Monthly Archives: September 2012

More Examples of Lie Groups: 3-Sphere as a Lie group, The 3-Dimensional Heisenberg Group

3-Sphere $S^3$ as a Lie Group

Consider 4 elements $1,i,j,k$ satisfying the following relation
\begin{align*}
1^2&=1,\ i^2=j^2=k^2=-1,\\
1i&=i1=i,\ 1j=j1=j,\ 1k=k1=k,\\
ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j.
\end{align*}
Let $\mathbb{H}$ be the algebra spanned by $1,i,j,k$ over $\mathbb{R}$
$$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$
Then $\mathbb{H}\cong\mathbb{R}^4$ as a vector space over $\mathbb{R}$. Define $||q||$ of $q=a1+bi+cj+dk\in\mathbb{H}$ by
$$||q||^2:=q\bar q=a^2+b^2+c^2+d^2,$$
where $\bar q=a1-bi-cj-dk$.

Set $S^3=\{q\in\mathbb{H}: ||q||=1\}$. Then $S^3$ is a unit sphere in $\mathbb{R}^4$. $S^3$ is closed under the multiplication of $\mathbb{H}$. So, $S^3$ is a group. In fact, it is a Lie group.

The 3-Dimensional Heisenberg Groups

Set
$$G=\left\{(x,y,z):=\begin{pmatrix}
1 & y & z\\
0 & 1 & x\\
0 & 0 & 1
\end{pmatrix}: x,y,z\in\mathbb{R}\right\}.$$
Define a multiplication in $G$ by
\begin{align*}
(x,y,z)\cdot (a,b,c)&=\begin{pmatrix}
1 & y & z\\
0 & 1 & x\\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
1 & b & c\\
0 & 1 & a\\
0 & 0 & 1
\end{pmatrix}\\
&=\begin{pmatrix}
1 & y+b & z+ya+c\\
0 & 1 & x+a\\
0 & 0 & 1
\end{pmatrix}\\
&=(x+a,y+b,z+ya+c)\in G.
\end{align*}
The identity element is $(0,0,0)$ and that $(x,y,z)^{-1}=(-x,-y,xy-z)$. $G$ is a Lie subgroup of $\mathrm{GL}(3,\mathbb{R})$.

Let $\gamma:(-\epsilon,\epsilon)\longrightarrow G$ be a regular curve in $G$ such that $\gamma(0)=(0,0,0)=\begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 &1 \end{pmatrix}$.  Then                                        \begin{align*}\dot\gamma(t)&=\frac{\partial\gamma}{\partial x}\frac{dx}{dt}+\frac{\partial\gamma}{\partial y}\frac{dy}{dt}+\frac{\partial\gamma}{\partial z}\frac{dz}{dt}\\
&=\begin{pmatrix}0 & 0 & 0\\0 & 0 & 1\\0 & 0&0\end{pmatrix}\frac{dx}{dt}+\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\frac{dy}{dt}+\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}\frac{dz}{dt}.\end{align*}
Let
$$\mathfrak{e}_1:=\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix},\mathfrak{e}_2:=\begin{pmatrix}0 & 1 & 0\\ 0 & 0 & 0\\0 & 0 & 0\end{pmatrix},\mathfrak{e}_3:=\begin{pmatrix}0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{pmatrix}.$$
Then $\mathfrak{e}_1,\mathfrak{e}_2,\mathfrak{e}_3$ generate the Lie algebra $\mathfrak{g}$ of $G$. The Lie algebra $\mathfrak{g}$ is called Heisenberg algebra. Note the commutation relation
$$[\mathfrak{e}_1,\mathfrak{e}_3]=[\mathfrak{e}_2,\mathfrak{e}_3]=0,\ [\mathfrak{e}_1,\mathfrak{e}_2]=-\mathfrak{e}_3$$
which resembles the commutation relation $[\hat p,\hat x]=-i\hbar$ in quantum mechanics.

