For any (unital and commutative) ring $R$ we can define the special linear group as the kernel of the determinant, that is $$ 0 \to \operatorname{SL_n}(R) \to \operatorname{GL_n}(R) \xrightarrow{\det} R^\times \to 0 $$ is an exact sequence.
Of course, when $R = \mathbb{R}$ or $\mathbb{C}$, this means that we can think of this group as the linear automorphisms of $R^n$ preserving the canonical measure and orientation (by the change of variable formula and the fact that the derivative of a linear transformation is itself).
Is there any way to describe $\operatorname{SL_n}(R)$ in this manner for a general ring?
Perhaps we want to think of an action of $\operatorname{SL_n}(R)$ on the affine $n$-space over $R$ (maybe even using the language of group schemes) and think on some invariant it preserves. Even better would be to find an algebraic variety (if $R$ is a field) or a scheme which realizes $\operatorname{SL_n}(R)$ as its group of automorphisms. Are there any results on this?
An object left invariant by $SL_n(R)$ is just what you'd think if you already know orientation forms: Define the exterior powers $\bigwedge^\ast M$ of an $R$-module as a quotient of the tensor powers of $M$ as usual. If $M$ is a free $R$-module, say of dimension $n$, then $\bigwedge^n M$ is a free, one-dimensional $R$-module with basis $b_1 \wedge \ldots \wedge b_n$ where $(b_1,\ldots,b_n)$ is a any basis of $M$. And you could go ahead and call that an "$R$-orientation of $M$" or something similar. With this definition, all $R$-linear maps $F: M\to M$ satisfy what you'd expect them to satisfy: $$(\bigwedge^n F)(b_1\wedge\ldots\wedge b_n) = \det(F)\cdot(b_1\wedge \ldots \wedge b_n)$$
In other words: $SL_n(R)$ is precisely the automorphism group of $M$ taken together with any orientation of $M$.