Proof. The implications

$$ \begin{align*}(1)\Rightarrow(2)\Rightarrow(3),\quad(1)\Rightarrow(4)\Rightarrow(5)\Rightarrow(6) \end{align*} $$

are straightforward. For $(3)\Rightarrow (1)$ , see [Reference Rao34, Proposition 1.1] (cf. [Reference Johnson and Wilson15, Proof of Theorem 4.16]).

Assume that $(6)$ holds. Then by [Reference Johnson and Wilson15, Theorem 4.8 and Remark 4.9], one has that the natural map $\mathrm P(n)^{\mathrm {top}}_*(X)\rightarrow \mathrm {CK}(n)^{\mathrm {top}}_*(X)$ is surjective. In particular, $(6)\Rightarrow (3)$ . This finishes the proof of equivalence (1)–(6). Moreover, we see that the induced surjection

$$ \begin{align*}\mathrm P(n)^{\mathrm{top}}_{*}(X)\otimes_{\mathbb F_2[v_n,v_{n+1},\ldots]}\mathbb F_2[v_n]\twoheadrightarrow\mathrm{CK}(n)^{\mathrm{top}}_*(X) \end{align*} $$

is in fact an isomorphism by (2) and (5) comparing the dimensions. Therefore, $(6)\Rightarrow (7)$ . Finally, for $(4)\Rightarrow (8)$ , see [Reference Adams1, Lemma 13.9].

Proof. The integral cohomology of $K/T$ is a free module with a basis given by classes of Schubert varieties; in particular, it is concentrated in even degrees. Since $\mathrm {MU}^*(*)$ and $\mathrm {CK}(n)_{\mathrm {top}}^*(*)$ are also even graded, the Atiyah–Hirzebruch spectral sequences

$$ \begin{align*}\mathrm H^p(K/T,\,A_q(*))\Longrightarrow A^{p+q}(K/T) \end{align*} $$

collapse at the second page for $A=\mathrm {MU}$ , $\mathrm {CK}(n)_{\mathrm {top}}$ . In particular, the natural map

$$ \begin{align*}\mathrm{MU}^*(K/T)\rightarrow\mathrm{CK}(n)_{\mathrm{top}}(K/T)\end{align*} $$

is surjective.

The Bruhat decomposition gives a cellular decomposition for $K/T$ ; therefore, the natural map $\Omega ^*(G/B)\rightarrow \mathrm {MU}^*(K/T)$ is an isomorphism by [Reference Hornbostel and Kiritchenko12, Theorem 6.1]. Then the induced map $\mathrm {CK}(n)^*(G/B)\rightarrow \mathrm {CK}(n)_{\mathrm {top}}^*(K/T)$ is a surjective map between free $\mathbb {F}_2[v_n]$ -modules of the same rank (cf. [Reference Calmès, Petrov and Zainoulline4, Lemma 13.7], Proposition 2.4); therefore, it is also an isomorphism.

Taking the localization by the powers of $v_n$ , we conclude that

$$ \begin{align*}\mathrm{K}(n)^*(G/B)\rightarrow\mathrm{K}(n)^*_{\mathrm{top}}(K/T) \end{align*} $$

is an isomorphism as well.

Proof. For (1), see [Reference Petrov and Semenov29, Lemma 7.2]. The proof of [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Theorem 5.1] can be repeated verbatim to obtain (2). For (3), see [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Proposition 5.2].

Proof. For $m\leq 2^{n+1}$ , see [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Theorem 6.10]. For $m\geq 2^{n+1}+1$ , we will argue by induction.

In the notation of Lemma 3.4, consider arbitrary lifts $\widetilde e_i$ of the elements ${e_i\in \mathrm {CK}(n)^*(\mathrm {SO}_{m-2})}$ to $\mathrm {CK}(n)^*(\mathrm {SO}_m)$ . These lifts together with x generate $\mathrm {CK}(n)^*(\mathrm {SO}_m)$ as an algebra.

First, we will construct a map from R to $\mathrm {CK}(n)^*(\mathrm {SO}_m)$ sending $e_i$ to $\widetilde e_i$ for $i<s$ , and sending $e_s$ to x.

Using Lemma 3.4 (2) we have $v_nx=0$ and $x^2=0$ .

We claim that $v_n\widetilde e_i=0$ for $2^n\leq i<s$ . Using the inductive assumption, we obtain $v_n\widetilde e_i=xz$ for some $z\in \mathrm {CK}(n)^*(\mathrm {SO}_m)$ . Decompose z as a linear combination of monomials in $\widetilde e_i$ and x. Since $v_nx=0$ , we can assume that this linear combination has $\mathbb F_2$ -coefficients, rather than just $\mathbb F_2[v_n]$ -. Moreover, since $x^2=0$ , we can assume that this linear combination consists only of monomials in $\widetilde e_i$ (and not in x). However, the image of $xz$ in $\mathrm {Ch}^*(\mathrm {SO}_m)$ is $0$ . In other words, the image of z in $\mathrm {Ch}^*(\mathrm {SO}_m)$ is a linear combination in $e_1,\ldots ,e_{s-1}$ which is annihilated by $e_s$ . This can only happen if the image of z in $\mathrm {Ch}^*(\mathrm {SO}_m)$ is zero. In other words, z is divisible by $v_n$ , but this implies that $xz=0$ .

Similarly, one shows that $\widetilde e_i^2-\widetilde e_{2i}=0$ for $2i<s$ . Finally, we have to show that $\widetilde e_{s/2}^{\,2}=x$ for s even. By induction, we know that $\widetilde e_{s/2}^{\,2}=xz$ for some z. Then $\widetilde e_{s/2}^{\,2}-x=x(z-1)$ , and we can argue for $z-1$ as above to obtain the claim.

In other words, the map from R to $\mathrm {CK}(n)^*(\mathrm {SO}_m)$ is well defined (and surjective). Then $\mathrm {CK}(n)^*(\mathrm {SO}_m)=R/(f_i)$ for some $f_1,\ldots ,f_k\in R$ . However, the images of $f_i$ in $\mathrm {CK}(n)^*(\mathrm {SO}_{m-2})$ should also be $0$ ; therefore, $f_i=e_sg_i$ for some $g_i$ by induction. We can again assume that $g_i$ are polynomials in $e_i$ for $i<s$ with $\mathbb F_2$ -coefficients. However, $f_i$ should equal $0$ modulo $v_n$ , since we can specialize $\mathrm {CK}(n)^*(\mathrm {SO}_m)$ to $\mathrm {Ch}^*(\mathrm {SO}_m)$ . This again implies that $g_i$ are divisible by $v_n$ , and therefore, $f_i=0$ .

Proof. Let $G=K_{\mathbb C}$ denote the corresponding reductive group. We conclude by Proposition 2.3 that $\mathrm {CK}(n)_{\mathrm {top}}^*(K)$ is a free $\mathbb F_2[v_n]$ -module. Consider the diagram

Take an element $x\in \mathrm {CK}(n)^*(G)$ which is not divisible by $v_n$ . Then the injectivity of the bottom horizontal arrow implies that the image of x in $\mathrm {CK}(n)_{\mathrm {top}}^*(K)$ cannot be zero. Since $\mathrm {CK}(n)_{\mathrm {top}}^*(K)$ cannot have a $v_n$ -torsion, we conclude that the top horizontal arrow is injective.

