Proof. If $R=A\langle S\rangle =A\langle x_s,x_s^* \ | \ s\in S\rangle $ is the free A-algebra with involution on a set S, then clearly $A[\operatorname {SpRep}_{(R,*)}^{\square ,2d}]$ is equal to the polynomial algebra $A[M_{2d}^S]=A[\mathbb X_{h,k}^{(s)}\ | \ 1\le h,k\le 2d,\ s\in S]$ and $\rho ^{u}(x_s)=\mathbb X^{(s)}=(\mathbb X_{h,k}^{(s)})_{h,k}$ .

For a general A-algebra with involution R, there is a presentation $R=A\langle S\rangle /I$ for some involution-stable two-sided ideal I of $A\langle S\rangle $ . Then $\rho ^{u}(I)$ generates a two-sided ideal in $M_{2d}(A[M_{2d}^S])$ which, as any two-sided ideal in a matrix algebra, is of the form $M_{2d}(J)$ with J an ideal of $A[M_{2d}^S]$ . Thus, the universal map for R is given by

By the universal property, $M_{2d}(A[M_{2d}^S]/J)$ is independent of the presentation of R.

Proof. We know that Azumaya algebras of symplectic type are classified by $\check H^1_{\operatorname {\acute {e}t}}(A, \operatorname {PGSp}_d)$ (see [Reference KnusKnu91, III. §8.5]). It is then sufficient to find an étale trivialisation of the cohomology class associated to $(\mathcal {A}, \sigma )$ . The orthogonal case is treated similarly.

Proof. We first observe, that symplectic vector bundles are the same in the Zariski and in the étale topology. This follows from the equivalence of categories [Sta23, 03DX], which is also used in the proof of Hilbert’s Theorem 90 [Sta23, 03P7] in the case of line bundles.

The group scheme $\operatorname {Sp}_{2d}$ defines an étale group sheaf $\operatorname {Sp}_{2d} : \operatorname {Sch}_{/X} \to \operatorname {Grp}$ . The bijection between étale symplectic vector bundles and étale $\operatorname {Sp}_{2d}$ -torsors is now the standard one: take an étale symplectic vector bundle $(\mathcal V,\sigma )$ to the étale $\mathcal I som$ -sheaf

$$ \begin{align*}\mathcal I som((\mathcal V,\sigma), (\mathcal O_X^{2d}, \mathrm{std}))(U) := \mathrm{Isom}((\mathcal V,\sigma)|_U, (\mathcal O_X^{2d}, \mathrm{std})|_U)\end{align*} $$

with $\operatorname {Sp}_{2d}$ -action induced by the standard action on $\mathcal O_X^{2d}$ . It follows directly from local triviality of $(\mathcal V,\sigma )$ , that $\mathcal I som((\mathcal V,\sigma ), (\mathcal O_X^{2d}, \mathrm {std}))$ is an $\operatorname {Sp}_{2d}$ -torsor.

Let $\mathcal T$ be an étale $\operatorname {Sp}_{2d}$ -torsor seen as an étale sheaf on $\operatorname {Sch}_{/X}$ . The group sheaf $\operatorname {Sp}_{2d}$ acts diagonally on $\mathcal T \times \mathcal O_X^{2d}$ . Take $\mathcal T$ to the étale sheaf quotient $\mathcal T \times ^{\operatorname {Sp}_{2d}} \mathcal O_X^{2d} := (\mathcal T \times \mathcal O_X^{2d})/\operatorname {Sp}_{2d}$ , which by local triviality of $\mathcal T$ is again easily seen to be an étale symplectic vector bundle.

By the same argument using Lemma 2.5, we see that the groupoid of Azumaya algebras with symplectic involution is equivalent to the groupoid of étale $\operatorname {PGSp}_{2d}$ -torsors.

Proof. By [Sta23,003Z], it is enough to show that the functors induce equivalences of fibre groupoids. For each Y-scheme $t \colon T \to Y$ , the stack $[\operatorname {SpRep}_{(R,*)}^{\square ,2d}/\operatorname {Sp}_{2d}]$ parametrises pairs

$$ \begin{align*} (f \colon \mathcal G \to T, ~\mathcal G \to \operatorname{SpRep}_{(R,*)}^{\square,2d}) \in [\operatorname{SpRep}_{(R,*)}^{\square,2d}/\operatorname{Sp}_{2d}](T), \end{align*} $$

where $\mathcal G$ is an étale $\operatorname {Sp}_{2d}$ -torsor over T and $\mathcal G \to \operatorname {SpRep}_{(R,*)}^{\square ,2d}$ is an $\operatorname {Sp}_{2d}$ -equivariant map of S-schemes.

