Lie Groups III Examples Top Down March 19, 2025
In each case we have two equivalent approaches:
Top Down : We can start from the definition of the Lie group itself, for
example the orghogonal Lie group O(n) is defined as all transformations that
leave the Euclidean metric invariant. Then we derive the Lie algebra and the
commutation relations.
Bottom Up : Or we start from the Lie algebra, defined using the commutation
relations, and we derive the Lie group from it.
The first approach is in some ways simpler, but it requires to assume what
metric we are preserving. The second approach is more work to derive
everything, but it has less assumptions, the only assumption is how many
generators there are and their commutation relations. From that we derive the
Lie group and what metric it leaves invariant.
In this section we only do the top down approach. In the next section we do the
bottom up approach.
Euclidean group ISO(n) ¶ The Euclidean group ISO(n) is defined as all (nonlinear) transformations that
leave the metric d i a g ( 1 , 1 , … , 1 ) \mathrm{diag}(1, 1, \dots, 1) diag ( 1 , 1 , … , 1 )
It can be shown that the transformation must be linear, of the form A x + b Ax + b A x + b A A A
A T A = 1 . A^T A = 1\,. A T A = 1 . The matrix A A A b b b
I S O ( n ) = T ( n ) ⋊ O ( n ) . \mathrm{ISO}(n) = \mathrm{T}(n) \rtimes \mathrm{O}(n)\,. ISO ( n ) = T ( n ) ⋊ O ( n ) . O(n) ¶ From the previous section we can see that
the orthogonal Lie group is defined as:
O ( n ) = { X : X T X = 1 } . \mathrm{O}(n) = \{X: X^T X = 1\}\,. O ( n ) = { X : X T X = 1 } . It represents all matrices that leave the Euclidean metric invariant.
To find the Lie algebra, we write X = exp ( t x ) X = \exp(t x) X = exp ( t x )
( exp ( t x ) ) T exp ( t x ) = 1 , (\exp(tx))^T \exp(tx) = 1\,, ( exp ( t x ) ) T exp ( t x ) = 1 ,
exp ( t x T ) exp ( t x ) = 1 , \exp(tx^T) \exp(tx) = 1\,, exp ( t x T ) exp ( t x ) = 1 ,
exp ( t x T + t x ) = 1 , \exp(tx^T + tx) = 1\,, exp ( t x T + t x ) = 1 ,
exp ( t ( x T + x ) ) = 1 , \exp(t(x^T + x)) = 1\,, exp ( t ( x T + x )) = 1 , which is equivalent to:
t ( x T + x ) = 0 , t(x^T + x) = 0\,, t ( x T + x ) = 0 ,
x T + x = 0 . x^T + x = 0\,. x T + x = 0 . The corresponding Lie algebra is thus all antisymmetric matrices
n × n n \times n n × n
o ( n ) = { X : X T = − X } . \mathfrak{o}(n) = \{X: X^T = -X\}\,. o ( n ) = { X : X T = − X } . Taking a determinant of X T X = 1 X^T X = 1 X T X = 1
1 = det ( X T X ) = det ( X T ) det ( X ) = det ( X ) det ( X ) = det ( X ) 2 , 1 = \det(X^T X) = \det(X^T)\det(X) = \det(X)\det(X) = \det(X)^2\,, 1 = det ( X T X ) = det ( X T ) det ( X ) = det ( X ) det ( X ) = det ( X ) 2 ,
det ( X ) = ± 1 . \det(X) = \pm 1\,. det ( X ) = ± 1 . The group thus has two connected components:
O ( n ) ≅ { − 1 , 1 } × S O ( n ) . \mathrm{O}(n) \cong \{-1, 1\} \times \mathrm{SO}(n)\,. O ( n ) ≅ { − 1 , 1 } × SO ( n ) . Where the special orthogonal group SO(n) is:
S O ( n ) = { X : X T X = 1 , det X = 1 } . \mathrm{SO}(n) = \{X: X^T X = 1,\quad \det X = 1\}\,. SO ( n ) = { X : X T X = 1 , det X = 1 } . O(3) ¶ O ( 3 ) ≅ { − 1 , 1 } × S O ( 3 ) . \mathrm{O}(3) \cong \{-1, 1\} \times \mathrm{SO}(3)\,. O ( 3 ) ≅ { − 1 , 1 } × SO ( 3 ) . O ( 3 ) / { I , − I } ≅ S O ( 3 ) \mathrm{O}(3)/\{I,−I\} \cong \mathrm{SO}(3) O ( 3 ) / { I , − I } ≅ SO ( 3 ) SO(3) ¶ S O ( 3 ) = { X : X T X = 1 , det X = 1 } . \mathrm{SO}(3) = \{X: X^T X = 1,\quad \det X = 1\}\,. SO ( 3 ) = { X : X T X = 1 , det X = 1 } . s o ( 3 ) = { x : x T = − x } . \mathfrak{so}(3) = \{x: x^T = -x\}\,. so ( 3 ) = { x : x T = − x } . The s o ( 3 ) \mathfrak{so}(3) so ( 3 ) x i j = ϵ i j k θ k x_{ij}=\epsilon_{ijk}\theta^k x ij = ϵ ijk θ k
