Proof. Let us consider the line bundle $ q:\mathcal {O}_{}(1):=\det (E_n) \longrightarrow \mathrm{BGL}_n $ . We can identify the map $ \pi _0:\mathrm{BSL}_n \longrightarrow \mathrm{BGL}_n $ with the $ \mathbb G_m $ -principal bundle $ q\circ j:\mathcal {O}_{}(1)\setminus \left \{ 0 \right \} \longrightarrow \mathrm{BGL}_n $ where $ j: \mathcal {O}_{}(1) \setminus \left \{ 0 \right \} \hookrightarrow \mathcal {O}_{}(1) $ . By homotopy invariance we can also identify $ \mathrm A(\mathrm{BGL}_n)\simeq \mathrm A(\mathcal {O}_{}(1)) $ , thus we have the localization sequences:

(5)

(6) $$ \begin{align} \ldots \rightarrow \mathrm A^{a,b}\left( \mathrm{BGL}_n; \mathcal{O}_{}(-1) \right) \stackrel{}{\rightarrow} \mathrm A^{a,b}\left( \mathrm{BSL}_n; \pi_0^*\mathcal{O}_{}(-1)\right) \stackrel{\bar\partial_{a,b}}{\rightarrow} \mathrm A^{a-1,b-1}\left( \mathrm{BGL}_n\right) \stackrel{}{\rightarrow} \ldots \end{align} $$

By [Reference Levine20, lemma 4.3] the cup product with $ e(L) $ for any line bundle $ L $ is zero in any $ \eta $ -inverted $ \mathrm{SL} $ -oriented ring spectrum, so the long exact sequence (5) splits in short exact sequences:

$$\begin{align*}0 \rightarrow \mathrm A^{a,b}\left( \mathrm{BGL}_n \right) \stackrel{}{\rightarrow} \mathrm A^{a,b}\left( \mathrm{BSL}_n \right) \stackrel{\partial_{a,b}}{\rightarrow} \mathrm A^{a-1,b-1}\left( \mathrm{BGL}_n; det(E_n) \right) \rightarrow 0 \end{align*}$$

We want now to find a splitting for $ \partial $ and we already have a natural candidate $ \pi _1^* $ , in the notation used above the proposition. Let us consider the tautological section $ can: \mathcal {O}_{}(1)\rightarrow q^*\mathcal {O}_{}(1) $ . Using definition 1.35, the tautological section defines an element $ \langle t_{can} \rangle \in \mathrm A^{0,0}(\mathrm{BSL}_n, \pi ^*_0\mathcal {O}_{}(1))\simeq \mathrm A^{0,0}(\mathrm{BSL}_n, \pi ^*_0\mathcal {O}_{}(-1)) $ . The map $ \pi _1^* $ we constructed before was:

$$\begin{align*}\mathrm A^{\bullet}\left( \mathrm{BGL}_n; \mathcal{O}_{}(1) \right)\stackrel{\pi_0^*}{\longrightarrow} \mathrm A^{\bullet}\left( \mathrm{BSL}_n; \pi_0^*\mathcal{O}_{}(1)\right) \stackrel{\theta_*}{\longrightarrow} \mathrm A^{\bullet}\left( \mathrm{BSL}_n \right) \end{align*}$$

but $ \theta _* $ is just multiplication by $ \langle t_{can} \rangle $ . To show that the short exact sequence actually splits we need to prove that $ \partial \circ \pi _1^*$ is an isomorphism. As a first step we claim that for any $ x \in \mathrm A^{\bullet }(\mathrm{BGL}_n; \mathcal {O}_{}(1)) $ we have:

$$\begin{align*}\partial(\pi_1^*(x))=\bar\partial(\langle t_{can} \rangle)\cup x \end{align*}$$

Indeed:

$$\begin{align*}\partial(\pi_1^*(x))=\partial( \langle t_{can} \rangle \cup \pi_0^*(x) )= \partial( \langle t_{can} \rangle \cup q^*j^*(x) ) \end{align*}$$

and $ \partial $ is $ \mathrm A^{\bullet }(\mathrm{BGL}_n; \mathcal {O}_{}(1)) $ -linear, with $ \mathrm A^{\bullet }(\mathrm{BGL}_n; \mathcal {O}_{}(1)) $ acting on $ E^{\bullet }(\mathrm{BSL}_n) $ via multiplication through $ q^*j^* $ and on $ E^{\bullet }(\mathrm{BGL}_n; \mathcal {O}_{}(1)) $ just via multiplication, and this proves the claim.

