# Can elements in the orthogonal group of a non-split Azumaya algebra with an orthogonal involution have reduced norm -1?

Let $R$ be a connected (commutative) ring with $2\in R^\times$. Let $A$ be an Azumaya algebra over $R$ and let $\sigma:A\to A$ be an orthogonal involution. (This means that there is a faithfully flat ring extension $R\to R'$ such that $(A\otimes R',\sigma\otimes \mathrm{id}_{R'})$ is isomorphic to $(\mathrm{Mat}_{n\times n}(R'),\tau)$ as $R'$-algebras with involution, where $n=\deg A$ and $\tau$ is the transpose involution.)

Denote the Brauer class of $A$ by $[A]$ and call $A$ split if $A\cong \mathrm{Mat}_{n\times n}(R)$ for some $n$. Note that the conditions "$A$ is split" and "$[A]=0$" are equivalent when $R$ is semilocal, but not in general.

Let $O(A,\sigma)$ denote the orthogonal group of $(A,\sigma)$, i.e. $\{a\in A\,:\,\sigma(a)a=1\}$, and let $\mathrm{Nrd}:A\to R$ denote the reduced norm map. It is easy to see that $\mathrm{Nrd}$ defines a group homomorphism $$\mathrm{Nrd}:O(A,\sigma)\to\{\pm1\}.$$ This question concerns with the image of this map. My personal motivation in looking into this is that surjectivity is equivalent to saying that the non-neutral component of the group scheme $\mathbf{O}(A,\sigma)\to \mathrm{Spec}\, R$ has an $R$-point.

When $R$ is a field, this image is well-understood:

Theorem A. Suppose $R$ is a field $K$. Then $\mathrm{Nrd}(O(A,\sigma))=\{\pm 1\}$ if and only if $A$ is split.

Sketch of Proof. If $A=\mathrm{Mat}_{n\times n}(K)$, then $\sigma$ is adjoint to some non-degnerate symmetric bilinear form $b:K^n\times K^n\to K$. Now, any reflection relative to $b$ is an element of $O(A,\sigma)$ with determinant $-1$. Conversely, suppose there is $a\in O(A,\sigma)$ satisfying $\mathrm{Nrd}(a)=-1$. Working in $A\otimes \overline{K}\cong \mathrm{Mat}_{n\times n}(\overline{K})$, the fact that $a^{-1}=\sigma(a)$ means that the characteristic polynomials of $a$ and $a^{-1}$ are the same. Thus, the multiset of eigenvalues of $a$ is a disjoint union of sets of the form $\{1\}$, $\{-1\}$, $\{\lambda,\lambda^{-1}\}$. Since $\mathrm{Nrd}(a)$ is the product of the eigenvalues of $a$, the multiplicity of $-1$ must be ood, hence $(a+1)^n$ has odd rank for $n$ sufficiently large, and this is impossible if $A$ is non-split. $\square$

My question is whether a similar result can hold for a more general (connected) ring $R$. Before making that more specific, let us first note that Theorem A gives a number of positive results, for instance:

Theorem B. Suppose $R$ is a regular domain. Then $[A]\neq 0$ implies $\mathrm{Nrd}(O(A,\sigma))=\{1\}$. The converse holds when $R$ is semilocal.

Sketch of Proof. Let $K$ denote the fraction field of $R$. Since $R$ is regular, a theorem of Auslander and Goldman (Theorem 7.2 here) implies that $\mathrm{Br}(R)\to\mathrm{Br}(K)$ is injective, hence $A\otimes K$ is not split, and the first statement follows from Theorem A. As for the second statement, since $R$ is semilocal, $[A]=0$ implies $A$ is split. Now, with some work, one can argue as in the proof of Theorem A to prove $\mathrm{Nrd}(O(A,\sigma))=\{\pm1\}$. $\square$

The proof of Theorem B also applies to (connected) rings $R$ for which the map $$\mathrm{Br}(R)\to \prod_{{\mathfrak p} \in\mathrm{Spec} R} \mathrm{Br}(\mathrm{Frac}(R/\mathfrak p))$$ is injective. However, Ojanguren gave examples of rings failing to satisfy it. (I did not check it thoroughly, but it seems likely that Ojanguren's example can be made semilocal.)