x = ( 0 θ 3 − θ 2 − θ 3 0 θ 1 θ 2 − θ 1 0 ) = θ 1 J 1 + θ 2 J 2 + θ 3 J 3 , x =
\begin{pmatrix}
0 & \theta_3 & -\theta_2 \\
-\theta_3 & 0 & \theta_1 \\
\theta_2 & -\theta_1 & 0
\end{pmatrix} = \theta_1 J_1 + \theta_2 J_2 + \theta_3 J_3\,, x = ⎝ ⎛ 0 − θ 3 θ 2 θ 3 0 − θ 1 − θ 2 θ 1 0 ⎠ ⎞ = θ 1 J 1 + θ 2 J 2 + θ 3 J 3 , where we have three generators J i J_i J i i = 1 , 2 , 3 i=1, 2, 3 i = 1 , 2 , 3
J 1 = ( 0 0 0 0 0 1 0 − 1 0 ) . J_1 =
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{pmatrix}\,. J 1 = ⎝ ⎛ 0 0 0 0 0 − 1 0 1 0 ⎠ ⎞ . J 2 = ( 0 0 − 1 0 0 0 1 0 0 ) , J_2 =
\begin{pmatrix}
0 & 0 & -1 \\
0 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix}\,, J 2 = ⎝ ⎛ 0 0 1 0 0 0 − 1 0 0 ⎠ ⎞ , J 3 = ( 0 1 0 − 1 0 0 0 0 0 ) , J_3 =
\begin{pmatrix}
0 & 1 & 0 \\
-1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\,, J 3 = ⎝ ⎛ 0 − 1 0 1 0 0 0 0 0 ⎠ ⎞ , The matrices J i J_i J i s o ( 3 ) \mathfrak{so}(3) so ( 3 )
[ J i , J j ] = ϵ i j k J k . [J_i, J_j]=\epsilon_{ijk} J_k\,. [ J i , J j ] = ϵ ijk J k . The general element X X X
X = exp x = exp ( θ 1 J 1 + θ 2 J 2 + θ 3 J 3 ) . X = \exp x = \exp (\theta_1 J_1 + \theta_2 J_2 + \theta_3
J_3)\,. X = exp x = exp ( θ 1 J 1 + θ 2 J 2 + θ 3 J 3 ) . We can also write every element X X X
X = exp ( θ 1 ′ J 1 ) exp ( θ 2 ′ J 2 ) exp ( θ 3 ′ J 3 ) . X = \exp (\theta_1' J_1)\exp(\theta_2' J_2)\exp(\theta_3' J_3)\,. X = exp ( θ 1 ′ J 1 ) exp ( θ 2 ′ J 2 ) exp ( θ 3 ′ J 3 ) . That is a different parametrization and one can compute the relation between
θ i \theta_i θ i θ i ′ \theta_i' θ i ′
Note that X = exp t x X = \exp tx X = exp t x x x x x x x t x tx t x X = exp x X = \exp x X = exp x
We can change the basis, for example J z = J 3 J_z=J_3 J z = J 3 J ± = 1 2 ( J 1 ± i J 2 ) J_\pm={1\over2}(J_1\pm i J_2) J ± = 2 1 ( J 1 ± i J 2 )
[ J z , J + ] = J + , [J_z, J_+] = J_+\,, [ J z , J + ] = J + ,
[ J z , J − ] = − J − , [J_z, J_-] = -J_-\,, [ J z , J − ] = − J − ,
[ J + , J − ] = 2 J z . [J_+, J_-] = 2J_z\,. [ J + , J − ] = 2 J z . But the Lie algebras are equivalent.
Every Lie bracket is always equal to a linear combination of all the
generators, e.g.:
[ J 1 , J 2 ] = α J 1 + β J 2 + γ J 3 , [J_1, J_2] = \alpha J_1 + \beta J_2 + \gamma J_3\,, [ J 1 , J 2 ] = α J 1 + β J 2 + γ J 3 , because the Lie bracket lies in the same vector space and the generators form a
basis. In the above example we get α = β = 0 \alpha=\beta=0 α = β = 0 γ = 1 \gamma=1 γ = 1
[ J i , J j ] = f i j k J k , [J_i, J_j] = f_{ij}{}^k J_k\,, [ J i , J j ] = f ij k J k , where f i j k f_{ij}{}^k f ij k
The structure constants are C i j k = ϵ i j k C_{ij}{}^k = \epsilon_{ij}{}^k C ij k = ϵ ij k
g i j = C i k l C j l k = ϵ i k l ϵ j l k = − 2 δ i j . g_{ij} = C_{ik}{}^l C_{jl}{}^k = \epsilon_{ik}{}^l \epsilon_{jl}{}^k =
-2 \delta_{ij}\,. g ij = C ik l C j l k = ϵ ik l ϵ j l k = − 2 δ ij . We have:
det g i j = 8 ≠ 0 \det g_{ij} = 8 \neq 0 det g ij = 8 = 0 So the SO(3) group is semisimple.