We want now to prove that multiplication by $ \bar \partial (\langle t_{can} \rangle ) \in \mathrm A^{-1,-1}(\mathrm{BGL}_n) $ is an isomorphism:

$$\begin{align*}\bar \partial(\langle t_{can} \rangle) \cup \cdot : \mathrm A^{a,b}(\mathrm{BGL}_n; \mathcal{O}_{}(1)) \longrightarrow \mathrm A^{a-1,b-1}(\mathrm{BGL}_n; \mathcal{O}_{}(1)) \end{align*}$$

Using a Mayer-Vietoris argument on the finite approximation pieces $ {B}_m\mathrm{GL}_n $ we can reduce ourselves to a local computation, where we already know the result by [Reference Ananyevskiy5, Lemma 6.4]. Indeed given two open sets $ U, V $ of $ {B}_m\mathrm{GL}_n $ , we get a Nisnevich excision square related to $ \left \{ U,V, U\cup V \right \} $ and a Mayer-Vietoris exact sequence for any Nisnevich sheaf. In particular multiplication by an element $ u \in \mathrm A^{-1,-1}({B}_m\mathrm{GL}_n) $ gives us a map of Mayer-Vietoris sequences:

So to show that multiplication by $ \bar \partial (\langle t_{can}\rangle ) $ is an isomorphism, it is enough to show that $ \bar \partial (\langle t_{can}\rangle ) $ restricts to an invertible element when passing to open sets $ U_i $ covering $ {B}_m\mathrm{GL}_n $ . But in local coordinates $\bar \partial (\langle t_{can}\rangle ) $ restricts to $ \eta $ by [Reference Ananyevskiy5, Lemma 6.4] and thus we get our desired splitting for $ B_m\mathrm{GL}_n $ . This implies that for any $ k $ we have:

$$\begin{align*}\pi_m^*: \mathrm A^{k}(B_m\mathrm{GL}_n)\oplus \mathrm A^{k}(B_m\mathrm{GL}_n; \mathcal{O}_{}(1))\stackrel{\sim}{\longrightarrow} \mathrm A^{k}(B_m\mathrm{SL}_n) \end{align*}$$

where $ \pi _m^* $ is the map induced by $ \pi ^* $ at the finite level approximations. Since the isomorphism holds for any $ k $ , we have an equivalence of mapping spectra:

$$\begin{align*}\pi^*_m: \mathrm A(B_m\mathrm{GL}_n)\oplus \mathrm A^{}(B_m\mathrm{GL}_n; \mathcal{O}_{}(1))\stackrel{\sim}{\longrightarrow} \mathrm A^{}(B_m\mathrm{SL}_n) \end{align*}$$

where $ \mathrm A(B_mG)=\operatorname {\mathrm {\underline {Map}}}(\Sigma ^{\infty } B_mG, \mathrm A) $ (and similarly for the twisted version). But since $ \mathrm A(BG)=\operatorname {\mathrm {\underline {Map}}}(BG, \mathrm A)=\lim _m \operatorname {\mathrm {\underline {Map}}}(B_mG, \mathrm A) $ , from the equivalence of mapping spectra $ \pi _m^* $ , we get:

$$\begin{align*}\pi^*: \mathrm A(\mathrm{BGL}_n)\oplus \mathrm A^{}(\mathrm{BGL}_n; \mathcal{O}_{}(1))\stackrel{\sim}{\longrightarrow} \mathrm A^{}(\mathrm{BSL}_n) \end{align*}$$

again at the level of mapping spectra, and this gives us our claim.

For the explicit presentation of the image of $ \pi ^* $ the same argument as in the proof of [Reference Levine20, Proposition 4.1] will give us the result, using the statement of [Reference Ananyevskiy1, Theorem 10] generalised over any smooth $ \mathbb {k} $ -scheme $ \mathrm{S} $ in theorem 3.3 (we will actually generalise it to smooth NL-stacks).

Proof. Use the same exact proof as in proposition 2.1, replacing $ \mathrm{BSL}_n $ with $ BN $ and $ \mathrm{BGL}_n $ with $ BN_G $ and pulling back the appropriate maps along the morphisms $ BN \rightarrow \mathrm{BSL}_2 $ and $ BN_G \rightarrow \mathrm{BGL}_2 $ induced by $ \mathrm{N}\hookrightarrow \mathrm{SL}_2 $ and $ N_G \hookrightarrow \mathrm{GL}_2 $ respectively.