With that in mind, one could ask whether Theorem A holds for all semilocal rings.

Question 1. Suppose $R$ is semilocal and $A$ is non-split. Is true that $\mathrm{Nrd}(O(A,\sigma))=\{1\}$?

As for general rings $R$, it is not too difficult to find examples of $(A,\sigma)$ with $A$ being non-split but satisfying $[A]=0$ and $\mathrm{Nrd}(O(A,\sigma))=\{\pm 1\}$. However, one can still ask:

Question 2. Suppose $[A]\neq 0$. Is it true that $\mathrm{Nrd}(O(A,\sigma))=\{1\}$?

Question 3. Is it possible to have $\mathrm{Nrd}(O(A,\sigma))=\{1\}$ when $[A]=0$?

Question 4. Is it possible to have $\mathrm{Nrd}(O(A,\sigma))=\{1\}$ when $A$ is split?

Remark. Question 3 is related to the question of whether it is possible for the reduced norm map of an Azumaya algebra with a trivial Brauer class to be non-surjective.

Remark. The proof of the "if" part of theorem A cannot be applied in the context of question 4. The reason is that $\sigma$ need not be adjoint to a non-degenerate symmetric bilinear form $b:R^n\times R^n\to R$. Rather, one has to replace the range $R$ with a line-bundle. Moreover, even when $b$ exists, it need not be diagnoalizable, and so one cannot assert the existence of reflections.

In the end it turns out that:

• The answer to Question 1 is "Yes" (i.e., if $$A$$ is non-split and $$R$$ is semilocal, then all elements in $$O(A,\sigma)$$ have reduced norm $$1$$).

• The answer to Questions 3 and 4 is "Yes" (i.e. there is a ring $$R$$, a natural number $$n$$ and an orthogonal involution $$\sigma:\mathrm{M}_{n}(R)\to \mathrm{M}_n(R)$$ such that all elements in $$O(\mathrm{M}_n(R),\sigma)$$ have determinant $$1$$.)

Question 2 is still open.

This had turned out to be sufficiently nontrivial to deserve a short preprint on arxiv (at least in my opinion), but since I asked the question, I feel committed to sketch the proofs here.

Concerning Question 1, denote by $$SO(A,\sigma)$$ the kernel of the reduced norm map $$\mathrm{Nrd}:O(A,\sigma)\to \{\pm 1\}$$. The idea is to first prove the following result:

Theorem C. Let $$(A,\sigma)$$ be an Azumaya algebra with orthogonal involution over a semilocal ring $$R$$ and let $$k_1,\dots,k_t$$ denote the residue fields of $$R$$ at its maximal ideals. Then the specialization map $$SO(A,\sigma)\to \prod_{i=1}^t SO(A_{k_i},\sigma_{k_i})$$ is surjective.

Remark. The specialization map $$O(A,\sigma)\to \prod_{i=1}^t O(A_{k_i},\sigma_{k_i})$$ is not surjective in general.

Provided Theorem C holds, and one is given $$u\in O(A,\sigma)$$ with reduced norm $$-1$$, one can prove that $$A$$ is split as follows:

• Without loss of generality, $$d:=\deg A$$ is even. If it is odd, then $$[A]=0$$ because $$[A]$$ is $$d$$-torsion and $$2[A]=0$$ (because $$A\cong A^{\mathrm{op}}$$ via $$\sigma$$). Since $$R$$ is semilocal, $$A$$ is split.

• The specialization of $$u$$ to $$k_i$$ ($$i=1,\dots,t$$) has reduced norm $$-1$$, so $$A_{k_i}$$ must be split (Theorem A).

• Note that the proof of Theorem A shows that we can find $$u'_i\in O(A_{k_i},\sigma_{k_i})$$ which is a simple reflection, i.e., the characteristic polynomial of $$u'_i$$ is $$(X+1)(X-1)^{d-1}$$.

