Is the spin group in a metaplectic group? Is every spin group $Spin(n,R)$ over the reals contained in some metaplectic group $Mp(m,R)$ for some $m$ in such a way that the spin representation is obtained by restriction of the metaplectic representation?
 A: The fact is that because $Sp(n, \mathbb{R})$ is diffeomorphic to the product of the unitary group $U(n)$ and an Euclidean space. So, the fundamental group of $Sp(n, \mathbb{R})$ is $\mathbb{Z}$. So  $Sp(n, \mathbb{R})$ has unique double covering, which we denote by $Mp(n, \mathbb{R})$ and is called Metaplectic group. Note also that, the metaplectic group $Mp(2,\mathbb{R})$ is not a matrix group, so metaplectic group is a little bit complicate.
To have a better picture of metaplectic group we give a general definition for it. Let $(V, \omega)$ be a symplectic vector space with $dimV=2n$ over $\mathbb{F}$ (here $\mathbb{F}$ is a nonarchimedean local field of characteristic 0 and residual characteristic $p$) with associated symplectic group $Sp(V)$. The group $Sp(V)$ has a unique two-fold central extension $Mp(V)$ which is called the metaplectic group: 
$$0\to \{\pm 1\}\to Mp(V)\to Sp(V)\to 0.$$
So, we can write $Mp(V)=Sp(V)\oplus \{\pm 1\}$ with group law given by 
$$(g_1, \epsilon_1).(g_2, \epsilon_2) = \left(g_1g_2,\epsilon_1\epsilon_2c(g_1, g_2)\right)$$ for some 2-cocycle $c$ on $Sp(V)$ valued in $\{\pm 1\}$. 
Now, note that we have the natural embedding $GL(n,\mathbb{R})\hookrightarrow Sp(n,\mathbb{R})$ given by 
$$A\mapsto \left[\matrix{ A&0\cr0&A^{*-1}}\right]$$
where $A^*$ is the transpose of $A$.
So,  $GL(n,\mathbb{R})$ can be viewed as subgroup in $Sp(2n, \mathbb{R})$ as the subgroup that preserves the standard Lagrangian submanifold $\mathbb{R}^n\hookrightarrow \mathbb{R}^{2n}$.  Restriction of the metaplectic group extension along this inclusion defines the metalinear group $Ml(n,\mathbb{R})$. 
\begin{array}{lll}
  ML(n,\mathbb{R})  & \rightarrow &Mp(2n,\mathbb{R})\\
  \downarrow && \downarrow \\
 GL(n,\mathbb{R}) & \rightarrow & Sp(2n,\mathbb{R})
 \end{array}
For the definition of metalinear group, the quotient $ML(n,\mathbb{C}):=({\mathbb{C}\times SL(n,\mathbb{C})})/{2\mathbb{Z}}$ is called as complex metalinear group of dimension $n$. The elements of $ML(n,\mathbb{C})$ can be written as following forms
$$\overline{(m,B)}=\left\{\left(m+\frac{4\pi ik}{n},e^{-\frac{4\pi ik}{n}}B\right):k\in {\mathbb{Z}} , \right\}$$
where $B\in SL(n,\mathbb{C}) $
So, we will have a covering map 
$$\rho :ML(n,\mathbb{C})\to GL(n,\mathbb{C}),$$ 
$$\overline{(m,B)}\mapsto e^mB$$
See my expose in Lille, 2014
A: Let $\alpha : 1\to K \to Spin(m)\to SO(m)\to 1$ be the universal central extension, so that $K=\mathbf Z$ if $m= 2$ and $K=\mathbf Z/2$ if $m\geq 3$.
Let $\beta : 1\to \mathbf Z \to \tilde{Sp}(n)\to {Sp}(n)\to 1$ be the universal central extension.
Let $\gamma : 1\to \mathbf Z/2 \to {Mp}(n)\to {Sp}(n)\to 1$ be the metaplectic central extension.
I interpret the question as : can one find a morphism $f:SO(m)\to Sp(n)$ such that the pullback $f^*(\gamma)$ of $\gamma$ along $f$ be isomorphic to $\alpha$ ?
It seems that the answer is always no. Indeed, 


*

*for $m\geq 3$, $f^*(\gamma)$ is the image of $f^*(\beta)$, which is an extension of $SO$ by $\mathbf Z$ and hence is trivial.

*for $m=2$, it is impossible since $\mathbf Z\neq \mathbf Z/2$. Nevertheless, in this case, if $f$ is the inclusion $SO(2)\to SL(2)$, then $f^*(\beta)$ is isomorphic to $\alpha$.
