# A general element of U(2)

The group U(2) is the group of all $2\times 2$ matrices such that $U^\dagger U=I$. Evidently it has $4$ real parameters, and can be represented as:
$$U(2) = \{\begin{bmatrix}a&b\\ 0&d \end{bmatrix}:\,\{a,b,d\}\in\mathbb{C}\wedge|a|^2=|d|^2=1\}\cup\{\begin{bmatrix}a&0\\ c&d \end{bmatrix}:\,\{a,c,d\}\in\mathbb{C}\wedge|a|^2=|d|^2=1\}$$

I know that the Pauli matrices “generate” $SU(2)$ in the sense that I could write a general element of $SU(2)$ as $\exp(i\theta\mathbf{\hat{n}}\cdot\mathbf{\sigma})$ where $\theta\in\mathbb{R}$, $\mathbf{\hat{n}}\in S^2$ and $\mathbf{\sigma}$ is the vector of the Pauli matrices; Furthermore I can write a general element of $U(1)$ as $\exp(ix)$ where $x\in\mathbb{R}$.

To put things together, I know that $U(2)\simeq SU(2)\cup U(1)$. So does that mean I could write:

1. $U(2) =\{\exp(ix) \exp(i\theta\mathbf{\hat{n}}\cdot\mathbf{\sigma}) \,:\,\{x,\theta\}\in\mathbb{R}\wedge\mathbf{\hat{n}}\in S^2 \}$
or
2. $U(2) =\{\exp(ix)I+ \exp(i\theta\mathbf{\hat{n}}\cdot\mathbf{\sigma}) \,:\,\{x,\theta\}\in\mathbb{R}\wedge\mathbf{\hat{n}}\in S^2 \}$

Which one of these options is true, and how do I get to know that just from the prescription $U(2)\simeq SU(2)\cup U(1)$?

#### Solutions Collecting From Web of "A general element of U(2)"

Preliminary Note: For some reason, physicists and mathematicians have different notation for Lie groups and Lie algebras. For mathematicians, the Lie algebra corresponding to a matrix Lie group $G$ is the set
$$\mathcal{L}_{\mathrm{math}} \;=\;\{A\in\mathbb{C}^{n\times n} \mid \exp(tA) \in G\text{ for all }t\in\mathbb{R}\}.$$
For physicists, the corresponding Lie algebra is the set
$$\mathcal{L}_{\mathrm{phys}} \;=\; \{A\in\mathbb{C}^{n\times n} \mid \exp(itA) \in G\text{ for all }t\in\mathbb{R}\}.$$
These are related by the equation
$$\mathcal{L}_{\mathrm{math}} \;=\; \{i A \mid A \in \mathcal{L}_{\mathrm{phys}} \}.$$

Answer: The Lie algebra $\mathcal{L}_{\mathrm{phys}}$ associated to $U(2)$ is the set of all $2\times 2$ Hermitian matrices. Every such matrix has the form
$$a I \,+\, b_1\sigma_1 \,+\, b_2\sigma_2 \,+\, b_3\sigma_3$$
where $I$ is the $2\times 2$ identity matrix, $\sigma_1,\sigma_2,\sigma_3$ are the three Pauli matrices, and $a,b_1,b_2,b_3\in\mathbb{R}$. We can write this as
$$a I + \textbf{b} \cdot \boldsymbol{\sigma}$$
where $\textbf{b} = (b_1,b_2,b_3) \in \mathbb{R}^3$, and $\boldsymbol{\sigma}$ is the $3$-tuple $(\sigma_1,\sigma_2,\sigma_3)$ of matrices.

It is true that every element of $U(2)$ can be written as an exponential $e^{iH}$, where $H$ is a $2\times 2$ Hermitian matrix. (This sort of thing isn’t always true for Lie groups, but it is true in this case.) It follows that
$$U(2) \;=\; \{\exp(iaI + i\textbf{b}\cdot \boldsymbol{\sigma}) \mid a\in\mathbb{R}\text{ and }\textbf{b}\in\mathbb{R}^3\}.$$
Now, it is not true in general that $\exp(A+B) = \exp(A)\exp(B)$ when $A$ and $B$ are matrices. However, it is true as long as $A$ and $B$ commute. Since $iaI$ is a multiple of the identity matrix, it commutes with anything, so
$$U(2) \;=\; \{\exp(iaI) \exp(i\textbf{b}\cdot \boldsymbol{\sigma}) \mid a\in\mathbb{R}\text{ and }\textbf{b}\in\mathbb{R}^3\}.$$
It turns out that $\exp(iaI) = \exp(ia)I$, so it follows that
$$U(2) \;=\; \{\exp(ia) \exp(i\textbf{b}\cdot \boldsymbol{\sigma}) \mid a\in\mathbb{R}\text{ and }\textbf{b}\in\mathbb{R}^3\}.$$
This is essentially option (1) in your question.

It is possible to derive this result from a version of the reasoning you gave. In particular, let
$$H \;=\; \left\{\left.\begin{bmatrix}u & 0 \\ 0 & u\end{bmatrix} \;\right|\; u\in U(1) \right\}.$$
Then $H$ is a subgroup of $U(2)$ isomorphic to $U(1)$, and $U(2)$ is generated by the elements of $H$ together with the elements of $SU(2)$:
$$U(2) \;=\; \langle SU(2) \cup H\rangle.$$
Now, in general if you have two matrix groups it is not true that every element of $\langle G\cup H\rangle$ can be written as an element of $G$ multiplied by an element of $H$. (Instead, elements of $\langle G\cup H\rangle$ can be written as words involving elements of $G$ and elements of $H$.) However, it is true whenever every element of $G$ commutes with every element of $H$, which is why it works in this case.