Proof. If $ \mathrm{S}=\operatorname {\mathrm {Spec}}(\mathbb {k}) $ , then this is just [Reference Ananyevskiy1, Theorem 9]. For a general smooth $ \mathbb {k} $ -scheme $ \mathrm{S} $ , we have a map:

$$\begin{align*}{\varphi}:\mathrm A^{\bullet}(\mathrm{S})[p_1,\ldots, e] \longrightarrow \mathrm A^{\bullet}(SGr_S(2n,2k+1)) \end{align*}$$

sending $ p_i $ to $ p_i(T_1) $ and $ e$ to $ e(T_1) $ . The structure map $ \pi _S: \mathrm{S} \rightarrow \operatorname {\mathrm {Spec}}(\mathbb {k}) $ induces a pullback map:

$$\begin{align*}\pi_S^*:\mathrm A^{\bullet}(SGr_{\mathbb{k}}(2n,2k+1))\rightarrow \mathrm A^{\bullet}(SGr_S(2n,2k+1)) \end{align*}$$

The tautological bundle $ T_1 $ on $ SGr_S(2n,2k+1) $ is the pullback of the tautological bundle $ T_{1, \mathbb {k}} $ on $ SGr_{\mathbb {k}}(2n,2k+1) $ , hence the Pontryagin and Euler classes of $ T_{1} $ satisfy all the relations in $ J_{2m,2k+1} $ since they hold for the classes of $ T_{1,\mathbb {k}} $ (by [Reference Ananyevskiy1, Theorem 9]). This implies that $ \varphi $ passes to the quotient, that is, we get a map:

$$\begin{align*}\varphi: {\unicode{x0022}}{\mathrm A^{\bullet}(\mathrm{S})[p_1,p_2,\ldots, p_n, e]}{J_{2n, 2k+1}} \longrightarrow \mathrm A^{\bullet}( SGr_S(2n,2k+1)) \end{align*}$$

By [Reference Asok, Déglise and Nagel7, Theorem 4.2.9], we have Leray spectral sequences both for source and target of $ \varphi $ :

$$\begin{align*}E_1^{p,q}=\bigoplus_{s\in \mathrm{S}^{(p)}} {\unicode{x0022}}{\mathrm A^{q}(\kappa(s))[p_1,p_2,\ldots, p_n, e]}{J_{2n, 2k+1}} \Rightarrow {\unicode{x0022}}{\mathrm A^{p+q}(\mathrm{S})[p_1,p_2,\ldots, p_n, e]}{J_{2n, 2k+1}} \end{align*}$$

$$\begin{align*}'E_1^{p,q}=\bigoplus_{s \in \mathrm{S}^{(p)}} \mathrm A^{q}(SGr_{\kappa(s)}(2n,2k+1)) \Rightarrow \mathrm A^{p+q}(SGr_S(2n,2k+1)) \end{align*}$$

The map between spectral sequences by $ \varphi $ is an isomorphism by [Reference Ananyevskiy1, Theorem 9], hence $ \varphi $ is an isomorphism too as claimed.

Proof. Via the pullback map induced by the first projection $ p_1: \mathcal {X}\times _S SGr_S(2n,2k+1) $ , we get an $ \mathrm A^{\bullet }(\mathcal {X}) $ -algebra structure on $ \mathrm A^{\bullet }(\mathcal {X} \times _S SGr_S(2n,2k+1)) $ . Since the Pontryagin and Euler classes in $ \mathrm A^{\bullet }(\mathcal {X}\times _S SGr_S(2n,2k+1)) $ are the pullback of the respective classes in $ \mathrm A^{\bullet }(SGr_S(2n,2k+1)) $ , then there exists a unique map of $ \mathrm A^{\bullet }(\mathcal {X}) $ -algebras:

$$\begin{align*}\varphi_{\mathcal{X}}: {\unicode{x0022}}{\mathrm A^{\bullet}(\mathcal{X})[p_1,p_2,\ldots, p_n, e]}{J_{2n, 2k+1}} \longrightarrow \mathrm A^{\bullet}( \mathcal{X}\times_S SGr_{\mathrm{S}}(2n,2k+1)) \end{align*}$$

defined by sending $ p_i $ to $ p_i(T_1) $ and $ e $ to $ e(T_1) $ . Let $ \mathrm{x}: \mathrm{X}\rightarrow \mathcal {\mathrm{X}} $ be a NL-atlas and let $ \mathrm{X}^r_{\mathcal {X}}:=\check {C}_{r}(/ {X}{\mathcal {X}}) $ be the scheme at the $ r $ th level of the Čech nerve. By proposition 3.1 we have isomorphisms:

