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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.