How to calculate the dimension of the Lie algebra associated with the complex Lorentz group? Calculate the generators associated with the Lie algebra

Question

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?)

1. that the associative property is valid
2. that there exists an inverse $\text{inv}(U)*U=1$
3. 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:

1. Casimir operators
2. Cartan subalgebra

Answer 1

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 $k$.

Answer 2

Answer to next the question, "Is $\text{SU}(1,3)$ a group?"

By $\text{U} (1, 3)$ I denote the pseudo-unitary group of complex matrices $U$ satisfying the condition $U\,g\,U^\dagger = g$, where $g$ is a diagonal matrix with elements $g_{11} = 1$ and $g_{kk} = -1$ for $2 \leq k \leq 4$. Explicitly, $$ g = \begin{bmatrix} 1&0&0&0 \\ 0&-1&0&0 \\ 0&0&-1&0 \\ 0&0&0&-1 \end{bmatrix}. $$

Paraphrasing from [1],

A group is a non-empty set together with a binary operation on the set that combines any two elements of to form an element of the set such that the group axioms, are satisfied.

(a) We must show that the set $\text{U}(1,3)$, which is a non-empty set, together with the binary operation matrix multiplication on $\text{U}(1,3)$ combines any two elements $U_1$ and $U_2$ of $\text{U}(1,3)$ to form an element of $\text{U}(1,3)$, denoted $U_1\,U_2$.

To begin with observe that the product $U_1\,U_2$ is an element of $\text{U}(1,3)$: \begin{align} \left(U_1\,U_2\right)^\dagger g \left(U_1\,U_2\right) &= \left(U_2^\dagger\, U_1^\dagger\right) g\left(U_1\,U_2\right) && (\text{rule of matrix multiplication}) \\ &= U_2^\dagger\, U_1^\dagger \,g\, U_1\,U_2 && (\text{rule of matrix multiplication}) \\ &= U_2^\dagger\, g \, U_2 && (U_1~\text{satisfies OP's Eq. 1}) \\ &= g && (U_2~\text{satisfies OP's Eq. 1}) \end{align}

(b) First group axiom Next, we must show that for $U_1$, $U_2$, and $U_3$ in $\text{U}(1,3)$ one has $U_1 \left(U_2 \,U_3\right)= \left(U_1 \,U_2 \right)U_3$. Well, each of these three matrices are square matrices. Thus, the products $(U_1 \,U_2 )\,U_3$ and $U_1 (U_2 \,U_3)$ are defined; and one has the associative property $(U_1 \,U_2 )\,U_3=U_1 (U_2 \,U_3)$.

(c) Second group axiom Next, we must show that there exists an element $e$ in $\text{U}(1,3)$ such that, for every $U$ in $\text{U}(1,3)$, one has $e \, U = U$ and $U\, e = U $. Well, note that $e = \begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix}$ works. I can show that $ \begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix}$ is in $\text{U}(1,3)$ by writing \begin{align} \begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix}^\dagger g \begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix} &= \begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0 \\0&0&0&1\end{bmatrix}g\begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix} \\ &= g. \end{align} Further, one can see by matrix multiplication that $$\begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix} \, U = U\quad\text{and}\quad U\, \begin{bmatrix}1&0&0&0\\0&1&0&0\\ 0&0&1&0\\0&0&0&1\end{bmatrix} = U .$$

(d) Third group axiom We must show that for each $U_1$ in $\text{U}(1,3)$, there exists an element $U_2$ in $\text{U}(1,3)$ such that $U_1 \, U_2 = e$ and $U_2\, U_1 =e$. Well, from the OP's Equation 1, we get that \begin{align} \det\left( U_1^\dagger \,g\,U_1 \right) &= \det(g) \\ \det \left(U_1^\dagger \right) \det (g)\,\det(U_1) &= \det(g) \\ \det \left(U_1^\dagger \right) \det(U_1) &= 1. \end{align} So, $$ \tag{2} \left(\det U_1 \right)^\dagger \det{U_1} = 1.$$ Assuming falsely that $\det U_1 = 0$, gives $\left(\det U_1 \right)^\dagger \det{U_1} =0$. This leads to a contradiction with Equation 2. So, truly $\det U_1 \neq 0$. Since the determinant of $ U_1 $ is not equal to zero, therefore exists an element $U_2$ such that $U_2\,U_1 = e$ and $U_1\,U_2 = e$. By $U_1^{-1}$ I denote the inverse of $U_1$ . Further, I show that the inverse is in $\text{O}(1,3)$. Now, since $g = U\,g \, U^\dagger $, therefore \begin{align} \left(U^{-1}\right)\,g \left(U^\dagger \right)^{-1} & = U^{-1}\,U\,g \, \left(U^\dagger \right)\left(U^\dagger \right)^{-1} \\ \left(U^{-1}\right)\,g \left(U^\dagger \right)^{-1} & = g \\ \left(U^{-1}\right)\,g \left(U^{-1} \right)^\dagger & = g \end{align} In the last line, we see that $\left(U^{-1}\right) $ satisfies the OP's Equation 1. Thus, $\left(U^{-1}\right) \in \text{U}(1,3)$.