Using Lemma 2.7, we attach to $\mathcal G$ a symplectic vector bundle $(V, b)$ on T. Since $\mathcal G(\mathcal G)$ contains $\operatorname {id}_{\mathcal G}$ , $(V,b)$ is canonically trivialised over $\mathcal G$ . The composition

$$ \begin{align*}f^* t^*R \to (M_{2d}(\mathcal O_{\mathcal G}), \mathrm j) \to \operatorname{End}_{\mathcal O_{\mathcal G}}(f^*V, \sigma_b)\end{align*} $$

can be descended to a map $t^*R \to \operatorname {End}_{\mathcal O_{\mathcal G}}(V, \sigma _b)$ using $\operatorname {Sp}_{2d}$ -equivariance of $\mathcal G \to \operatorname {SpRep}_{(R,*)}^{\square ,2d}$ . The functor $\mathcal G \mapsto (V,b)$ realises the identification Lemma 2.7 between symplectic vector bundles and $\operatorname {Sp}_{2d}$ -torsors. In particular, it induces an equivalence between the groupoid of symplectic vector bundles and the groupoid of $\operatorname {Sp}_{2d}$ -torsors.

To show that the functor $[\operatorname {SpRep}_{(R,*)}^{\square ,2d}/\operatorname {Sp}_{2d}](T) \to \operatorname {SpRep}_{(R,*)}^{2d}(T)$ is an equivalence, we give a functor in the other direction. It is then formal to verify that this realises an equivalence of groupoids.

An object of $\operatorname {SpRep}_{(R,*)}^{2d}(T)$ is a triple $(V, b, \rho )$ as in Definition 2.6. We define an $\operatorname {Sp}_{2d}$ -torsor $\mathcal G$ over T by setting

$$ \begin{align*}\mathcal G(X) := \mathrm{Isom}_{\mathcal O_X}((x^*V, b), ~(\mathcal O_X^{\oplus 2d}, b_{\mathrm{std}}))\end{align*} $$

for all T-schemes $x \colon X \to T$ . Here $b_{\mathrm {std}}$ is the standard symplectic form and isomorphisms shall preserve the bilinear forms. The functor $\mathcal G$ is representable by a flat scheme $f \colon \mathcal G \to T$ of finite presentation over T [Reference YoucisYou, Theorem 3.24]. The identity map in $\mathcal G(\mathcal G)$ corresponds to an isomorphism $f^* V \xrightarrow {\sim } \mathcal O_{\mathcal G}^{\oplus 2d}$ compatible with b and $b_{\mathrm {std}}$ . The composition

$$ \begin{align*}(f^*t^* R, *) \overset{\rho}{\to} (\operatorname{End}_{\mathcal O_{\mathcal G}}(f^*V), \sigma_b) \xrightarrow{\sim} (\operatorname{End}_{\mathcal O_{\mathcal G}}(\mathcal O_{\mathcal G}^{\oplus 2d}), \mathrm j)\end{align*} $$

defines a representation in $\operatorname {SpRep}_{(R,*)}^{\square ,2d} (\mathcal G)$ , so we obtain a map $\mathcal G \to \operatorname {SpRep}_{(R,*)}^{\square ,2d}$ . The latter is $\operatorname {Sp}_{2d}$ -equivariant, for the action of $\operatorname {Sp}_{2d}$ realises a change of basis. We have constructed an object of $[\operatorname {SpRep}_{(R,*)}^{\square ,2d}/\operatorname {Sp}_{2d}](T)$ . The equivalence $[\operatorname {SpRep}_{(R,*)}^{\square ,2d}/\operatorname {PGSp}_{2d}] \xrightarrow {\sim } \overline {\operatorname {SpRep}}_{(R,*)}^{2d}$ follows by an analogous argument.

Proof. See [Reference ChenevierChe14, Lemma 1.21].

Proof. Let $\mathcal P := \mathbb {Z}[\tfrac {1}{2}][x_{ij} \ | \ 1\le i,j\le 2d]/I$ , where I is the ideal generated subject to the relations $X^{\mathrm j} = X$ , where X is a $2d \times 2d$ matrix defined by $X_{ij} := x_{ij}$ . Since $2$ is invertible, $\mathcal P$ is a polynomial algebra over $\mathbb {Z}[\tfrac {1}{2}]$ in $2d^2-d$ variables. Using the ring homomorphism $\mathcal P \rightarrow A, ~x_{ij}\mapsto m_{ij}$ , where $M=(m_{i,j})_{1\le i,j\le 2d}$ is as in the statement, and an injection $\mathcal P \hookrightarrow \mathbb C$ , we can assume that $A=\mathbb {C}$ .

By density, we may assume that $\operatorname {pf}_M(\lambda )=\prod _{i=1}^d(\lambda -\lambda _i)$ for mutually distinct $\lambda _i\in \mathbb {C}$ . Indeed, this is the locus in the affine space of symmetric matrices with respect to $\mathrm {j}$ where $\mathrm {disc}(\operatorname {pf}_M(\lambda ))\neq 0$ , which is non-empty since it contains the matrices of the form $M=\operatorname {diag}(a_1,\dots ,a_d,a_1,\dots ,a_d)$ with $a_i\neq a_j$ . Since $\operatorname {pf}_M(\lambda )^2$ is the characteristic polynomial of M, by the Cayley–Hamilton theorem, we can decompose $V=\mathbb {C}^{2d}$ as a direct sum of generalised eigenspaces

$$ \begin{align*} V=V_1\oplus \cdots \oplus V_d, \end{align*} $$

with $(M-\lambda _i)^2v=0$ for $v\in V_i$ . Let $b\colon V\times V\to \mathbb {C}$ be the standard symplectic bilinear form, we need to show that the above decomposition is orthogonal with respect to b. Let $v_i\in V_i$ and $v_j\in V_j$ for $i\neq j$ . If $v_i,v_j$ are both eigenvectors for M, then by symmetry of M, we have