• For $$i=1,\dots,t$$, set $$w_i=u_iu'_i\in SO(A_{k_i},\sigma_{k_i})$$. By Theorem C, there is $$w\in SO(A,\sigma)$$ such that $$w_{k_i}=w_i$$. Set $$u'=u^{-1}w$$. Then $$u'_{k_i}=u'_i$$ for all $$i$$ and $$\mathrm{Nrd}(u')=\mathrm{Nrd}(u^{-1})\mathrm{Nrd}(w)=-1$$.

• Let $$f\in R[X]$$ denote the (reduced) characteristic polynomial of $$u'$$ in $$A$$. An argument similar to the proof of Theorem A shows that $$f(-1)=0$$, so $$f=(X+1)g$$ for some $$g$$.

• Since $$u'_{k_i}=u_i$$, the image of $$g\in R[X]$$ in $$k_i[X]$$ is $$(X-1)^{d-1}$$. In particular, $$g(-1)\equiv (-2)^{d-1}$$ modulo every maximal ideal of $$R$$. Thus, $$(X-1)$$ and $$g(X)$$ generate the unit ideal in $$R[X]$$ (we assume $$2\in R^\times$$).

• At this point a standard argument shows that $$g(u')A\oplus (u'+1)A=A$$ and $$g(u')A$$ has $$R$$-rank $$\deg A$$, forcing $$[A]=0$$. Since $$R$$ is semilocal, $$A\cong \mathrm{M}_d(R)$$.

It remains to prove Theorem C. I will sketch the proof when $$R$$ is local with an algebraically closed residue field $$k$$.

Let $$S=\{a\in A_k\,:\,\sigma(a)=-a\}$$ and view it as a $$k$$-variety. The Cayley transform $$y\mapsto (1+y)(1-y)^{-1}$$ defines a birational isomorphism from $$S$$ to $$SO(A_k,\sigma_k)$$. Since the image of the Cayley transform contains a Zariski-open subset, $$SO(A_k,\sigma_k)$$ is generated by elements of the form $$s_y:=(1+y)(1-y)^{-1}$$ with $$y\in S\cap (1+A_k^\times)$$. It is easy to see that every $$s_y$$ lifts to an element of $$SO(A,\sigma)$$, so Theorem C holds.

With very mild modifications, this argument works when all residue fields of $$R$$ are infinite. What is more surprising is the much-more-delicate fact that it works in general.

Concerning Questions 3 and 4, loosely quoting from the preprint: "Take $$R$$ to be an integral domain admitting a non-principal invertible fractional ideal $$L$$. Define $$A=\left[\begin{array}{cc} R & L^{-1} \\ L & R\end{array}\right]$$ and let $$\sigma:A\to A$$ be the involution given by $$\left[\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right]^\sigma= \left[\begin{smallmatrix} d & b \\ c & a \end{smallmatrix}\right]$$. .... Straightforward computation shows that elements of $$O(A,\sigma)$$ of determinant $$-1$$ are of the form $$\left[\begin{smallmatrix} 0 & x^{-1} \\ x & 0 \end{smallmatrix}\right]$$, where $$x\in L$$. If such an element exists, then $$x^{-1}R\subseteq L^{-1}$$, or rather, $$L\subseteq xR$$. Since $$x\in L$$, this means that $$L=xR$$, contradicting our assumption that $$L$$ is not principal. Thus, $$\mathrm{Nrd}_{A/R}:O(A,\sigma)\to \{\pm 1\}$$ is not surjective.

If $$R\oplus L\cong M\oplus M$$ for some invertible fractional ideal $$M$$, then we also have $$A\cong \mathrm{End}_R(M\oplus M)\cong\mathrm{M}_2(R)$$. Such examples exist, e.g., take $$R$$ to be a Dedekind domain with class group containing an element $$[M]$$ of order $$4$$."

Concerning Question 2, if there is a counterexample, then by the positive answer to Question 1, it will have the remarkable property that $$[A]\neq 0$$, while $$[A_S]=0$$ for every semilocal $$R$$-ring $$S$$. I do not know if Azumaya algebras with this property can exist.