& =\prod_{j=1}^l\big(((a_jb_j)^{2^n-1}\otimes a_j)\cdot (b_j\otimes 1+1\otimes b_j)\big) + \prod_{j=1}^l\big( ((a_j)^{2^n}(b_j)^{2^n-1}\otimes 1) \cdot (b_j\otimes 1+1\otimes b_j) \big) \\ &= \big(\prod_{j=1}^l((a_jb_j)^{2^n-1}\otimes a_j)+\prod_{j=1}^l((a_j)^{2^n}(b_j)^{2^n-1}\otimes 1)\big) \cdot \prod_{j=1}^l (b_j\otimes 1+1\otimes b_j) \\&= \big(\prod_{j=1}^l(a_jb_j)^{2^n-1}(1 \otimes a_j)+\prod_{j=1}^l(a_jb_j)^{2^n-1}(a_j\otimes 1)\big) \cdot \prod_{j=1}^l (b_j\otimes 1+1\otimes b_j) \\&= (\prod_{j=1}^l(a_jb_j)^{2^n-1})\big(1 \otimes (\prod_{j=1}^la_j)+(\prod_{j=1}^la_j)\otimes 1\big) \cdot \prod_{j=1}^l (b_j\otimes 1+1\otimes b_j), \end{align*} $$

which belongs to $J_{\langle 1 \rangle }^{l+1}$ .

Proof of Theorem 3.4

By Remark 3.5, in degree $q=0$ , the map $u_n$ is an isomorphism for all $n\geq 0$ . Thus, let $q\geq 1$ . By Lemma 3.6, the map $u_0\colon \Omega ^q_k\longrightarrow J_{\langle 1\rangle }^q/J_{\langle 1\rangle }^{q+1}$ is an isomorphism for every $q\geq 1$ . Thus, let $n\geq 1$ , assume that $u_{n-1}$ is an isomorphism, and let us show that $u_{n}$ is an isomorphism. Since $u_{n}$ is a map of Witt complexes, it clearly hits all the generators of $J_{\langle 2^{n} \rangle }^q$ from Proposition 2.16, and it is therefore surjective. To see that it is injective, consider the commutative diagram with exact rows

where the top row is exact by [Reference CosteanuCos08, Lemma 3.5]. It then suffices to show that $u_{n}$ is injective when restricted to the image of the top left horizontal map $V^n+dV^n$ . Thus, let $a\in \Omega ^q_k$ and $b\in \Omega ^{q-1}_k$ , and suppose that $u_{n}(V^n(a)+dV^n(b))$ can be represented by an element of $J_{\langle 2^{n} \rangle }^{q+1}$ . We need to show that $c:=V^n(a)+dV^n(b)$ is divisible by $2$ in ${\mathrm {W}}_{\langle 2^{n}\rangle }\Omega ^q_k$ . Let us explicitly calculate $u_{n}(V^n(a)+dV^n(b))$ . Given a $2$ -basis $\{x_{i}\}_{i\in I}$ of k, let us write

$$\begin{align*}a=\sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }\lambda_\xi(dx)^\xi \ \ \ \ \ \ \ \mbox{and}\ \ \ \ \ \ \ \ b=\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} }\mu_\nu(dx)^\nu, \end{align*}$$

where $\lambda _\xi , \mu _\nu \in k$ and $(dx)^\xi =\prod _{i\in \xi }dx_i$ . Since u is a map of Witt complexes, we must have that

$$\begin{align*}u_{n}(V^n(a)+dV^n(b))=V^n(u_0(a))+dV^n(u_0(b))= \sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }V^n(\lambda_\xi\Delta(x)^\xi) + \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } dV^n(\mu_\nu \Delta(x)^\nu). \end{align*}$$

Recall from Proposition 2.10 that for every $x\otimes y\in k\otimes _S k$ , we have that $V^n(x\otimes y)$ is represented by $({\mathrm {tran}}(x\otimes y), 0)$ in $\pi _0\ {\mathrm {TRR}}^{n+1}(k;2)^{\phi \mathbb {Z}/2}$ , where ${\mathrm {tran}}(x\otimes y)=x\otimes y+y\otimes x$ . Thus, since $d=1+\sigma $ , we have

$$ \begin{align*} u_{n}(V^n(a)+dV^n(b))&= \sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }({\mathrm{tran}}(\lambda_\xi\Delta(x)^\xi), 0) + \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } ({\mathrm{tran}}(\mu_\nu \Delta(x)^\nu), {\mathrm{tran}}(\mu_\nu \Delta(x)^\nu)). \end{align*} $$

