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I want to calculate the higher homotopy groups of $\Sigma_g$ and $\mathbb{R}P^2\# \mathbb{R}P^2\#\cdots\# \mathbb{R}P^2$. But I haven't found the methods to calculate the homotopy groups of connected sum. So how to solve this question?

346699
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If $X$ is a path-connected topological space nice enough so that universal cover $\tilde{X}$ exists, then the covering map $p : \tilde{X} \to X$ induces an isomorphism $p_* : \pi_n(\tilde{X}, \tilde{x_0}) \to \pi_n(X, x_0)$ for all $n > 1$. This is a simple application of the functoriality of $\pi_n$ coupled with the lifting lemma (key point is $S^n$ for $n > 1$ is simply connected). You should work it out if you don't know this fact already.

Universal cover of $M_1 = T^2$ is just $\Bbb R^2$. For $g > 1$, it can be seen that it is the upper half plane $\Bbb H^2$ or the open unit disk: there is a triangulation of $\Bbb H^2$ by geodesic $4g$ sided polygons, and an appropriate isometry group of $\Bbb H^2$ acts so that the quotient identifies the sides of the fundamental polygon appropriately. Alternatively apply classification of $2$-manifolds: there are not too many simply connected noncompact $2$-manifolds out there (noncompact because $\pi_1(M_g)$ is infinite, so the cover has to be infinite sheeted).

In both cases, the universal cover is contractible. Thus, $\pi_n(M_g) \cong \pi_n(D^2) = 0$ for $n > 1$ whenever $g \neq 0$. The nonorientable surface with $n$ cross caps has the same universal cover as the orientable surface of genus $n - 1$ (why?), so from that you can similarly conclude. Note that for $g = 0$ we would end up computing $\pi_n(S^2)$ which is not quite entirely trivial so I'm not including it.

Balarka Sen
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    "Not quite entirely trivial". Understatement of the year. – Eric Wofsey Apr 14 '16 at 06:58
  • @EricWofsey Sarcasm was intended. The honest reason I didn't include it is because I don't really know how to compute homotopy groups of $S^2$. :) – Balarka Sen Apr 14 '16 at 07:05
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    You shouldn't feel bad about that; no one else does either. :) – Eric Wofsey Apr 14 '16 at 07:08
  • Oh, I wasn't aware of that. I knew computing homotopy groups of spheres is open but not that it's open itself for $S^2$. Thanks. – Balarka Sen Apr 14 '16 at 07:13
  • May I inquire you how to prove the cliam " The nonorientable surface with $n$ cross caps has the same universal cover as the orientable surface of genus $n-1$". You can give me a reference. Thanks. – 346699 Apr 15 '16 at 08:51
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    @user34669 Sure, the reason is that nonorientable surface with $n$ cross caps is double covered by the orientable surface of genus $n - 1$. You can see this explicitly: a nonorientable surface with $n$ cross caps is connected sum of $n$ copies of $\Bbb{RP}^2$ (so you remove a pair of disks from each pair of those projective planes and tube it up). $\Bbb{RP}^2$ are double covered by $S^2$. So $\Bbb{RP}^2 - D^2$ is covered by $S^2$ with two $D^2$'s removed. You should be able to construct the double cover now by tubing up the $n$ copies of twice punctured $S^2$'s appropriately. – Balarka Sen Apr 15 '16 at 08:59
  • It is in general true that any compact closed nonorientable manifold admits a double cover by an orientable manifold, but that's a bit more general thing which you don't really need to invoke right now. – Balarka Sen Apr 15 '16 at 09:01