Proof. Consider the standard presentation of $\mathcal {M}_{1,1}$ as the quotient $\left [ \left (\mathbb {A}^2 \setminus V\right ) / \mathbb {G}_m \right ]$, where V is the curve $27x^2 + 4y^3 = 0$ and $\mathbb {G}_m$ acts by $(x, y) \mapsto \left (xt^6 , yt^4\right )$. The curve V is universally homeomorphic to $\mathbb {A}^1$, so we have an exact sequence

$$ \begin{align*} 0 \rightarrow \mathrm{M}^{\bullet}(k) \rightarrow A^{0}(\mathbb{A}^2 \setminus V, \mathrm{M}) \rightarrow \mathrm{M}^{\bullet}(k) \rightarrow 0. \\[-15pt]\end{align*} $$

The boundary of the element $\left \lbrace 27x^2 + 4y^3 \right \rbrace \in \mathrm {H}^{1}(k(x,y), \mu _\ell )$ at V is $1$, and it is unramified everywhere else, so it belongs to $A^{0}\left (\mathbb {A}^2 \setminus V, \mathrm {H}_{\mathbb {Z}/\ell \mathbb {Z}}\right )$. In particular, the submodule $\left \lbrace 27x^2 + 4y^3 \right \rbrace \cdot \mathrm {M}^{\bullet }(k)$ maps bijectively to $\mathrm {M}^{\bullet }(k) = A^{0}(V,\mathrm {M})$, splitting the exact sequence. Thus

$$ \begin{align*} A^{0}\left(\mathbb{A}^2 \setminus V, \mathrm{M}\right) = \mathrm{M}^{\bullet}(k)\oplus \left\lbrace 27x^2 + 4y^3 \right\rbrace \cdot \mathrm{M}^{\bullet}(k). \end{align*} $$

We have to understand which of these elements glue to invariants of $\mathcal {M}_{1,1}$. This is equivalent to checking whether the two pullbacks through

$$ \begin{align*} \mathrm{pr}_1, \mathrm{m}: \left(\mathbb{A}^2 \setminus V\right) \times \mathbb{G}_m = \left(\mathbb{A}^2 \setminus V\right) \times_{\mathcal{M}_{1,1}} \left(\mathbb{A}^2 \setminus V\right) \rightrightarrows \mathbb{A}^2 \setminus V \end{align*} $$

coincide, where $\mathrm {m}$ denotes the multiplication map. By Lemma 2.22 we have

$$ \begin{align*} A^{0}\left(\left(\mathbb{A}^2 \setminus V\right) \times \mathbb{G}_m, \mathrm{M}\right) = A^{0}\left(\mathbb{A}^2 \setminus V, \mathrm{M}\right) \oplus t \cdot A^{0}\left(\mathbb{A}^2 \setminus V, \mathrm{M}\right) \end{align*} $$

and $\mathrm {m}^*\left \lbrace 27x^2 + 4y^3 \right \rbrace = 12\lbrace t \rbrace + \left \lbrace 27x^2 + 4y^3 \right \rbrace $. Then an element $ \left \lbrace 27x^2 + 4y^3 \right \rbrace \cdot \alpha $ is unramified if and only if $\alpha \in \mathrm {M}^{\bullet }(k)_{12}$.

Proof. The formula for the cohomological Brauer group follows immediately by applying Theorem 3.1 to $\mathrm {M}=\mathrm {H}_{\mu _{\ell }^{\vee }}$ for increasing $\ell $. Then Lemmas 2.18 and 2.19 allow us to conclude immediately.

Proof. This is the content of [Reference Pirisi22, Thm. 4.16] in the case where the coefficients are in étale cohomology. The general case can be obtained in the exact same way, as an immediate consequence of Theorem 2.10 and Corollary 2.12.

Proof. Let V be a finite-dimensional vector space and let $\mathrm {Gr}_m(V)$ be the Grassmannian of m-subspaces of V. We are going to prove that

$$ \begin{align*} \mathrm{CH}^{i}(\mathrm{Gr}_m(V))\otimes \mathrm{M}^{\bullet}(k) \simeq A^{i}(\mathrm{Gr}_m(V),\mathrm{M}) \end{align*} $$

for every $i\geq 0$ and $0<m\leq \mathrm {dim}(V)$. This will readily imply the proposition, because of the isomorphisms

$$ \begin{align*} \mathrm{CH}^i(\mathrm{Gr}_m(V))\simeq \mathrm{CH}^i_{\textrm{GL}_{\textrm{m}}}(\mathrm{Spec}(k)),\qquad A^i(\mathrm{Gr}_m(V),\mathrm{M})\simeq A^i_{\textrm{GL}_{\textrm{m}}}(\mathrm{Spec}(k),\mathrm{M}), \end{align*} $$

for vector spaces V of sufficiently high dimension (the proof in [Reference Edidin and Graham10, Subsec. 3.2] for classical Chow rings holds verbatim).

Let $\mathrm {Fl}_m(V)$ be the scheme of complete flags of length m in V. Observe that $\mathrm {Fl}_m(V)$ can be constructed as a tower of projective bundles both over $\mathrm {Spec}(k)$ and $\mathrm {Gr}_m(V)$.

If the Chow ring of a smooth scheme X is generated by Chern classes of vector bundles, thanks to Proposition 2.21 there is a well-defined multiplication morphism

$$ \begin{align*} \mathrm{CH}^i(X)\otimes \mathrm{M}^{\bullet}(k) \rightarrow A^i(X,\mathrm{M}). \end{align*} $$

Moreover, if $f:Y\to X$ is a flat morphism of relative constant dimension from a smooth scheme whose Chow ring is also generated by Chern classes, it is easy to check that the pullback of a product is equal to the product of pullbacks.

Applying this to the morphism $\mathrm {Fl}_m(V)\to \mathrm {Gr}_m(V)$, we obtain the following commutative diagram:

(4.1)

The fact that $\mathrm {Fl}_m(V)$ is a tower of projective bundles over a point, combined with an iterated application of [Reference Pirisi21, Prop. 2.4], shows that the right vertical morphism in the diagram is an isomorphism.

Define a splitting s of the bottom horizontal arrow of diagram (4.1) as follows: at each level of the tower of projective bundles $\mathrm {Fl}_m(V)\to \mathrm {Gr}_m(V)$, multiply r times with the appropriate hyperplane section (here r is the dimension of the fibre at that level) and then take the push-forward to the level below. In the same way, we can construct a splitting $s'$ at the level of Chow groups.

The pullback $f^*:\mathrm {CH}^{*}(\mathrm {Gr}_m(V))\to \mathrm {CH}^{*}(\mathrm {Fl}_m(V))$ is then split injective, and from this it is easy to deduce that the top horizontal arrow in diagram (4.1) is injective, and hence the left vertical arrow must be injective as well.

