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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.

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    $\begingroup$ When working with automorphic representations, you're in good shape if the representations show up in the cohomology of a Shimura variety that satisfies Deligne's axioms. Not every connected reductive group gives rise to such a situation---but if $F$ is totally real and $E$ CM then this is one of the cases where everything works. Because of this you can attach Galois representations to certain automorphic forms on these groups. If Deligne's axioms fail you may not know where to start. $\endgroup$
    – Kevin Buzzard
    Commented Apr 13, 2011 at 19:32
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    $\begingroup$ If you take $F$ to be totally real and $F$ CM, i.e. a totally imaginary quadratic extension of $F$, then the restriction of scalars $G:=R_{F/\mathbb{Q}} \mathrm{SU}(V)$ is a group of Hermitian type, i.e. it acts on a Hermitian symmetric domain $X$. If you choose a congruence subgroup $\Gamma\subset G(\mathbb{Q})$, then $X/\Gamma$ is a Shimura variety. $\endgroup$
    – Mikhail Borovoi
    Commented Apr 13, 2011 at 19:37
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    $\begingroup$ From a representation theoretic point of view (in automorphic forms), you don't need to, and most papers I look at (not that I look at unitary groups often) don't require this. E.g., see Gross-Gan-Prasad: math.ucsd.edu/~wgan/work8-3.pdf. $\endgroup$
    – Kimball
    Commented May 2, 2011 at 2:47

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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.

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.

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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.

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  • $\begingroup$ As in this answer, the restriction is mainly assure that the archimedean behavior is as expected, namely, classical (archimedean) unitary groups. Since (as it seems to turn out) the archimedean components of automorphic repns play a disproportionately large role, this is a reasonable distinction for automorphic purposes. For example, by the Hasse principle for hermitian forms, in more than 2 dimensions, compactness of arithmetic quotients of such groups depends most directly on the behavior at archimedean places. For more-purely algebraic-group purposes, it might be somewhat less significant. $\endgroup$
    – paul garrett
    Commented Jun 26, 2016 at 22:23

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