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.