Unitary groups over number fields - MathOverflow

Unitary groups over number fields

2011-04-13T19:10:35Z

When defining unitary groups over number fields, one usually takes $F$ to be a totally real number field, $E$ a CM quadratic extension of $F$, and $V$ a hermitian space attached to $E/F$. Then $U(V)$ is simply the isometry group for the hermitian form attached to $V$.

My question is: why does one take $F$ to be totally real and $E$ CM? The definition makes perfect sense without this, but most references (at least in the context of automorphic forms) make these assumptions.

Answer by Jon Yard for Unitary groups over number fields

2011-05-01T20:44:28Z

If $E$ is not CM, then the action of complex conjugation on $E$ depends on how it is embedded into $\mathbb{C}$. In particular, it could have different real subfields depending on which embedding you are using. When $E$ is CM, so that it is a totally imaginary quadratic extension of the totally real subfield $F$, then it has a unique complex conjugation that commutes with all automorphisms of $E$, such that $F$ contains precisely the elements that are fixed by complex conjugation in all embeddings. This lets you talk about unitarity and hermiticity for the abstract field $E$, and not just for some particular embedding, which I would imagine could be problematic if not impossible to say anything useful about.

Or, to put it another way, there can be more than one way to consistently define a complex conjugation on your field, unless it is CM, in which case there is a unique way to define it.

Answer by Emerton for Unitary groups over number fields

2011-05-01T21:40:04Z

Just to add to Jon Yard's answer: when one defines a unitary group as described in the OP, the group that one gets after extending scalars from $F$ to $F_v = \mathbb R$ for each archimedean place $v$ of $F$ is indeed a unitary group in the usual sense, i.e. the set of complex matrices preserving some non-degenerate Hermitian form.

If one were to apply the defintion in a more general setting, say to a quadratic extension $E$ over $F$ which is not a CM extension of a totally real field, the groups at infinity could end up being $GL_n(\mathbb R)$ or $GL_n(\mathbb C)$, which are not unitary groups.