$$ \begin{align*}\lambda_i b(v_i,v_j)=b(Mv_i,v_j)=b(v_i,Mv_j)=\lambda_jb(v_i,v_j),\end{align*} $$

which implies that $b(v_i,v_j)=0$ . Now if $v_i$ is an eigenvector, and $v_j$ is a generalised eigenvector, then a similar calculation also gives $b(v_i,v_j)=0$ . Finally, if $v_i,v_j$ are both generalised eigenvectors, then by the previous case, $b((M-\lambda _i)v_i,v_j)=0$ . This implies that $b(Mv_i,v_j)=\lambda _ib(v_i,v_j)$ . Analogously we also have $b(Mv_i,v_j)=\lambda _jb(v_i,v_j)$ which implies the desired vanishing.

For dimension reasons, we have $\dim (V_i)=2$ for all i, we are then reduced to the $2$ -dimensional case. We conclude by noting that if $M\in M_2(\mathbb {C})$ satisfies $M^{\mathrm {j}}=M$ , then M is scalar.

Proof. From $(\mathcal A, \sigma )$ we obtain a descent datum of quasi-coherent involutive $\mathcal O_Y$ -algebras on the small étale site of Y: by Lemma 2.5 there is an étale covering $f \colon \tilde Y \to Y$ and an isomorphism $\psi \colon (M_{2d}(\mathcal O_{\tilde S}), \mathrm j) \cong (f^* \mathcal A, \sigma )$ . So we obtain an isomorphism $\varphi \colon (M_{2d}(\mathcal O_{\tilde Y \times _Y \tilde Y}), \mathrm j) \to (M_{2d}(\mathcal O_{\tilde Y \times _Y \tilde Y}), \mathrm j)$ satisfying the axioms of a descent datum, as in [Sta23, 023V]. To define $\mathrm {Prd}$ by effectivity of étale descent for maps of vector bundles, we have to give a morphism of descent data from $\varphi $ to the descent datum of $\mathcal O_Y$ induced by the covering $\tilde Y \to Y$ . We define $\mathrm {Prd}_{\tilde Y} \colon M_{2d}(\mathcal O_{\tilde Y})^+ \to \mathcal O_{\tilde Y}$ by $M \mapsto (-1)^{d(d-1)/2}\cdot \operatorname {Pf}(MJ)$ using the Leibniz formula for $\operatorname {Pf}$ . To check that $\mathrm {Prd}_{\tilde Y}$ gives a morphism of descent data and to prove at the same time that $\mathrm {Prd}$ satisfies the desired property stated in the Proposition (which also implies uniqueness), we have to show that for an arbitrary scheme T and an arbitrary isomorphism of Azumaya algebras $\alpha \colon (M_{2d}(\mathcal O_T), \mathrm j) \to (M_{2d}(\mathcal O_T), \mathrm j)$ over $\mathcal O_T$ , we have $\mathrm {Prd}_T \circ \alpha = \mathrm {Prd}_T$ where $\mathrm {Prd}_T \colon M_{2d}(\mathcal O_T)^+ \to \mathcal O_T$ is defined by $M \mapsto (-1)^{d(d-1)/2}\cdot \operatorname {Pf}(MJ)$ . For that we may assume that $T = \operatorname {Spec}(A)$ for a local ring A. By the Skolem-Noether theorem (which holds over local rings, see e.g. [Reference Aljadeff, Giambruno, Procesi and RegevAGPR20, Remark 3.4.19]), $\alpha $ is given by conjugation by a matrix in $\operatorname {GSp}_{2d}(A)$ . To show $\mathrm {Prd}_T \circ \alpha = \mathrm {Prd}_T$ is now entirely formal, hence we may also assume that A is an integral domain. We have $(\mathrm {Prd}_T \circ \alpha )^2 = \det \circ \alpha = \det = \mathrm {Prd}_T^2$ , hence $\mathrm {Prd}_T \circ \alpha = \pm \mathrm {Prd}_T$ . It follows that $\mathrm {Prd}_T \circ \alpha = \mathrm {Prd}_T$ , since $\operatorname {GSp}_{2d}$ is a connected group scheme over A.

Proof. All desired properties follow from Proposition 3.14 and étale descent.

Proof. Since $P(1) = P'(1) = 1$ , we have $\mathcal T^P_0 = \mathcal T^{P'}_0 = 1$ . By comparing the coefficients $\mathcal T_i^P$ and $\mathcal T_i^{P'}$ and the coefficients $\Lambda _i^D$ using $\chi ^D(\cdot ,t)|_{R^+} = \chi ^P(\cdot , t)^2 = \chi ^{P'}(\cdot , t)^2$ we obtain

$$ \begin{align*}\Lambda_{i}^D|_{R^+} = \sum_{j=0}^i \mathcal T_{j}^P \mathcal T_{i-j}^P = \sum_{j=0}^i \mathcal T^{P'}_{j} \mathcal T^{P'}_{i-j}\end{align*} $$

for $1 \leq i \leq 2d$ and $\mathcal T^P_{d} = P$ and $\mathcal T^{P'}_{d} = P'$ . For $i=0$ , we know, that $1 = \Lambda _0 = {\mathcal T^P_0}^2 = {\mathcal T^{P'}_0}^2$ .