By choosing $i\in \xi $ , we can write

$$ \begin{align*} {\mathrm{tran}}(\lambda_\xi\Delta(x)^\xi)&={\mathrm{tran}}(\lambda_\xi\Delta(x_i)\Delta(x)^{\xi\setminus i})={\mathrm{tran}}(\lambda_\xi\Delta(x_i))\Delta(x)^{\xi\setminus i}={\mathrm{tran}}(\lambda_\xi\otimes x_i+\lambda_\xi x_i\otimes 1)\Delta(x)^{\xi\setminus i} \\&= (\lambda_\xi\otimes x_i+\lambda_\xi x_i\otimes 1+x_i\otimes \lambda_\xi+1\otimes \lambda_\xi x_i)\Delta(x)^{\xi\setminus i}=\Delta(\lambda_\xi)\Delta(x_i)\Delta(x)^{\xi\setminus i} \\&=\Delta(\lambda_\xi)\Delta(x)^{\xi}, \end{align*} $$

where the second equality holds since $\Delta (x)^{\xi \setminus i}$ is fixed by the involution. Similarly, ${\mathrm {tran}}(\mu _\nu \Delta (x)^\nu )=\Delta (\mu _\nu )\Delta (x)^{\nu }$ (which is obvious in the case where $q=1$ and $\nu =\emptyset $ ). Thus, we find that

$$ \begin{align*} u_{n}(V^n(a)+dV^n(b))&= \sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }(\Delta(\lambda_\xi)\Delta(x)^{\xi}, 0) + \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} } (\Delta(\mu_\nu)\Delta(x)^{\nu}, \Delta(\mu_\nu)\Delta(x)^{\nu}) \\&= \big( \sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }\Delta(\lambda_\xi)\Delta(x)^{\xi} + \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} }\Delta(\mu_\nu)\Delta(x)^{\nu} , \sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} }\Delta(\mu_\nu)\Delta(x)^{\nu} \big), \end{align*} $$

and this element is by assumption in $J_{\langle 2^{n} \rangle }^{q+1}$ . Let us analyse the two components separately, starting from the second one. As observed above Lemma 3.8, these components must in fact belong to $J_{\langle 1 \rangle }^{q+1}$ , and therefore,

$$\begin{align*}\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} }\Delta(\mu_\nu)\Delta(x)^{\nu} \in J_{\langle 1 \rangle}^{q+1}. \end{align*}$$

By Lemma 3.7, $V(\mu _\nu (dx)^\nu )$ vanishes in the de Rham Witt complex modulo $2$ , and therefore so does

$$\begin{align*}dV^n(b)=dV^{n-1}(\sum_{ \begin{smallmatrix} \nu\subset I \\ |\nu|=q-1 \end{smallmatrix} }V(\mu_\nu (dx)^\nu)). \end{align*}$$

Our original element $c=V^n(a)+dV^n(b)$ is then equal to $V^n(a)$ in the de Rham-Witt complex modulo $2$ , and the map $u_n$ sends this element to

$$\begin{align*}u_n(c)=V^n(a)=\big( \sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }\Delta(\lambda_\xi)\Delta(x)^{\xi} , 0 \big). \end{align*}$$

Moreover, this element is by assumption in $J_{\langle 2^{n} \rangle }^{q+1}$ . By applying Lemma 3.8, the first component $\sum _{ \begin {smallmatrix} \xi \subset I \\ |\xi |=q \end {smallmatrix} }\Delta (\lambda _\xi )\Delta (x)^{\xi }$ in fact belongs to $J_{\langle 1 \rangle }^{q+2}$ . Again by Lemma 3.7, we find that

$$\begin{align*}c=V^n(a)=V^{n-1}V(\sum_{ \begin{smallmatrix} \xi\subset I \\ |\xi|=q \end{smallmatrix} }\lambda_\xi(dx)^{\xi})=0\end{align*}$$

in $({\mathrm {W}}_{\langle 2^{n}\rangle }\Omega ^q_k)/2$ , proving that $u_n$ is injective.