Q.E.D

Bibliography

[1] https://en.wikipedia.org/wiki/Group

[2]

Answer 3

Answer to next the question, "Is $\text{SU}(1,3)$ a Lie group?"

There appears to be many ways and several conditions to determine the matter (cf, [1]). That said, I can wrap my head around the following definition from [2]:

Definition: [Matrix Lie group]A matrix Lie group, which is also called a linear Lie group, is a closed subgroup of GL$(n;\mathbb{C})$.

I have showed in a accompanying answer to this question that $\text{U}(1,3)$ is a group. Further, $\text{U}(1,3)\subset \text{GL}(4, \mathbb{C} ).$ What remains to show is that the limit of a sequence of psuedo-unitary matrices is itself a psuedo-unitary matrix.

Proof

Let us take the sequence $\left(A_m\right)$ of matrices all of which are in U(1,3). I start with a proof by induction [3], and finish by taking a limit.

The base case: I start with
\begin{align} A_1 = \prod_{j=1}^1 U_{j}. \end{align} By construction $U_1 \in \text{U}(1,3) $.

The inductive step: I show that if the statement holds for any given case $m =k$, then it also hold for the next case $m = k + 1$. Observe. Now, \begin{align} \left(A_{k+1}\right) \,g\, \left(A_k\right)^\dagger &= \left(A_k\,U_{k+1} \right) \,g\, \left(A_k\,U_{k+1}\right)^\dagger && (\text{by construction}) \\ &= \left(A_k\,U_{k+1} \right) \,g\, \left(U_{k+1}\right)^\dagger \left(A_k \right)^\dagger && (\text{rule of matrix multiplication}) \\ &= A_k\,U_{k+1} \,g\, U_{k+1} ^\dagger\, A_k ^\dagger && (\text{rule of matrix multiplication}) \\ &= A_k \,g\, A_k ^\dagger && (U_{k+1}~\text{satisfies}~U_{k+1}\,g\,U_{k+1}^\dagger = g) \\ &= g && ( A_k~\text{satisfies}~ A_k\,g\, A_k^\dagger = g) \end{align} By induction, I conclude that a finite sequence of pseudo-unitary matrices is pseudo-unitary.

Finally, given $A = \lim_{m\to\infty}\left(A_m\right)$, we have that $A \in \text{U}(1,3) $. This is true because the relation $A \,g\,A ^\dagger = g$ is preserved under taking limits.

Q.E.D.

Bibliography

[1] https://en.wikipedia.org/wiki/Closed-subgroup_theorem

[2] Hall, "Lie Groups, Lie Algebras, and Representations," pp, 3-4,6, 2000.

[3] https://en.wikipedia.org/wiki/Mathematical_induction

Answer 4

Answer to the questions, "What is is the dimension of the associated Lie algebra?"

I begin with a definition from [1]

Definition. Let $\mathcal{G}$ be a matrix Lie group. The Lie algebra of $\mathcal{G}$, denoted $\mathfrak{g}$, is the set of all matrices $X$ such that $e^{tX}$ is in $\mathcal{G}$ for all real numbers $t$.

Adapting from [1], I reason as follows. Recall that a matrix $U$ is pseudo unitary if, and only if, $u\,g\,U^\dagger = g$. Thus, $e^{tX}$ is psuedo unitary if, and only if, $$ e^{tX}\,g\,\left(e^{tX}\right)^\dagger = g$$ for all real numbers $t$. Or, alternatively, if, and only if,
$$ g\,\left(e^{tX}\right)^\dagger = e^{-tX}\,g \tag{1}$$ for all real numbers $t$. Since Eq. 1 holds for all $t$, then by differentiating at $t=0$, we see that $$ g\,X ^\dagger = -X \,g $$ is the necessary condition for Eq. 1 to hold.

Thus, the Lie algebra of $U(1,3)$, which is denoted by $\mathfrak{u}(1,3)$, is the space of all $n\times n $ complex matrices $X$ such that $ g\,X ^\dagger = -X \,g \tag{2} $.

The dimension of each and every $n\times n $ invertible complex matrices is $2\,n^2$. Yet, as given by Eq. 2, each and every $n\times n $ matrix in $\mathfrak{u}(1,3)$ has $n^2$ constraints imposed upon them. Thus, the dimension of $\mathfrak{u}(1,3)$ is $n^2$ (i.e., $2\,n^2-n^2$ ).

Bibliography

[1] Hall, "Lie Groups, Lie Algebras, and Representations," pp. 39-40, 2000.