Finally, since $\mathrm {CK}(n)_{\mathrm {top}}^*(K)$ cannot have a $v_n$ -torsion, and the natural map

$$ \begin{align*}\mathrm{CK}(n)_{\mathrm{top}}^*(K)\rightarrow\mathrm{K}(n)_{\mathrm{top}}^*(K)\end{align*} $$

coincides with the localization, we conclude that this map is injective. This implies that

$$ \begin{align*}\mathrm{K}(n)^*(G)\rightarrow\mathrm{K}(n)_{\mathrm{top}}^*(K) \end{align*} $$

is also injective.

Proof. First, assume that $m\leq 2^{n+1}$ . Under this condition, the Atiyah–Hirzebruch spectral sequence $\mathrm H^p\big (K;\,\mathrm K(n)_q^{\mathrm {top}}(*)\big )\Rightarrow \mathrm K(n)_{\mathrm {top}}^{p+q}(K)$ collapses by [Reference Nishimoto27, Theorems 2.4 and 3.2]. Therefore, the claim follows from Lemma 3.6.

Next, consider the case $m>2^{n+1}$ , and assume that the natural map from the algebraic Morava K-theory $\mathrm K(n)^*(G)$ to the topological one $\mathrm K(n)^*_{\mathrm {top}}(K)$ is not injective. Consider the natural inclusion of $K_0=\mathrm {SO}(2^{n+1})$ for m even or $K_0=\mathrm {SO}(2^{n+1}-1)$ for m odd into K, and denote $G_0=(K_0)_{\mathbb C}$ . Consider the following diagram

where the left vertical arrow is an isomorphism by [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Theorem 5.1]. Then a simple diagram chase provides a contradiction.

Finally, assume that the natural map from $\mathrm {CK}(n)^*(G)$ to $\mathrm {CK}(n)^*_{\mathrm {top}}(K)$ is not injective, and take an element x in the kernel. Since x maps to $0$ in $\mathrm {Ch}^*(G)$ , we conclude that $x=v_ny$ for some y. However, since the image of x in $\mathrm {K}(n)^*(G)$ is also $0$ , we conclude that x (and therefore y) is a $v_n^{\mathbb Z}$ -torsion. However, using Theorem 3.5, we conclude that if y is a $v_n^{\mathbb Z}$ -torsion, then in fact, $v_ny=0$ . This finishes the proof.

Proof. For (1), see [Reference Rao35, Proposition 3.1], for (2), see [Reference Rao35, Proposition 2.4 and Lemma 3.4], for (3), see [Reference Rao35, Lemma 3.4], for (4), see [Reference Rao35, Proposition 5.8]. To obtain (5), it suffices to show that the commutators of all generators are $0$ modulo $\beta _{2i-1}$ . This is shown in [Reference Rao35, Lemmas 3.2–3.4, Corollary 5.3, Propositions 5.4–5.5, Lemma 5.7].

Proof. Consider the map

(4.3)

dual to the natural inclusion $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})\hookrightarrow \mathrm {CK}(n)_{\mathrm {top}}^*(\mathrm {SO}({2^{n+1}-1}))$ , and denote

$$ \begin{align*}\overline\gamma_t(\beta_{2i-1})=\varphi\big(\gamma_t(\beta_{2i-1})\big). \end{align*} $$

Using Propositions 2.4 and 4.6, we conclude that $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ is generated over $\mathbb F_2[v_n]$ by the monomials $g_1\ldots g_k$ , where $g_i$ lie in

$$ \begin{align*}\{\overline\gamma_{2^s}(\beta_{2i-1})\mid1\leq i\leq2^{n-1},\ 1\leq s\leq k_i\}. \end{align*} $$

In fact, comparing the ranks, we conclude that these monomials form a basis. Moreover, by Proposition 4.6, we conclude that $\overline \beta _{2i-1}$ are divisible by $v_n$ . Let

$$ \begin{align*}\overline\beta_{2i-1}=v_n^aP+v_n^{a+1}R \end{align*} $$

in $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ , where P is not divisible by $v_n$ (i.e., the image of P in $\mathrm {Ch}^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ is nonzero). Since (4.3) is a morphism of algebras and co-algebras, and $\beta _{2i-1}$ is primitive by Theorem 4.4, we conclude that P is primitive modulo $v_n$ . This implies that P coincides with a sum of $\overline \gamma _2(\beta _{2j-1})$ for some $1\leq j\leq 2^{n-1}$ modulo $v_n$ by Corollary 4.7. Comparing the gradings

$$ \begin{align*}2i-1=2(2j-1)-a(2^n-1), \end{align*} $$

we conclude that $a=1$ and P is congruent just to $\overline \gamma _2(\beta _{2j-1})$ modulo $v_n$ (rather than to a sum of several $\overline \gamma _2(\beta _{2j-1})$ ). Moreover, $2j-1\geq 2^{n-1}-1$ (under this condition, $k_j=1$ ), $i\leq j$ (in particular, $i\leq 2^{n-1}$ ), and $i=j$ implies that $i=2^{n-1}$ .

Take the largest i such that $\overline \beta _{2i-1}\neq 0$ and let

$$ \begin{align*}\overline\beta_{2i-1}=v_n\overline\gamma_2(\beta_{2j-1})+v_n^{2}R. \end{align*} $$

Since $k_j=1$ , we conclude that

$$ \begin{align*}\gamma_2(\beta_{2j-1})^2=[\gamma_2(\beta_{2j-1}),\,\beta_{2j-1}]\cdot\beta_{2j-1}+v_n\gamma_2(\beta_{2i-1}) \end{align*} $$

by Theorem 4.4 (4). However,

$$ \begin{align*}\beta_{2i-1}^2=0 \end{align*} $$

by Theorem 4.4 (3). We do not know yet that $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ is commutative; however, all commutators are divisible by $v_n$ in $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ , and we obtain

$$ \begin{align*}0=\overline\beta_{2i-1}^{\,2}=v_n^2\,\overline\gamma_2(\beta_{2j-1})^2+v_n^4R'. \end{align*} $$

For $i<2^{n-1}$ , we have $\overline \beta _{2j-1}=0$ (by the choice of i), and therefore,

$$ \begin{align*}0=v_n^3\,\overline\gamma_2(\beta_{2i-1})+v_n^4R', \end{align*} $$

but the image of $\gamma _2(\beta _{2i-1})$ in $\mathrm {Ch}^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ is nonzero. This provides a contradiction.