To show surjectivity, observe that if $\alpha $ is an element in $A^{*}(\mathrm {Gr}_m(V),\mathrm {M})$, we have

$$ \begin{align*} \alpha=s(f^*(\alpha))=s\left(\sum \xi_i \cdot \beta_i\right)=\sum s'(\xi_i) \cdot \beta_i \end{align*} $$

for some $\xi _i$ in $\mathrm {CH}^{*}(\mathrm {Fl}_m(V))$ and $\beta _i$ in $\mathrm {M}^{\bullet }(k)$. Therefore the left vertical morphism of equation (4.1) is surjective and this concludes the proof.

Proof. We have $\mathrm {B}\mu _\ell =\left [\left (\mathbb {A}^1\setminus \{0\}\right )/\mathbb {G}_m\right ]$, where the action is defined as $\lambda \cdot x=\lambda ^\ell x$.

The localisation long exact sequence for the $\mathbb {G}_m$-equivariant embedding $\{0\}\hookrightarrow \mathbb {A}^1$ reads as follows:

$$ \begin{align*} \dotsb \to A^i_{\mathbb{G}_m}\left(\mathbb{A}^1,\mathrm{M}\right)\to A^i_{\mathbb{G}_m}\left(\mathbb{A}^1\setminus\{0\},\mathrm{M}\right) \to A^i_{\mathbb{G}_m}(\{0\},\mathrm{M}) \to A^{i+1}_{\mathbb{G}_m}\left(\mathbb{A}^1,\mathrm{M}\right) \to \dotsb. \end{align*} $$

By Proposition 4.2, we have

$$ \begin{align*} A_{\mathbb{G}_m}^i\left(\mathbb{A}^1,\mathrm{M}\right)\simeq A_{\mathbb{G}_m}^i(\{0\},\mathrm{M}) \simeq \mathrm{M}^{\bullet}(k) s^i. \end{align*} $$

Moreover by Proposition 2.21 the push-forward $i_*:A^i_{\mathbb {G}_m}(\{0\},\mathrm {M}) \to A^{i+1}_{\mathbb {G}_m}\left (\mathbb {A}^1,\mathrm {M}\right )$ corresponds to multiplication by $[\{0\}]=\ell s$.

Proposition 4.2 assures us that multiplying by s defines an injective morphism, hence we deduce

$$ \begin{align*} \mathrm{ker}(i_*)=\mathrm{M}^{\bullet}(k)_{\ell}\cdot s^i,\qquad \mathrm{im}(i_*)=\mathrm{M}^{\bullet}(k)\cdot \ell s^{i+1}. \end{align*} $$

It follows that for every $i\geq 0$ we have the following short exact sequence:

$$ \begin{align*} 0\to (\mathbb{Z}\left[s\right]/\ell s)\otimes\mathrm{M}^{\bullet}(k) \to A_{\mathbb{G}_m}^i\left(\mathbb{A}^1\setminus\{0\},\mathrm{M}\right) \to \mathrm{M}^{\bullet}(k)_{\ell}\cdot s^i \to 0. \end{align*} $$

Recall from Lemma 2.22 that there is a splitting

$$ \begin{align*} A^i(\{0\},\mathrm{M})\rightarrow A^i\left(\mathbb{A}^1\setminus\{0\},\mathrm{M}\right) \end{align*} $$

given by multiplication by the element $\{t\} \in A^0\left (\mathbb {A}^1\setminus \{0\},\mathrm {K}_{\mathrm {M}}^1\right )$.

If we consider $\{t\} \cdot \tau $, where $\tau \in \mathrm {M}^{\bullet }(k)$ is an $\ell $-torsion element, this element is invariant for the $\mathbb {G}_m$-action, as the action sends $\lbrace t\rbrace $ to $\left \lbrace \lambda ^\ell t\right \rbrace =\ell \lbrace \lambda \rbrace + \lbrace t \rbrace $. Thus it glues to an element $\gamma \in A^0_{\mathbb {G}_m}\left (\mathbb {A}^1\setminus \{0\}, \mathrm {M}\right )$, splitting the exact sequence for $i=0$.

Combining this splitting with multiplication by $s^i$, we obtain a splitting for every codimension i, proving our claim.

Proof. Note that if $\mathbb {G}_m$ acts trivially on X, then $A^i_{\mathbb {G}_m}(X,\mathrm {M})=A^i\left (X\times \mathbb {P}^r_k,\mathrm {M}\right )$ for some $r>i$, showing that

$$ \begin{align*} A^{*}_{\mathbb{G}_m}(X,\mathrm{M})=A^{*}(X,\mathrm{M})\left[s\right]= \mathrm{CH}^{*}_{\mathbb{G}_m}(\mathrm{Spec}(k))\otimes A^{*}(X,\mathrm{M}). \end{align*} $$

Using this, we can just repeat the proof of the previous Proposition 4.3 using the $\ell $-twisted action of $\mathbb {G}_m$ on $X \times \mathbb {G}_m$ and the compatibility of the pullback

$$ \begin{align*} A^{*}_{\mathbb{G}_m}(X,\mathrm{M})\rightarrow A^{*}_{\mathbb{G}_m}(X\times \mathbb{G}_m,\mathrm{M}) \end{align*} $$

with the long exact sequence.

Proof. Let $\mathrm {M}$ be an odd torsion cycle module. Using the description of $A^{*}_{\mu _2}(\mathrm {Spec}(k),\mathrm {M})$ given by Proposition 4.3, we see that every element of codimension $>0$ is of both $2$-torsion and $\ell $-torsion, hence $0$. We deduce

(4.2)$$ \begin{align} A^{*}_{\mathrm{O}_{\mathrm {1}}}(\mathrm{Spec}(k),\mathrm{M})\simeq A^{*}_{\mu_2}(\mathrm{Spec}(k),\mathrm{M})\simeq \mathrm{M}^{\bullet}(k). \end{align} $$

Define $C\subset V\setminus \{0\}$ as the vanishing locus of q. Let B be the complement of C in $V\setminus \{0\}$. All these schemes are $\mathrm {O}_{\mathrm n}$-invariant.

We can compute $A_{\mathrm {O}_{\mathrm n}}^{*}(B,\mathrm {M})$ as follows: consider the morphism $B\to \mathbb {G}_m$ induced by q and form the Cartesian square

where the bottom morphism is the square map. We have $\widetilde {B}\simeq Q\times \mathbb {G}_m$, where $Q\subset B$ is the fibre over the unit in $\mathbb {G}_m$, and moreover both horizontal arrows are $\mu _2$-torsors. Therefore

(4.3)$$ \begin{align} A_{\mathrm{O}_n\times\mu_2}^{*}(Q\times\mathbb{G}_m,\mathrm{M})\simeq A_{\mathrm{O}_n}^{*}(B,\mathrm{M}). \end{align} $$

Let $\mathcal {L}$ be the $(\mathrm {O}_{\mathrm n}\times \mu _2)$-equivariant line bundle over Q such that the equivariant $\mathbb {G}_m$-torsor $Q\times \mathbb {G}_m\to Q$ is the complement in $\mathcal {L}$ of the zero section. The associated localisation exact sequence is

(4.4)

Via the identification $A^{*}_{\mathrm {O}_{\mathrm n}\times \mu _2}(\mathcal {L},\mathrm {M})\simeq A^{*}_{\mathrm {O}_{\mathrm n}\times \mu _2}(Q,\mathrm {M})$, we see that the last map is given by multiplication by $c_1^{\mathrm {O}_{\mathrm n}\times \mu _2}(\mathcal {L})$.

