How can I demonstrate that the set of matrices, $U$, that are consistent with Equation 1 form a group?
\begin{align}
\tag{1}
U^\dagger
\begin{bmatrix}
1&0&0&0
\\
0&-1&0&0
\\
0&0&-1&0
\\
0&0&0&-1
\end{bmatrix}
U
=
\begin{bmatrix}
1&0&0&0
\\
0&-1&0&0
\\
0&0&-1&0
\\
0&0&0&-1
\end{bmatrix},
\end{align}
where the $\dagger$ indicates conjugate transpose.
I guess I have to show the 3 main properties of a group:(But, how?)
- that the associative property is valid
- that there exists an inverse $\text{inv}(U)*U=1$
- and identity element. $U*I=U$
In the case it is a group. Is it a Lie group? Which is the dimension of the associated Lie algebra? I also have to find the generators of the Lie algebra associated with that group.
Note that the we are working in the complex space $\Bbb C$ in 4 dimensions.This is not the classical Lorentz group of all the books because it is not in $\Bbb R^4$.
So far I have found that this group of matrices is called complex Lorentz group.(Is it a subgroup of the pseudo-unitary group (3,1). In this case we are working with a signature diag(-1,1,1,1).Physics convention.
And the last question is if there is any relation between this Lie algebra with:
- Casimir operators
- Cartan subalgebra
Best Answer
A standard method of proving this is a Lie group is by constructing the homomorphism $\Phi:GL_n(\mathbb{C}) \to Herm(n)$ given by
$$ \Phi(A) \;\; =\;\; AMA^\dagger. $$
Here the set $Herm(n)$ is the $n\times n$ Hermitian matrices which is a vector space of real dimension $n^2$, and of course $\dim GL_n(\mathbb{C}) = 2n^2$. We want to prove that $\Phi$ is a smooth equivariant function. This will show that $\Phi$ is of constant rank. If this occurs then the inverse image $\Phi^{-1}(M)$ will be equivalent to the Lorentz group in question and the theory of Lie groups shows us that this is a Lie subgroup of $GL_n(\mathbb{C})$ of dimension $2n^2 - \text{rank}(\Phi)$.
Let $GL_n(\mathbb{C})$ act on itself by left multiplication and let it act on the set of Hermitian matrices as $A\cdot H = AHA^\dagger$. Then the function $\Phi$ is equivariant since:
$$ \Phi(AB) \;\; =\;\; ABM(AB)^\dagger \;\; =\;\; ABMB^\dagger A^\dagger \;\; =\;\; A\cdot \Phi(B). $$
Equivariant maps are of constant rank, hence all we need to do is find the rank of the map $\Phi$. We have freedom to choose the base point, hence we can compute $\text{rank }d\Phi_I$ at the identity. Observe that if we let $\gamma:(-\epsilon, \epsilon)$ be a curve where $\gamma(0) = I$ and $\gamma'(0) = B$ then we find that
\begin{eqnarray*} d\Phi_I(B) & = & \left. \frac{d}{dt} \right |_{t=0}\Phi(\gamma(t)) \\ & = & \left . \frac{d}{dt} \right |_{t=0} \gamma(t)M\gamma(t)^\dagger \\ & = & MB^\dagger + BM \;\; =\;\; BM + (BM)^\dagger \end{eqnarray*}
hence showing that $d\Phi_I \subseteq Herm(n)$. To show that it is surjective, we can acknowledge that $T_IGL_n(\mathbb{C}) = M_n(\mathbb{C})$, so if we let $P$ be any Hermitian matrix, then $\frac{1}{2}PM \in M_n(\mathbb{C})$ will get sent to $P$:
$$ d\Phi_I\left (\frac{1}{2}PM\right ) \;\; =\;\; \frac{1}{2}(PM)M + \left (\frac{1}{2}PMM\right)^\dagger \;\; =\;\; \frac{1}{2}P + \frac{1}{2}P \;\; =\;\; P $$
hence $d\Phi_I$ has full rank which is $n^2$. This proves the complex Lorentz group is a Lie group of dimension $2n^2 - n^2 = n^2$. Hence, in the case the $n=4$ this would 16-dimensional.
Generators
In general, we can at best expect to find local generators for a Lie group, and these can be discerned from the Lie algebra. If $\mathfrak{g}$ is the Lie algebra of a Lie group $G$, then letting $A_1, \ldots, A_k$ be a basis of $\mathfrak{g}$, then we can find local generators of the form $\exp(tA_i)$, thus the issue in discerning the generators comes down to finding the Lie algebra $\mathfrak{l}_n(\mathbb{C})$ of $\Lambda_n(\mathbb{C})$ and picking a basis. Note that we can uncover the Lie algebra structure by looking at the equation above for $d\Phi_I(B) = BM + MB^\dagger$. We know this must be zero given the equation $AMA^\dagger = M$. What we find here is:
$$ BM + MB^\dagger =0 \;\; \Longrightarrow \;\; B^\dagger = -MBM, $$
which if you multiply this out essentially gives you the equation: $$ \left [ \begin{array}{cc} \alpha^* & \overline{v^T} \\ \overline{u^T} & \tilde{B}^\dagger \\ \end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc} -\alpha & u \\ v & - \tilde{B} \\ \end{array} \right ] $$
where we find $\alpha \in \mathbb{C}, \; u,v \in \mathbb{C}^3$, and $\tilde{B} \in M_3(\mathbb{C})$. The relationships $\alpha^* = - \alpha$ tells us that $\alpha \in i\mathbb{R}$. Also that $u = \overline{v^T}$ restricts the structure of the matrix to just one vector in $\mathbb{C}^3$. Lastly that $\tilde{B}^\dagger = - \tilde{B}$ is that $\tilde{B} \in \mathfrak{u}(3)$ the Lie algebra of $3\times 3$ skew-Hermitian matrices. This tells us that
$$ \mathfrak{l}_n(\mathbb{C}) \;\; \cong\;\; i\mathbb{R} \oplus \mathbb{C}^{n-1} \oplus \mathfrak{u}(n-1) $$
You can verify this Lie algebra is 16-dimensional. My suggestion is to pick a basis $\{A_1, \ldots, A_{16}\}$ of this vector space and compute $\exp(tA_k)$ for each value of $k$.