Lie Brackets (for $n\times n$ Matrices)

Definition. Given any two $n\times n$ matrices $A$ and $B$ define
$$[A,B]:=AB-BA.$$
$[\ ,\ ]$ is a bilinear operator on the Lie algebra $\mathfrak{gl}(n)$ of the general linear group $\mathrm{GL}(n)$ and is called Lie bracket. Note that with $[\ ,\ ]$, $\mathfrak{gl}(n)$ becomes an algebra over $\mathbb{R}$ or $\mathbb{C}$. So Lie algebra is actually an algebra. Lie bracket plays an important role in physics, especially in quantum mechanics. Physicists call it commutator.

Proposition. For any $n\times n$ matrices $A$, $B$ and $C$, the following identity holds
$$[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0.$$
This identity is called the Lie identity or Jacobi identity.

Lie algebra can be studied from purely algebraic point of view without knowing its relationship with Lie group. The algebraists’ definition of Lie algebra is:

Definition. A Lie algebra $\mathfrak{g}$ is an algebra over $\mathbb{R}$ or $\mathbb{C}$ with a bilinear vector product $[\ ,\ ]:\mathfrak{g}\times\mathfrak{g}\longrightarrow\mathfrak{g}$ satisfying the Jacobi identity.

In case you are interested, there are a couple of good books on algebraic approach of Lie algebra. They are

Hans Samelson, Notes on Lie Algebra, 2nd Edition, Springer 1990

J. E. Humphreys, Introduction to Lie Algebras and Representation Theory, Springer 1973

Proposition. $\mathfrak{so}(n)$ is a Lie algebra with $[\ ,\ ]$.

Proof. In here we have shown that $\mathfrak{so}(n)$ is the set of all $n\times n$ skew-symmetric matrices. It suffices to show that $\mathfrak{so}(n)$ is closed under $[\ ,\ ]$.

For any $A,B\in\mathfrak{so}(n)$,
\begin{align*}
{}^t[A,B]&={}^t(AB-BA)\\
&={}^t(AB)-{}^t(BA)\\
&={}^tB{}^tA-{}^tA{}^tB\\
&=BA-AB\\
&=-[A,B].
\end{align*}
Hence $[A,B]\in\mathfrak{so}(n)$.

The Lorentz Group

Let $\mathbb{L}^4$ be $\mathbb{R}^4$ with the Lorentzian inner product $\langle\ ,\ \rangle$ defined by
$$\langle v,w\rangle=-v^0w^0+v^1w^1+v^2w^2+v^3w^3$$
for $v={}^t(v^0,v^1,v^2,v^3),w={}^t(w^0,w^1,w^2,w^3)\in\mathbb{R}^4$. In particular,
$$||v||^2=-(v^0)^2+(v^1)^2+(v^2)^2+(v^3)^2.$$
$\mathbb{L}^4$ is called the Minkowski $4$-spacetime or simply 4-spacetime. In physics, $\mathbb{L}^4$ is commonly denoted by $\mathbb{R}^{3+1}$. Another common notation for $\mathbb{L}^4$ in differential geometry is $\mathbb{R}^4_1$.

A linear transformation $A:\mathbb{L}^4\longrightarrow\mathbb{L}^4$ is called a Lorentz transformation if it is a Lorentzian isometry i.e. a Lorentzian inner product preserving map. Note that $\langle v,w\rangle$ can be written in matrix form as
$$\langle v,w\rangle={}^tv(g_{ij})w,$$
where
$$(g_{ij})=\begin{pmatrix}
-1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix}.$$
The matrix $(g_{ij})$ is called a Lorentzian metric tensor. The set of all Lorentz transformations forms a Lie group called Lorentz group and is denoted  by $\mathrm{O}(3,1)$:
$$\mathrm{O}(3,1)=\{A\in\mathrm{GL}(\mathbb{L}^4): {}^tA(g_{ij})A=(g_{ij})\}.$$
$\mathrm{O}(3,1)$ contains conventional rotations such as one in the $x^1-x^2$ plane
$$\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & \cos\theta & -\sin\theta & 0\\
0 & \sin\theta & \cos\theta & 0\\
0 & 0 & 0 & 1
\end{pmatrix},\ 0\leq\theta<2\pi$$
plus Lorentz boots which may be regarded as rotation between space and time directions. An example of boosts is
$$\begin{pmatrix}
\cosh\phi & \sinh\phi & 0 &  0\\
\sinh\phi & \cosh\phi & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{pmatrix},\ -\infty<\phi<\infty.$$
Lorentz transformations leave the origin (present) fixed due to linearity. The set of all isometries of $\mathbb{L}^4$ constains Lorentz transformations and translations. It is a Lie group called Poincaré group.

