Special unitary group

Definition
Suppose $$L$$ is a separable quadratic extension of a field $$K$$ and $$\sigma:L \to L$$ is the unique automorphism of $$L$$ that fixes $$K$$ pointwise. The special unitary group of degree $$n$$ for this quadratic extension, denoted $$SU(n,L)$$ (if the extension being referred to is understood) is defined as the subgroup of the special linear group $$SL(n,L)$$ comprising those matrices on which the transpose-inverse map gives the same result as the entry-wise application of $$\sigma$$.

$$SU(n,L) = \{ A \in SL(n,L) \mid \sigma(A) = (A^t)^{-1} \}$$

Here, $$\sigma(A)$$ is the matrix obtained by applying $$\sigma$$ to each of the entries of $$A$$.

Alternatively, we can define it as the intersection of the unitary group and the special linear group, both viewed as subgroups of the general linear group:

$$SU(n,L) = U(n,L) \cap SL(n,L)$$

For the real and complex numbers
The most typical usage of the term special unitary group is in the context where $$K$$ is the field of real numbers, $$L$$ is the field of complex numbers, and the automorphism $$\sigma$$ is complex conjugation. In this case, the group $$SU(n,\mathbb{C})$$ is the subgroup of the special linear group $$SL(n,\mathbb{C})$$ comprising those matrices whose complex conjugate equals the transpose-inverse. When it's understood that we are working over the complex numbers, this group is sometimes just denoted $$SU(n)$$.

For a finite field
If $$K$$ is the (unique up to isomorphism) finite field of size a prime power $$q$$, there is a unique quadratic extension $$L$$ of $$K$$, and this extension is separable. The extension field is the finite field (unique up to isomorphism) of order $$q^2$$. The automorphism $$\sigma$$ is the map $$x \mapsto x^q$$. The special unitary group for this extension may be denoted $$SU(n,q)$$ (the more standard choice) or $$SU(n,q^2)$$ (a less standard choice). Note that due to the ambiguity of notation, it is important to understand from context what exactly is meant.

Note that, if we denote this group by $$SU(n,q)$$, then, somewhat confusingly:

$$SU(n,q) = U(n,q) \cap SL(n,q^2)$$