$$\begin{align*}\varphi_{\mathrm{X}^r_{\mathcal{X}}}: {\unicode{x0022}}{\mathrm A^{p,q}(\mathrm{X}^r_{\mathcal{X}})[p_1,\ldots,p_n,e]}{J_{2n,2k+1}} \stackrel{\sim}{\longrightarrow} \mathrm A^{p+q}(\mathrm{X}^r_{\mathcal{X}}\times_S SGr_S(2n,2k+1)) \end{align*}$$

for each $ r $ and for each bi-degree $ (p,q) $ . This implies that we have an equivalence of mapping spectra:

(7) $$ \begin{align} \operatorname{\mathrm{\underline{Map}}}(\mathrm{X}^r_{\mathcal{X}}, A_{J_{2n,2k+1}}) \stackrel{\sim}{\longrightarrow} \operatorname{\mathrm{\underline{Map}}}(SGr_{\mathrm{X}^r_{\mathcal{X}}}(2n,2k+1), \mathrm A) \end{align} $$

where $ A_{J_{2n,2k+1}} $ is the ring spectrum representing $ {\unicode{x0022}} {\mathrm A^{\bullet }(-)[p_1,\ldots ,p_n,e]}{J_{2n,2k+1}} $ . The equivalence (7), by [Reference D’Angelo12, Remark 3.38], implies that we have:

$$ \begin{align*} \mathrm A_{J_{2n,2k+1}}(\mathcal{X})&\simeq \lim_{r \in \Delta} \mathrm A_{J_{2n,2k+1}}(\mathrm{X}^r_{\mathcal{X}})\simeq\\ & \simeq \lim_{r \in \Delta} \operatorname{\mathrm{\underline{Map}}}(\mathrm{X}^r_{\mathcal{X}}, A_{J_{2n,2k+1}}) \simeq\\ &\simeq \lim_{r \in \Delta} \operatorname{\mathrm{\underline{Map}}}(SGr_{\mathrm{X}^r_{\mathcal{X}}}(2n,2k+1), \mathrm A) \simeq\\ &\simeq \lim_{r \in \Delta} \mathrm A(SGr_{\mathrm{X}^r_{\mathcal{X}}}(2n,2k+1))\simeq \\ & \simeq \mathrm A(\mathcal{X}\times_S SGr_S(2n,2k+1)) \end{align*} $$

In other words:

$$\begin{align*}\varphi_{\mathcal{X}}: {\unicode{x0022}}{\mathrm A^{\bullet}(\mathcal{X})[p_1,p_2,\ldots, p_n, e]}{J_{2n, 2k+1}} \longrightarrow \mathrm A^{\bullet}( \mathcal{X}\times_S SGr_{\mathrm{S}}(2n,2k+1)) \end{align*}$$

is an equivalence as claimed.

Proof. Let us start with the even case (the only case we are actually interested in for future applications). We know that the ind-scheme $ \mathrm{BSL}_{n,\mathrm{S}}=\underset {k}{\mathrm {colim}}\ SGr_{\mathrm{S}}(n, 2k+1) $ , since the system made of $ SGr_{\mathrm{S}}(n,2k+1) $ is cofinal inside the system made by all special linear Grassmannians. Then theorem 3.2 tells us that the system $ \left \{ \mathrm A^{\bullet }(\mathcal {X} \times _S SGr_{\mathrm{S}}(n,2k+1)) \right \} $ satisfies the Mittag-Leffler condition. Hence by [Reference D’Angelo12, Proposition 3.33], applied to the spectrum $ \mathrm A_{\mathcal {X}} $ as constructed in [Reference D’Angelo13, Lemma 3.17], we have:

$$\begin{align*}A^{\bullet}( \mathcal{X}\times_S \mathcal{B}\mathrm{SL}_{n,\mathrm{S}}) \simeq \lim_{k} \mathrm A^{\bullet}(\mathcal{X} \times SGr_{\mathrm{S}}(n, 2k+1)) \end{align*}$$

By theorem 3.2 this gives us the isomorphism:

$$\begin{align*}\varphi_{\mathcal{X}}: {\mathrm A^{\bullet}(\mathcal{X}){[\![} p_1,p_2,\ldots, p_{n-1}, e{]\!]}} \longrightarrow \mathrm A^{\bullet}( \mathcal{X}\times_S \mathcal{B}\mathrm{SL}_{2n,\mathrm{S}}) \end{align*}$$

For the odd case, it is enough to use the identification $ \mathrm A^{\bullet }(SGr_S(2n+1, 2k+1))\simeq \mathrm A^{\bullet }(SGr_S(2k-2n,2k+1)) $ as pointed out in [Reference Ananyevskiy1, Remark 14] and hence reduce to the same argument used in the even case.

