The most common matrix representation of $SU(2)$ is given by $$ \begin{pmatrix} a & b\\ b^* & -a^*\\ \end{pmatrix} $$ where $a,b\in\mathbb{C}$. If we denote real components by the subscript $r$, imaginary by $i$, one can clearly write this as

$$ \begin{pmatrix} a & b\\ -b^* & a^*\\ \end{pmatrix} = a_r\begin{pmatrix} 1 & 0\\ 0 & 1\\ \end{pmatrix} +b_r\begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix} +b_i\begin{pmatrix} 0 & i \\ i & 0 \\ \end{pmatrix} +a_i\begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix}, $$ so the Lie algebra should be obtained by discarding the coefficient of the identity matrix and identifying the remaining traceless matrices as the generators of $\mathfrak{su}(2)$. We quickly realize these are $i\sigma_a$, where $a = 1,2,3$ denote the Pauli spin matrices. Many books however define the standard basis of $\mathfrak{su}(2)$ as $(i/2)\sigma_a$. While this would normally be simply a trivial convention, R. Gilmore, on page 107 of "Lie Groups, Physics, and Geometry", seems to use this to show the $SU(2)\rightarrow SO(3)$ cover is 2:1 by splitting $\mathfrak{su}(2)$ into the form $\mathfrak{su}(2)\ni X = S_1 + (ia_i/2)\sigma_3$ where $S_1 = (ib_i/2)\sigma_1 + (ib_r/2)\sigma_2$ and exponentiating $S_1$ and $(ia_i/2)\sigma_3$ individually to give

$$SU(2) \ni U = \exp(S_1)\times\begin{pmatrix} e^{ia_i/2} & 0\\ 0 & e^{-ia_i/2}\\ \end{pmatrix}.$$

Using a similar splitting of $\mathfrak{so}(3)$ into two components (an $SO(3)/SO(2)$ quotient and $SO(2)$) which are individually exponentiated, he concludes the quotients $SO(3)/SO(2)$ and $SU(2)/U(1)$ are both topologically $S^2$, thus bijective, and the thus the cover must be 2:1 because putting $\theta \mapsto \theta + 2\pi$ is a complete rotation in $SO(2)$, while $a_i \mapsto a_i + 4\pi$ is a complete rotation in the given representation of $U(1)$. I've been wanting to emulate this argument but can't do so without the factor of $1/2$. My questions are thus as follows:

(1) where does the $1/2$ come from? Is it just convention or something else?

(2) Is there any easy way to argue the cover is $2:1$ without using the $1/2$?

(3) I thought that generally the whole Lie algebra had to be exponentiated to map back onto the Lie group. How does separating the Lie algebra and exponentiating the terms individually make sense in the context of the BCH formula? Does this have to do with irreducible representations?

  • 2
    $\begingroup$ Can you add a bit about what exactly he does for $\mathfrak{so}_3$, in particular what basis he uses for that? The argument sounds sketchy since that Lie algebra is actually isomorphic to $\mathfrak{su}_2$. $\endgroup$ Commented Jun 22, 2019 at 20:27
  • $\begingroup$ @TorstenSchoeneberg thanks for the comment. Honestly, I posted this question years ago, during my master's thesis. Evidently the system has "made it active" again, but I no longer have much interest in this. Actually, I don't even have a copy of that book anymore. $\endgroup$
    – JMJ
    Commented Jun 22, 2019 at 20:46
  • $\begingroup$ @JMJ Then you should delete the question to avoid further efforts of readers. $\endgroup$
    – Paul Frost
    Commented Jul 22, 2020 at 10:14
  • $\begingroup$ I’m voting to close this question because the OP comments that he has lost interest in the question. $\endgroup$
    – Paul Frost
    Commented Jul 24, 2020 at 9:42

1 Answer 1


Evidently this is due to the basis with the $1/2$ giving the "fundamental representation" of $\mathfrak{su}(2)$, e.g. according to




But I don't know enough about representation theory to make a good argument about how that guarantees the covering map is 2:1 for $SU(2) \rightarrow SO(3)$ regardless of the representation.

  • 1
    $\begingroup$ could you add the essential parts of the links to your answer? $\endgroup$
    – user190080
    Commented Mar 31, 2016 at 18:47
  • $\begingroup$ top of the SU(2) section in the first link while I read the whole of the second link. I asked this question and happened upon something resembling an answer, but really would like an expert in representation theory to expand on it in their own answer. $\endgroup$
    – JMJ
    Commented Apr 1, 2016 at 3:56

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