I would like to understand precisely the structure of unitary groups. 


Let $F$ be a global number field, $E$ a quadratic extension of $F$, and $U$ a unitary group on $E$ (*i.e.* the group of automorphism preserving an hermitian form on an $E$-vector space), say of rank $n$ (*i.e.* the hermitian form is on a squared $n$-dimensional vector space).


What can be said for the local groups, that is for the $U\_v$ where $v$ goes through the places of $F$? More precisely:


* are the $U\_v$ also unitary groups, or can they be isomorphic to $GL\_n(F\_v)$?
* is there only a finite number of possibility for $U\_v$?
* what are the obstructions for a unitary group $G$ to have prescribed local groups $G\_v$?


Thanks in advance !



Understanding the structure of unitary groups
As you already seem to know, unitary groups are classified by separable quadratic field extensions (in fact one should really work with separable quadratic algebras, i.e. also $F\times F$, corresponding to the ''trivial'' unitary group $\mathrm{GL}\_n$).


Thus your questions can essentially be reduced to problems about quadratic field extensions.


* *are the $U\_v$ also unitary groups, or can they be isomorphic to $GL\_n(F\_v)$?*


You need to be a bit careful, since, as I already point out, you should really view $GL\_n$ itself as a unitary group. But you really want to ask whether $U\_v$ can be isomorphic to a general linear group. As already explained in the comments, and following the classification, this happens iff $v$ is split in $E/F$.


* *is there only a finite number of possibility for $U\_v$?*


You again need to be careful with the phrasing of this question, as it is not clear what you mean, since it is not possible to compare unitary groups over different fields. However, over a given local field $F\_v$, there are only finitely many isomorphism classes of unitary groups of given dimension. This follows from the fact that there are only finitely many quadratic extensions of a given local field. This is nothing special about unitary groups, since, more generally, over any local field $F\_v$ there are only finitely many isomorphism classes of reductive algebraic groups of given dimension (this can be proved using the classification of reductive algebraic groups over algebraically closed fields combined with the fact that $F\_v$ admits only finitely many field extensions of given degree).


* *what are the obstructions for a unitary group $G$ to have prescribed local groups $G\_v$?*


This is the most interested question. The obvious obstruction is of course the dimension of the groups, so we assume the local groups $G\_v$ have the same dimension. Also one cannot hope for a positive answer in general when you consider infinitely many places, so we suppose that we are given only finitely many places $v$ and finitely many groups $G\_v$. Then there the answer is: there is no obstruction! There is always some unitary group $G$ which realises the given $G\_v$. This follows from the [Grunwald-Wang theorem](https://en.wikipedia.org/wiki/Grunwald%E2%80%93Wang_theorem), which implies that you can always realise any given finite collection of quadratic algebras over $F\_v$ by some quadratic extension (though of course this elementary case does not require the full force of Grunwald-Wang). 


Note that this is an important place where one works with quadratic field extensions; this approximation property fails for field extensions in general. You can read more about this in Section IX.2 of the book "Cohomology of Number Fields" by Neukirch, Schmidt, Wingberg.