By induction over the above equations and using $2 \in A^{\times }$ , we obtain $\mathcal T_i' = \mathcal T_i$ for all $0 \leq i \leq d$ , in particular $P' = P$ .

Proof. For a fixed y as in the statement, consider the polynomial laws $Q_1\colon R\otimes _A B\to B, \ x\mapsto P(xyx^*)$ and $Q_2\colon R\otimes _A B\to B, \ x\mapsto D(x)P(y)$ . Then, it is clear that $Q_1^2=Q_2^2$ , and so evaluating at the formal power series ring $B[[t]]$ , we have

$$ \begin{align*}\big(Q_{1}(tx-t+1)-Q_{2}(tx-t+1)\big)\big(Q_{1}(tx-t+1)+Q_{2}(tx-t+1)\big)=0.\end{align*} $$

The evaluation of the second summand at $t=0$ gives $2P(y)$ , thus $Q_{1}(tx-t+1)+Q_{2}(tx-t+1)$ is a non zero divisor. And so, $Q_{1}(tx-t+1)=Q_{2}(tx-t+1)$ whose evaluation at $t=1$ gives the result.

Proof. The fact that $xy\in R^+\otimes _A B$ is immediate. Now we introduce the commuting elements $1+t_1x,1+t_2y\in R^+\otimes _A B[t_1,t_2]$ , and the polynomials

$$ \begin{align*}Q_x=P_B(1+t_1x), \quad Q_y=P_B(1+t_2y), \quad Q_{xy}=P_B((1+t_1x)(1+t_2y))\end{align*} $$

in $B[t_1,t_2]$ . The $Q_x$ is a polynomial in $t_1$ of degree at most d whose coefficient of $t^d$ is $P_B(x)$ . Similarly $Q_y$ is a polynomial in $t_2$ of degree at most d whose coefficient of $t^d$ is $P_B(y)$ , and $Q_{xy}$ is a polynomial in $t_1,t_2$ whose coefficient of $t_1^dt_2^d$ is $P_{B}(xy)$ . Thus, to prove the statement, it suffices to show the equality $Q_xQ_y=Q_{xy}$ , which can be checked inside the power series ring $B[[t_1,t_2]]$ .

Note that for every power series $g\in B[[t_1,t_2]]^\times $ with $g(0,0)\in B^\times $ and every square root $f_0\in B^\times $ of $g(0,0)$ , there exists a unique power series $f\in B[[t_1,t_2]]^\times $ with $f(0,0)=f_0$ such that $f^2=g$ . This can be seen by considering the power series expansion of the square root function at $1$ . Using this fact, the equality $Q_{xy}^2=Q_{x}^2Q_{y}^2$ (coming from multiplicativity of D), and $Q_{xy}(0,0)=Q_{y}(0,0)Q_{x}(0,0)=1$ , we get that $Q_xQ_y=Q_{xy}$ as desired.

Proof. Since D is $*$ -invariant, it follows that $\ker (D)$ is a $*$ -ideal. Using [Reference ChenevierChe14, Lemma 1.19] we have that

$$ \begin{align*}\ker(D)= \left\{r\in R \ | \ \forall B \in \operatorname{CAlg}_A, \ \forall m\in R\otimes_A B, \ \forall i \ge 1, \ \Lambda_i(rm)=0 \right\}\end{align*} $$

By Proposition 3.16, we know that P can be expressed as a polynomial in the $\Lambda _i$ , thus to show that $r\in \ker (D)\cap R^+$ is in $\ker (P)$ , it suffices to show that $\Lambda _i(r\otimes b + m) = \Lambda _i(m)$ for all commutative A-algebras B, $b\in B$ and $m\in R^+\otimes _A B$ . But this follows from the definition of the $\Lambda _i$ and the definition of $\ker (D)$ .

Since $2 \in R^{\times }$ , we have a surjection $R^+ \twoheadrightarrow (R/I)^+$ and $(R/I)^+$ is identified with $R^+/(I \cap R^+)$ . Since $I \cap R^+ \subseteq \ker (D) \cap R^+ \subseteq \ker (P)$ , P descends to a well-defined A-polynomial law $\overline P \colon (R/I)^+ \to A$ satisfying the desired properties.

Proof. As in [Reference ChenevierChe14, §2.1], $D_1D_2$ is a determinant law of dimension $2(d_1+d_2)$ and one checks, that it is a $*$ -determinant law. Similarly $P_1P_2 \colon R^+ \to A$ is homogeneous of degree $d_1+d_2$ . Further $(P_1P_2)^2 = D_1|_{R^+}D_2|_{R^+}$ and $(P_1P_2)(1) = 1$ . This proves the claim for weak symplectic determinant laws.

