# Is the metaplectic group not a matrix group - counterexample

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.

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Perhaps you could include details of your proposed counterexample? – MTS Jan 16 '13 at 5:27
This question is written in an argumentative tone of voice, which risks turning people off. I recommend that you revise it, taking into consideration advice from mathoverflow.net/howtoask. Among other things, you could post your putative faithful finite-dimensional representation. At the minimum, writing up your construction carefully will probably lead you to find an error. – Theo Johnson-Freyd Jan 16 '13 at 5:29
Sorry about that. I hope this is better for everyone. – Jules Berracasa Jan 16 '13 at 5:45
@Alexander Schlering: You're just constructed the disconnected double cover of the symplectic group inside of the disconnected Pin group. This has nothing to do with the metaplectic group, and is not useful. Be more attentive to the importance of checking connectedness. – user29720 Jan 16 '13 at 6:04
-1 for persisting despite Theo's explanation (and any number of sources which state that the metaplectic group is not a matrix group) – Yemon Choi Jan 16 '13 at 7:26

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.