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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 defenetion 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 $Mp(V)=Sp(V)\oplus \mathbb{Z}/2\mathbb{Z}$ 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

$$A\mapsto \left[\matrix{ A&0\cr0&A^{*-1}}\right]$$

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

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

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 defenetion 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 $Mp(V)=Sp(V)\oplus \mathbb{Z}/2\mathbb{Z}$ 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

$$A\mapsto \left[\matrix{ A&0\cr0&A^{*-1}}\right]$$

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

$$\rho :ML(n,\mathbb{C})\to GL(n,\mathbb{C}),$$ $$\overline{(m,B)}\mapsto e^mB$$

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 defenetion 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 $Mp(V)=Sp(V)\oplus \mathbb{Z}/2\mathbb{Z}$ 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

$$A\mapsto \left[\matrix{ A&0\cr0&A^{*-1}}\right]$$

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

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 defenetion 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 $Mp(V)=Sp(V)\oplus \mathbb{Z}/2\mathbb{Z}$ 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

$$A\mapsto \left[\matrix{ A&0\cr0&A^{*-1}}\right]$$

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}

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 defenetion 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 $Mp(V)=Sp(V)\oplus \mathbb{Z}/2\mathbb{Z}$ 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

$$A\mapsto \left[\matrix{ A&0\cr0&A^{*-1}}\right]$$

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

$$\rho :ML(n,\mathbb{C})\to GL(n,\mathbb{C}),$$ $$\overline{(m,B)}\mapsto e^mB$$