Now for the last statement of the theorem, it is enough to show that:

$$\begin{align*}\mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_1,\mathrm{S}}\times \mathcal{B}\mathrm{SL}_{k_2,\mathrm{S}})\simeq \mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_1,\mathrm{S}})\widehat{\otimes}_{\mathrm A^{\bullet}(\mathrm{S})} \mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_2,\mathrm{S}}) \end{align*}$$

and then iterate. Let us suppose that $ k_1=2n $ and $ k_2=2m+1 $ . Denote the Pontryagin and Euler classes of $ \mathcal {B}\mathrm{SL}_{k_1,\mathrm{S}} $ by $ p_{i,1}, e_{1} $ and denote the Pontryagin classes of $ \mathcal {B}\mathrm{SL}_{k_2,\mathrm{S}} $ by $ p_{i,2} $ . Then by what we just proved, for $ \mathcal {X}= \mathcal {B}\mathrm{SL}_{k_1,\mathrm{S}} $ , we have:

$$ \begin{align*} \mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_1,\mathrm{S}}\times \mathcal{B}\mathrm{SL}_{k_2,\mathrm{S}})&\simeq \mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_1,\mathrm{S}}){[\![} p_{1,2},\ldots,p_{m,2}{]\!]} \simeq \\ & \simeq \left( \mathrm A^{\bullet}(\mathrm{S}){[\![} p_{1,1}, \ldots, p_{n-1,1},e_1{]\!]} \right){[\![} p_{1,2},\ldots,p_{m,2}{]\!]} \simeq \\ &\simeq \mathrm A^{\bullet}(\mathrm{S}){[\![} p_{1,1}, \ldots, p_{n-1,1},e_1{]\!]} \widehat{\otimes}_{\mathrm A^{\bullet}(\mathrm{S})} \mathrm A^{\bullet}(\mathrm{S}){[\![} p_{1,2}, \ldots, p_{m,2}{]\!]} \simeq \\ & \simeq \mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_1,\mathrm{S}}) \widehat{\otimes}_{\mathrm A^{\bullet}(\mathrm{S})} \mathrm A^{\bullet}(\mathcal{B}\mathrm{SL}_{k_2,\mathrm{S}}) \end{align*} $$

as we wanted. All the other cases are totally analogous and then we can iterate the computations for the general case $ \prod _{j=0}^{s}\mathcal {B}\mathrm{SL}_{k_j} $ .

Proof. Under the identifications $ \mathrm{BGL}_{2n,\mathrm{S}}\simeq \mathrm{BGL}_{2n+1,\mathrm{S}}\simeq \mathrm{BSL}_{2r+1} $ of [Reference Haution17, Theorem 6.3.3, Theorem 6.3.7] in $ \mathrm {SH}(\mathrm{S})[\eta ^{-1}] $ , the corollary follows from theorem 3.3.

Proof. The proposition follows by a Mayer-Vietoris argument from the untwisted cases in theorem 3.3 and corollary 3.4.

Notation 4.4. To distinguish between Thom classes coming from (twisted) symplectic bundles and the ones coming from $ \mathrm{SL} $ -vector bundles, we will denote the latter as $ th^{\mathrm{SL}}(\cdot ) $ .

Proof. For a rank 2 vector bundle $ V $ over some $ \mathrm{X} \in \mathbf {Sch}_{/ {}{\mathrm{S}}} $ , we get the canonical identification $ \omega _{can}: \Lambda ^2V \longrightarrow \det (V) $ . Then we define:

$$\begin{align*}\mathrm{th}(V, can):=\mathrm{th}^{\mathrm{SL}}(V) \in \mathbb E^{4,2}(\mathrm{Th}_{\mathrm{X}}\left( V \right); \det{}^{-1}(V)) \end{align*}$$

where $ \mathrm {th}^{\mathrm{SL}}(V) $ is the $ \mathrm{SL} $ -Thom class living in the twisted cohomology as defined in Definition 1.30, and where we implicitly used the fact that:

$$\begin{align*}\det{}^{-1}(V)\simeq \det(V)\otimes \left( \det{}^{-1}(V) \right)^{\otimes 2} \end{align*}$$

since for $\mathrm{SL}$ -oriented theories we have a natural equivalence $\mathbb E(-; - \otimes L^{\otimes })\simeq \mathbb E(-; -)$ .