Finally, let $i=2^{n-1}$ . Then

$$ \begin{align*}0=\overline\beta_{2^n-1}^{\,2}=v_n^2\,\overline\gamma_2(\beta_{2^n-1})^2+v_n^4R' =v_n^3\,\overline\gamma_2(\beta_{2^n-1})+v_n^2\,[\overline\gamma_2(\beta_{2^n-1}),\,\overline\beta_{2^n-1}]\cdot\overline\beta_{2^n-1}+v_n^4R', \end{align*} $$

and using again the fact that all commutators are divisible by $v_n$ and $\overline \beta _{2^n-1}=v_n\overline \gamma _2(\beta _{2^n-1})+v_n^2R$ , we obtain

$$ \begin{align*}0=v_n^3\,\overline\gamma_2(\beta_{2^n-1})+v_n^4R", \end{align*} $$

which again provides a contradiction.

Therefore, all $\overline \beta _{2i-1}$ are equal to zero, and $\varphi $ factors through the quotient modulo the ideal generated by $\beta _{2i-1}$ . Comparing the ranks, we obtain the claim.

Proof. This follows from Theorem 4.4 (5).

Proof. The identity (4.4) and the formula

(4.6) $$ \begin{align} \Delta(\gamma_t(\alpha_{2i-1}))=\sum_{s=0}^t\gamma_s(\alpha_{2i-1})\otimes\gamma_{t-s}(\alpha_{2i-1}) \end{align} $$

for $1\leq t\leq 2^{k_i-1}$ follow from Theorem 4.4. Using this and the fact that $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ is a Hopf algebra, we can prove (4.6) for all $1\leq t<2^{k_i}$ , and obtain the first claim. Finally, (4.5) follows from Theorem 4.4 (4) and Corollary 4.9.

Proof. We will denote by $\Big \langle \ ,\ \Big \rangle $ the natural pairing of $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})$ with $\mathrm {CK}(n)^*(\mathrm {SO}_{2^{n+1}-1})^\vee $ .

Let $e_{2i-1}$ denote the dual elements to $\alpha _{2i-1}$ . Then the first claim follows from Proposition 4.10. To compute $\Delta (e_{2^n-1})$ , observe that

$$ \begin{align*}\Big\langle\Delta(e_{2^n-1}),\,\gamma_{t_1}(\alpha_1)\ldots\gamma_{t_{2^n-1}}(\alpha_{2^n-1})\otimes\gamma_{s_1}(\alpha_1)\ldots\gamma_{s_{2^n-1}}(\alpha_{2^n-1})\Big\rangle \end{align*} $$

can only be nonzero for the basis elements $1\otimes \alpha _{2^n-1}$ , $\alpha _{2^n-1}\otimes 1$ and $\alpha _{2^n-1}\otimes \alpha _{2^n-1}$ in the right-hand side of the pairing $\Big \langle \ ,\,\Big \rangle $ by Proposition 4.10. Similarly, for odd j, we see that $2^{n-1}-1+j$ is even; that is, if $2j-1=(2i-1)2^{k_i}-2^{n}+1$ , then $k_i>1$ . This also implies that

$$ \begin{align*}\Big\langle\Delta(e_{2j-1}),\,\gamma_{t_1}(\alpha_1)\ldots\gamma_{t_{2^n-1}}(\alpha_{2^n-1})\otimes\gamma_{s_1}(\alpha_1)\ldots\gamma_{s_{2^n-1}}(\alpha_{2^n-1})\Big\rangle \end{align*} $$

can only be nonzero for the basis elements $1\otimes \alpha _{2j-1}$ , $\alpha _{2j-1}\otimes 1$ and $\gamma _{2^{k_i-1}}(\alpha _{2i-1})\otimes \gamma _{2^{k_i-1}}(\alpha _{2i-1})$ , where $2j-1=(2i-1)2^{k_i}-2^{n}+1$ by Proposition 4.10.

However, if j is even, then $k_i=1$ , that is, $\alpha _{2i-1}^2=v_n\alpha _{2j-1}$ , and

$$ \begin{align*} &\Big\langle\Delta(e_{2j-1}),\,\alpha_{2i-1}\gamma_{2^{k_h-1}}(\alpha_{2h-1})\otimes\gamma_{2^{k_h-1}}(\alpha_{2h-1})\Big\rangle=v_n^2\qquad\qquad\qquad\\ &\qquad\qquad=\Big\langle\Delta(e_{2j-1}),\,\gamma_{2^{k_h-1}}(\alpha_{2h-1})\otimes\gamma_{2^{k_h-1}}(\alpha_{2h-1})\alpha_{2i-1}\Big\rangle \end{align*} $$

for $2i-1=(2h-1)2^{k_h}-2^{n}+1$ . Let us denote $2j-1=2^n-1-2k$ . Then we have $2i-1=2^{n}-1-k$ , and $2h-1=2^{n}-1-\frac k2$ . Moreover, if k is divisible by $4$ , we see that h is even and we can continue this process by induction.

Denoting $\langle t\rangle =2^n-1-t$ , we obtain formula (4.7).

Proof. First, assume that m is odd. We know that the natural map

$$ \begin{align*}\mathrm{Ch}^*(\mathrm{SO}_{2^{n+1}-1})\rightarrow\mathrm{Ch}^*(\mathrm{SO}_{m}) \end{align*} $$

coincides with taking a quotient modulo $e_k$ for $k>\lfloor \frac {m-1}{2}\rfloor $ . Therefore, the corresponding map on the connective Morava K-theories

$$ \begin{align*}\mathrm{CK}(n)^*(\mathrm{SO}_{2^{n+1}-1})\rightarrow\mathrm{CK}(n)^*(\mathrm{SO}_{m}) \end{align*} $$

is surjective by Nakayama’s Lemma (see [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Lemma 5.5]), and the images of $e_k$ become divisible by $v_n$ for $k>\lfloor \frac {m-1}{2}\rfloor $ .

Denote the image of $e_i$ in $\mathrm {CK}(n)^*(\mathrm {SO}_{m})$ by $\overline e_i$ and let

$$ \begin{align*}\overline e_k=v_n^aP+v_n^{a+1}R \end{align*} $$

for some P not divisible by $v_n$ . We will show that such an equation is impossible by formal reasons; therefore, in fact $\overline e_k=0$ for $k>\lfloor \frac {m-1}{2}\rfloor $ .