The action of $\mathrm {O}_{\mathrm n}\times \mu _2$ on Q is transitive: its stabiliser is $\mathrm {O}_{\mathrm{n-1}}\times \mu _2$, hence $[Q/\mathrm {O}_{\mathrm n}]\simeq B(\mathrm {O}_{\mathrm{n-1}}\times \mu _2)$ and

(4.5)$$ \begin{align} A_{\mathrm{O}_{\mathrm{n}}\times\mu_2}^{*}(Q,\mathrm{M})\simeq A_{\mathrm{O}_{n-1}\times\mu_2}^{*}(\mathrm{Spec}(k),\mathrm{M}) \simeq A_{\mathrm{O}_{\mathrm{n-1}}}^{*}(\mathrm{Spec}(k),\mathrm{M}), \end{align} $$

where the last isomorphism is a consequence of the triviality of $A_{\mu _2}^{*}(\mathrm {Spec}(k),\mathrm {M})$.

By Proposition 2.21, we have

$$ \begin{align*}c_1^{\mathrm{O}_{\mathrm n}\times\mu_2}(\mathcal{L})(\alpha)=c_1^{\mathrm{O}_{\mathrm{n}}\times\mu_2}(\mathcal{L})(1)\cdot\alpha \end{align*} $$

for every $\alpha $ in $A^{*}_{\mathrm {O}_{\mathrm n}\times \mu _2}(Q,\mathrm {M})$, where $c_1^{\mathrm {O}_{\mathrm n}\times \mu _2}(\mathcal {L})$ is regarded as an element inside $A^1_{\mathrm {O}_{\mathrm n}\times \mu _2}\left (Q,\mathrm {H}_{\mathbb {Z}/\ell \mathbb {Z}}\right )$.

Observe that from formula (4.5), it follows that the part of $A^1_{\mathrm {O}_{\mathrm n}\times \mu _2}\left (Q,\mathrm {H}_{\mathbb {Z}/\ell \mathbb {Z}}\right )$ of cohomological degree $0$ is equal to $\mathbb {Z}/\ell \mathbb {Z} \cdot c_1/\langle 2\cdot c_1 \rangle =0$. Therefore, from sequence (4.4) we deduce the exact sequence

$$ \begin{align*} 0\to A^i_{\mathrm{O}_{\mathrm{n}}\times\mu_2} (Q,\mathrm{M}) \to A^i_{\mathrm{O}_{\mathrm{n}}\times\mu_2} (Q\times\mathbb{G}_m,\mathrm{M}) \to A^i_{\mathrm{O}_n\times\mu_2}(Q,\mathrm{M})\to 0. \end{align*} $$

Combining this with the content of formulas (4.5) and (4.3), we get

(4.6)$$ \begin{align} A^{*}_{\mathrm{O}_n}(B,\mathrm{M})\simeq A^{*}_{\mathrm{O}_{n-1}}(\mathrm{Spec}(k),\mathrm{M})\oplus A^{*}_{\mathrm{O}_{n-1}}(\mathrm{Spec}(k),\mathrm{M})\cdot\beta \end{align} $$

as $\mathrm {H}^{\bullet }_{\mathbb {Z}/\ell \mathbb {Z}}(k)$-modules, where $\beta $ has codimension $0$ and cohomological degree $1$.

Next we compute $A_{\mathrm {O}_{\mathrm n}}^{*}(C,\mathrm {M})$, where C is the vanishing locus of q in $V\setminus \{0\}$. The action of $\mathrm {O}_{\mathrm n}$ on C is transitive: the stabiliser is given by the semidirect product of $\mathrm{O}_{\mathrm n-2}$ with a subgroup H of the group of affine transformations of a vector space W. More precisely, the group H is isomorphic to W via the natural projection of $\mathrm {Aff}(V)$ on translations. It follows from [Reference Pirisi23, Lemma 1.7] that

(4.7)$$ \begin{align} A_{\mathrm{O}_n}^{*}(C,\mathrm{M})\simeq A_{\mathrm{O}_{\mathrm n-2}\ltimes H}^*(\mathrm{Spec}(k),\mathrm{M})\simeq A_{\mathrm{O}_{\mathrm n-2}}^*(\mathrm{Spec}(k),\mathrm{M}). \end{align} $$

Next we compute $A^{*}_{\mathrm{O}_{\mathrm {2}}}(\mathrm {Spec}(k),\mathrm {M})$. First observe that $O_0=\{id\}$ and $\mathrm{O}_{\mathrm {1}}\simeq \mu _2$. Consider the push-forward morphism

$$ \begin{align*} i_*:A^{*}_{\mathrm{O}_{\mathrm {2}}}(C,\mathrm{M})\simeq \mathrm{M}^{\bullet}(k) \rightarrow A^{*+1}_{\mathrm{O}_{\mathrm {2}}}(V\setminus\{0\},\mathrm{M}). \end{align*} $$

The isomorphism on the left side comes from formula (4.7). By the projection formula, $i_*\alpha =\alpha \cdot i_*(1)$, and $i_*(1)$ is in degree $0$ and codimension $1$. We have $A^1_{\mathrm{O}_{\mathrm {2}}}(V\setminus \{0\},\mathrm {M})\simeq A^1_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm {M})$, which is in turn isomorphic to $A^1_{\mathrm{O}_{\mathrm {2}}}(\mathrm {Spec}(k),\mathrm {M})$.

The piece of the latter group of cohomological degree $0$ turns out to be equal to $\mathrm {CH}^1_{\mathrm{O}_{\mathrm {2}}}(\mathrm {Spec}(k))\otimes M^0(k)$, which vanishes because it is of both $2$-torsion and $\ell $-torsion. We deduce that $i_*=0$.

This last remark, applied to the localisation exact sequence for $C\subset V\setminus \{0\}$, implies that the following short sequence is exact:

(4.8)$$ \begin{align} 0\to A^{*}_{\mathrm{O}_{\mathrm {2}}}(V\setminus\{0\},\mathrm{M})\to A^{*}_{\mathrm{O}_{\mathrm {2}}}(B,\mathrm{M}) \to A^{*}_{\mathrm{O}_{\mathrm {2}}}(C,\mathrm{M}) \to 0. \end{align} $$

By formula (4.7), the group on the right is equal to $\mathrm {M}^{\bullet }(k)$ in codimension $0$ and is $0$ in higher codimension. By formula (4.6), the group in the middle is equal to $A^{*}_{\mu _2}(\mathrm {Spec}(k),\mathrm {M})\oplus A^{*}_{\mu _2}(\mathrm {Spec}(k),\mathrm {M})\cdot \beta $ in codimension $0$ and vanishes in higher codimension, where $\beta $ has cohomological degree $1$.

