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Let $K$ be a non-archimedean local field of characteristic $>2$. Consider the double cover metaplectic extension of symplectic groups $p: Mp_{2n}(K)\rightarrow Sp_{2n}(K)$.

Q1: Let $H_{2n+1}(K)$ be a 2n+1 dimensional Heisenberg group. Are $Mp_{2n}(K)$ and $H_{2n+1}(K)\rtimes Mp_{2n}(K)$ always non-linear groups? and why?

Q2: Consider the Jacobi group $H_{2(n-1)+1}(K)\rtimes Sp_{2(n-1)}(K)$ as a subgroup in $Sp_{2n}(K)$. Is the following true $p^{-1}(H_{2(n-1)+1}(K)\rtimes Sp_{2(n-1)}(K))=H_{2(n-1)+1}(K)\rtimes Mp_{2(n-1)}(K)$?

In "Representations of Metaplectic Groups" (see http://www.math.nus.edu.sg/~matgwt/iccm-offprint.pdf) Wee Teck Gan consider $K$ has characteristic ZERO. On page 157 it states that "As a set, we may write $Mp_{2n}(K)=Sp_{2n}(K)\times \{+,-1\}$ So I would like to guess that the covering map $p$ is just the projection on the first coordinate. If this is the case, Q2 has affirmative answer as least for characteristic zero case. On page 158 in the last paragraph it states that "Mp(Wn) is not a linear group". I have no argument for this fact.

Happy New Year!

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    $\begingroup$ Concerning your Q1, over any sort of reasonable field a symplectic group of rank at least 2 is "simply connected" in the sense of algebraic groups (which coincides with the usual notion over the real or complex fields). So a non-split extension should typically not be a linear algebraic group. $\endgroup$
    – Jim Humphreys
    Commented Jan 2, 2016 at 21:24
  • $\begingroup$ @jim: not over the reals ($SL_2({\mathbb R})$ is not simply connected) but over the complex numbers $\endgroup$
    – Venkataramana
    Commented Jan 3, 2016 at 0:55
  • $\begingroup$ @Venkataramana: Yes, it gets more complicated over the reals, but your example just has rank 1. What is true for symplectic groups of higher rank? Anyway, the complex case is the crucial one here. $\endgroup$
    – Jim Humphreys
    Commented Jan 3, 2016 at 1:30
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    $\begingroup$ @jim: the symplectic group over ${\mathbb R}$ is not simply connected (has $\mathbb Z$ as fundamental group) but over $\mathbb C$, it is simply connected. $\endgroup$
    – Venkataramana
    Commented Jan 3, 2016 at 2:37
  • $\begingroup$ As far as I know, metaplectic groups over reals are not linear algebraic groups, because linear algebraic groups have finitely many topological components (proved by Whitney '57) and in this case metaplectic groups contain $\mathbb{Z}$ as a central subgroup (center has infinitely many components ). Metaplectic groups over complex number are linear, in fact, metaplectic Groups over complex number are equal to symplectic groups. However, I am interested in function fields case. Wee Teck Gan is interested in $Q_p$ case. $\endgroup$
    – m07kl
    Commented Jan 3, 2016 at 13:02

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