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:

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

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

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

  • $\begingroup$ See page 10 of mat.univie.ac.at/~cap/files/wisser.pdf . The associated Lie algebra isomorphism is a real consequence, by the way, of the exceptional isomorphism $B_2 \cong C_2$ of semisimple (complex) Lie algebras, which itself has numerous beautiful consequences. $\endgroup$ Commented Jan 15, 2015 at 12:34
  • $\begingroup$ At a glance the Lie algebras look to be isomorphic, possibly up to a reversal of the roles of $r, s$. $\endgroup$ Commented Jan 15, 2015 at 12:52
  • $\begingroup$ Are you familiar with the classification of real semisimple Lie algebras? $\endgroup$ Commented Jan 15, 2015 at 12:57
  • $\begingroup$ No, i'm not familiar, but please explain in those terms that I would learn this topic. I hate to be abother, but could you please tell me how could you tell, a glance, this would be isomorphic? $\endgroup$
    – user208059
    Commented Jan 15, 2015 at 12:58
  • $\begingroup$ Are you familiar with how to think of quaternionic matrices as complex ones in the "usual" way? $\endgroup$ Commented Jan 15, 2015 at 12:59

1 Answer 1


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

Sketched proof:

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

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

  3. The symplectic group over the complex numbers is $$\begin{align} Sp(2n;\mathbb{C})~:=~& \left\{ M\in{\rm Mat}_{2n\times 2n}(\mathbb{C}) \left| M^t\Omega M =\Omega \right. \right\}\cr ~\cong~&\left\{ M\in{\rm Mat}_{2n\times 2n}(\mathbb{C}) \left| M^t(\eta\otimes\omega) M =(\eta\otimes\omega) \right. \right\}\cr ~=:~&Sp(2p,2q;\mathbb{C}), \cr \Omega ~:=~&{\bf 1}_{n \times n}\otimes \omega~=~-\Omega^t, \cr \omega ~:=~&\mathrm{i}\sigma_2~=~-\omega^t , \cr \Omega^2~=~&-{\bf 1}_{2n \times 2n}.\end{align}\tag{6}$$ The proof that we can use $\eta\otimes\omega$ 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 $$\begin{align} Sp(p,q)~:=~&U(2p,2q) \cap Sp(2p,2q;\mathbb{C})\cr ~=~&\left\{ M\in U(2p,2q) \left| M^t(\eta\otimes\omega) M =(\eta\otimes\omega) \right. \right\} \cr ~=~&\left\{ M\in U(2p,2q) \left| \overline{M}\Omega = \Omega M \right. \right\}, \cr (\eta\otimes {\bf 1}_{2 \times 2}) \Omega~=~& \Omega(\eta\otimes {\bf 1}_{2 \times 2}).\end{align}\tag{7} $$

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

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

  7. The first isomorphism in Proposition A now follows from the fact that symplectic matrices has unit determinant.


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 $$\begin{align} {\rm Re~tr} (x)~=~&\frac{1}{2} {\rm tr} (x+x^{\dagger})\cr ~=~&\frac{1}{2} {\rm tr} (\Phi(x)), \cr x~\in~&{\rm Mat}_{2\times 2}(\mathbb{H}),\end{align}\tag{14} $$ 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 $$\begin{align} V~=~& \left\{ x\in {\rm Mat}_{2\times 2}(\mathbb{H}) \left| x^{\dagger\eta}=x~~\wedge~~ {\rm Re~tr} (x)=0 \right. \right\} \cr ~=~&\left\{\left. \begin{pmatrix} r & q \cr \pm\bar{q} & -r \end{pmatrix}\in {\rm Mat}_{2\times 2}(\mathbb{H}) \right| r\in\mathbb{R}~~\wedge~~q\in\mathbb{H}\right\},\end{align}\tag{16}$$ 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 $\rho:G\times V \to V$ given by conjugation $$ \rho(g)x~:=~gxg^{-1}~=~gxg^{\dagger\eta}, \qquad g~\in ~G, \qquad x~\in~ V, \tag{18}$$ which is length preserving, i.e. $g$ is a (pseudo)orthogonal transformation. In other words, there is a Lie group homomorphism $\rho: G\rightarrow O(V)$, 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 $$so(4,\mathbb{C})~:=~\left\{ A\in{\rm Mat}_{4\times 4}(\mathbb{C}) \left| A^t = -A \right. \right\}\tag{21}$$ denote the set of antisymmetric complex $4\times 4$ matrices, endowed with the Pfaffian $$\begin{align} {\rm Pf}(A)~=~&\frac{1}{8}\sum_{\mu,\nu,\lambda,\sigma=1}^4\epsilon_{\mu\nu\lambda\sigma}A^{\mu\nu} A^{\lambda\sigma}\cr ~=~&A^{12}A^{34} + A^{31}A^{24} + A^{23}A^{14}, \cr \epsilon_{1234}~=~&1.\end{align}\tag{22}$$

