Contextualization
While researching a line of investigation in physics I came across matrices with a certain structure. I want to characterize the structure. Specifically, I want to determine if infinitesimal generators of the Lie algebra exist; if they exist, to determine one representation of them; and the to determine the Lie bracket.
From what I understand I have to first determine if the matrices form a Lie group. Secondly, I have to determine the tangent vector [80], third I have to form the commutators with the elements of the Lie algebra.
Problem
Let $S$ be the set $$ S = \left\{\begin{pmatrix} \exp (i\varphi_1) & 0 \\ 0 & \exp (i\varphi_2) \end{pmatrix} : \varphi_1 \in \mathbb{R}\ /\ 2\pi\mathbb{Z}~~\text{and}~~ \exp (2i\varphi_1) =\exp (2i\varphi_2)\right\}. $$ Given the group comprised of the set $S$ with matrix multiplication as the group operation, determine (1) if this forms a Lie group; and, if it is a Lie group, (2) determine thee infinitismal generators and the Lie bracket.
Personal Thoughts
Preliminarily, I have to show that I am working with a group---namely the group formed by the set $S$ with matrix mutlplication as the group operation. The set contains the identity element and every element has an inverse in $S$. The only non-trivial item to check for is closure. One can verify that $$ \begin{pmatrix} \exp (i\theta_1) & 0 \\ 0 & \exp (i\theta_1) \end{pmatrix} \begin{pmatrix} \exp (i\theta_2) & 0 \\ 0 & \exp (i\theta_2+i\pi) \end{pmatrix} = \begin{pmatrix} \exp (i[\theta_1+\theta_2]) & 0 \\ 0 & \exp (i[\theta_1+\theta_2+\pi]) ) \end{pmatrix} , $$ is a member of the group since $$ \exp (2i[\theta_1+\theta_2]) = \exp (2i[\theta_1+\theta_2+\pi]) ) . $$
Next, to show that this is a Lie group, I have to check for smoothness. I think that what I have to show is that both $$ \frac{d^n}{d\varphi_1^n}\begin{pmatrix} \exp (i\varphi_1) & 0 \\ 0 & \exp (i\varphi_1) \end{pmatrix} \quad \text{and} \quad \frac{d^n}{d\varphi_1^n}\begin{pmatrix} \exp (i\varphi_1) & 0 \\ 0 & -\exp (i\varphi_1) \end{pmatrix} $$ are differentiable for $n = 1,2,3,\ldots$. If this is so, then yes. I do have a Lie group.
In fact, I believe that this a disconnected Lie group [20,30]. First, there is an identity component of the group $S_0$ that contains the identity element of the group. However, there is one, and only one, additional connected component that exists that does not contain the identity. This component is not a subgroup, but rather a coset of the identity component. This coset of the identity group can be found, for example, by multiplying the identity subgroup by the fixed element $c$ in $S$ that given by $$ c = \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}. $$
Next, I wish to find the generators [10]. I am honestly not sure how many parameters there are in this Lie group. That is because the parameter $\varphi_1$ and $\varphi_2$ are interdependent. To try find the generators I follow along with the thinking of Greg Graviton in [80] along with the comments therein. Let $A\in S$. To find the Lie algebra, take a smooth path $A(\varphi_1)$ with $A(0) = I$. One must note that to find the Lie algebra we must first focus on the identity subgroup whereby $A$ is also a member of $S_0$. To first order approximation, $A$ can be written as the Taylor expansion
$$A(t) = I + \varphi_1\, H + \mathcal O(\varphi_1^2) ,$$
where the tangent vector, $H$,
$$
H \equiv \left. \frac{d A(\varphi_1)}{d\varphi_1}\right|_{\varphi_1 = 0}
=
\left. \frac{d \begin{pmatrix} \exp (i\varphi_1) & 0 \\ 0 & \exp (i\varphi_1) \end{pmatrix}
}{d\varphi_1}\right|_{\varphi_1 = 0}
=
i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} .
$$
However, there is a problem. This alone does not generate all the elements in $S$. Thus, I suppose that to find the generator of the one and only coset, I should multiply $H$ by the same fixed element $c$ in $S$ that I used before. Thus, the second generator is
$$
\begin{pmatrix}
1&0
\\
0&-1
\end{pmatrix}.
$$
What I find is that that generators of the Lie algebra can be written (non-uniquely) by the elements $g_1$ and $g_2$, where $$ \boxed{ g_1 = i \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \quad \textrm{and}\qquad g_2 = i \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} .} $$ It is worth noting that if, instead of differentiating with respect to $\vartheta_1$, I were to use $\vartheta_2$, I could get the same (non-unique) representation of the generators.
Lastly, to find the Lie bracket, all that is to write is that $$ \left[ g_1 , g_2\right] = 0. $$
Explicit Question
Is this a one-parameter Lie group or a two parameter Lie group?
Steps that are in doubt
I am in doubt about whether or not I have to use the element $c$ to obtain a second generator of the Lie group. Please verify this.
Bibliography
[20] https://en.wikipedia.org/wiki/Identity_component
[30] Is a Lie group connected? https://en.wikipedia.org/wiki/Identity_component
[40] https://en.wikipedia.org/wiki/Coset
[80] How do you find the Lie algebra of a Lie group (in practice)?
