most recent 30 from http://mathoverflow.net 2013-06-19T14:01:00Z http://mathoverflow.net/feeds/question/119037 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119037/is-the-metaplectic-group-not-a-matrix-group-counterexample Jules Berracasa 2013-01-16T04:53:35Z 2013-01-16T17:15:00Z

<p>Is the statement below false?</p> <p>"The metaplectic group Mp2(R) is not a matrix group: it has no faithful finite-dimensional representations."</p> <hr> <p>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.</p> <p>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<br> [a+b,c-d;c+d,a-b] &lt;---> [a,-bi,-ci,-d;bi,a,d,-ci;ci,-d,a,bi;d,ci,-bi,a] </p> <p>Since Spin(4,C) will double cover SO(4,C), we could have a connected double cover of Sp(2,R).</p> <hr> <p>Note: The proposed "example" is false due to submitted answer. Thanks.</p>

http://mathoverflow.net/questions/119037/is-the-metaplectic-group-not-a-matrix-group-counterexample/119040#119040 Theo Johnson-Freyd 2013-01-16T05:46:58Z 2013-01-16T05:46:58Z

<p>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.</p> <p>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.</p>