Proof. Let $\nu ^\ast (k)$ and be $\epsilon ^\ast (k)$ be respectively the equaliser and coequaliser of the projection map and the inverse Cartier operator

The map R of the de Rham-Witt complex induces multiplicative maps $\nu _{dRW/2}^\ast (k;2)\to \nu ^\ast (k)$ and $\epsilon _{dRW/2}^\ast (k;2)\to \epsilon ^\ast (k)$ , which are isomorphisms by the proof of [Reference Clausen, Mathew and MorrowCMM21, Proposition 2.26]. Moreover, by [Reference KatoKat82, Theorem (2)], the graded ring associated to the fundamental ideal I of the symmetric Witt group is isomorphic to $\nu ^\ast (k)$ , and the graded module $I^\ast {\mathrm {W}}^q(k)/I^{\ast +1} {\mathrm {W}}^q(k)$ to $\epsilon ^\ast (k)$ . Thus, since [Reference KatoKat82, Theorem (1)] and Corollary 2.5 identify ${\mathrm {W}}^s(k)$ and $\pi _0{\mathrm {TCR}}(k;2)^{\phi \mathbb {Z}/2}$ , as rings, with the equaliser of

and ${\mathrm {W}}^q(k)$ and $T_{-1}$ , as modules, with their coequaliser, it suffices to show that I and K correspond to the same ideal under these identifications.

For the symmetric Witt group, the isomorphism with the equaliser is given by the unique additive map that sends the rank $1$ form $\langle a\rangle $ , with $a\in k^\times $ , to $a^{-1}\otimes a$ . For $\pi _0{\mathrm {TCR}}(k;2)^{\phi \mathbb {Z}/2}$ , it is induced by the map

$$\begin{align*}\pi_0{\mathrm{TCR}}(k;2)^{\phi\mathbb{Z}/2}\stackrel{R}{\longrightarrow}\pi_0{\mathrm{TRR}}^{2}(k;2)^{\phi\mathbb{Z}/2}\stackrel{c}{\longrightarrow} \pi_0({\mathrm{THR}}(k)^{\phi\mathbb{Z}/2})^{C_2}, \end{align*}$$

followed by the identification of the target with $(k\otimes _S k)^{C_2}$ from Proposition 2.3. Let us note that, after including the fixed points into $k\otimes _Sk$ , this is the map

$$\begin{align*}\pi_0{\mathrm{TCR}}(k;2)^{\phi\mathbb{Z}/2}\stackrel{R}{\longrightarrow} \pi_0{\mathrm{THR}}(k)^{\phi\mathbb{Z}/2}\cong k\otimes_Sk, \end{align*}$$

where the isomorphism is from [Reference Dotto, Moi, Patchkoria and ReehDMPR21, Theorem 5.1] (there it is stated for the fixed points, but since the transfer map of k is zero, the isomorphism descends to the geometric fixed points). Thus, in order to compare I and K, it suffices to show that, under the isomorphism of Corollary 2.5, the restriction map from $\pi _0{\mathrm {TCR}}(k;2)^{\phi \mathbb {Z}/2}$ to $(\pi _0{\mathrm {TC}}(k;2))^{\mathbb {Z}/2}/Im(1+w)$ sends $a^{-1}\otimes a$ to $1$ . As we do not have a good handle of $\pi _0{\mathrm {TC}}(k;2)$ for a general field k, we found ourselves unable to prove this by direct calculation.

Instead, we can employ the existence of a trace map of $\mathbb {Z}/2$ -equivariant spectra ${\mathrm {tr}}\colon {\mathrm {KR}}(k)\to {\mathrm {TCR}}(k;2)$ which lifts the trace map ${\mathrm {K}}(k)\to {\mathrm {TC}}(k;2)$ from [Reference Bökstedt, Hsiang and MadsenBHM93]. This trace map is constructed in the forthcoming paper [Reference Harpaz, Nikolaus and ShahHNS21] in the setting of Poincaré $\infty $ -categories. For the purpose of our paper, we content ourselves with giving a point-set construction of this trace map in the case of rings with involution, as carried out in Proposition A.1 below. In fact, we need very little from this trace map: since ${\mathrm {tr}}\colon {\mathrm {KR}}(k)\to {\mathrm {TCR}}(k;2)$ is a map of $\mathbb {Z}/2$ -equivariant spectra and ${\mathrm {W}}^{s}(k)\cong \pi _0{\mathrm {KR}}(k)^{\phi \mathbb {Z}/2}$ , it induces a commutative square

where the bottom map is induced by the usual trace map from [Reference Bökstedt, Hsiang and MadsenBHM93]. The composite on the top row sends the rank $1$ form $\langle a\rangle $ to $a^{-1}\otimes a$ , as proved in Proposition A.1. It follows that the top trace map must be an isomorphism, and since the bottom map is a ring homomorphism and therefore injective, the respective vertical kernels I and K are then isomorphic.