Let $k=2^t(2^n-1-2j)$ and let $\langle t\rangle $ denote $2^n-1-t$ . Then

$$ \begin{align*}\Delta(e_k)=\Delta(e_{\langle 2j\rangle}^{2^t})=e_{\langle 2j\rangle}^{2^t}\otimes1+1\otimes e_{\langle 2j\rangle}^{2^t}+\sum_i v_n^{2^t(i+1)}\,e_{\langle j/2^i\rangle}^{2^t}\otimes e_{\langle j/2^i\rangle}^{2^t}\,S_i, \end{align*} $$

and therefore,

(4.8) $$ \begin{align} \overline e_k\otimes1+1\otimes\overline e_k+\sum_iv_n^{2^t(i+1)}\,\overline e_{\langle j/2^i\rangle}^{\,2^t}\otimes\overline e_{\langle j/2^i\rangle}^{\,2^t}\,\overline S_i=v_n^a\Delta(P)+v_n^{a+1}\Delta(R). \end{align} $$

For $k=2^n-1$ , we obtain

$$ \begin{align*}\overline e_{2^n-1}\otimes1+1\otimes\overline e_{2^n-1}+v_n^{}\,\overline e_{2^n-1}\otimes\overline e_{2^n-1}=v_n^a(P\otimes1+1\otimes P)+v_n^{a+1}S, \end{align*} $$

which implies that the image of P in $\mathrm {Ch}^*(\mathrm {SO}_{m})$ is primitive and therefore coincides with $\sum e_{2h-1}^{2^{d_h}}$ for some $0\leq d_h<\lfloor \log _2\left (\frac {m-1}{2h-1}\right )\rfloor $ (cf. [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Lemma 6.1 a)]). Then

$$ \begin{align*}\overline e_{2^n-1}=v_n^a\sum\overline e_{2h-1}^{\,2^{d_h}}+v_n^{a+1}R', \end{align*} $$

and comparing the gradings, we have

$$ \begin{align*}2^n-1=a(1-2^n)+2^{d_h}(2h-1) \end{align*} $$

for each h. However, $a\geq 1$ and $2^{d_h}(2h-1)\leq \frac {m-1}{2}$ ; therefore, the above identity is impossible. This shows that $\overline e_{2^n-1}=0$ .

Take the largest k such that

$$ \begin{align*}\overline e_k=\overline e_{[2j]}^{2^t}\neq0. \end{align*} $$

Then by (4.8), we conclude that

$$ \begin{align*}\overline e_k\otimes1+1\otimes\overline e_k=v_n^a\Delta(P)+v_n^{a+1}\Delta(R), \end{align*} $$

and the image of P in $\mathrm {Ch}^*(\mathrm {SO}_{m})$ is primitive. Then

$$ \begin{align*}\overline e_{k}=v_n^a\sum\overline e_{2h-1}^{\,2^{d_h}}+v_n^{a+1}R' \end{align*} $$

as above, and

$$ \begin{align*}k=a(1-2^n)+2^{d_h}(2h-1) \end{align*} $$

for each h. But again, $a\geq 1$ , $k>\lfloor \frac {m-1}{2}\rfloor $ and $2^{d_h}(2h-1)\leq \frac {m-1}{2}$ ; therefore, the above identity is impossible.

This shows that all $\overline e_k=0$ for $k>\lfloor \frac {m-1}{2}\rfloor $ . Comparing the ranks, we obtain the claim for odd m (cf. [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Theorem 6.10]).

Now assume that m is even. As we already remarked, we can assume that $k=\mathbb C$ ; in particular, we can consider the natural map $\mathrm {SO}_{m-1}\rightarrow \mathrm {SO}_m$ induces a surjection on Chow rings

$$ \begin{align*}\mathrm{CH}^*(\mathrm{SO}_m)\rightarrow\mathrm{CH}^*(\mathrm{SO}_{m-1}) \end{align*} $$

(in fact, an isomorphism) (cf., for example, [Reference Pittie31, Section 2]), and therefore, a surjection on $\mathrm {CK}(n)^*$ by Nakayama’s Lemma [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Lemma 5.5]. Comparing the ranks, we conclude that

$$ \begin{align*}\mathrm{CK}(n)^*(\mathrm{SO}_m)\rightarrow\mathrm{CK}(n)^*(\mathrm{SO}_{m-1}) \end{align*} $$

is an isomorphism of Hopf algebras (cf. [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Theorem 6.10]).

Finally, we will show that $e_1$ coincides with the first Chern class of a certain bundle. Let $e_1^{\mathrm {CH}}$ denote a generator of $\mathrm {CH}^1(\mathrm {SO}_m;\,\mathbb Z)$ , and $e_1'=c_1^{\mathrm {CK}(n)}(e_1^{\mathrm {CH}})$ . The images of $e_1$ and $e_1'$ coincide in $\mathrm {Ch}^*(\mathrm {SO}_m)$ , and

$$ \begin{align*}\Delta(e_1')=\mathrm{FGL}_{\mathrm{CK}(n)}(e_1'\otimes1,\,1\otimes e_1'); \end{align*} $$

see the proof of [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Lemma 6.12]. Since $e_{2i-1}^{2^n}=0$ in $\mathrm {CK}(n)^*(\mathrm {SO}_m)$ for all i, we conclude that $(e_1')^{2^n}=0$ , and therefore,

$$ \begin{align*}\widetilde\Delta(e_1')=v_n\,(e_1')^{2^{n-1}}\otimes(e_1')^{2^{n-1}} \end{align*} $$

by, for example, [Reference Petrov and Semenov28, Subsection 2.3]. Let

$$ \begin{align*}e_1'-e_1=v_n^aP+v_n^{a+1}R \end{align*} $$

for some P not divisible by $v_n$ . Again, we will show that the above equation is impossible by formal reasons.

Indeed, one has

$$ \begin{align*} \Delta(e_1')&=(e_1+v_n^aP+v_n^{a+1}R)\otimes1+1\otimes(e_1+v_n^aP+v_n^{a+1}R)+\\ &\quad+v_n(e_1+v_n^aP+v_n^{a+1}R)^{2^{n-1}}\otimes(e_1+v_n^aP+v_n^{a+1}R)^{2^{n-1}}\\ &=\Delta(e_1)+v_n^a(P\otimes1+1\otimes P)+v_n^{a+1}S. \end{align*} $$

However,

$$ \begin{align*}\Delta(e_1')=\Delta(e_1+v_n^aP+v_n^{a+1}R)=\Delta(e_1)+v_n^a\,\Delta(P)+v_n^{a+1}\Delta(R). \end{align*} $$

This implies that P becomes primitive modulo $v_n$ ; that is,

$$ \begin{align*}e_1'-e_1=v_n^a\sum e_{2h-1}^{\,2^{d_h}}+v_n^{a+1}R', \end{align*} $$

and

$$ \begin{align*}1=a(1-2^n)+2^{d_h}(2h-1), \end{align*} $$

where $a\geq 1$ and $2^{d_h}(2h-1)\leq \frac {m-1}{2}\leq 2^n-1$ . Therefore, the above identity is impossible and $e_1'=e_1$ .

Proof. For $m\leq 2^{n+1}$ , the claim follows from Proposition 4.12. Assume that $m>2^{n+1}$ and denote $m_0=2^{n+1}-1$ for m odd, and $m_0=2^{n+1}$ for m even.

The claim about the algebra structure is proven in Theorem 3.5. Moreover, from the proof of Theorem 3.5, it follows that one can take any lift of $e_1\in \mathrm {CK}(n)^*(\mathrm {SO}_{m_0})$ as a generator $e_1\in \mathrm {CK}(n)^*(\mathrm {SO}_{m})$ . Therefore, we can assume that $e_1=c_1^{\mathrm {CK}(n)}(e_1^{\mathrm {CH}})$ is the first Chern class of the generator $e_1^{\mathrm {CH}}$ of $\mathrm {CH}^1(\mathrm {SO}_m;\mathbb Z)$ .