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

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

  4. Endow $V\!j$ with the indefinite quadratic form $$\begin{align} ||xj||^2 ~=~& {\rm Pf}(\Phi(xj))\cr ~=~& \frac{1}{2}{\rm Re~tr} (x^2)-\frac{1}{2}{\rm Re~tr} (x)^2\cr ~=~&rs \pm |q|^2 .\end{align}\tag{27} $$

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


  1. Paul Garrett, Sporadic isogenies to orthogonal groups, 2015.
  • $\begingroup$ Notes for later: The complexification $u(p,q)_{\mathbb{C}}:=u(p,q)\otimes_{\mathbb{R}}\mathbb{C} \cong gl(n,\mathbb{C})$; $sp(2n,\mathbb{C})_{\mathbb{C}}:=sp(2n,\mathbb{C})\otimes_{\mathbb{R}}\mathbb{C} \cong sp(2n,\mathbb{C})\oplus sp(2n,\mathbb{C})$. Hence $U(p,q)_{\mathbb{C}} \cong GL(n,\mathbb{C})$ and $Sp(p,q)_{\mathbb{C}} \cong Sp(2n,\mathbb{C})$. $\endgroup$
    – Qmechanic
    Commented Apr 2, 2023 at 12:13
  • $\begingroup$ Notes for later: $gl(n,\mathbb{H})_{\mathbb{C}} \cong gl(2n,\mathbb{C})$. Hence $GL(n,\mathbb{H})_{\mathbb{C}} \cong GL(2n,\mathbb{C})$. $\endgroup$
    – Qmechanic
    Commented Apr 2, 2023 at 12:41
  • $\begingroup$ Notes to self: Helgason defines $Sp(n,\mathbb{H}):=\{x\in{\rm Mat}_{n\times n}(\mathbb{H})\mid x^{\dagger}(j{\bf 1}_{n\times n})x=j{\bf 1}_{n\times n}\}$ $=\{x\in{\rm Mat}_{n\times n}(\mathbb{H})\mid x^tx={\bf 1}_{n\times n}\}$ $=\{M\in{\rm Mat}_{2n\times 2n}(\mathbb{C})\mid M^tM={\bf 1}_{2n\times 2n},~\overline{M}\Omega=\Omega M\}$. $\quad sp(n,\mathbb{H})=\{x\in{\rm Mat}_{n\times n}(\mathbb{H})\mid x^t=-x\}$ $=\{x\in{\rm Mat}_{n\times n}(\mathbb{H})\mid x^{\dagger}=jxj\}$ $=\{x\in{\rm Mat}_{n\times n}(\mathbb{H})\mid (jx)^{\dagger}=jx\}$. $\quad\dim_{\mathbb{R}}sp(n,\mathbb{H})=n(2n\!-\!1)$. $\endgroup$
    – Qmechanic
    Commented Apr 6, 2023 at 13:56
  • $\begingroup$ Signature is not important: $\quad D~:=~{\rm diag}(\underbrace{1,\ldots,1}_p,\underbrace{i,\dots, i}_q)$. $\quad x^{\prime}=DxD^{-1}$. $\quad Sp(n,\mathbb{H})\cong Sp(p,q;\mathbb{H}):=\{x\in{\rm Mat}_{n\times n}(\mathbb{H})\mid x^{\dagger}j\eta x = j\eta\}$. $\endgroup$
    – Qmechanic
    Commented May 6, 2023 at 14:42
  • $\begingroup$ Notes for later: $\quad sp(2,\mathbb{H})\cong su(2)\oplus sl(2,\mathbb{R})$, where $su(2)$ is spanned by 3 compact/anti-Hermitian ${\bf 1}_{2\times 2}\otimes \omega$, $\omega\otimes\sigma_x$, $\omega\otimes\sigma_z$; while $sl(2,\mathbb{R})$ is spanned by 1 compact/anti-Hermitian $\omega\otimes {\bf 1}_{2\times 2}$ and 2 non-compact/Hermitian $\sigma_z\otimes i\omega$, $\sigma_x\otimes i\omega$. $\endgroup$
    – Qmechanic
    Commented May 8, 2023 at 11:04

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