Mfrhtyj

i'm interested in the spin group $Spin(4,1)$ wich correspond to the symplectic group $Sp(1,1)$. The only source that I could find about it was wikipedia (http://en.wikipedia.org/wiki/Spin_group). It seems that there is no general definition about what a indefinite symplectic group is. It would be helpful if someone could provide me a text (or a proof) where I could find how come $Spin(4,1)=Sp(1,1)$. Thanks alot to everyone.

For example Cornwell (Group Theory in Physics Vol II,page 392) says:

$Sp(r,,s)={Aintextbf{GL}(n;,mathbb C): A^T,J,A=J::{rm and}:: A^dagger,G,A=G}$, where $r+s=n/2$ and

$J=left(
begin{array}{clc}
0 && I_{ntimes n}\
-I_{ntimes n}
end{array}right),hspace{.5cm} G=left(
begin{array}{clclc}
-I_{rtimes r}&&0&&0&&0\
0&&I_{stimes s}&&0&&0\
0&&0&&-I_{rtimes r}&&0\
0&&0&&0&&I_{stimes s}
end{array}
right)$

And there is another definition tha involves quaternions... so, which one?

Proposition A:
$$U(p,q;mathbb{H})~cong~Sp(p,q).tag{1}$$

Sketched proof:

1. The indefinite unitary group over the quaternions is
   $$U(p,q;mathbb{H})~:=~ left{
   xin{rm Mat}_{ntimes n}(mathbb{H}) left| x^{dagger}eta x =eta right. right},qquad n~=~p+q,$$
   $$ ~=~left{
   xin{rm Mat}_{ntimes n}(mathbb{H}) left| x^{daggereta} x ={bf 1}_{n times n} right. right},qquad x^{daggereta}~:=~eta x^{dagger}eta,$$
   $$ eta~:=~{rm diag}(underbrace{+1,ldots,+1}_p,underbrace{-1,dots, -1}_q),qquad eta^2~=~{bf 1}_{n times n} tag{2} . $$
   Let us mention for completeness that the corresponding Lie algebra
   $$u(p,q;mathbb{H})~:=~ left{
   xin{rm Mat}_{ntimes n}(mathbb{H}) left| x^{daggereta}=- x right. right} tag{3} $$
   of anti-Hermitian matrices has real dimension
   $$dim_{mathbb{R}}u(p,q;mathbb{H})~=~4frac{n(n-1)}{2}+3n~=~n(2n+1).tag{4} $$




2. The indefinite unitary group over the complex numbers is
   $$U(2p,2q)~:=~ left{
   Min{rm Mat}_{2ntimes 2n}(mathbb{C}) left| M^{dagger}(etaotimes {bf 1}_{2 times 2} ) M =etaotimes {bf 1}_{2 times 2} right. right}. tag{5}$$




3. The symplectic group over the complex numbers is
   $$Sp(2n;mathbb{C})~:=~ left{
   Min{rm Mat}_{2ntimes 2n}(mathbb{C}) left| M^tOmega M =Omega right. right}$$
   $$~cong~left{
   Min{rm Mat}_{2ntimes 2n}(mathbb{C}) left| M^t(etaotimesomega) M =(etaotimesomega) right. right}, $$
   $$ Omega ~:=~{bf 1}_{n times n}otimes omega~=~-Omega^t, qquad
   omega ~:=~mathrm{i}sigma_2~=~-omega^t , qquadOmega^2~=~-{bf 1}_{2n times 2n}.tag{6}$$
   The proof that we can use $etaotimesomega$ rather than $Omega$ as symplectic unit follows from index relabelling of rows (and the corresponding index relabelling of columns).




4. The indefinite symplectic group is defined as
   $$Sp(p,q)~:=~U(2p,2q) cap Sp(2n;mathbb{C}), $$
   $$~=~left{ Min U(2p,2q) left| M^t(etaotimesomega) M =(etaotimesomega) right. right} $$
   $$~=~left{ Min U(2p,2q) left| overline{M}Omega = Omega M right. right}, qquad eta Omega~=~ Omegaeta.tag{7} $$




5. The condition
   $$overline{M}Omega ~=~ Omega M tag{8} $$
   
   is the condition
   $$ overline{m}_{ij}omega ~=~omega m_{ij}, qquad
   i,j~in~{1,ldots,n},tag{9} $$
   that each of the $2times 2$ blocks $m_{ij}$ in
   $$M~=~begin{pmatrix} m_{11} &ldots & m_{1n} cr
   vdots &ddots & vdots cr
   m_{n1} &ldots & m_{nn}end{pmatrix}~in~{rm Mat}_{2ntimes 2n}(mathbb{C})tag{10}$$
   can be identified with a quaternion, cf. e.g. my Phys.SE answer here. In other words, there is an $mathbb{R}$-algebra isomorphism
   $$ Phi:~~{rm Mat}_{ntimes n}(mathbb{H})~~longrightarrow left{
   Min{rm Mat}_{2ntimes 2n}(mathbb{C}) left| overline{M} Omega = Omega M right. right}.tag{11} $$
   $Phi$ is actual a star algebra isomorphism
   $$ Phi(x^{dagger})~=~Phi(x)^{dagger}, qquad x~in~{rm Mat}_{ntimes n}(mathbb{H}).tag{12}$$




6. Conclude the group isomorphism
   $$U(p,q;mathbb{H})~stackrel{Phi_|}{cong}~Sp(p,q).tag{13}$$
   $Box$

Proposition B:

1. $G:=U(2; mathbb{H})$ is (the double cover of) the special orthogonal group $SO(5).$




2. $G:=U(1,1; mathbb{H})$ is (the double cover of) the restricted de Sitter group $SO^+(1,4).$

Sketched proof:

1. Since quaternions are non-commutative, care should be taken when defining
   trace (& determinant). The reduced trace is defined as
   $$ {rm Re~tr} (x)~=~frac{1}{2} {rm tr} (x+x^{dagger})~=~frac{1}{2} {rm tr} (Phi(x)), qquad x~in~{rm Mat}_{2times 2}(mathbb{H}),tag{14} $$
   which is cyclic,
   e.g. because
   $$ {rm Re}(qq^{prime})~=~{rm Re}(q^{prime}q), qquad q,q^{prime}~in~mathbb{H}.tag{15} $$




2. There is a bijective isometry from $mathbb{R}^5$ [or $mathbb{R}^{1,4}$] to the $mathbb{R}$-inner product space
   $$ V~=~ left{ xin {rm Mat}_{2times 2}(mathbb{H}) left| x^{daggereta}=x~~wedge~~ {rm Re~tr} (x)=0 right. right} $$
   $$~=~left{left. begin{pmatrix} r & q cr pmbar{q} & -r end{pmatrix}in {rm Mat}_{2times 2}(mathbb{H}) right| rinmathbb{R}~~wedge~~qinmathbb{H}right},tag{16}$$
   equipped with the (indefinite) quadratic form
   $$ || x||^2~:=~ frac{1}{2}{rm Re~tr} (x^2)~=~r^2 pm |q|^2 , tag{17}$$
   where $pm$ corresponds to the two cases in Proposition B.




3. There is a group action $rho:Gtimes V to V$ given by conjugation
   $$ rho(g)x~:=~gxg^{-1}~=~gxg^{daggereta}, qquad g~in ~G, qquad x~in~ V, tag{18}$$
   which is length preserving, i.e. $g$ is a (pseudo)orthogonal transformation.
   In other words, there is a Lie group homomorphism
   $rho: Grightarrow O(V)$, where
   $$rho(pm {bf 1}_{2 times 2})~=~{bf 1}_V.tag{19}$$
   $Box$

Proposition C:
$$G~:=~SL(2;mathbb{H})~:=~Phi^{-1}(SL(4;mathbb{C})) tag{20}$$
is (the double cover of) the special orthogonal group $SO^+(1,5).$

Sketched proof:

1. Let
   $$so(4,mathbb{C})~:=~left{
   Ain{rm Mat}_{4times 4}(mathbb{C}) left| A^t = -A right. right}tag{21}$$
   denote the set of antisymmetric complex $4times 4$ matrices,
   endowed with the Pfaffian
   $${rm Pf}(A)~=~frac{1}{8}sum_{mu,nu,lambda,sigma=1}^4epsilon_{munulambdasigma}A^{munu} A^{lambdasigma}
   ~=~A^{12}A^{34} + A^{31}A^{24} + A^{23}A^{14}, $$
   $$ epsilon_{1234}~=~1.tag{22}$$




2. If we define a "transposed" quaternion as
   $$ q^t~:=~q|_{q^2to-q^2}~=~-jbar{q}j, qquad q~in~mathbb{H}, tag{23}$$
   then
   $$ Phi(x)^t~=~Phi(x^t), qquad x~in~{rm Mat}_{2times 2}(mathbb{H}). tag{24}$$




3. Let the set of "antisymmetric" quaternion $2times 2$ matrices be
   $$so(2;mathbb{H})~:= left{ xin{rm Mat}_{2times 2}(mathbb{H}) mid x^t=-x right}~=~Phi^{-1}(so(4,mathbb{C}))~=~V({bf 1}_{2 times 2}otimes j) ,tag{25} $$
   where
   $$ V~:=~u(2;mathbb{H})~:= left{ xin{rm Mat}_{2times 2}(mathbb{H}) mid x^{dagger}=x right} $$
   $$~=~left{left. begin{pmatrix} r & q cr bar{q} & s end{pmatrix} in{rm Mat}_{2times 2}(mathbb{H}) right| r,sinmathbb{R}, ~qinmathbb{H} right}
   ~cong~mathbb{R}^{1,5}tag{26} $$
   is the real vector space of Hermitian quaternion $2times 2$ matrices.




4. Endow $V$ with the indefinite quadratic form
   $$ ||x||^2 ~=~ {rm Pf}(Phi(x({bf 1}_{2 times 2}otimes j)))~=~ frac{1}{2}{rm Re~tr} (x^2)-frac{1}{2}{rm Re~tr} (x)^2~=~rs pm |q|^2 .tag{27} $$




5. There is a group action $rho:Gtimes V to V$
   $$ rho(g)x~:=~gxg^{dagger}~=~-gx({bf 1}_{2 times 2}otimes j)g^t({bf 1}_{2 times 2}otimes j), qquad g~in ~G, qquad x~in~ V, tag{28}$$
   which is length preserving, i.e. $g$ is a (pseudo)orthogonal transformation.
   In other words, there is a Lie group homomorphism
   $rho: Grightarrow O(V)$, where
   $$rho(pm {bf 1}_{2 times 2})~=~{bf 1}_V.tag{29}$$
   $Box$

References:

1. Paul Garrett, Sporadic isogenies to orthogonal groups, 2015.