I am a bit confused regarding the possible constructions/realizations of symplectic groups. Basically I am looking for the following:

A linear algebraic group $\mathbb{G}$ defined over $\mathbb{Q}$ such that $\mathbb{G}(\mathbb{C})$ is (isomorphic to) the group $\mathrm{Sp_{2m}}(\mathbb{C})$ while $\mathbb{G}(\mathbb{R})$ is compact (and possibly $\mathbb{G}$ split at as many finite primes as possible). Does such a group exist and if so what does it look like?

I am especially interested in the case $m=3$ and would like such a realization of the symplectic group for some explicit computations of algebraic modular forms.

  • 3
    $\begingroup$ All simply connected semisimple real algebraic groups are definable over the rationals. Here the simplest is just the stabilizer of the standard positive definite hermitian form on the $H^n$ where $H$ are the usual quaternions (with the standard basis). $\endgroup$
    – YCor
    Jun 23, 2015 at 14:02
  • $\begingroup$ PS: I guess that a semisimple group over $\mathbf{Q}$ will be automatically be split over $\mathbf{Q}_p$ for $p$ ranging over a positive density set of primes. $\endgroup$
    – YCor
    Jun 23, 2015 at 21:38
  • $\begingroup$ I guess that the standard quaternion algebra $H=\mathbb{Q}(i,j)$ with $i^2=-1,\ j^2=-1,\ ji=-ij$ ramifies exactly at $\infty$ and 2. It follows that the unitary group $G=SU(H^n,F)$ of the hermitian form $F(x)=x_1 \bar{x}_1+\dots+x_n\bar{x}_n$ from YCor's first comment splits at every prime $p$ except $\infty$ and maybe at $p=2$. It does not split at 2 because it cannot be nonsplit at one place only by the reciprocity law (I mean the Hasse-Brauer-Noether theorem). $\endgroup$ Jun 24, 2015 at 4:10
  • $\begingroup$ @YCor: Moreover, any $\mathbb{Q}$-form of ${\rm Sp}_{2m}$ splits at almost all primes $p$ (because it is an inner form of a split $\mathbb{Q}$-group). $\endgroup$ Jun 24, 2015 at 4:18
  • $\begingroup$ Thank you for your comments, they were very helpful. I came across the construction via the quaternions earlier but was somehow thrown off the tracks by a a confusing remark. $\endgroup$ Jun 24, 2015 at 11:11

1 Answer 1


Yes, and you can make it split away from whatever even finite set $S$ of places of $\mathbf{Q}$ you wish that contains the archimedean place, and such a form of ${\rm{Sp}}_{2n}$ is uniquely determined as well. This is seen via Galois cohomology and class field theory (a not exactly constructive method, but Springer's book on linear algebraic groups has a rather explicit description of twisted forms of classical groups in terms of quaternion algebras and other linear-algebraic data, so it can make explicit various constructions provided abstractly by Galois cohomology).

The main point is that for any connected semisimple group $G$ that is simply connected and has no nontrivial automorphisms of its Dynkin diagram (such as your question for type C), the automorphism variety of $G$ is connected, which is to say is exactly the adjoint central quotient $G^{\rm{ad}}$, and hence forms of that type can be studied via the cohomology of the adjoint quotient, to which Hasse Principle, etc. may be applied. (The book of Platonov-Rapinchuk provides proofs of the statements about cohomology of semisimple groups made below.)

Let's now focus on the case of simply connected groups of type C over a field $k$, as in your question. Such groups are exactly the twisted forms of the split form $G_0 := {\rm{Sp}}_{2n}$, so the set of isomorphism classes over $k$ is naturally in bijection with ${\rm{H}}^1(k, G_0^{\rm{ad}}) = {\rm{H}}^1(k, G_0/\mu)$ where $\mu = \mu_2$ is the center of $G_0$. By the Hasse Principle for adjoint groups, the natural map $$c:{\rm{H}}^1(k, G_0/\mu) \rightarrow \prod_v {\rm{H}}^1(k_v, G_0/\mu)$$ is injective (as a map of sets, not just that its kernel as a map of pointed sets is trivial). Moreover, by the arithmetic theory of connected semisimple groups over local fields (including the archimedean case), the natural connecting map $$\delta_v: {\rm{H}}^1(k_v, G_0/\mu) \rightarrow {\rm{H}}^2(k_v,\mu) = {\rm{Br}}(k_v)[2]$$ is bijective.

Since ${\rm{Br}}(k_v)[2]$ has order 2 when $k_v \ne \mathbf{C}$ and is trivial when $k_v = \mathbf{C}$, we see that a $k$-form $G$ of $G_0$ is split at $v$ precisely when its class $[G] \in {\rm{H}}^1(k, G_0/\mu)$ has image $c([G])$ with trivial $v$-component. Likewise, if $k_v = \mathbf{R}$ then $G$ is anisotropic at $v$ (equivalently, $G(k_v)$ is compact) precisely when $[G]_v$ is nontrivial.

It is a general fact that the connecting map $\delta:{\rm{H}}^1(k,G_0/\mu) \rightarrow {\rm{H}}^2(k,\mu) = {\rm{Br}}(k)[2]$ is surjective for any number field $k$, and by class field theory elements of ${\rm{Br}}(k)[2]$ are specified exactly by choosing an even finite set $S$ of non-complex places of $k$ (corresponding to a unique quaternion division algebra over $k$ split exactly away from $S$). Hence, given any such $S$ there exists a $k$-form $G$ of $G_0$ that is split away from $S$ and is non-split at all places in $S$, with $G(k_v)$ compact for all real places in $S$. Such a $k$-form $G$ is unique, due to the injectivity of $c$ above.


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