For $A\in\mathrm{O}(3,1)$, $\det A=\pm 1$. Those Lorentz transformations with determinant 1 are spatial orientation (parity) preserving transformations. They form a Lie subgroup of $\mathrm{O}(3,1)$ and is denoted by $\mathrm{SO}(3,1)$:
$$\mathrm{SO}(3,1)=\{A\in\mathrm{O}(3,1): \det A=1\}.$$
$\mathrm{SO}(3,1)$ has two connected components. ($\mathrm{O}(3,1)$ has four connected components.) The identiy componenent of $\mathrm{SO}(3,1)$ is denoted by $\mathrm{SO}^+(3,1)$. $\mathrm{SO}^+(3,1)$ is the group of both time orientation and parity preserving Lorentz transformations.

Orthogonal Group $\mathrm{O}(n)$ and Symmetry

Denote by $\langle\ ,\ \rangle$ the standard Euclidean inner product in $\mathbb{R}^n$. Then for any $v,w\in\mathbb{R}^n$,
$$\langle v,w\rangle={}^tvw.$$

Definition. A bijective map $A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is said to be an isometry if it preserves the inner product $\langle\ ,\ \rangle$ i.e. for any vectors $v,w\in\mathbb{R}^n$
$$\langle Av,Aw\rangle=\langle v,w\rangle.$$
Isometries are usually called symmetries in physics.

The set of all isometries of $\mathbb{R}^n$ forms a group with composition $\circ$. The group is denoted by $\mathrm{E}(n)$ and called the Euclidean motion group. The orthogonal group $\mathrm{O}(n)$ is a subgroup of $\mathrm{E}(n)$ and that it is a group of all isometries of $\mathbb{R}^n$ which leaves the origin fixed. In other words,

Theorem. $\mathrm{O}(n)$ is the group of all linear isometries of $\mathbb{R}^n$.

Proof. Let $A\in\mathrm{O}(n)$. Then
\begin{align*}
\langle Av,Aw\rangle&=\langle v,{}^tAAw\rangle\\
&=\langle v,w\rangle
\end{align*}
since ${}^tAA=I$. Hence, $A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry. Conversely, if $A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry, then for any $v,w\in\mathbb{R}^n$
\begin{align*}
\langle Av,Aw\rangle=\langle v,w\rangle&\Longrightarrow\langle{}^tAAv,w\rangle=\langle v,w\rangle\\
&\Longrightarrow\langle {}^tAAv-v,w\rangle=0.
\end{align*}
Since $w$ is arbitrary,
$$({}^tAA-I)v=0.$$
Since $v$ is also arbitrary,
$${}^tAA=I$$
i.e. $A$ is an orthogonal matrix.

Remark. For any $v,w\in\mathbb{R}^n$,
$$||v+w||^2=||v||^2+2\langle v,w\rangle+||w||^2$$
where $||v||=\sqrt{\langle v,v\rangle}$, the Euclidean norm of $v$. Plugging in $Av$ and $Aw$ for $v$ and $w$ respectively, we obtain
$$\langle Av,Aw\rangle=\frac{1}{2}(||A(v+w)||^2-||Av||^2-||Aw||^2).$$
This equation tells that a linear isomorphism $A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry if and only if it preserves the norm  $||\cdot||$.

Let $A\in\mathrm{O}(n)$. Since $\det A=\det {}^tA$, $\det A=\pm 1$. This implies that $\mathrm{O}(n)$ has two connected components. One that contains orthogonal matrices whose determinant is $1$, i.e. $\mathrm{SO}(n)$, and the other that contains orthogonal matrices whose determinant is $-1$. The identity component $\mathrm{SO}(n)$ of $\mathrm{O}(n)$ is the group of all linear isometries that preserve orientation. Clearly $\mathrm{SO}(n)$ is a normal subgroup of $\mathrm{O}(n)$.

The Lie groups $\mathrm{O}(n)$ and $\mathrm{SO}(n)$ are compact.