From formula (4.2) we know that $A^{*}_{\mu _2}(\mathrm {Spec}(k),\mathrm {M})$ is trivial, hence the short exact sequence (4.8) implies

(4.9)$$ \begin{align} A^{*}_{\mathrm{O}_{\mathrm {2}}}(V\setminus\{0\},\mathrm{M})\simeq \mathrm{M}^{\bullet}(k). \end{align} $$

Now consider the localisation exact sequence

(4.10)$$ \begin{align} \dotsb\to A^i_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm{M}) \to A^i_{\mathrm{O}_{\mathrm {2}}}(V\setminus\{0\},\mathrm{M}) \to A^{i-1}_{\mathrm{O}_{\mathrm {2}}}(\{0\},\mathrm{M}) \to A^{i+1}_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm{M})\to\dotsb \end{align} $$

induced by the open embedding $V\setminus \{0\}\hookrightarrow V$.

Observe that $A^0_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm {M})\simeq A^0_{\mathrm{O}_{\mathrm {2}}}(V\setminus \{0\})$ because the codimension of $\{0\}$ in V is $>1$. Moreover, from formula (4.9) we know that $A^i_{\mathrm{O}_{\mathrm {2}}}(V\setminus \{0\},\mathrm {M})=0$ for $i>0$. Observe also that the push-forward induced by the closed embedding $\{0\}\hookrightarrow V$ coincides with multiplication by $c_2^{\mathrm{O}_{\mathrm {2}}}(V)=:c_2$.

Hence from the long exact sequence (4.10) we readily deduce the following:

• • $A^0_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm {M})\simeq \mathrm {M}^{\bullet }(k)$.

• • $A^i_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm {M})=0$ for i odd.

• • $A^{i+2}_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm {M})\simeq c_2\cdot A^i_{\mathrm{O}_{\mathrm {2}}}(V,\mathrm {M})$.

Therefore

(4.11)$$ \begin{align} A^{*}_{\mathrm{O}_{\mathrm {2}}}(\mathrm{Spec}(k),\mathrm{M})\simeq\mathrm{M}^{\bullet}(k)[c_2]\simeq \mathrm{CH}_{\mathrm{O}_{\mathrm {2}}}^*(\mathrm{Spec}(k))\otimes\mathrm{M}^{\bullet}(k). \end{align} $$

Next we prove that the same result holds for $O_3$. As before, we start by computing $A_{\mathrm{O}_{\mathrm{3}}}^{*}(V\setminus \{0\},\mathrm {M})$.

The triviality of $A^{*}_{\mu _2}(\mathrm {Spec}(k),\mathrm {M})$ together with formula (4.7) implies that $A_{\mathrm{O}_{\mathrm{3}}}^{*}(C,\mathrm {M})\simeq \mathrm {M}^{\bullet }(k)$. The same argument used for $\mathrm{O}_{\mathrm {2}}$ shows that the push-forward morphism $i_*:A^{*}_{\mathrm{O}_{\mathrm{3}}}(C,\mathrm {M})\to A^{*}_{\mathrm{O}_{\mathrm{3}}}(V\setminus \{0\},\mathrm {M})$ is zero.

Plugging this information into the localisation exact sequence induced by the open embedding $B\hookrightarrow V\setminus \{0\}$, we deduce that $A^i_{\mathrm{O}_{\mathrm{3}}}(V\setminus \{0\},\mathrm {M})\simeq A^i_{\mathrm{O}_{\mathrm{3}}}(B,\mathrm {M})$ for $i>0$ and that the following is exact:

$$ \begin{align*} 0\to A^0_{\mathrm{O}_{\mathrm{3}}}(V\setminus\{0\},\mathrm{M}) \to A^0_{\mathrm{O}_{\mathrm{3}}}(B,\mathrm{M}) \to A^0_{\mathrm{O}_{\mathrm{3}}}(C,\mathrm{M})\simeq \mathrm{M}^{\bullet}(k) \to 0. \end{align*} $$

By formula (4.6) together with formula (4.11), we have

$$ \begin{align*} A_{\mathrm{O}_{\mathrm{3}}}^{*}(B,\mathrm{M})\simeq \mathrm{M}^{\bullet}(k)[c_2] \oplus \mathrm{M}^{\bullet}(k)[c_2] \cdot \beta, \end{align*} $$

with $\beta $ of codimension $0$ and cohomological degree $1$. We deduce that

$$ \begin{align*} A^{*}_{\mathrm{O}_{\mathrm{3}}}(V\setminus\{0\},\mathrm{M})\simeq \mathrm{M}^{\bullet}(k)[c_2]\oplus\mathrm{M}^{\bullet}(k)[c_2]\cdot\gamma, \end{align*} $$

where $\gamma $ has codimension $2$ and cohomological degree $1$.

Observe that the push-forward morphism $A^{*}_{\mathrm{O}_{\mathrm{3}}}(\{0\},\mathrm {M})\to A^{*+3}_{\mathrm{O}_{\mathrm{3}}}(V,\mathrm {M})$ coincides with multiplication by $c_3(1)$, which is both a $2$-torsion and an $\ell $-torsion element, hence zero.

Consider now the open embedding $V\setminus \{0\}\hookrightarrow V$. By what we have just proved, the following short sequences are exact for every i:

$$ \begin{align*} 0\to A^i_{\mathrm{O}_{\mathrm{3}}}(V,\mathrm{M}) \to A^i_{\mathrm{O}_{\mathrm{3}}}(V\setminus\{0\},\mathrm{M})\to A^{i-2}_{\mathrm{O}_{\mathrm{3}}}(\{0\},\mathrm{M})\to 0. \end{align*} $$

We immediately deduce the following:

• • $A^0_{\mathrm{O}_{\mathrm{3}}}(V,\mathrm {M})\simeq \mathrm {M}^{\bullet }(k)$.

• • $A^i_{\mathrm{O}_{\mathrm{3}}}(V,\mathrm {M})=0$ for i odd.

• • $A^{2i}_{\mathrm{O}_{\mathrm{3}}}(V,\mathrm {M})\simeq \mathrm {M}^{\bullet }(k)\cdot c_2^i$.

Therefore

(4.12)$$ \begin{align} A^{*}_{\mathrm{O}_{\mathrm{3}}}(\mathrm{Spec}(k),\mathrm{M})\simeq \mathrm{M}^{\bullet}(k)[c_2]\simeq \mathrm{CH}^{*}_{\mathrm{O}_{\mathrm{3}}}(\mathrm{Spec}(k))\otimes\mathrm{M}^{\bullet}(k), \end{align} $$

which concludes the proof of the lemma.