It is not difficult to check that this class satisfies the desired relations, see for example [Reference Levine and Raksit23, §3] for a detailed account of these classes (in the case $ \mathrm{X} \in \mathbf {Sm}_{/ {}{\mathrm{S}}} $ ).

Proof. The map we are looking for is given by:

$$\begin{align*}\begin{array}{@{}cccc} (1,\zeta,\zeta^2,\ldots,\zeta^{n-1}): &\bigoplus_{j=0}^{n-1}\mathbb E_Z(\mathrm{X}; L^{\otimes -j}) & \longrightarrow & \mathbb E_{\pi^{-1}(Z)}(\mathrm H\mathbb{P}^n_L)\\ & (a_0,\ldots, a_{n-1}) & \mapsto & \sum_{0}^{n-1} a_j\zeta^{j} \end{array} \end{align*}$$

Find opens that trivialise $ L $ over $ \mathrm{X} $ and use them with a Mayer-Vietoris argument to reduce to the untwisted case proved in [Reference Panin and Walter25, Theorem 8.2].

Construction 4.11. Let $ (E,\omega ^{L}) $ be an $ L $ -twisted symplectic bundle of rank $ 2r $ : we now want to construct a twisted version of the quaternionic flag bundle. First consider:

$$\begin{align*}\pi_1: \mathrm{HFlag}_X^{L}(1, r-1; E):=\mathrm{HGr}^{L}_X(2; E) \longrightarrow \mathrm{X} \end{align*}$$

Then over $ \mathrm {HFlag}_X^{L}(1, r-1; E) $ we have the tautological rank 2 twisted bundle $ U_{1,1} $ (the one classified by the identity map of $ \mathrm {HGr}^{L}_X(2; E) $ ). The bundle $ U_{1,1} $ is a sub-bundle of $ \pi _1^*E $ , therefore we can consider:

$$\begin{align*}U_{1,1}^{\perp}:= \left \{ w\in E{\mid} \ \omega^{L}(w,u)=0 \ \forall \ u \in U_{1,1} \right \} \end{align*}$$

that is the tautological sub-bundle of rank $ 2r-2 $ over $ \mathrm {HFlag}_X^{L}(1, r-1; E) $ . Let us rename $ U_{1,1}^{\perp } $ as:

$$\begin{align*}W_{1}:=U_{1,1}^{\perp} \end{align*}$$

This gives us a decomposition:

$$\begin{align*}(\pi_1^*E, \pi_1^*\omega^{L})\simeq \left(U_{1,1}, \omega_{1,1}\right) \perp \left(W_1, \omega_{W_1}\right) \end{align*}$$

where $ \omega _{1,1}:=\left. { \pi _1^*\omega ^{L}}\right|_{U_{1,1}} $ and $ \omega _{W_1}:=\left. { \pi _1^*\omega ^{L}}\right|_{W_1} $ .

Now we iterate; to ease the notation we will drop the upper and lower scripts from the twisted hyperbolic Grassmannians $ \mathrm {HGr} $ . Consider the natural map:

$$\begin{align*}p_2: \mathrm{HFlag}_X^L(1^2,r-2; E):= \mathrm{HGr}(2; W_1) \longrightarrow \mathrm{HFlag}_X^{L}(1, r-1; E) \end{align*}$$

and denote by $ U_{2,2} $ the universal tautological rank 2 bundle over $ \mathrm {HFlag}_X^L(1^2,r-2; E) $ . This bundle is a natural sub-bundle of $ p_2^*W_1 $ and again, naming $ W_2:=U_{2,2}^{\perp } $ , we get:

$$\begin{align*}(p_2^*W_1, p_2^*\omega_{W_1})\simeq \left(U_{2,2}, \omega_{2,2}\right) \perp \left(W_2, \omega_{W_2}\right) \end{align*}$$

where $ \omega _{2, 2} $ and $ \omega _{W_2} $ are the restrictions of $ p_2^*\omega _{W_1} $ to $ U_{2,2} $ and $ W_2 $ respectively. In particular, if $ \pi _2: \mathrm {HFlag}_X^L(1^2,r-2; E)\rightarrow \mathrm{X} $ is the map given by $ \pi _1\circ p_2 $ , we have the following decomposition:

$$\begin{align*}(\pi_2^*E, \pi_2^*\omega^{L})\simeq p_2^*\left(U_{1,1}, \omega_{1,1}\right) \perp \left(U_{1,2}, \omega_{1,1}\right) \perp \left(U_{2,2}, \omega_{2,1}\right) \end{align*}$$

Notice that on $ \mathrm {HFlag}_X^L(1^2,r-2; E) $ we have three tautological bundles, namely $ U_{1,2}:=p_2^*U_{1,1} $ , $ U_{2,2} $ and $ W_2 $ .