Translation-Rotation Group SE(3) ¶ Also called the special Euclidean group.
The group decomposes as:
S E ( 3 ) = T ( 3 ) ⋊ S O ( 3 ) . \mathrm{SE}(3) = \mathrm{T}(3) \rtimes \mathrm{SO}(3)\,. SE ( 3 ) = T ( 3 ) ⋊ SO ( 3 ) . The group of all translations T T T R R R P 1 P_1 P 1 P 2 P_2 P 2 P 3 P_3 P 3 J 1 J_1 J 1 J 2 J_2 J 2 J 3 J_3 J 3 P i P_i P i T T T R R R
R T R − 1 = T ′ R T R^{-1} = T' RT R − 1 = T ′ So T T T invariant subgroup . Furthermore for all T T T T ′ T' T ′
[ T , T ′ ] = 0 [T, T'] = 0 [ T , T ′ ] = 0 So T T T abelian invariant subgroup .
The commutation relations of the generators:
[ J i , J j ] = ϵ i j k J k , [J_i, J_j] = \epsilon_{ijk} J_k\,, [ J i , J j ] = ϵ ijk J k ,
[ J i , P j ] = ϵ i j k P k , [J_i, P_j] = \epsilon_{ijk} P_k\,, [ J i , P j ] = ϵ ijk P k ,
[ P i , P j ] = 0 . [P_i, P_j] = 0\,. [ P i , P j ] = 0 . One way to represent these is using 4x4 matrices as follows:
J 1 = ( 0 0 0 0 0 0 1 0 0 − 1 0 0 0 0 0 0 ) , J_1 =
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\,, J 1 = ⎝ ⎛ 0 0 0 0 0 0 − 1 0 0 1 0 0 0 0 0 0 ⎠ ⎞ , J 2 = ( 0 0 − 1 0 0 0 0 0 1 0 0 0 0 0 0 0 ) , J_2 =
\begin{pmatrix}
0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\,, J 2 = ⎝ ⎛ 0 0 1 0 0 0 0 0 − 1 0 0 0 0 0 0 0 ⎠ ⎞ , J 3 = ( 0 1 0 0 − 1 0 0 0 0 0 0 0 0 0 0 0 ) , J_3 =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\,, J 3 = ⎝ ⎛ 0 − 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 ⎠ ⎞ , P 1 = ( 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 ) , P_1 =
\begin{pmatrix}
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\,, P 1 = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 ⎠ ⎞ , P 2 = ( 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 ) , P_2 =
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\,, P 2 = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 ⎠ ⎞ , P 3 = ( 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 ) . P_3 =
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}\,. P 3 = ⎝ ⎛ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ⎠ ⎞ . SL(2, C) ¶ The special linear Lie group can be defined as 2 × 2 2\times2 2 × 2
S L ( 2 , C ) = { X : det X = 1 } . \mathrm{SL}(2,\mathbb{C}) = \{X: \det X = 1\}\,. SL ( 2 , C ) = { X : det X = 1 } . To find the Lie algebra, we use:
det exp x = exp T r x . \det\exp x = \exp \mathrm{Tr}\,x\,. det exp x = exp Tr x . We write X = exp ( t x ) X = \exp(t x) X = exp ( t x )
1 = det X = det exp ( t x ) = exp T r t x = exp ( t T r x ) , 1 = \det X = \det\exp(t x)=\exp\mathrm{Tr}\, tx=\exp (t\,\mathrm{Tr}\, x)\,, 1 = det X = det exp ( t x ) = exp Tr t x = exp ( t Tr x ) , that is equivalent to:
t T r x = 0 , t\,\mathrm{Tr}\, x = 0\,, t Tr x = 0 ,
T r x = 0 . \mathrm{Tr}\, x = 0\,. Tr x = 0 . The corresponding Lie algebra is thus all n × n n \times n n × n
s l ( 2 , C ) = { x : T r ( x ) = 0 } . \mathfrak{sl}(2,\mathbb{C}) = \{x: \mathrm{Tr}(x) = 0\}\,. sl ( 2 , C ) = { x : Tr ( x ) = 0 } . One can choose a basis of this space as:
H = ( 1 0 0 − 1 ) , H = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\,, H = ( 1 0 0 − 1 ) ,
X = ( 0 1 0 0 ) , X = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\,, X = ( 0 0 1 0 ) ,
Y = ( 0 0 1 0 ) . Y = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\,. Y = ( 0 1 0 0 ) . The Lie brackets are:
[ H , X ] = 2 X , [H, X] = 2X\,, [ H , X ] = 2 X ,
[ H , Y ] = − 2 Y , [H, Y] = -2Y\,, [ H , Y ] = − 2 Y ,
[ X , Y ] = H . [X, Y] = H\,. [ X , Y ] = H . SU(2, C) ¶ The special unitary Lie group can be defined as 2 × 2 2\times2 2 × 2
S U ( 2 , C ) = { X : X † X = 1 , det X = 1 } . \mathrm{SU}(2,\mathbb{C}) = \{X: X^\dagger X = 1, \quad \det X = 1\}\,. SU ( 2 , C ) = { X : X † X = 1 , det X = 1 } . Using the above derivations, the corresponding Lie algebra is
all 2 × 2 2 \times 2 2 × 2
s u ( 2 , C ) = { X : X H = − X , T r ( X ) = 0 } . \mathfrak{su}(2,\mathbb{C}) = \{X: X^H=-X, \quad \mathrm{Tr}(X) = 0\}\,. su ( 2 , C ) = { X : X H = − X , Tr ( X ) = 0 } . One can choose a basis of this space as:
u 1 = + i σ 1 = ( 0 i i 0 ) , u_1 = +i\sigma_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\,, u 1 = + i σ 1 = ( 0 i i 0 ) ,
u 2 = − i σ 2 = ( 0 − 1 1 0 ) , u_2 = -i\sigma_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\,, u 2 = − i σ 2 = ( 0 1 − 1 0 ) ,
u 3 = + i σ 3 = ( i 0 0 − i ) . u_3 = +i\sigma_3 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}\,. u 3 = + i σ 3 = ( i 0 0 − i ) . The matrices are a representation of quaternions.
The Lie brackets are:
[ u 1 , u 2 ] = 2 u 3 , [u_1, u_2] = 2u_3\,, [ u 1 , u 2 ] = 2 u 3 ,
[ u 2 , u 3 ] = 2 u 1 , [u_2, u_3] = 2u_1\,, [ u 2 , u 3 ] = 2 u 1 ,
[ u 3 , u 1 ] = 2 u 2 . [u_3, u_1] = 2u_2\,. [ u 3 , u 1 ] = 2 u 2 . One can pass to the complexified Lie algebra:
s u ( 2 , C ) + i s u ( 2 , C ) = s l ( 2 , C ) . \mathfrak{su}(2,\mathbb{C}) + i\,\mathfrak{su}(2,\mathbb{C})=
\mathfrak{sl}(2,\mathbb{C})\,. su ( 2 , C ) + i su ( 2 , C ) = sl ( 2 , C ) . The complexified Lie algebra is spanned by three elements X, Y, H:
H = 1 i u 3 = ( 1 0 0 − 1 ) , H = {1\over i} u_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\,, H = i 1 u 3 = ( 1 0 0 − 1 ) ,
X = 1 2 i ( u 1 − i u 2 ) = ( 0 1 0 0 ) , X = {1\over 2i}(u_1-i u_2) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\,, X = 2 i 1 ( u 1 − i u 2 ) = ( 0 0 1 0 ) ,
Y = 1 2 i ( u 1 + i u 2 ) = ( 0 0 1 0 ) . Y = {1\over 2i}(u_1+i u_2) =\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\,. Y = 2 i 1 ( u 1 + i u 2 ) = ( 0 1 0 0 ) . The Lie brackets are:
[ H , X ] = 2 X , [H, X] = 2X\,, [ H , X ] = 2 X ,
[ H , Y ] = − 2 Y , [H, Y] = -2Y\,, [ H , Y ] = − 2 Y ,
[ X , Y ] = H . [X, Y] = H\,. [ X , Y ] = H . We have { I , − I } \{I,−I\} { I , − I }
S U ( 2 ) / { I , − I } ≅ S O ( 3 ) . \mathrm{SU}(2)/\{I,−I\} \cong \mathrm{SO}(3)\,. SU ( 2 ) / { I , − I } ≅ SO ( 3 ) . O(4) ¶ ISO(3) ¶ ISO(3) / O(3) is the 3D Euclidean space.
We can derive the metric diag(1, 1, 1).
Surface of the sphere ¶ [P,P]=J
[J,P]=P
[J,J]=J
ISO(4) ¶ Lorentz O(3,1) ¶ Poincare ISO(3,1) ¶ Galilei ¶ Carroll ¶