Proof of Proposition 4.5. Observe that $\mathrm {PGL}_2\simeq \mathrm {SO}_3$ and $O_3\simeq \mathrm {SO}_3\times \mu _2$. The result is then a direct consequence of Proposition 4.4 and Lemma 4.6.

Proof. By the compatibility of the morphism of cycle modules with the boundary map, the ramification of $\alpha '$ at any point x of codimension $1$ has to belong to the kernel of $\mathrm {K}_{\mathrm {Mil}}^{\bullet }(k(x)) \rightarrow \mathrm {K}_{\ell }^{\bullet }(k(x))$, which means it is a multiple of $\ell $, so $\alpha ' \cdot x$ is unramified as $\partial _v(\alpha ' \cdot x)=\partial _v(\alpha ')\cdot x = 0$. By the same reasoning, note that if $\alpha ''$ is another inverse image of $\alpha $, the difference $\alpha ' - \alpha ''$ has to be a multiple of $\ell $, concluding the proof.

Proof. For the sake of simplicity, we give a proof in the nonequivariant setting. The general case can be deduced in the same way using equivariant approximation.

Let $\alpha $ be an element of $A^0(X,\mathrm {K}_{\ell })\subset \mathrm {K}_{\ell }^{\bullet }(k(X))$ and let $\beta $ be an element of $A^0_G(X,\mathrm {M})_\ell $. For every inverse image $\alpha '$ of $\alpha $ in $\mathrm {K}_{\mathrm {Mil}}^{\bullet }(k(X))$, we have a well-defined product $\alpha '\cdot \beta $ in $\mathrm {M}^{\bullet }(k(X))$: this product is unramified on X because the ramification of $\alpha '$ at any point of codimension $1$ is a multiple of $\ell $, just as in the proof of the Lemma 5.1.

By the same reasoning, the definition of the product does not depend on the choice of an inverse image $\alpha '$. The compatibility with the pullback follows from the compatibility of the product $A^0(X,\mathrm {K}_{\mathrm {Mil}})\otimes A^0(X,\mathrm {M}) \to A^0(X,\mathrm {M})$.

Proof. Consider the $2$-dimensional $\mu _\ell $-representation V given by diagonal matrices. The representation is free outside of $0$, so $V/\mu _\ell $ is an equivariant approximation of $\mathrm {B}(\mu _\ell )$, and $V^i/\mu _\ell ^i$ is an equivariant approximation of $\mathrm {B}\left (\mu _\ell ^i\right )$.

Our claim is true for $r=1$ by Proposition 4.3. Assume by induction that it is true for $r-1$; then it follows immediately by applying Proposition 4.4 to $V^{r-1}/\mu _\ell ^{r-1}$.

Proof. Obviously the first point implies the second. We claim that the converse is also true. We have a commutative square

The left-hand vertical map is surjective by hypothesis. The bottom horizontal map is surjective as well; it can be easily seen from the structure of $\mathrm {Inv}^{\bullet }\left (\mathrm {B} \mu _l^r, \mathrm {M}\right )$ that the G-invariant elements are generated by those in the form

$$ \begin{align*} \alpha_{J} + \alpha_{g_1(J)} + \dotsb + \alpha_{g_m(J)}, \end{align*} $$

where $\lbrace J, g_1(J), \dotsc , g_m(J)\rbrace $ is the orbit of the subset $J \subset \lbrace 1, \dotsc , r\rbrace $, and these elements belong to the image of the bottom horizontal map. Then the right-hand vertical map is injective (by hypothesis) and surjective. This implies that the map on nonnormalised cohomological invariants is surjective as well, proving our claim.

Proof. The fact that all normalised invariants are of $2$-torsion is an obvious consequence of the inclusion $\mathrm {Inv}^{\bullet }(\mathrm {B} \mathrm {O}_{\mathrm n},\mathrm {M}) \subset \mathrm {Inv}^{\bullet }\left (\mathrm {B} \mu _2^n,\mathrm {M}\right )$. Note that moreover, we have $\mathrm {Inv}^{\bullet }(\mathrm {BO}_n,\mathrm {M}) \subset \mathrm {Inv}^{\bullet }\left (\mathrm {B}\mu _2^n, \mathrm {M}\right )^{\mathrm {S}_{\textrm{n}}}$, by the gluing conditions, and $\mathrm {Inv}^{\bullet }\left (\mathrm {B}\mu _2^n, \mathrm {M}\right )^{\mathrm {S}_{\textrm{n}}}$ is exactly the group in our formula. The description of the cohomological invariants with coefficients in a $\mathrm {K}_2$ is obtained in [Reference Gille and Hirsch15, Sec. 23]; the description for an arbitrary cycle module then follows by Lemma 5.4 as $\mathrm {S}_{\textrm{n}}$ acts on $\mu _2^n$ by permuting the components.

Proof. This is a combination of [Reference Garibaldi, Merkurjev and Serre16, Thm. 7, Cor. 8].

Proof. This is immediate from Lemma 5.3.

Proof. We start by proving (1). Let $H\simeq (\mathbb {Z}/2\mathbb {Z}) ^{\times m}$ be the subgroup of $\mathrm {S}_{\textrm{n}}$ generated by the transpositions $(1,2)$, $(3,4),\dotsc , (n-1,n)$ for n even and by $(1,2),\dotsc , (n-2,n-1)$ for n odd. Observe that if $m=\left \lfloor {n/2}\right \rfloor $, the normaliser of H is a copy of $\mathrm {S}_{\textrm{m}}$, which acts on H by permuting its generators. By Theorem 5.7, the pullback of invariants along the morphism $\mathrm {B} H\to \mathrm {B} \mathrm {S}_n$ induces an injective morphism

$$ \begin{align*} \mathrm{Inv}^{\bullet}(\mathrm{B} \mathrm{S}_n,\mathrm{M})\longrightarrow \mathrm{Inv}^{\bullet}(\mathrm{B} H,\mathrm{M})^{\mathrm {S}_{\textrm{m}}} \end{align*} $$

for all $\mathrm {M}$. Then by Lemma 5.4 we only need to prove the case of $\mathrm {M}=\mathrm {K}_2$. This is done in in [Reference Gille and Hirsch15, Thm. 25.6.(1)].

To prove (2), note that for n odd, we have $\mathrm {SO}_n = \mathrm {O}_{\mathrm n} \times \mu _2$. Then we can combine Proposition 4.3 with Proposition 5.5 to conclude that

$$ \begin{align*} \mathrm{Inv}^{\bullet}(\mathrm{B} \mathrm{SO}_n,\mathrm{M}) = \mathrm{Inv}^{\bullet}(\mathrm{B} \mathrm{O}_n,\mathrm{M})/\alpha_1 \cdot \mathrm{Inv}^{\bullet}(\mathrm{B} \mathrm{O}_n,\mathrm{M}), \end{align*} $$

where the multiplication is given by seeing $\alpha _1$ as an element of $\mathrm {Inv}^{\bullet }(\mathrm {B} \mathrm {O}_{\mathrm n},\mathrm {K}_2)$. The multiplicative structure of $\mathrm {Inv}^{\bullet }(\mathrm {B} \mathrm {O}_{\mathrm n},\mathrm {K}_2)$ (see [Reference Gille and Hirsch15, Remark 17.4]) immediately shows that the resulting submodule is generated by the odd-degree Stiefel–Whitney classes. Our claim follows.