Proceeding again in the same way, we can inductively construct:

$$\begin{align*}p_k: \mathrm{HFlag}_X^L(1^k,r-k; E):=\mathrm{HGr}(2; W_{k-1})\longrightarrow \mathrm{HFlag}_X^L(1^{k-1},r-k+1; E) \end{align*}$$

together with a natural map:

$$\begin{align*}\pi_k: \mathrm{HFlag}_X^L(1^k,r-k; E) \longrightarrow \mathrm{X} \end{align*}$$

Over $ \mathrm {HFlag}_X^L(1^k,r-k; E) $ we have the following (inductively defined) tautological bundles:

$$\begin{align*}(U_{i,k}, \omega_{i,k}):=p_k^*(U_{i,k-1}, \omega_{i,k-1})\ \ \ \ i=1,\ldots, k-1 \end{align*}$$

$$\begin{align*}(U_{k,k}, \omega_{k,k}) \end{align*}$$

$$\begin{align*}(W_k, \omega_{W_k}):=(U_{k,k}^{\perp}, \omega_{U_{k,k}^{\perp}} )\end{align*}$$

where $ U_{k,k} $ is the tautological rank 2 bundle of $ \mathrm {HGr}(2; W_{k-1}) $ . By construction, we have:

$$\begin{align*}\pi_k^*(E,\omega^{L})\simeq \displaystyle{\perp}_{i=1}^{k} \left( U_{i,k}, \omega_{i,k} \right) \perp \left( W_k, \omega_{W_k} \right) \end{align*}$$

Proof. Recall that we have:

$$\begin{align*}1 \rightarrow \mathrm{SL}_2^{\times r} \longrightarrow \mathrm{GL}_{2,\det}^{\times r} \stackrel{\det(-)}{\longrightarrow} \mathbb G_m \rightarrow 1 \end{align*}$$

The bundle $ \mathcal {O}_{}(1) $ is the tautological bundle of $ B\mathbb G_m $ pulled back to $ B \mathrm{GL}_{2,\det }^{\times r} $ , and we have that $ \mathrm{BSL}_2^{\times r}\simeq \mathcal {O}_{}(1)\setminus \left \{ 0 \right \} $ . Then, we can identify $ \mathrm{BSL}_2^{\times r} \longrightarrow \mathrm{BGL}_{2,\det }^{\times r} $ with the natural map $ \mathcal {O}_{}(1)\setminus \left \{ 0 \right \} \rightarrow \mathrm{BGL}_{2,\det }^{\times r} $ given by the composition of the open immersion $ j: \mathcal {O}_{}(1)\setminus \left \{ 0 \right \} \hookrightarrow \mathcal {O}_{}(1) $ with the projection map from $ \mathcal {O}_{}(1) $ . From here, the same argument used to prove the first part of proposition 2.1 applies verbatim replacing $ \mathrm{BSL}_n $ with $ \mathrm{BSL}_2^{\times r} $ and $ \mathrm{BGL}_n $ with $ \mathrm{BGL}_{2,\det }^{\times r} $ , and we are done.

Proof. The first claim is just an iterated application of the Twisted Projective Bundle Theorem 4.9. For the remaining claim, we already noticed that $ q^*E \simeq \bigoplus _{i=1}^{r} \left ( \mathcal {U}_i, \omega ^{L}_i \right ) $ . Now these $ (\mathcal {U}_i, \omega ^{L}_i) $ are twisted symplectic bundles of rank 2, but that means they are just rank 2 vector bundles together with isomorphisms $ \omega ^{L}_i: \det \left (\mathcal {U}_i\right ) \stackrel {\sim }{\rightarrow } \left. {L}\right|_{\mathcal {U}_i} $ . Therefore, we get a natural map:

$$\begin{align*}p: \mathrm{HFlag}^{L}_X(E,\omega^{L}) \longrightarrow \mathcal{B}\mathrm{GL}_{2,\det}^{\times r} \end{align*}$$

