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As the question title suggests, what is the quickest self-contained way of computing $\pi_1(\text{SU}(2))$ and $\pi_1(\text{SO}(3))$?

Santiago
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Jakob W
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1 Answers1

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The quickest way to compute $\pi_1(SU(2))$ by itself is to observe that it is diffeomorphic to $S^3$, which is easily seen to be simply connected.

The quickest way to compute $\pi_1(SO(3))$ by itself is, as Mariano notes, to observe that it acts on $S^2$ with stabilizer $SO(2)$, obtain a fibration and use the long exact sequence in homotopy groups.

If you complain that method is not self-contained, you can instead do the following: Observe that $SU(2)$ double covers $SO(3)$, so $\pi_1(SO(3))$ must be the group of order two. In my opinion this is slightly more work, but can be done with elementary differential topology.

Max
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    Wikipedia gives a nice way to obtain the topology of $SO(3) \cong \Bbb RP^3$ directly: Consider the solid ball in $\Bbb R^3$ of radius $\pi$ (that is, all points of $\Bbb R^3$ of distance $\pi$ or less from the origin). Given the above, for every point in this ball there is a rotation, with axis through the point and the origin, and rotation angle equal to the distance of the point from the origin. The identity rotation corresponds to the point at the center of the ball... – Ben Grossmann Aug 29 '16 at 23:48
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    ...Rotation through angles between $0$ and $-\pi$ correspond to the point on the same axis and distance from the origin but on the opposite side of the origin. The one remaining issue is that the two rotations through $\pi$ and through $-\pi$ are the same. So we identify (or "glue together") antipodal points on the surface of the ball. After this identification, we arrive at a topological space homeomorphic to the rotation group. Source – Ben Grossmann Aug 29 '16 at 23:48
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    Note: For a diffeomorphism $SU(2) \to S^3$, take $$ \DeclareMathOperator{\re}{Re} \DeclareMathOperator{\im}{Im} f\pmatrix{a&b\c&d} = (\re a, \im a, \re b, \im b) $$ verifying that this map is injective and bicontinuous is a nice undergraduate exercise. – Ben Grossmann Aug 29 '16 at 23:56
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    @Omnomnomnom: One does not even need to prove bi-continuity. Simply use the fact that a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism. – Berrick Caleb Fillmore Aug 30 '16 at 00:36