I want to demonstrate that the manifold of $SU(2)/U(1)$ is a 2-sphere. In a text-book I've found this way of solution, where there are some unclear points.

Let to be $g= a\mathbb{1} + i b_j\sigma_j$ the generic element of $SU(2)$ (with $b_j\sigma_j$ I mean a sum over $j=1,2,3$), with $a^2+b_jb^j=1$. Now, let define a subgroup $U(1)$ of $SU(2)$ by the generic element on $\sigma_3$-direction $h=a\mathbb{1} + ix_3\sigma_3 \in U(1)$ (why is the coefficient of $\mathbb{1}$ the same in the two cases?) with $a^2+x_3^2=1$. Now, we write $b_j = x_3\xi_j$ (why?) and the condition about the determinant returns $\xi_j\xi^j = 1$ that is the 2-sphere.

I don't understand why we don't calculate the left coset $gU(1)$ and how to deduce the isomorphism with the 2-sphere...


1 Answer 1


That is indeed not a valid demonstration, since substituting $b_j=x_3\xi_j$ is not well-defined when $x_3=0$.

  • $\begingroup$ Thanks Jesper. What's the correct demonstration ? $\endgroup$
    – apt45
    Jan 21, 2015 at 20:07
  • 1
    $\begingroup$ This is the Hopf fibration, see for example the explanation in Wikipedia. $\endgroup$ Jan 22, 2015 at 9:22

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