classifying $ ( \left \{ \mathcal {U}_i \right \} , L, \left \{ \omega _i \right \} ) $ (cf. remark 4.15). The universal bundle $ {} {1.25}{\mathcal{D}}_r $ of $ \mathcal {B}\mathrm{GL}_{2,\det }^{\times r} $ is given by an $ r $ -tuple of rank $ 2 $ bundles $ \mathcal {V}_i $ , together with isomorphisms $ \rho _i: \det (\mathcal {V}_i)\stackrel {\sim }{\rightarrow }\mathcal {O}_{}(1) $ , where $ \mathcal {O}_{}(1) $ is the pullback of the tautological bundle over $ \mathcal {B}\mathbb G_m $ . We can regard the $ \mathcal {V}_i $ as twisted symplectic bundles, with twisted symplectic forms given by $ \omega _{i,can}: \Lambda ^2 \mathcal {V}_i\simeq \det (\mathcal {V}_i) \stackrel {\rho _i}{\rightarrow } \mathcal {O}_{}(1) $ . Then by construction of $ p $ , we get that:

$$\begin{align*}p^*(\mathcal{V}_i, \omega_{i,can})\simeq (\mathcal{U}_i, \omega_i^{L}) \end{align*}$$

If we prove the claim of our theorem for the bundle $ \bigoplus (\mathcal {V}_i, \omega _{i,can}) $ , then, via the pullback map $ p^* $ , we get the relation we want for $ q^*(E,\omega ^{L}) $ . This means that it is enough to prove the following:

$$\begin{align*}b_t^{\mathcal{O}_{}(1)}\left(\bigoplus (\mathcal{V}_i, \omega_{i,can})\right)=\prod b_t^{\mathcal{O}_{}(1)}((\mathcal{V}_i, \omega_{i,can})) \end{align*}$$

By proposition 4.16, we have an injective map:

(9) $$ \begin{align} \mathfrak i_1^*:\mathrm A^{\bullet}\left( \mathcal{B}\mathrm{GL}_{2,\det}^{\times r}; \mathcal{O}_{}(1)\right) \hookrightarrow \mathrm A^{\bullet}\left( \mathcal{B}\mathrm{SL}_2^{\times r} \right) \end{align} $$

induced by $ \mathfrak i_0: \mathcal {B}\mathrm{SL}_2^{\times r} \rightarrow \mathcal {B}\mathrm{GL}_{2,\det }^{\times r} $ . For each $ i $ , the pullback $ \mathfrak i_0^* \mathcal {V}_i $ is isomorphic to the universal rank 2 $ \mathrm{SL} $ -vector bundle $ \mathcal {W}_i $ of $ \mathcal {B}\mathrm{SL}_2^{\times r} $ . Recall that the map $ \mathfrak i_1^* $ was given by the pullback map $ \mathfrak i_0^* $ , together with the inverse of the Thom isomorphism associated to the determinant of the $ \mathrm{SL} $ -vector bundles $ \mathfrak j^*\mathcal {V}_i $ . By injectivity of (9), we can therefore reduce ourselves to prove:

$$\begin{align*}b_t\left( \bigoplus (\mathcal{W}_i, \psi_{i,can}) \right)=\prod (\mathcal{W}_i, \psi_{i,can}) \end{align*}$$

with $ \psi _{i,can} $ the canonical symplectic form given by their $ \mathrm{SL} $ -structure. But this amounts to the usual Cartan Sum formula for (untwisted) symplectic bundles, and therefore we can conclude by [Reference Panin and Walter25, Theorem 10.5].

Proof. Using the twisted symplectic splitting principle, we can just follow the same steps as in [Reference Panin and Walter25, Theorem 10.5].

Proof of 4.19.

It is enough to prove the theorem for $ \mathrm{X}=\mathcal {B}\mathrm{GL}_2^{\times 3} $ , but again by the Künneth formula 3.5 and proposition 2.1 we can reduce to the $ \mathcal {B}\mathrm{SL}_2^{\times 3} $ case. Taking the finite level approximation $ \mathcal {B}_m\mathrm{SL}_2 $ we can use Ananyevskiy’s result [Reference Ananyevskiy4, Lemma 8.2] and since Witt theory for $ \mathcal {B}\mathrm{SL}_2 $ satisfies the Mittag-Leffler condition by [Reference Ananyevskiy1, Theorem 9] (or by proposition 3.1), taking the limit of the theories on the finite level approximation we get the desired result for $ {B}\mathrm{SL}_2 $ .