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Standard Basis of $SU(2)$--where does the 1/2 come from?

Mathematics Asked by JMJ on December 24, 2020

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

$$
begin{pmatrix}
a & b\
-b^* & a^*\
end{pmatrix}
=
a_rbegin{pmatrix}
1 & 0\
0 & 1\
end{pmatrix}
+b_rbegin{pmatrix}
0 & 1 \
-1 & 0 \
end{pmatrix}
+b_ibegin{pmatrix}
0 & i \
i & 0 \
end{pmatrix}
+a_ibegin{pmatrix}
i & 0 \
0 & -i \
end{pmatrix},
$$
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 $isigma_a$, 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

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

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 $theta mapsto theta + 2pi$ is a complete rotation in $SO(2)$, while $a_i mapsto a_i + 4pi$ 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?

One Answer

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

https://en.wikipedia.org/wiki/Pauli_matrices

and

https://en.wikipedia.org/wiki/Representation_theory_of_SU(2),

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.

Answered by JMJ on December 24, 2020

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