Why isn't $S^2$ a lie group?

Question

Why isn't $S^2$ a lie group? The only spheres that are Lie groups are $S^0$,$S^1$ and $S^3$. What about these spheres makes them special so that they are the only spheres that can be given a continuous group structure?

On the 2-sphere, what group operation would you put ? On the circle, it's induced by complex multiplication, and on the 3-sphere, by quaternionic multiplication. — Anthony, Mar 29 '21 at 00:13

score 8 · Accepted Answer · edited Mar 29 '21 at 00:17

8

A lie group is a parallelizable manifold, in particular there exists a vector field which does not vanish. This implies that the Euler characteristic is zero. The Euler characteristic of the sphere is $2$

edited Mar 29 '21 at 00:17

Thomas Andrews

186,215

answered Mar 29 '21 at 00:11

Tsemo Aristide

89,587

2

Also see "hairy ball theorem". – Tristan Duquesne Mar 29 '21 at 00:18
Incidentally much more is true (and in fact even more than that), but that takes a lot more work. – Noah Schweber Mar 29 '21 at 00:19
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Since this question was closed as a duplicate, and the older question has only a wrong answer, this answer should be moved there. – Toby Bartels Mar 29 '21 at 01:07

Why isn't $S^2$ a lie group?

1 Answers1