Now suppose that $\operatorname {CH}(P_i) \subseteq \ker (D_i)$ for $i=1,2$ . We will show that $\operatorname {CH}(P_1P_2) \subseteq \ker (D_1D_2)$ . Let $P := P_1P_2$ and $D := D_1D_2$ . Let $n \geq 1$ be an integer. Let $r_1, \dots , r_n \in R^+$ . For the purpose of this proof, we write $\mathbf t \cdot \mathbf r := t_1r_1 + \dots + t_nr_n \in R[t_1, \dots , t_n]$ . Recall Definition 3.3 of the functions $\chi _{\alpha }^{P_i}$ , where $\alpha \in \mathbb N_0^n$ with $\sum \nolimits _{j=1}^n \alpha _j = d_i$ . The equation

$$ \begin{align*}\chi^P(\mathbf t \cdot \mathbf r, \mathbf t \cdot \mathbf r) = \chi^{P_1}(\mathbf t \cdot \mathbf r, \mathbf t \cdot \mathbf r) \chi^{P_2}(\mathbf t \cdot \mathbf r, \mathbf t \cdot \mathbf r)\end{align*} $$

in $R[t_1, \dots , t_n]$ implies

$$ \begin{align*}\chi_{\alpha}^P(r_1, \dots, r_n) = \sum_{\alpha' + \alpha" = \alpha} \chi_{\alpha'}^{P_1}(r_1, \dots, r_n) \chi_{\alpha"}^{P_2}(r_1, \dots, r_n)\end{align*} $$

by comparing the coefficients of $t^{\alpha }$ . By [Reference ChenevierChe14, Lemma 1.19] we need to check that $D(1 + \chi _{\alpha }^P(r_1, \dots , r_n)r) = 1$ for all $r {\kern-1pt}\in{\kern-1pt} R{\kern-1pt}\otimes _A{\kern-1pt} B$ , so it suffices to check that ${D_i(1 {\kern-1pt}+{\kern-1pt} \chi _{\alpha }^P(r_1, \dots , r_n)r) {\kern-1pt}={\kern-1pt} 1}$ for all $r \in R\otimes _A B$ . This follows from [Reference ChenevierChe14, Lemma 1.19], since

$$ \begin{align*}\sum_{\alpha' + \alpha" = \alpha} \chi_{\alpha'}^{P_1}(r_1, \dots, r_n) \chi_{\alpha"}^{P_2}(r_1, \dots, r_n) \in \ker(D_1)\cdot\ker(D_2)\subseteq \ker(D_1) \cap \ker(D_2).\end{align*} $$

Proof. Let I be the ideal of $\operatorname {Sym}_A(\Gamma _A^d(R^+))$ generated by the element $[1]^d-1$ . Then the ring $\operatorname {Sym}_A(\Gamma _A^d(R^+))/I$ represents the functor which associates to a commutative A-algebra B the set of homogeneous polynomial laws P on $R^+$ of degree d such that $P(1)=1$ . Using the isomorphism

$$ \begin{align*} \Gamma_A(R^+\times R^+) &\xrightarrow{\sim} \Gamma_A(R^+)\otimes_A \Gamma_A(R^+) \\ [(r_1,r_2)]^i &\mapsto \sum_{p+q=i} [r_1]^p\otimes [r_2]^q \end{align*} $$

we get a morphism of A-modules

$$ \begin{align*}\widetilde{\varphi}\colon \Gamma_A^{2d}(R^+)\xrightarrow{\Gamma_A(\Delta)} \Gamma_A(R^+\times R^+)\twoheadrightarrow \Gamma_A^d(R^+)\otimes_A \Gamma_A^d(R^+)\rightarrow \operatorname{Sym}_A(\Gamma_A^{d}(R^+))/I.\end{align*} $$

For $[r_1]^{i_1}\cdots [r_m]^{i_m}\in \Gamma _A^{2d}(R^+)$ with $i_1+\cdots +i_m=d$ , it is given by

$$ \begin{align*}\widetilde{\varphi}([r_1]^{i_1}\cdots [r_m]^{i_m})=\sum_{p_1,q_1,\dots,p_m,q_m} ([r_1]^{p_1}\cdots[r_m]^{p_m})\odot ([r_1]^{q_1}\cdots [r_m]^{q_m}),\end{align*} $$

where the sum runs over the integers $p_j,q_j$ satisfying $p_j+q_j=i_j$ and $p_1+\cdots +p_m=q_1+\cdots +q_m=d$ . Therefore, we get a morphism of A-algebras $\varphi \colon \operatorname {Sym}_A(\Gamma _A^{2d}(R^+))\to \operatorname {Sym}_A(\Gamma _A^{d}(R^+))/I$ .

From this it follows that, the canonical map $\Gamma _A^{2d}(R^+)\to \Gamma _A^{2d}(R)$ induces a morphism of commutative A-algebras $\operatorname {Sym}_A(\Gamma _A^{2d}(R^+))\to \Gamma _A^{2d}(R)^{\text {ab}}$ . Then we can take the representing ring for weak symplectic determinant laws to be

$$ \begin{align*}A[\operatorname{w-SpDet}_{(R,*)}^{2d}]=(\Gamma_A^{2d}(R)^{\text{ab}}/*)\otimes_{\operatorname{Sym}_A(\Gamma_A^{2d}(R^+)),\varphi} \operatorname{Sym}_A(\Gamma_A^{d}(R^+))/I.\end{align*} $$

Here $\Gamma _A^{2d}(R)^{\text {ab}}/*$ is the quotient of $\Gamma _A^{2d}(R)^{\text {ab}}$ by the ideal generated by $\gamma -\gamma ^*$ for $ \gamma \in \Gamma _A^{2d}(R)^{\text {ab}}$ .