Denote $H=\mathrm {CK}(n)^*(\mathrm {SO}_{m})$ , $I=(e_{2^n},e_{2^n+1},\ldots ,e_s)\trianglelefteq H$ , and

$$ \begin{align*}H_0=\mathrm{CK}(n)^*(\mathrm{SO}_{m_0})=H/I. \end{align*} $$

Then one also has

$$ \begin{align*}H\otimes_{\mathbb F_2[v_n]}H/(H\otimes_{\mathbb F_2[v_n]}I+I\otimes_{\mathbb F_2[v_n]}H)=H_0\otimes_{\mathbb F_2[v_n]}H_0. \end{align*} $$

Observe that all generators of I have degree at least $2^n$ and are annihilated by $v_n$ . This implies that $H\otimes _{\mathbb F_2[v_n]}I+I\otimes _{\mathbb F_2[v_n]}H$ does not have nonzero elements of degree less than $2^n$ .

For $0<k<2^{n-1}$ , we conclude by Proposition 4.11 that

$$ \begin{align*}\widetilde\Delta(e_{\langle 2k\rangle})-\sum_{i=0}^{\nu_2(k)}v_n^{i+1}\,e_{\langle k/2^{i}\rangle}\otimes e_{\langle k/2^{i}\rangle}\,\prod_{j=0}^{i-1}\left(e_{\langle k/2^{j}\rangle}\otimes1+1\otimes e_{\langle k/2^{j}\rangle}\right) \end{align*} $$

is a (homogeneous) element of $H\otimes _{\mathbb F_2[v_n]}I+I\otimes _{\mathbb F_2[v_n]}H$ of degree $\langle 2k\rangle <2^n$ . Therefore, this element is $0$ . Similarly, $\widetilde \Delta (e_{2^n-1})-v_n\,e_{2^{n}-1}\otimes e_{2^{n}-1}=0$ .

It remains to compute $\Delta (e_{2k-1})$ for $k>2^{n-1}$ . Let $I'=(e_{2k-1},e_{2k},\ldots ,e_s)$ and $H':=H/I'=\mathrm {CK}(n)^*(\mathrm {SO}_{m'})$ for $m'=4k-3$ for m odd or $m'=4k-2$ for m even. Observe that $\Delta (e_{2k-1})$ lies in the kernel of the map

$$ \begin{align*}H\otimes_{\mathbb F_2[v_n]}H\rightarrow H'\otimes_{\mathbb F_2[v_n]}H', \end{align*} $$

and the latter kernel coincides with $H\otimes _{\mathbb F_2[v_n]}I'+I'\otimes _{\mathbb F_2[v_n]}H$ . However, $\Delta (e_{2k-1})$ is homogeneous of degree $2k-1$ , and the only nonzero element of $I'$ of degree less than or equal to $2k-1$ is $e_{2k-1}$ itself. This implies that $\Delta (e_{2k-1})$ is a linear combination of $1\otimes e_{2k-1}$ and $e_{2k-1}\otimes 1$ , which finishes the proof.

Proof. The first claim is obvious. To get 2), one only has to observe that

$$ \begin{align*}H\otimes H/(I\otimes H+H\otimes I)\cong (H/I)\otimes(H/I) \end{align*} $$

is free, and use 1).

Proof. Since $ (2j-1)2^y=(2k-1)2^z+(2^n-1)x\geq 2^n, $ we have

$$ \begin{align*}2^{y+1}\geq\frac{2^{n+1}}{2j-1}\geq\frac{m}{2j-1}>\frac{m-1}{2j-1}. \end{align*} $$

In other words, $y+1>\left \lfloor \mathrm {log}_2\left (\frac {m-1}{2j-1}\right )\right \rfloor $ contradicting the assumptions.

Proof. Let $\overline {(\ )}$ denote the image modulo $v_n$ . Since $\overline I$ is also a bi-ideal, we conclude by Lemma 5.4 that

$$ \begin{align*}\overline I=\big(\overline e_1^{2^{a_1}},\,\overline e_3^{2^{a_2}},\ldots,\overline e_{2r-1}^{2^{a_r}}\big) \end{align*} $$

for some $0\leq a_i\leq k_i$ . We claim that

$$ \begin{align*}I=\big(e_1^{2^{a_1}},\,e_3^{2^{a_2}},\ldots,e_{2r-1}^{2^{a_r}}\big) \end{align*} $$

with the same indices $a_k$ . In fact, it is enough to show that $e_k^{2^{a_k}}\in I$ for all k, and comparing dimensions, we will get the claim.

Assume that $e_k^{2^{a_k}}\not \in I$ for some k, fix an arbitrary lift $E_{2k-1}$ of $\overline e_{2k-1}^{2^{a_{k}}}$ to an element of I, and write

(5.1) $$ \begin{align} E_{2k-1}=e_{2k-1}^{2^{a_k}}+v_n^aP \end{align} $$

for $\overline P\neq 0\in \overline H$ . We will show that it is possible to find another lift $E_{2k-1}'\in I$ of $\overline e_{2k-1}^{2^{a_k}}$ with greater a in the decomposition (5.1). Continuing this process, we will eventually show that $P=0$ by the degree reasons.

Case 1. First, we will show that if $\overline e_{2k-1}^{\,2^e}\in \overline I$ for some $e>0$ , then in fact, $e_{2k-1}^{\,2^e}\in I$ .

Let $E_{2k-1}$ denote a pre-image of $\overline e_{2k-1}^{2^{e}}$ in I, and write

(5.2) $$ \begin{align} E_{2k-1}=e_{2k-1}^{2^{e}}+v_n^aP \end{align} $$

for $\overline P\neq 0\in \overline H$ . Observe that $e_{2k-1}^{2^{e}}$ is primitive in H (see Example 4.2). Then, applying $\widetilde \Delta $ to (5.2), we get

$$ \begin{align*}v_n^a\widetilde\Delta(P)=\widetilde\Delta(E_{2k-1}) \in I\otimes H+H\otimes I \end{align*} $$

since I is a bi-ideal. By Lemma 5.3, we conclude that

$$ \begin{align*}\widetilde\Delta(P) \in I\otimes H+H\otimes I, \end{align*} $$

that is, the class of $\overline P$ is primitive in $\overline H/\,\overline I$ , and therefore, $\overline P$ is congruent modulo $\overline I$ to $\sum _{i\in \mathcal {I}} \overline e_{2i-1}^{2^{s_i}}$ for some $0\leq s_i<a_i$ and some index set $\mathcal {I}\subseteq \{1,\ldots ,r\}$ (see [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Lemma 6.1 a)]). Comparing the codimensions, we obtain an equation

$$ \begin{align*}(2k-1)2^{e}=(1-2^n)a+(2i-1)2^{s_i}. \end{align*} $$

However, this equation has no solutions by Lemma 5.5. Therefore, $\mathcal {I}=\emptyset $ and, thus, $\overline P\in \overline I$ .

