Is the metaplectic group not a matrix group - counterexample

Question

Is the statement below false?

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

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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).

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

Answer 1

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.

Answer 2

The answer of Theo seems to be right. Nevertheless, it is not explained that "the Mp(2,R)-representation that gave rise to V factors through SL(2,C) and on the other hand Mp(2,R) -> SL(2,C) factors through SL(2,R)". This cannot be true in general. E.g. in the case of SO(n) and Spin(n), i.e., when SL(2,R) is replaced by SO(n) and Mp(2,R) by Spin(n). Note that Spin(n,C) is (as SL(2,C)) simply connected as well (so topologically, it works). I think a bit incomplete part of the answer - for me - can be in factorizations related to the complexifications. (SO(n) is a compact real of SO(n,C), but SL(n,R) is split real of SL(n,C) - this should explain the difference bteween the SO and Sp-cases).