Mixing
A description of the compact symplectic group
$begingroup$
Let $mathrm{Sp}(2m;mathbb{C})={Xinmathrm{GL}(2m;mathbb{C});X^tOmega X=Omega}$, where $Omega=begin{bmatrix}0& I_m\ -I_m& 0end{bmatrix}$, $I_m$ is $mtimes m$ identity matrix.
The compact symplectic group is defined by $mathrm{Sp}(m)=mathrm{Sp}(2m;mathbb{C})cap U(2m)$. The following statement is from Kobayashi's Differential Geometry of Complex Vector Bundles pp.45.
By simple calculation we obtain
$$mathrm{Sp}(m)=left{begin{bmatrix} A& B\-bar{B}&bar{A}end{bmatrix};bar{A}^tA+B^tbar{B}=I_m,bar{A}^tB=B^tbar{A}right}.$$
However I cannot obtain this form via computation. If we write $Xinmathrm{Sp}(m)$ as $X=begin{bmatrix} A& B\C&Dend{bmatrix}$, then from $X^tOmega X=Omega$ and $bar{X}^tX=I$, we obtain
$$A^tC=C^tAquad B^tD=D^t Bquad A^tD=C^tB+I_mquad D^tA=B^tC+I_m$$ $$ bar{A}^tB=-bar{C}^tDquadbar{B}^tA=-bar{D}^tCquadbar{A}^tA+bar{C}^tC=I_mquadbar{B}^tB+bar{D}^tD=I_m.$$
The third and fourth equations are equivalent, fifth and sixth equations are equivalent. How to deduce $C=-bar{B}$ and $D=bar{A}$?
I appreciate it if you can give a proof or a counterexample.
linear-algebra matrices symplectic-linear-algebra classical-groups
asked Feb 2 at 17:18
NienzNienz
1108
$endgroup$
add a comment |
$begingroup$
Let $mathrm{Sp}(2m;mathbb{C})={Xinmathrm{GL}(2m;mathbb{C});X^tOmega X=Omega}$, where $Omega=begin{bmatrix}0& I_m\ -I_m& 0end{bmatrix}$, $I_m$ is $mtimes m$ identity matrix.
The compact symplectic group is defined by $mathrm{Sp}(m)=mathrm{Sp}(2m;mathbb{C})cap U(2m)$. The following statement is from Kobayashi's Differential Geometry of Complex Vector Bundles pp.45.
By simple calculation we obtain
$$mathrm{Sp}(m)=left{begin{bmatrix} A& B\-bar{B}&bar{A}end{bmatrix};bar{A}^tA+B^tbar{B}=I_m,bar{A}^tB=B^tbar{A}right}.$$
However I cannot obtain this form via computation. If we write $Xinmathrm{Sp}(m)$ as $X=begin{bmatrix} A& B\C&Dend{bmatrix}$, then from $X^tOmega X=Omega$ and $bar{X}^tX=I$, we obtain
$$A^tC=C^tAquad B^tD=D^t Bquad A^tD=C^tB+I_mquad D^tA=B^tC+I_m$$ $$ bar{A}^tB=-bar{C}^tDquadbar{B}^tA=-bar{D}^tCquadbar{A}^tA+bar{C}^tC=I_mquadbar{B}^tB+bar{D}^tD=I_m.$$
The third and fourth equations are equivalent, fifth and sixth equations are equivalent. How to deduce $C=-bar{B}$ and $D=bar{A}$?
I appreciate it if you can give a proof or a counterexample.
linear-algebra matrices symplectic-linear-algebra classical-groups
asked Feb 2 at 17:18
NienzNienz
1108
$endgroup$
add a comment |
$begingroup$
Let $mathrm{Sp}(2m;mathbb{C})={Xinmathrm{GL}(2m;mathbb{C});X^tOmega X=Omega}$, where $Omega=begin{bmatrix}0& I_m\ -I_m& 0end{bmatrix}$, $I_m$ is $mtimes m$ identity matrix.
The compact symplectic group is defined by $mathrm{Sp}(m)=mathrm{Sp}(2m;mathbb{C})cap U(2m)$. The following statement is from Kobayashi's Differential Geometry of Complex Vector Bundles pp.45.
By simple calculation we obtain
$$mathrm{Sp}(m)=left{begin{bmatrix} A& B\-bar{B}&bar{A}end{bmatrix};bar{A}^tA+B^tbar{B}=I_m,bar{A}^tB=B^tbar{A}right}.$$
However I cannot obtain this form via computation. If we write $Xinmathrm{Sp}(m)$ as $X=begin{bmatrix} A& B\C&Dend{bmatrix}$, then from $X^tOmega X=Omega$ and $bar{X}^tX=I$, we obtain
$$A^tC=C^tAquad B^tD=D^t Bquad A^tD=C^tB+I_mquad D^tA=B^tC+I_m$$ $$ bar{A}^tB=-bar{C}^tDquadbar{B}^tA=-bar{D}^tCquadbar{A}^tA+bar{C}^tC=I_mquadbar{B}^tB+bar{D}^tD=I_m.$$
The third and fourth equations are equivalent, fifth and sixth equations are equivalent. How to deduce $C=-bar{B}$ and $D=bar{A}$?
I appreciate it if you can give a proof or a counterexample.
linear-algebra matrices symplectic-linear-algebra classical-groups
asked Feb 2 at 17:18
NienzNienz
1108
$endgroup$
Let $mathrm{Sp}(2m;mathbb{C})={Xinmathrm{GL}(2m;mathbb{C});X^tOmega X=Omega}$, where $Omega=begin{bmatrix}0& I_m\ -I_m& 0end{bmatrix}$, $I_m$ is $mtimes m$ identity matrix.
The compact symplectic group is defined by $mathrm{Sp}(m)=mathrm{Sp}(2m;mathbb{C})cap U(2m)$. The following statement is from Kobayashi's Differential Geometry of Complex Vector Bundles pp.45.
By simple calculation we obtain
$$mathrm{Sp}(m)=left{begin{bmatrix} A& B\-bar{B}&bar{A}end{bmatrix};bar{A}^tA+B^tbar{B}=I_m,bar{A}^tB=B^tbar{A}right}.$$
However I cannot obtain this form via computation. If we write $Xinmathrm{Sp}(m)$ as $X=begin{bmatrix} A& B\C&Dend{bmatrix}$, then from $X^tOmega X=Omega$ and $bar{X}^tX=I$, we obtain
$$A^tC=C^tAquad B^tD=D^t Bquad A^tD=C^tB+I_mquad D^tA=B^tC+I_m$$ $$ bar{A}^tB=-bar{C}^tDquadbar{B}^tA=-bar{D}^tCquadbar{A}^tA+bar{C}^tC=I_mquadbar{B}^tB+bar{D}^tD=I_m.$$
The third and fourth equations are equivalent, fifth and sixth equations are equivalent. How to deduce $C=-bar{B}$ and $D=bar{A}$?
I appreciate it if you can give a proof or a counterexample.
linear-algebra matrices symplectic-linear-algebra classical-groups
linear-algebra matrices symplectic-linear-algebra classical-groups
asked Feb 2 at 17:18
NienzNienz
1108
asked Feb 2 at 17:18
NienzNienz
1108
edited Mar 13 at 14:04
Nienz
asked Feb 2 at 17:18
NienzNienz
1108
asked Feb 2 at 17:18
NienzNienz
1108
asked Feb 2 at 17:18
NienzNienz
1108
1108
add a comment |
add a comment |
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