Take a pre-image $Q\in I$ of $\overline P$ . Then $P-Q$ is divisible by $v_n$ , (i.e., there exist $P'\in H$ such that $P-Q=v_nP'$ ). Therefore,

$$ \begin{align*}E_{2k-1}':=E_{2k-1}-v_n^aQ=e_{2k-1}^{2^{e}}+v_n^aP-v_n^aQ=e_{2k-1}^{2^{e}}+v_n^{a+1}P'\in I \end{align*} $$

(i.e., we obtain a pre-image of $\overline e_{2k-1}^{\,2^{e}}$ in I with a greater a than in (5.2)). Continuing this process, we will eventually show that $e_{2k-1}^{2^{e}}\in I$ .

Case 2. By Case 1, we can assume that $a_k=0$ . We use the notation $\langle t\rangle =2^n-1-t$ from Proposition 4.12, and denote $\langle 2j\rangle =2k-1$ . We can assume, moreover, that k is the largest possible (i.e., j is the least possible) such that $a_k=0$ and $e_{2k-1}\not \in I$ .

Our aim is to prove that $\overline P\in \overline I\otimes \overline H+\overline H\otimes \overline I$ and use the same argument as in Case 1 to find a presentation as (5.1) with a greater a.

Applying $\widetilde \Delta $ to (5.1), we obtain

$$ \begin{align*}v_n\,e_{\langle j\rangle}\otimes e_{\langle j\rangle}+v_n^2R+v_n^a\widetilde\Delta(P) \in I\otimes H+H\otimes I, \end{align*} $$

where $v_n^2R=\widetilde \Delta (e_{\langle 2j\rangle })-v_ne_{\langle j\rangle }\otimes e_{\langle j\rangle }$ is determined by Theorem 4.13.

Subcase 2.1. For $j=0$ (i.e., $2k-1=2^n-1$ ), observe that $R=0$ . Using that $E_{2^n-1}\otimes E_{2^n-1}\in I\otimes I$ , we obtain by (5.1) that

$$ \begin{align*}v_n^a\widetilde\Delta(P)+v_n^{a+1}R'\in I\otimes H+H\otimes I \end{align*} $$

for $R'=e_{2^n-1}\otimes P+P\otimes e_{2^n-1}+v_n^{a}P\otimes P$ . As above, this implies that $\overline P$ becomes primitive modulo $\overline I$ ; however, $\overline H/\,\overline I$ does not have primitive elements of the required degree by Lemma 5.5. This implies that $\overline P\in \overline I$ , and therefore, we can find a new lift $E_{2^n-1}'$ for $\overline e_{2^n-1}$ with greater a in the decomposition (5.1). Continuing this process, we will eventually prove that $e_{2^n-1}\in I$ .

Subcase 2.2. Now consider the case $j>0$ . The proof in this case is divided in three steps.

Step 1. First, we show that that $\overline e_{\langle j\rangle }\in \overline I$ . Indeed, if $a>1$ , we obtain that $\overline e_{\langle j\rangle }\otimes \overline e_{\langle j\rangle }\in \overline I\otimes \overline H+\overline H\otimes \overline I$ by Lemma 5.3, and therefore, $\overline e_{\langle j\rangle }\in \overline I$ . If $a=1$ , we similarly obtain that $\overline e_{\langle j\rangle }\otimes \overline e_{\langle j\rangle }+\widetilde \Delta (\overline P)\in \overline I\otimes \overline H+\overline H\otimes \overline I$ .

Consider the natural basis of $\overline H$ consisting of the monomials in $\overline e_{2i-1}$ . Then the part of this basis divisible by $\overline e_{2i-1}^{\,2^{a_i}}$ for some i is a basis of $\overline I$ , and the tensor product $\overline H\otimes \overline H$ has a natural basis consisting of decomposable tensors of basis elements.

Observe that $\widetilde \Delta (\overline P)$ cannot have a monomial $\overline e_{\langle j\rangle }\otimes \overline e_{\langle j\rangle }$ in its decomposition as a sum of basis elements (this follows from the facts that $\overline e_{2i-1}$ are primitive, and ${2n\choose n}\equiv 0\ \mod 2$ ). Therefore, we obtain that $\overline e_{\langle j\rangle }\otimes \overline e_{\langle j\rangle }\in \overline I\otimes \overline H+\overline H\otimes \overline I$ , and $\overline e_{\langle j\rangle }\in \overline I$ .

Step 2. Now we will show that in fact $e_{\langle j\rangle }\in I$ . Recall that we chose k in such a way that $2k-1=\langle 2j\rangle $ is the largest possible number such that $a_k=0$ and $e_{2k-1}\not \in I$ . Since $\overline e_{\langle j\rangle }\in \overline I$ , we conclude that $\overline e_{\langle j\rangle }$ is equal to $\overline e_{2i-1}^{\,2^{a_i+e}}$ for some i and $e\geq 0$ . However, if $a_i+e>0$ , we already proved that $e_{2i-1}^{\,2^{a_i+e}}\in I$ (see Case 1). But if $a_i=0=e$ , we know that $e_{2i-1}\in I$ by the choice of k. In both cases, we obtain that $e_{\langle j\rangle }\in I$ .

Step 3. Finally, we show that $e_{\langle 2j\rangle }\in I$ . Since $e_{\langle j\rangle }\in I$ by Step 2, we conclude that

$$ \begin{align*}\widetilde\Delta(e_{\langle2j\rangle})\in e_{\langle j\rangle}\otimes H+H\otimes e_{\langle j\rangle}\subseteq I\otimes H+H\otimes I \end{align*} $$

by Theorem 4.13. Consequently, applying $\widetilde \Delta $ to (5.1), we have

$$ \begin{align*}v_n^a\widetilde\Delta(P)\in I\otimes H+H\otimes I, \end{align*} $$

and therefore, $\widetilde \Delta (P)\in I\otimes H+H\otimes I$ by Lemma 5.3. As above, this implies that $\overline P$ becomes primitive modulo $\overline I$ ; however, $\overline H/\,\overline I$ does not have primitive elements of the required degree by Lemma 5.5. This implies that $\overline P\in \overline I$ , and therefore, we can find a new lift $E_{2k-1}'$ for $\overline e_{2k-1}$ with greater a in the decomposition (5.1). Continuing this process, we will eventually prove that $e_{2k-1}=e_{\langle 2j\rangle }\in I$ . This finishes the proof of Subcase 2.2.

We proved that $e_{2i-1}^{2^{a_i}}\in I$ , and it remains to observe that they clearly generate I since it is true modulo $v_n$ and I is saturated.

Proof. It is easy to see that $I_0$ is also a bi-ideal of $H_0$ as an image of I.