To prove the last statement, we just note that in characteristic different from $2$ we have an isomorphism $\mathrm {PGL}_2 \simeq \mathrm {SO}_3$.

Proof. The formulas for the cohomological Brauer group are immediate from the description of the generalised cohomological invariants given in Proposition 5.10.

For $\mathrm {B}\mathrm {S}_n$, we know that the Brauer group is equal to the cohomological Brauer group thanks to Proposition 2.2.

For $\mathrm {B}\mathrm {PGL}_2$, note that the universal conic $\mathcal {C}=\left [P^1/\mathrm {PGL}_2\right ]$ induces a nonzero element $\lbrace \mathcal {C} \rbrace $ in the Brauer group which is trivial when pulled back to the base field. This implies that $\lbrace \mathcal {C} \rbrace = w_2 + \alpha _0 $, where $\alpha _0$ belongs to $\mathrm {Br}(k)$, showing that $\mathrm {Br}(\mathrm {B}\mathrm {PGL}_2)=\mathrm {Br}'(\mathrm {B}\mathrm {PGL}_2)$.

Proof. This is proven in [Reference Arsie and Vistoli2, 4.7].

Proof. This is [Reference Pirisi21, Prop. 3.3].

Proof. This is a combination of [Reference Di Lorenzo and Pirisi8, Prop. 3.1- Cor. 3.2].

Proof. For $G=\mathrm {GL}_2$, the statement is proved in [Reference Pirisi21, Prop. 4.3]. For $G=\mathrm {PGL}_2$, see [Reference Di Lorenzo5, Prop. 5.2], albeit there is a mistake in the statement of the result, where $4(n-2)h_n$ should be replaced by $(4n-2)h_n$ (note that the n in the statement of [Reference Di Lorenzo5, Prop. 5.2] would be $n/2$ in our lemma’s notation).

Proof. We have

$$ \begin{align*} A^*_{G}\left(P^1,\mathrm{M}\right) \simeq A^{*}_{H}(\mathrm{Spec}(k),\mathrm{M}), \end{align*} $$

where H is the stabiliser of a point in $P^1$, which is isomorphic to a semidirect product $\mathbb {G}_m \ltimes \mathbb {G}_a$. Then by [Reference Pirisi23, Reference Di Lorenzo and Pirisi8] we have

$$ \begin{align*} A^{*}_{H}(\mathrm{Spec}(k),\mathrm{M})\simeq A^{*}_{\mathbb{G}_m}(\mathrm{Spec}(k),\mathrm{M}), \end{align*} $$

which allows us to conclude immediately.

Proof. As $A^{0}_{G}\left (\Delta ^n_{1},\mathrm {M}\right )$ is isomorphic to $A^{0}_{G}\left (\Delta ^n_{\left [ 1,2 \right ]},\mathrm {M}\right ) \left (\text {because}\ \Delta ^n_{3}\ \text {has codimension}\ 2\ \text {in}\ \Delta ^n_{1}\right )$, we can compute it using the following exact sequence:

$$ \begin{align*} 0 \rightarrow A^{0}_{G}\left(\Delta^n_{\left[ 1,2\right]},\mathrm{M}\right) \rightarrow A^{0}_{G}\left(\Delta^n_{\left[1\right]},\mathrm{M}\right) \xrightarrow{\partial} A^{0}_{G}\left(\Delta^n_{\left[2\right]},\mathrm{M}\right). \end{align*} $$

We want to prove that the kernel of $\partial $ is equal to $\mathrm {M}^{\bullet }(k)$. This will then imply that the image of $A^{0}_{G}\left (\Delta ^n_{\left [1,2\right ]},\mathrm {M}\right )$ must be equal to $\mathrm {M}^{\bullet }(k)$, thus showing that $A^{0}_{G}\left (\Delta ^n_{\left [1,2\right ]},\mathrm {M}\right )=\mathrm {M}^{\bullet }(k)$.

The map $P^{n-2}_{\left [0,1\right ]} \times P^{1} \xrightarrow {\pi } \Delta ^n_{\left [1,2\right ]} $ yields the following commutative diagram with exact columns:

The second horizontal map is an isomorphism by Proposition 2.23, because $\pi $ is a universal homeomorphism when restricted to $\Delta ^n_{\left [ 1\right ]}$ (Proposition 6.2).

The kernel of $\partial _{1}$ is $\mathrm {M}^{\bullet }(k)$, because $A^{0}_{G}\left (P^{n-2}_{\left [0,1\right ]} \times P^1,\mathrm {M}\right )=\mathrm {M}^{\bullet }(k)$, as we have $A^{0}_{G}\left (P^{n-2} \times P^1,\mathrm {M}\right )=\mathrm {M}^{\bullet }(k)$ by the projective bundle formula and $\Delta ^{n-2}_{2} \times P^r$ has codimension $2$.

We claim that the third horizontal map is injective, implying that the kernel of $\partial $ must be equal to $\mathrm {M}^{\bullet }(k)$ as well.

Let $\psi $ be the map from $P^{n-4}_{\left [0\right ]} \times P^{1} \times P^1$ to $P^{n-4}_{\left [0\right ]} \times P^{2}$ sending $(f,g,h)$ to $(f,gh)$. We have a commutative diagram:

where $\pi _1$ and $\pi _2$ are defined respectively by $(f,g,h) \rightarrow \left (fg^2,h\right )$ and $(f,g) \rightarrow \left (fg^2\right )$. The maps $\pi _1$ and $\pi _2$ are universal homeomorphisms, so the push-forward maps $(\pi _1)_*,(\pi _2)_*$ are isomorphisms. Then if we prove that $\psi _*$ is injective, $\pi _*$ will be injective too. Consider this last diagram:

The pullbacks along $p_1$ and $p_2$ are both surjective, the first due to Lemma 7.3 and the second due to the projection formula, implying that the pullback along $\psi $ is surjective. We have $\psi _* ( \psi ^* \alpha ) = \mathrm {deg}(\psi ) \alpha $ by the projection formula. Then, as the degree of $\psi $ is $2$, $\psi _*$ is injective.

Proof. We begin by computing the invariants of $\left [ P^{2g+2}_{\mathrm {sm}}/ G \right ]$.