Now let $(D^{\mathrm {w}},P^{\mathrm {w}})$ be the universal weak symplectic determinant law on $(R,*)$ . The universal ring $A[\operatorname {SpDet}_{(R,*)}^{2d}]$ is the quotient of $A[\operatorname {w-SpDet}_{(R,*)}^{2d}]$ by $\chi _\alpha ^{D^{\mathrm {w}}}(rr_1,\dots ,rr_n)$ for every $r\in \operatorname {CH}(P^{\mathrm {w}})$ , every $r_1,\dots ,r_n\in R$ , and every ordered tuple of integers $\alpha $ (see Definition 3.3).

Proof. Applying the Artin-Wedderburn theorem to R, we see that $R\cong \prod _{i=1}^s R_i^{\prime }$ , where $R^{\prime }_i$ is a central simple algebra over a field $K_i$ . This product decomposition corresponds to a unique set of orthogonal central primitive idempotents $e_1, \dots , e_s \in R$ with $e_1 + \dots + e_s = 1$ . The involution $*$ defines a bijection $* \colon \{e_1, \dots , e_s\} \to \{e_1, \dots , e_s\}$ . It follows that there is a partition $\{1, \dots , s\}=I_0\sqcup I_1\sqcup I_2$ such that $e_i^* = e_i$ if and only if $i\in I_0$ , and $e_i^*= e_{i^{\prime }}$ for $i\in I_1$ if and only if $i^{\prime }\in I_2$ . Since $e_iR$ is $*$ -stable for all $i\in I_0$ , and $(e_i+e_i^*)R$ is $*$ -stable for all $i\in I_1$ , we obtain $*$ -stable k-algebras $R_i$ with $R_i := R^{\prime }_i$ if $i\in I_0$ and $R_i := R_i^{\prime } \times R_{i^{\prime }}^{\prime }$ if $i\in I_1$ with $e_i^*=e_{i^{\prime }}$ . In the latter case, the involution $*$ induces an isomorphism $R_{i^{\prime }}^{\prime }\cong (R_i^{\prime })^{\mathrm {op}}$ . This provides us with the desired decomposition. The rest of the proposition is deduced from the discussion in Section 2.2 (see also [Reference Knus, Merkurjev, Rost and TignolKMRT98, Chapter I, Proposition 2.20]) for the cases (I-II-IIIa), and the case (IIIb) is immediate.

Proof. This Proposition follows from [Reference ChenevierChe14, Theorem 2.16] and Proposition 3.29.

Proof. Let $(D,P)$ be a symplectic determinant of $(R,*)$ over $\overline k$ . By Proposition 3.31 there is a decomposition

$$ \begin{align*}(R/\ker(D), *) \cong \prod_{i=1}^s (R_i, \sigma_i)\end{align*} $$

where the $R_i$ are $K_i$ -algebras of the form described in Example 3.30 for some extension field $K_i/\overline {k}$ .

From the description of $K_i$ , we see that every element of $K_i$ is algebraic over $\overline {k}$ , therefore $K_i = \overline k$ for all i and we have the following three cases:

1. (I) $(R_i, \sigma _i) \cong (M_{2n_i}(\overline k), \mathrm j)$ . We let $\rho _i \colon (R, *) \to (M_{2n_i}(\overline k), \mathrm j)$ be the corresponding symplectic representation.

2. (II) $(R_i, \sigma _i) \cong (M_{n_i}(\overline k), \top )$ . We let

   $$ \begin{align*} \rho_i \colon (R,*) &\to (M_{2n_i}(\overline k), \mathrm j) \\ r &\mapsto \begin{pmatrix} \pi_i(r) & 0 \\ 0 & \pi_i(r) \end{pmatrix} \end{align*} $$

3. (III) $(R_i, \sigma _i) \cong (M_{n_i}(\overline k) \times M_{n_i}(\overline k), \mathrm {swap})$ . We let

   $$ \begin{align*} \rho_i \colon (R, *) &\to (M_{2n_i}(\overline k), \mathrm j) \\ r &\mapsto \begin{pmatrix} \operatorname{pr}_1(\pi_i(r)) & 0 \\ 0 & \operatorname{pr}_2(\pi_i(r))^{\top} \end{pmatrix} \end{align*} $$

In these three cases $(D_i, P_i)$ is of the form $(\det \circ \rho _i, \operatorname {pf} \circ \rho _i)$ . In particular $(D,P)$ is of the form $(\det \circ \rho , \operatorname {pf} \circ \rho )$ , where $\rho = \bigoplus _{i=1}^s \rho _i$ . Since R surjects onto the $R_i$ , the $\rho _i$ are semisimple and thus $\rho $ is semisimple.

To prove that the map is injective, let us consider two semisimple representations $\rho _1$ and $\rho _2$ of R over $\overline k$ of dimension $2d$ that have the same symplectic determinant. By [Reference ChenevierChe14, Theorem 2.12], $\rho _1$ and $\rho _2$ are conjugated by an element $g\in \operatorname {GL}_{2d}(\overline {k})$ . We need to show that we can take $g\in \operatorname {Sp}_{2d}(\overline {k})$ .