Let $J_0$ denote the saturation of $I_0$ , that is,

$$ \begin{align*}J_0=\{x\in H\mid\exists\,i\geq0,\ v_n^ix\in I_0\}.\end{align*} $$

For $x\in J_0$ , one has

$$ \begin{align*}v_n^i\Delta(x)=\Delta(v_n^ix)\in I_0\otimes H+H\otimes I_0\subseteq J_0\otimes H+H\otimes J_0, \end{align*} $$

and therefore, $J_0$ is a bi-ideal by Lemma 5.3. Then, by Proposition 5.6,

$$ \begin{align*}J_0=\big(e_1^{2^{a_1}},\,e_3^{2^{a_2}},\ldots,e_{2r-1}^{2^{a_r}}\big). \end{align*} $$

We will show that in fact $e_{2i-1}^{2^{a_i}}\in I_0$ .

Since $v_n^te_{2i-1}^{2^{a_i}}\in I_0$ , there exist homogeneous elements $h_k\in H_0$ and

$$ \begin{align*}x_k\in\mathrm{Im}\big(\mathrm{CK}(n)^*(E)^+\rightarrow H\rightarrow H_0\big) \end{align*} $$

such that $v_n^te_{2i-1}^{2^{a_i}}=\sum x_kh_k$ . However, since the codimension of $e_{2i-1}^{2^{a_i}}$ is less than or equal to $2^n-1$ , we can assume that for each k, either $x_k$ or $h_k$ has degree less than $0$ (if the degree of $x_kh_k$ is $0$ , use additionally that $e_{2i-1}\in H^+_0$ ). By a theorem of Levine–Morel, the algebraic cobordism is generated by the elements of non-negative degrees [Reference Levine and Morel19, Theorem 1.2.14]. This implies that either $x_k$ or $h_k$ is divisible by $v_n$ . In other words, there exist $h_k'\in H_0$ and $x_k'\in \mathrm {Im}\big (\mathrm {CK}(n)^*(E)^+\rightarrow H\rightarrow H_0\big )$ such that

$$ \begin{align*}v_n^te_{2i-1}^{2^{a_i}}=\sum v_nx_k'h_k'. \end{align*} $$

Since $H_0$ does not have $v_n$ -torsion, we conclude that $ v_n^{t-1}e_{2i-1}^{2^{a_i}}\in I_0. $ Continuing this process, we obtain the claim.

Proof. For $m\leq 2^{n+1}$ , the claim is proven in Proposition 5.6. We assume that $m>2^{n+1}$ and argue by induction.

We denote the reduction modulo $v_n$ by $\,\overline {\phantom {x}}$ . Let $ \overline I=(\overline e_1^{\,2^{a_1}},\ldots ,\overline e_{2r-1}^{\,2^{a_{r}}}) $ according to Lemma 5.4. Let $s=\lfloor \frac {m-1}{2}\rfloor $ .

Case 1. Assume that $e_s\neq e_{2i-1}^{2^{a_{i}}}$ for all i. Since $I/e_s$ is a bi-ideal in $H/e_s$ such that $(I/e_s)_0$ is saturated, using the induction, we have

$$ \begin{align*}I/e_s=(e_1^{2^{a_1}},\ldots,e_{2r-1}^{2^{a_r}}). \end{align*} $$

Then there exist $E_1,\ldots ,E_r\in I$ such that $E_i-e_{2i-1}^{2^{a_i}}=e_s\cdot x_i$ . Since $e_s^2=0$ and $v_ne_s=0$ , we can assume that $x_i$ are polynomials with $\mathbb F_2$ -coefficients in $e_1,\ldots ,e_{s-1}$ . However,

$$ \begin{align*}\overline e_s\cdot\overline x_i=\overline e_{2i-1}^{\,2^{a_i}}-\overline E_i\in\overline I, \end{align*} $$

and this can only happen if $\overline x_i\in \overline I$ . Then $ x_i-\sum e_{2j-1}^{2^{a_j}}\cdot y_{ij} = v_nz_i $ for some $y_{ij}$ , $z_i\in H$ . Then one has

$$ \begin{align*} e_{2i-1}^{2^{a_i}}=E_i-e_s\left(v_nz+\sum e_{2j-1}^{2^{a_j}}\cdot y_{ij}\right)\\=E_i-e_s\left(\sum (E_j-e_sx_j)\cdot y_{ij}\right)=E_i-e_s\left(\sum E_j\cdot y_{ij}\right)\in I. \end{align*} $$

It remains to prove that $I\subseteq (e_1^{2^{a_1}},\ldots ,e_{2r-1}^{2^{a_r}})$ . Take $x\in I$ . Then $x-\sum e_{2i-1}^{2^{a_i}}\, x_{i}=e_sz$ for some $x_i$ , $z\in H$ . Again, we can assume that z is a polynomial with $\mathbb F_2$ -coefficients in $e_1,\ldots ,e_{s-1}$ . But $\overline e_s\,\overline z\in \overline I$ , and therefore, $\overline z\in \overline I$ (i.e., $z-\sum e_{2j-1}^{2^{a_j}}\, y_{j}=v_nw$ for some $y_j$ , $w\in H$ ). Therefore,

$$ \begin{align*}x=\sum e_{2i-1}^{2^{a_i}}\, x_{i}+e_s\left(\sum e_{2j-1}^{2^{a_j}}\, y_{j}\right), \end{align*} $$

which proves the claim.

Case 2. Assume that $e_s= e_{2i_0-1}^{2^{a_{i_0}}}$ for some $i_0$ . Again, $I/e_s$ is a bi-ideal in $H/e_s$ such that $(I/e_s)_0$ is saturated. Then using the induction, we have

$$ \begin{align*}I/e_s=(e_1^{2^{a_1}},\ldots,\widehat e_s,\ldots,e_{2r-1}^{2^{a_r}}), \end{align*} $$

where $\widehat e_s$ means that we excluded $e_s=e_{2i_0-1}^{2^{a_{i_0}}}$ from $e_1^{2^{a_1}},\ldots ,e_{2r-1}^{2^{a_r}}$ . As above, we can find $E_i\in I$ such that $E_i-e_{2i-1}^{2^{a_i}}=e_sx_i$ for $i\neq i_0$ and some $x_i\in H$ , and $e_s+v_ny\in I$ for some $y\in H$ . Since the image of $e_s+v_ny$ coincides with the image of $v_ny$ in $I_0$ , and the latter ideal is saturated, we conclude that the image of y lies in $I_0$ . Then there exist $y_i\in H$ such that the image of $Y=\sum e_{2i-1}^{2^{a_i}}\,y_i$ in $H_0$ coincides with the image of y. Obviously, the image of $Y'=\sum E_i\,y_i$ also coincides with the image of Y. But the kernel of $H\rightarrow H_0$ is annihilated by $v_n$ ; in other words,

$$ \begin{align*}v_n(y-Y')=0\in H. \end{align*} $$

But then $e_s=(e_s+v_ny)-v_nY'\in I$ . Therefore, $e_{2i-1}^{2^{a_i}}=E_i-e_s\,x_i\in I$ .

Finally, it remains to show that $I\subseteq (e_1^{2^{a_1}},\ldots ,e_{2r-1}^{2^{a_r}})$ . Let $x\in I$ . Then

$$ \begin{align*}x-\sum e_{2i-1}^{2^{a_i}}\,z_i=e_s\,w \end{align*} $$

for some $z_i$ , $w\in H$ by induction. The claim follows.

