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The symplectic group is defined as

$$\text{Sp}(2n,F)=\{M\in M_{2n\times 2n}(F) : M^T\Omega M=\Omega\},$$

where

$$\Omega =\left( \begin{matrix}0&I_n\\-I_n&0\end{matrix}\right).$$

Is the symplectic group over the rationals $\text{Sp}(2n,\mathbb Q)$ dense on the symplectic group over the reals $\text{Sp}(2n,\mathbb R)$?

I am aware that the rational orthogonal group is dense on the real orthogonal group. There appears to be at least two ways to show this, but neither seem obvious to extend to the case of symplectic matrices.

  1. By the Cartan–Dieudonné theorem it is possible to show that the rational Householder matrices are dense on the reals Density of orthogonal matrices with rational coefficients

  2. By the Cayley formula https://mathoverflow.net/questions/90070/existence-of-rational-orthogonal-matrices

One possible method I have been considering is using the fact that the real symplectic group can be written in terms of generators

$$\text{Sp}(2n,\mathbb R)=D(n)\cup N(n) \cup \{\Omega\}$$ where

$${\displaystyle {\begin{aligned}D(n)&=\left\{\left.{\begin{bmatrix}A&0\\0&(A^{T})^{-1}\end{bmatrix}}\,\right|\,A\in \operatorname {GL} (n,\mathbb {R} )\right\}\\[6pt]N(n)&=\left\{\left.{\begin{bmatrix}I_{n}&B\\0&I_{n}\end{bmatrix}}\,\right|\,B\in \operatorname {Sym} (n)\right\}\end{aligned}}}$$.

Is it sufficient to show that there exists groups $D(n,\mathbb Q)$ and $N(n,\mathbb Q)$ which are each dense in $D(n)$ and $N(n)$ respectively? If so, what is the reasoning for this?

Cameron
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    I mean this is true by the strong approximation theorem, isn't it? For example, Theorem 2.3 here: http://library.msri.org/books/Book61/files/70rapi.pdf – Alex Youcis Aug 11 '22 at 21:44
  • @Alex Youcis (1) should be straight-forward to demonstrate (as the real symplectic group is simply connected). For (2): I believe it should also be a simply connected group (although I am not sure what it means to be absolutely simply connected). But how would you prove $G(\mathcal O(S))$ is infinite? – Cameron Aug 12 '22 at 10:25
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    WHy do you have to show that? In this case $S={\infty}$ and you just need to show that $G(\mathbb{R})$ is non-compact, but the symplectic group is non-compact. – Alex Youcis Aug 12 '22 at 14:03
  • @AlexYoucis Okay great! I therefore just need to show that the symplectic group is connected and "absolutely almost simple". I believe it is according to https://encyclopediaofmath.org/wiki/Symplectic_group But I am a bit confused as to how to come to the conclusion that it is simple, from the property of the center of the group – Cameron Aug 15 '22 at 10:09

1 Answers1

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For any field $\mathbb{F}$, the symplectic group $Sp(2n,\mathbb{F})$ is generated by symplectic transvections. These are maps of the form $f_{\alpha,u}$, where $\alpha \in \mathbb{F}$, $u \in \mathbb{F}^{2n}$, and $f_{\alpha,u}(v) = v + \alpha B(v,u) u$ for all $v \in \mathbb{F}^{2n}$. Here $B$ is the alternating bilinear form used to define $Sp(2n,\mathbb{F})$.

It seems to me this should be enough to show that $Sp(2n,\mathbb{Q})$ is dense in $Sp(2n,\mathbb{R})$: for $\alpha \in \mathbb{R}$ and $u \in \mathbb{R}^{2n}$, you can find $\alpha' \in \mathbb{Q}$ and $u' \in \mathbb{Q}^{2n}$ such that $f_{\alpha',u'}$ is arbitrarily close to $f_{\alpha,u}$.

spin
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  • why is it enough to show that its generators are dense? Isn't it possible (in general) that to build some target rational symplectic matrix we would need many of these generators, each contributing another error? Or how do we mathematically show that this isn't the case? – Cameron Aug 16 '22 at 07:24
  • Well, say every generator is in the closure of $Sp(2n,\mathbb{Q})$. Since closure of a subgroup is a subgroup, the closure must be all of $Sp(2n,\mathbb{R})$. – spin Aug 16 '22 at 08:37
  • I don't understand the last part of your sentence. If we denote $H={f_{u,v}}$ the set of generators in the rationals. $H\le Cl(Sp(2n,\mathbb Q)\le Sp(2n,\mathbb R)$. How does this imply $Cl(Sp(2n,\mathbb Q)= Sp(2n,\mathbb R)$? – Cameron Aug 16 '22 at 09:21
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    I mean that every generator of $Sp(2n,\mathbb{R})$ is in the closure of $Sp(2n,\mathbb{Q})$. – spin Aug 16 '22 at 09:43
  • Okay, so do you mean that the closure of the set of generators H is equal to the real symplectic matrices? i.e. $Cl(H)=Sp(2n,\mathbb R)$? Which would mean $Cl(H)\le Cl(Sp(2n,\mathbb Q))\le Sp(2n,\mathbb R)$. If so, why is $Cl(H)=Sp(2n,\mathbb R)$? Otherwise, what do you mean? – Cameron Aug 16 '22 at 10:03
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    Denote by $S_{\mathbb{Q}}$ generators in the rationals, and by $S_{\mathbb{R}}$ generators in the reals. Then $Cl(S_{\mathbb{Q}})$ contains $S_{\mathbb{R}}$. Hence $Cl(Sp(2n,\mathbb{Q}))$ contains $S_{\mathbb{R}}$, hence the result. – spin Aug 16 '22 at 11:59
  • Okay great, last question: how do you show that $f_{\alpha',u'}$ is arbitrarily close to $f_{\alpha,u}$? I guess you need to show $||f_{a,u}(v)-f_{a',u'}||=||\alpha B(v,u)u-\alpha' B(v,u')u'||$ is arbitrarily small, but I don't see how to simplify this – Cameron Aug 16 '22 at 15:51
  • Just express the difference in terms of $\alpha-\alpha'$ and $u-u'$, for example – spin Aug 17 '22 at 01:43