Could any one give me hint for this one?

how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$,

well, Is it the same: there is a 2-fold covering map from $SU(2)$ to $SO(3)$? what is that map will be?


Recall that we have a map $\Phi : SU(2) \to SO(3)$ that sends a matrix $A \in SU(2)$ to the linear endomorphism $\Phi_A$ on $\Bbb{R}^3$ defined as follows: recall that the Lie algebra $\mathfrak{su}(2)$ has basis given by

$$ E_1 = \left(\begin{array}{cc} 0 & i \\ i & 0 \end{array}\right),\ \ \ E_2 = \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right),\ \ \ E_3 = \left(\begin{array}{cc} i & 0 \\ 0 & -i \end{array}\right).$$

This basis is orthonormal with respect to inner product $\langle U,V\rangle = \frac{1}{2}\textrm{Tr}(UV^*)$ on $\mathfrak{su}(2)$, so we can identify $\mathfrak{su}(2)$ and $\Bbb{R}^3$ as an inner product space. Now, for any $U \in \mathfrak{su}(2)$ we define $\Phi_A(U)=AUA^{-1}$. Then $\Phi_A$ preserves the given inner product on $\mathfrak{su}(2)$, which means $\Phi_A$ is an orthogonal linear mapping of $\mathfrak{su}(2)\simeq \Bbb{R}^3$, by the above identification. Thus, $\mathrm{Im}\ \Phi\subset O(3)$.

Recall that an arbitrary element of $SU(2)$ looks like

$$A = \left(\begin{array}{cc} \alpha & -\overline{\beta} \\ \beta & \overline{\alpha} \end{array}\right)$$

for $\alpha,\beta \in \Bbb{C}$ satisfying $|\alpha|^2 + |\beta|^2 = 1$. Now it is easy to see that $SU(2)$ is homeomorphic to $S^3$, which means in particular that $SU(2)$ is connected. Therefore its image under the continuous group homomorphism $\Phi$ is necessarily contained in the identity component of $O(3)$, which is $SO(3)$.

Again appealing to the coordinate expression for $SU(2)$ given above, one can directly compute the kernel of $\Phi$, that is, find the matrices $A$ in $SU(2)$ such that $AE_iA^{-1} = E_i$ for $i = 1,2,3$. Eventually, one finds $\textrm{Ker}\ \Phi= \{\pm I\}$.

We now show that $\Phi$ is surjective. Suppose you know that $\Phi$ is a covering map. Then there is $U$ open about the identity $I \in SO(3)$ and $V$ open about the identity $I$ in $SU(2)$ such that $U$ is homeomorphic to $V$.

Let $\langle U \rangle$ denote the subgroup generated by the set $U$. Then $\langle U \rangle$ must be contained in $\textrm{Im} \Phi$ that is a subgroup. However because $U$ is open it is also closed using an argument via cosets and connectedness of $SO(3)$ now implies that $SO(3) \subseteq \textrm{Im}\Phi$. It follows that $\Phi$ is surjective and so we have proven that

$$SU(2)/\{ \pm I\} \cong SO(3).$$

$\hspace{6in} \square$

I should say that there is also another argument to prove that $\Phi$ is surjective. Namely, we already know that the associated Lie algebra homomorphism

$$\phi : \mathfrak{su}(2) \longrightarrow \mathfrak{so}(3)$$

is an isomorphism, so it would suffice to prove that the map

$$\textrm{exp} : \mathfrak{so}(3) \longrightarrow SO(3)$$

is surjective.

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  • $\begingroup$ Yes, it's better now (you want the matrices to exponentiate into SU(2)). The choice of $A$ should reflect the choice of basis in $\mathfrak{su}(2)$. Now it's consistent. To prove surjectivity, you can use that SO(3) is connected and that the quotient map sends a small neighborhood of the identity in SU(2) to a small neighborhood of the identity of SO(3), so the image is an open subgroup of the latter, hence everything by connectedness. $\endgroup$ – t.b. Aug 25 '12 at 9:02
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    $\begingroup$ What do you mean by "If SO(3) is abelian". It's certainly not abelian... Questions on surjectivity of the exponential map tend to be much harder than your first argument. Since the exponential map is a diffeomorphism in a neighborhood of $0$, the group generated by the image of the exponential map must be everything (this is basically your first argument again). $\endgroup$ – t.b. Aug 25 '12 at 13:18

$SU(2)$ is isomorphic to unit quaternions. In this this article you can find a way to represent every quaternion with a rotation of $\mathbb{R}^3$, namely an element of $SO(3)$, and for every rotation of $\mathbb{R}^3$ there's a quaternion that represents it. So you get a surjective function from quaternions to $SO(3)$, this function should be a homomorphism. Composing you get a surjective homomorphism from $SU(2)$ to $SO(3)$. Check if the kernel is $\{1,-1 \}$.

Maybe this can help you. Check for details :)

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    $\begingroup$ When you say quaternions you mean unit quaternions. $\endgroup$ – t.b. Aug 25 '12 at 8:43
  • $\begingroup$ Yes, thank you. $\endgroup$ – Lorban Dec 8 '16 at 18:59

The connection between SO(3) and SU(2) is particularly well and simply described, using quaternions as Lorban suggested above, in the following article, https://en.wikipedia.org/wiki/Rotation_group_SO(3)?wteswitched=1#Using_quaternions_of_unit_norm

The 2:1-nature of the map between quaternions and SO(3) results from the use of conjugation, i. e. p ↦ $q p q^{-1}$ , to describe the rotation of p $\in\Bbb{R}^3$, by q $\in\Bbb{H}$ and the fact that both "q" and "-q" define the same rotation of p.

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  • $\begingroup$ This is an old question which already has an accepted answer which seems to be pretty clear. I'm not sure what the link that you give above adds to the conversation, and the brief explanation you provide does little to clarify. Perhaps you could expand on those ideas a bit, and discuss (in greater detail) what can be found on the linked Wikipedia page and how it applies? $\endgroup$ – Xander Henderson May 25 '18 at 18:45
  • $\begingroup$ My point was to add a link, following Lorban's suggestion, highlighting the exceeding clarity that the use of quaternions brings to the comparison of SU(2) with SO(3) - and the discussion of rotations in general. $\endgroup$ – FG98337 May 25 '18 at 19:18

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