First consider the case of $\mathrm {M}=\mathrm {K}_{\ell }$. Consider the exact sequence

We have $A^{0}_G\left (\Delta ^{2g+2}_{1},\mathrm {K}_{\ell }\right )=\mathrm {K}_{\ell }^{\bullet }(k)$, and the push-forward map is just multiplication by $\left \lbrace \Delta ^{2g+2}_{1} \right \rbrace $. Assume first that $\ell $ divides $2g+1$, so that the image of $\left \lbrace \Delta ^{2g+2}_{1} \right \rbrace $ is zero. Then if we pick an inverse image $\gamma $ of $1$, the submodule $\gamma \cdot \mathrm {K}_{\ell }^{\bullet }(k)$ maps isomorphically to $A^{0}_G\left (\Delta _{1,n},\mathrm {K}_{\ell }\right )$, splitting the exact sequence.

Now let $\ell $ be arbitrary and let $\ell '$ be the greatest common divisor of $\ell $ and $2g+1$. Let $\gamma \in A^0_G\left (P^{2g+2}_{\mathrm {sm}},\mathrm {K}_{\ell '}\right )$ be as before.

Then for any $x \in \mathrm {K}_{\ell }^{\bullet }(k)_{\ell '}$, the element $\gamma \cdot x$ belongs to $A^0\left (P^{2g+2}_{\mathrm {sm}},\mathrm {K}_{\ell }\right )$, and moreover the boundary map sends $\gamma \cdot \mathrm {K}_{\ell }^{\bullet }(k)_{\ell '}$ to $\mathrm {K}_{\ell }^{\bullet }(k)_{\ell '} \subset A^0\left (\Delta ^{2g+2}_{1},\mathrm {K}_\ell \right )$, which is exactly the kernel, splitting the exact sequence.

For an arbitrary $\ell $-torsion cycle module $\mathrm {M}$, take $\gamma $ as before. We have an inclusion $\gamma \cdot \mathrm {M}^{\bullet }(k)_{\ell '}\subseteq A^{0}_G\left (P^{2g+2}_{\mathrm {sm}},\mathrm {M}\right )$, as these elements are clearly unramified. On the other hand, using the fact that $A^{0}_G\left (\Delta ^{2g+2}_{1},\mathrm {K}_{\ell }\right )=\mathrm {M}^{\bullet }(k)$, it is easy to see that we must have $A^{0}_G\left (P^{2g+2}_{\mathrm {sm}},\mathrm {M}\right ) = \gamma \cdot \mathrm {M}^{\bullet }(k)_{\ell '}$.

The next step consists of verifying that the $\mathbb {G}_m$-torsor ${\mathcal {H}}_g \rightarrow \left [ P^{2g+2}_{\mathrm {sm}}/ G \right ]$ does not generate any new invariant. This can be done as in the proof of Theorem 7.9. It is equivalent to proving that the first Chern class of the associated line bundle $\mathcal {L}$ does not annihilate any element of $A^{0}_G\left (P^{2g+2}_{\mathrm {sm}},\mathrm {M}\right )$; as the class is injective on the copy of $\mathrm {M}^{\bullet }(k)$ coming from the base field, we only have to check it on the submodule $\gamma \cdot \mathrm {M}^{\bullet }(k)$. This is done exactly as in the proof of Theorem 7.9, by taking the boundary of the elements and verifying that it is nonzero.

Proof of Lemma 7.3. First note that if G is $\mathrm {GL}_2$ or in the nonequivariant case, the statement is obvious by the projective bundle formula.

Now consider the case where $G=\mathrm {PGL}_2$. We have

$$ \begin{align*} A^0_{G}\left(P^n_{\mathrm{sm}}\times P^{1} \times P^1,\mathrm{M}\right) \simeq A^{0}_{H}\left(P^{n}_{\mathrm{sm}}\times P^{1},\mathrm{M}\right), \end{align*} $$

where H is the stabiliser of a point in $P^1$, which is isomorphic to a semidirect product $\mathbb {G}_m \ltimes \mathbb {G}_a$. As $\mathrm {H}$ is a special group, the pullback

$$ \begin{align*} A^{0}_{H}\left(P^{n}_{\mathrm{sm}}\times P^{1},\mathrm{M}\right) \rightarrow A^{0}\left(P^{n}_{\mathrm{sm}}\times P^{1}, \mathrm{M}\right)=A^{0}\left(P^{n}_{\mathrm{sm}}, \mathrm{M}\right) \end{align*} $$

has to be injective. We claim that the pullback $A^0_G\left (P^n_{\mathrm {sm}},\mathrm {M}\right ) \rightarrow A^0\left (P^n_{\mathrm {sm}},\mathrm {M}\right )$ is surjective, which would allow us to conclude immediately.

One can use the same techniques used to prove Lemma 7.4 to easily show that when $p \neq 2$, the nonequivariant group $A^{0}\left ( \Delta ^n_{1},\mathrm {M}\right )$ is equal to $\mathrm {M}^{\bullet }(k)$, and thus $A^{0}\left (P^n_{\mathrm {sm}},\mathrm {M}\right )$ either is $\mathrm {M}^{\bullet }(k)$ or is generated by $1$ and an element in degree $1$ corresponding to the equation for $\Delta _{1}^{n}$, multiplied by the submodule of $\mathrm {M}^{\bullet }(k)$ which annihilates the class of $\Delta _{1}^{n}$ in $A^1(P^{n},\mathrm {M})$.

In the latter case, consider the following commutative diagram induced by the pullback from equivariant to nonequivariant Chow groups with coefficients:

Both the top and the bottom horizontal map are isomorphisms, as one can see using the fact that the groups on the top row are $M(k)$ and the groups on the bottom row are given by $c_1^G\left (\mathcal {O}_{P^{n}}(-1)\right )\cdot \mathrm {M}^{\bullet }(k)$.

Moreover, $A^{0}\left (P^{n}_{\mathrm {sm}},\mathrm {M}\right )$ is generated as by $1 \cdot \mathrm {M}^{\bullet }(k)$ and an element $\alpha \cdot \mathrm {M}^{\bullet }(k)$, where $\alpha $ is an element such that $\partial (\alpha )=1$ in $A^{0}_{G}\left (\Delta ^n_{1},\mathrm {M}\right )$, with no additional relations.

The third horizontal map maps $1 \in A^{0}_{G}\left (\Delta _{1}^n,\mathrm {M} \right )$ to $1 \in A^{0}\left (\Delta _{1}^n,\mathrm {M}\right )$, which shows that an element $\tau $ of degree $0$ maps to $0$ in the equivariant group $ A^{1}_{G}\left (P^{n},\mathrm {M}\right )$ if and only if it maps to $0$ in $ A^{1}(P^n,\mathrm {M})$. Then there must be an element

$$ \begin{align*} \alpha' \in A^{0}_{G}\left(P^{n}_{\mathrm{sm}},\mathrm{M}\right) \end{align*} $$

which maps to $\alpha \in A^{0}\left (P^{n}_{\mathrm {sm}},\mathrm {M}\right )$, and thus the pullback $A^{0}_{G}\left (P^{n}_{\mathrm {sm}},\mathrm {M}\right ) \rightarrow A^0\left (P^{n}_{\mathrm {sm}},\mathrm {M}\right )$ is surjective. This concludes the proof.