Since the product of copies of the symplectic group embeds diagonally in a symplectic group up to conjugation, it suffices to check this for direct summands of $\rho _1$ and $\rho _2$ . We can match the irreducible symplectic subrepresentations of $\rho _1$ and $\rho _2$ . An irreducible subrepresentation of $\rho _1$ , which is contained in an indecomposable symplectic subrepresentation of $\rho _1$ that is not irreducible, is mapped into an indecomposable symplectic subrepresentation of $\rho _2$ that is also not irreducible. Thus, we can assume that $\rho _1$ and $\rho _2$ are indecomposable as symplectic representations.

We distinguish two cases:

1. (a) $\rho _1$ and $\rho _2$ are irreducible as representations. In this case, they are both surjective onto $M_{2d}(\overline {k})$ , so that $(M \mapsto g M g^{-1}) \in \operatorname {Aut}((M_{2d}(\overline {k}),\mathrm {j}))=\operatorname {PGSp}_{2d}(\overline {k})$ .

2. (b) The representations are of the form $\rho _i=\rho _{i,1}\oplus {\rho _{i,2}}$ with $\rho _i(r^\sigma )=(\rho _{i,2}(r)^{*},\rho _{i,1}(r)^{\top })$ . There exist $g_1,g_2\in \operatorname {GL}_{d}(\overline {k})$ such that $\rho _{1,1}=g_1\rho _{2,1}g_1^{-1}$ and $\rho _{1,2}=g_2\rho _{2,2}g_2^{-1}$ . The compatibility of the representations with the involution implies that $g_2=(g_1^{\top })^{-1}$ , and so $\operatorname {diag}(g_1,g_2)=\operatorname {diag}(g_1,(g_1^{\top })^{-1})\in \operatorname {Sp}_{2d}(\overline {k})$ .

Proof. By Lemma 3.21, we may assume that $\ker (D) = 0$ and that R is a finitely generated k-algebra. We know by Theorem 3.32 that there is a symplectic representation $\rho _{\overline k} \colon (R \otimes _k {\overline k},*) \to (M_{2d}(\overline k), \mathrm j)$ with $D \otimes _k {\overline k} = \det \circ \rho _{\overline k}$ and $P \otimes _k {\overline k} = \operatorname {pf}\circ \rho _{\overline k}|_{(R \otimes _k {\overline k})^+}$ . Then the image $\rho _{\overline k}(R) \subseteq M_{2d}(\overline k)$ is as a k-subalgebra generated by finitely many matrices in $M_{2d}(\overline k)$ . Hence, there is a finite field extension $k^{\prime }/k$ such that $\rho _{\overline k}(R) \subseteq M_{2d}(k^{\prime })$ , and so we get a symplectic representation $\rho \colon (R \otimes _k {k^{\prime }}, \sigma \otimes \operatorname {id}_{k^{\prime }}) \to (M_{2d}(k^{\prime }), \mathrm j)$ .

It remains to prove the last equality of the corollary. For every commutative $k^{\prime }$ -algebra B we obtain a diagram

By the functorialities of D, $\det $ and the base changes of $\rho $ , we know that every square commutes. The bottom triangle commutes by our choice of $\rho $ . The vertical maps are all injective and so it follows that the top triangle commutes, hence the equality $D \otimes _k k'=\det \circ \rho $ . The other equality follows from Proposition 3.16.

Proof. Since every element $r\in R$ can be written as $r=r_1+r_2$ with $r_1^*=r_1$ and $r_2^*=-r_2$ and that $r_1$ and $r_2$ are roots of $\chi ^P(r_1,t)$ and $\chi ^P(r_2^2,t^2)$ , we get that r is integral over A of degree $\le 2d^2$ . Therefore by [Reference ProcesiPro73, Proposition 3.22], we get that R is a PI (Polynomial Identity) A-algebra (we invite the reader to consult [Reference Wang-EricksonWE13, 1.2.2] for the definition of a PI algebra). Hence by [Reference ProcesiPro73, Theorem 2.7], we get that R is a finite A-module.

Proof. Ad (1)+(2)+(3). Following the argument in [Reference ChenevierChe14, Lemma 2.4 (1)] using [Reference ChenevierChe14, Lemma 2.2 (iii)], it is straightforward to see that $(D_e,P_e)$ defines a weak symplectic determinant law. Also part (2) follows easily. So let us show that if $\operatorname {CH}(P)\subseteq \ker (D)$ , then $\operatorname {CH}(P_e)\subseteq \ker (D_e)$ .