Proof of Corollary 6.3.

According to Theorem 6.1, there exists a $\mathrm K(n)$ -motive $\mathcal R$ such that the $\mathrm K(n)$ -motive of X decomposes as

$$ \begin{align*} \mathcal{M}_{\mathrm K(n)}(X)\simeq\bigoplus_{i\in\mathcal{I}}\mathcal{R}\{i\} \end{align*} $$

for some multiset of integers $\mathcal {I}$ ; moreover, the rank of $\mathcal R$ equals the rank of $H^*_{\mathrm K(n)}(q)$ .

By Corollary 5.2, we conclude that

$$ \begin{align*}H^*_{\mathrm K(n)}(q)=H^*_{\mathrm{Ch}}(q)[v_n^{\pm1}]/(\overline e_i\mid i\geq 2^n)=\mathrm K(n)^*(\mathrm{SO}_m), \end{align*} $$

where the last equality follows from the assumption on the J-invariant of q.

We can now compare the ranks of $\mathrm K(n)^*(\mathrm {SO}_m)$ and $\mathrm K(n)^*(\overline X)$ to determine the cardinality of $\mathcal I$ . By [Reference Geldhauser, Lavrenov, Petrov and Sechin9, Theorem 6.13], the rank of $\mathrm K(n)^*(\mathrm {SO}_m)$ equals $2^{\lfloor \frac {m-1}{2}\rfloor }$ for $m\leq 2^{n+1}$ , and $2^{2^n-1}$ for $m>2^{n+1}$ . However, the rank of $\mathrm K(n)^*(\overline X)$ always equals $2^{\lfloor \frac {m-1}{2}\rfloor }$ . This implies that $|\mathcal I|=1$ for $m\leq 2^{n+1}$ , and $|\mathcal I|=2^{\lfloor \frac {m-1}{2}\rfloor -2^n+1}$ for $m>2^{n+1}$ .

It remains to use that $\mathcal R$ is indecomposable for $m\leq 2^{n+1}-2$ , and that $\mathcal R$ has two indecomposable summands of the same rank for $m>2^{n+1}-2$ by Theorem 6.2.

Proof. Let e be an idempotent in $\mathrm {K}(n)^*(\mathrm {SO}_{2^{n+1}-1})^{\vee }$ . Decompose it in the basis of monomials

$$ \begin{align*}\gamma_{t_1}(\alpha_{1})\gamma_{t_3}(\alpha_{3})\ldots\gamma_{t_{2^{n-1}}}(\alpha_{2^n-1}), \end{align*} $$

where $0\leq t_i<2^{k_i}$ , and observe that

$$ \begin{align*}\big(\gamma_{t_1}(\alpha_{1})\gamma_{t_3}(\alpha_{3})\ldots\gamma_{t_{2^{n-1}}}(\alpha_{2^n-1})\big)^2= \begin{cases} v_n^k\,\alpha_{2h_1-1}\ldots\alpha_{2h_k-1},&\text{if}\ t_i\in\{0,2^{k_i-1}\}\ \forall\,i,\\ 0,&\text{elsewhere}. \end{cases} \end{align*} $$

Since $e^2=e$ , we conclude that e can contain only monomials of the form

$$ \begin{align*}\alpha_{2h_1-1}\ldots\alpha_{2h_k-1} \end{align*} $$

(with some coefficients from $\mathbb F_2[v_n^{\pm 1}]$ ) in its decomposition. However,

$$ \begin{align*}\big(\alpha_{2h_1-1}\ldots\alpha_{2h_k-1}\big)^2= \begin{cases} v_n^k\,\alpha_{2j_1-1}\ldots\alpha_{2j_k-1},&\text{if}\ k_{h_i}=1\ \forall\,i,\\ 0,&\text{elsewhere}. \end{cases} \end{align*} $$

The condition $k_{h_i}=1$ is equivalent to $2^{n-2}<h_i\leq 2^{n-1}$ , and the fact $e^2=e$ implies that e contains only monomials $\alpha _{2h_1-1}\ldots \alpha _{2h_k-1}$ with $2^{n-2}<h_i\leq 2^{n-1}$ in its decomposition. Recall that

$$ \begin{align*}\alpha_{2h_i-1}^2=v_n\alpha_{2j_i-1} \end{align*} $$

for $2j_i-1=2(2h_i-1)-2^n+1$ , and $2j_i-1$ cannot be greater than $2h_i-1$ for $2^{n-2}<h_i\leq 2^{n-1}$ . Moreover, $2j_i-1=2h_i-1$ can only happen for $2h_i-1=2^n-1$ . Then assume that e contains a monomial $\alpha _{2h_1-1}\ldots \alpha _{2h_k-1}$ in its decomposition distinct from $1$ and $\alpha _{2^n-1}$ , and take the greatest such monomial in the lexicographical order. Then $e^2$ cannot contain this monomial, which provides a contradiction.

Then e is a linear combination of $1$ and $\alpha _{2^n-1}$ , and it remains to observe that $e^2=e$ implies that $1$ can only have the coefficient $0$ or $1$ , and $\alpha _{2^n-1}$ can only have the coefficient $0$ or $v_n^{-1}$ in the decomposition of e.

Proof. By Proposition 6.6, $H^*_{\mathrm K(n)}(q)^\vee $ contains non-trivial idempotents if and only if $\alpha _{2^n-1}\in H^*_{\mathrm K(n)}(q)^\vee $ (i.e., $e_{2^n-1}$ does not map to zero in $H^*_{\mathrm K(n)}$ ). The latter exactly means that $2^n-1\not \in J(q)$ .

Proof of Theorem 6.2.

According to [Reference Petrov and Semenov29, Theorem 5.7], there is a one-to-one correspondence between motivic decompositions of $\mathcal R$ in the category of $\mathrm K(n)$ -motives and direct sum decompositions of $H^*_{\mathrm K(n)}(q)^{\vee }$ as a module over itself.

However, $H^*_{\mathrm K(n)}(q)^{\vee }$ is indecomposable as a module over itself for $m\leq 2^{n+1}-2$ or $2^n-1\in J(q)$ by Corollary 6.8; therefore, $\mathcal R$ is also indecomposable in these cases.

If $m\geq 2^{n+1}-1$ and $2^n-1\not \in J(q)$ , then the only nontrivial idempotents in $H^*_{\mathrm K(n)}(q)^{\vee }$ are $v_n^{-1}\alpha _{2^n-1}$ and $1-v_n^{-1}\alpha _{2^n-1}$ . The corresponding modules $H_1=v_n^{-1}\alpha _{2^n-1}H^*_{\mathrm K(n)}(q)^{\vee }$ and $H_2=(1-v_n^{-1}\alpha _{2^n-1})H^*_{\mathrm K(n)}(q)^{\vee }$ have equal ranks, but they are not isomorphic as $H^*_{\mathrm K(n)}(q)^{\vee }$ -modules ( $\alpha _{2^n-1}$ acts as the identity on the first one, and as zero on the second one).