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

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?

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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. Identify $\Bbb{R}^3$ with the Lie algebra $\mathfrak{su}(2)$ that has basis consisting of matrices

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

Then given any $U \in \mathfrak{su}(2)$, we have $\Phi_A(U) =AUA^{-1}$. Now with respect to the inner product $\langle A,B\rangle = \frac{1}{2}\textrm{Tr}(AB)$ on $\mathfrak{su}(2)$ $\Phi_A$ is an orthogonal linear transformation. Now because $\Phi$ is a continuous group homomorphism it must map connected sets to connected sets, so that in particular $\textrm{Im} \Phi$ is necessarily contained in $SO(3)$. The fact that $SU(2)$ is connected comes from the fact that as a topological space it is homeomorphic to $S^3$. Now we already know that a general 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 $|\alpha|^2 + |\beta|^2 = 1$. You can use this to compute the kernel of $\Phi$, namely find those matrices $A$ in $SU(2)$ such that $AE_iA^{-1} = E_i$ for $i = 1,2,3$. If you bash this out you will get that the kernel is precisely $\{\pm I\}$. 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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@t.b. Is it better now? Not sure about why we need to replace $A$ with its transpose. Also, I am not sure how to prove that $\Phi$ is surjective. –  fpqc Aug 25 '12 at 8:56
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. –  t.b. Aug 25 '12 at 9:02
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). –  t.b. Aug 25 '12 at 13:18

Hint:

The lead paragraphs of this Wiki page.

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$SU(2)$ is isomorphic to 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 $\{1,-1 \}$.