Note that for a commutative A-algebra B, and $r\in (eR^+e \oplus (1-e)R^+(1-e))\otimes _A B$ , $\chi ^{P_e}(er,t)=P_{e,B[t]}(te-er+1-e)$ . Since $te-er+1-e$ and $t(1-e)-(1-e)r+e$ commute and their product is equal to $t-r$ , we get by Lemma 3.20 that

$$ \begin{align*} \chi^P(r,t) &=P_{B[t]}(r-t)=P_{B[t]}(te-er+1-e)P_{B[t]}(t(1-e)-(1-e)r+e)\\ &=\chi^{P_e}(er,t)\chi^{P_{1-e}}((1-e)r,t). \end{align*} $$

Now for $r\in eR^+e\otimes B$ , we apply the above equation to r and $r+1-e$ to get that

(4.1) $$ \begin{align} \chi^P(r,t)=\chi^{P_e}(r,t)t^{d_{1-e}} \quad \text{ and }\quad \chi^P(r+1-e,t)=\chi^{P_e}(r,t)(t-1)^{d_{1-e}}. \end{align} $$

Now since the ideal generated by $t^{d_{1-e}}$ and $(t-1)^{d_{1-e}}$ in $B[t]$ is $B[t]$ itself, we get that there exist polynomials $Q,S\in B[t]$ such that

$$ \begin{align*}\chi^{P_e}(r,t)=Q(t)\chi^{P}(r,t)+ S(t)\chi^{P}(r+1-e,t).\end{align*} $$

Now consider $r_1,\dots ,r_n\in (eRe)^+$ . Noting that $(r_1t_1+\cdots +r_nt_n+1-e)^k=(r_1t_1+\cdots +r_nt_n)^k+(1-e)$ in $R[t_1,\dots ,t_{n+1}]$ , we get that

$$ \begin{align*}\chi^{P}(r+1-e,r)= \chi^{P}(r+1-e,r+1-e)+ b(1-e),\end{align*} $$

for some $b\in B$ . Therefore, we get that

$$ \begin{align*} & \chi^{P_e}(r_1 t_1+\dots+r_nt_n,r_1t_1+\dots+r_nt_n)\\ &= Q(r_1t_1+\dots+r_nt_n)\chi^{P}(r_1t_1+\dots+r_nt_n,r_1t_1+\dots+r_nt_n) \\ &\quad +S(r_1t_1+\dots+r_nt_n)(\chi^{P}(r_1t_1+\dots+r_nt_n+1-e,r_1t_1+\dots+r_nt_n+1-e)+b(1-e)). \end{align*} $$

Therefore for $\alpha =(\alpha _1,\dots ,\alpha _{n})\in {\mathbb {N}}^{n}$ non-zero, the coefficient in front of $t_1^{\alpha _1}\cdots t_n^{\alpha _n}$ in the above equation can be written as a linear combination of elements in $\operatorname {CH}(P)$ . This gives that $\operatorname {CH}(P_e)\subseteq \operatorname {CH}(P)$ , and so $\operatorname {CH}(P_e)\subseteq \operatorname {CH}(D_e)$ . We also immediately get $(3)$ from the inclusion $\operatorname {CH}(P_e)\subseteq \operatorname {CH}(P)$ .

Ad (4). As $e^2 = e$ , we have $\chi ^P(e,e) - P(e) \in Ae$ . Since $(D,P)$ is symplectic Cayley–Hamilton, we have $\chi ^P(e,e)=0$ . Assuming that $P(e) = 1$ , there is some $a \in A$ such that $ae = 1$ and hence $1 = ae = ae^2 = e$ . For the second claim, assume that $(D_e,P_e)$ is of dimension $0$ . This means that $D_e$ is constantly $1$ and also $P_e$ is constantly $1$ . So $P(1-e) = 1$ and $e=0$ follows from the previous claim. For the last assertion, set $e_{s+1} := 1- (e_1 + \dots + e_s)$ . The dimension of $D_{e_i}$ is $\leq 2d$ for all $1 \leq i \leq s+1$ and by (2) the dimensions sum up to $2d$ . But this implies that the dimension of $D_{e_{s+1}}$ is $0$ and thus $e_{s+1}=0$ .

Ad (5). We let $e=f+f^*$ , which is a symmetric idempotent. Suppose $r \in fRf$ , and set $x = tf-r+(1-f)$ . Using the equality $ff^* = f^*f = 0$ , we obtain

$$ \begin{align*} xx^* &= (f(t-r-1)f + 1) \cdot (f^*(t-r^*-1)f^* + 1) \\ &= et - (r+r^*) + 1-e, \end{align*} $$

so that $P_{A[t]}(xx^*) = P_{e, A[t]}(e t- (r+r^*)) = \chi ^{P_{e}}(r+r^*, t)$ . By Lemma 3.19 applied to x and $y=1$ , we have

$$ \begin{align*}\chi^{D_f}(r, t) = D_{A[t]}(tf-r+(1-f)) = D_{A[t]}(x) = P_{A[t]}(xx^*) = \chi^{P_{e}}(r+r^*, t).\end{align*} $$

Let us call this polynomial $q \in A[t]$ . As $rr^* = r^*r = 0$ , we have $q(r+r^*) = q(r) + q(r^*)$ . As $(D,P)$ is symplectic Cayley–Hamilton, then by $(3)$ , so is $(D_{e},P_{e})$ . Therefore, we find that $q(r+r^*) = 0$ . Further $q(r) \in fRf$ and $q(r^*) \in f^*Rf^*$ and since $fRf \cap f^*Rf^* = 0$ , we have $q(r)=q(r^*) = 0$ . So indeed $D_f$ is Cayley–Hamilton.