Proof. Consider the exact sequence

$$ \begin{align*} A^{0}_G\left(P^n,\mathrm{K}_{2^r}\right) \hookrightarrow A^{0}_G\left(P^n_{\mathrm{sm}},\mathrm{K}_{2^r}\right) \xrightarrow{\partial} A^{0}_G\left(\Delta^n_{1},\mathrm{K}_{2^r}\right) \subset A^{0} _G\left(P^{n-2}_{\mathrm{sm}}\times P^1, \mathrm{K}_{2^r}\right). \end{align*} $$

First consider the case $r=1$. We will proceed by induction on n. If we assume the result for $n-2$, we have that $ A^{0} _G\left (P^{n-2}_{\mathrm {sm}}\times P^1, \mathrm {K}_{2^r}\right )=\mathrm {Inv}^{\bullet }(\mathrm {B} \mathrm {S}_{n-2},\mathrm {K}_{2})$. Note that if $G=\mathrm {PGL}_2$, the invariant $w_2$ in $A^{0} _G\left (P^{n-2}_{\mathrm {sm}}\times P^1, \mathrm {K}_{2^r}\right )$ is killed by the $P^1$. The cokernel of the first is surjective on the normalised invariants of $\mathrm {S}_{\textrm{n}}$. Thus we have a map

$$ \begin{align*} \mathrm{Inv}^{\bullet}(\mathrm{B} \mathrm{S}_n,\mathrm{K}_{2}) / \mathrm{K}_{2}^{\bullet}(k) \rightarrow \mathrm{Inv}^{\bullet}(\mathrm{B} \mathrm{S}_{n-2},\mathrm{K}_{2})_2 \end{align*} $$

which lowers degree by $1$. Comparing the generators of the image with the generators of $A^{0} _G\left (P^{n-2}_{\mathrm {sm}}\times P^1, \mathrm {K}_{2^r}\right )$ degree by degree shows that the map is surjective, and in particular there cannot be any additional element in $A^{0}_G\left (P^n_{\mathrm {sm}},\mathrm {K}_{2}\right )$, yielding the result.

Now pick $r> 1$. We will proceed by induction on n even. If we assume the result for $n-2$, we have that $ A^{0} _G\left (P^{n-2}_{\mathrm {sm}}\times P^1, \mathrm {K}_{2^r}\right )=\mathrm {Inv}^{\bullet }(\mathrm {B} \mathrm {S}_{n-2})$. Note that if $G=\mathrm {PGL}_2$, the invariant $w_2$ is killed by the $P^1$. Moreover, note that as the image of $\left [ \Delta _{1,n} \right ]$ is divisible by $2$ but not by $4$, the intersection of the kernel of $A^{0}_{G}\left (\Delta _{1,n},\mathrm {K}_{2^r}Z\right ) \rightarrow A^{1}_G\left (P^n,\mathrm {K}_{2^r}\right )$ and the subring of constant invariants of $\mathrm {S}_{n-2}$ is exactly the $2$-torsion.

On the other hand, the cokernel of the inclusion of $A_G^0\left (P^n,\mathrm {K}_{2^r}\right )$ into the group $A_G^0\left (P^n_{\mathrm {sm}},\mathrm {K}_{2^r}\right )$ is isomorphic to the group of normalised invariants of $\mathrm {S}_{\textrm{n}}$. Thus we have a map

$$ \begin{align*} \mathrm{Inv}^{\bullet}\left(\mathrm{B} \mathrm{S}_n,\mathrm{K}_{2^r}\right) / \mathrm{K}_{2^r}^{\bullet}(k) \rightarrow \mathrm{Inv}^{\bullet}\left(\mathrm{B} \mathrm{S}_{n-2},\mathrm{K}_{2^r}\right)_2 \end{align*} $$

which lowers degree by $1$. We claim this map is surjective. Let

$$ \begin{align*} \alpha = \tau_0 + \alpha_{1}\tau_1 + \dotsb +\alpha_{n/2-1}\tau_{n/2-1}, \end{align*} $$

with $\tau _0, \dotsc , \tau _{n/2-1} \in \mathrm {K}_{2^r}^{\bullet }(k)_2$ elements of $\mathrm {Inv}^{\bullet }\left (\mathrm {B} \mathrm {S}_{n-2},\mathrm {K}_{2^r}\right )$. Let $\zeta _{1}, \dotsc , \zeta _{n}$ be elements of $A^0_{G}\left (P^{n}_{\mathrm {sm}}, \mathrm {K}_{2}\right )$ such that $\partial (\zeta _1)=1, \dotsc , \partial (\zeta _n)=\alpha _{n/2-1}$.

Then by Lemma 5.1 the element

$$ \begin{align*} \zeta = \zeta_{1} \tau_0 + \dotsb + \zeta_{n/2} \tau_{n/2-1} \end{align*} $$

belongs to $A^0_{G}\left (P^{n}_{\mathrm {sm}}, \mathrm {K}_{2^r}\right )$. Now note that by the compatibility of the boundary map with the morphism of cycle modules $\mathrm {K}_{2^r} \rightarrow \mathrm {K}_{2}$, the restriction modulo $2$ of $\partial (\zeta _i)$ is equal to $\alpha _{i-1}$. But then $\alpha _{i-1}-\partial (\zeta _{i})$ is a multiple of $2$, proving that $\partial (\zeta )=\alpha $. This concludes the proof.

Proof. Consider the exact sequence

$$ \begin{align*} A^{0}_G(P^n,\mathrm{M}) \hookrightarrow A^{0}_G\left(P^n_{\mathrm{sm}},\mathrm{M}\right) \xrightarrow{\partial} A^{0}_G\left(\Delta^n_{1},\mathrm{M}\right) \subset A^{0} _G\left(P^{n-2}_{\mathrm{sm}}\times P^1, \mathrm{M}\right). \end{align*} $$

As in the Lemma 7.6, we proceed by induction on even n. If the result is true for $n-2$, we reduce to showing that the map

$$ \begin{align*} \partial:\alpha_1\cdot \mathrm{M}^{\bullet}(k)_2 \oplus \dotsb \oplus \alpha_{n/2}\cdot \mathrm{M}^{\bullet}(k)_2 \rightarrow \mathrm{M}^{\bullet}(k)_2 \oplus \dotsb \oplus \alpha_{n/2-1}\cdot\mathrm{M}^{\bullet}(k)_2 \end{align*} $$

is surjective. This is done exactly as in the Lemma 7.6, using the structure of an $A^{0}_G\left (P^n_{\mathrm {sm}},\mathrm {K}_{\ell }\right )$-module of $A^{0}_G\left (P^n_{\mathrm {sm}},\mathrm {M}\right )$.

Proof. The first formula is proven in [Reference Edidin and Fulghesu9, Lemma 3.2] and the second in [Reference Fulghesu and Viviani13, Eq. 3.2].