Is the statement below false?


"The metaplectic group Mp2(R) is not a matrix group: it has no faithful finite-dimensional representations."




---


Possible "counterexample":
Sp(2n,R) is a subgroup of O(4n,C) (or O(2n,2n) if you prefer).
So the Clifford algebraic Pin group will contain a double cover. The double cover will definitely be disconnected if Sp(2n,R) is not a subgroup of SO(4n,C). It should be connected if it is entirely in the Spin subgroup of the Pin group.


Consider the case of Sp(2,R). If we have a 2x2 real matrix with determinant 1, we can establish an isomorphism in SO(4,C) as follows: a^2-b^2-c^2+d^2 = 1  

[a+b,c-d;c+d,a-b] <---> [a,-bi,-ci,-d;bi,a,d,-ci;ci,-d,a,bi;d,ci,-bi,a] 


Since Spin(4,C) will double cover SO(4,C), we could have a connected double cover of Sp(2,R).




---


Note: The proposed "example" is false due to submitted answer. Thanks.



Is the metaplectic group not a matrix group - counterexample
Keep in mind that any finite-dimensional representation of a Lie group determines a finite-dimensional representation of its Lie algebra, and for a connected Lie group the induced Lie algebra representation determines the Lie group representation.


However, every finite-dimensional representation of $\operatorname{Lie}(\mathrm{Mp}(2,\mathbb R)) = \mathfrak{sl}(2,\mathbb R)$ comes from a representation of $\mathrm{SL}(2,\mathbb R)$, and so does not come from a faithful rep of $\mathrm{Mp}(2,\mathbb R)$. One way to see this by directly classifying all finite-dimensional $\mathfrak{sl}(2,\mathbb R)$ representations, which is not too difficult. A better way is to observe that any $\mathfrak{sl}(2,\mathbb R)$-representation $V$ embeds in an $\mathfrak{sl}(2,\mathbb C)$-representation $V \otimes \mathbb C$, but $\mathrm{SL}(2,\mathbb C)$ is simply connected, so $V \otimes \mathbb C$ is a representation of $\mathrm{SL}(2,\mathbb C)$, and so the $\mathrm{Mp}(2,\mathbb R)$-representation that gave rise to $V$ factors through $\mathrm{SL}(2,\mathbb C)$, and on the other hand the map $\mathrm{Mp}(2,\mathbb R) \to \mathrm{SL}(2,\mathbb C)$ factors through $\mathrm{SL}(2,\mathbb R)$ and is not faithful.


