4

Concretely, I'm working with the spaces:

$S^n$, $\mathbb{C}P^n$ and $\mathbb{H}P^n$. I need to conclude that $T^1(M)$ is simply connected for all those manifolds $M$ I listed (with the exception of $S^2$).

Now, I know that $T^1S^2 \cong SO(3) $, hence $T^1S^2$ is not simply connected.

For the rest, what I can do is use the Gysin Sequence for those spaces and arrive at the fact that their first homology groups are trivial. But this is not enough for me to say that they are simply connected.

Do I need to take another route entirely to prove this, or can I circumvent this issue with some theorem that guarantees that $H_1=0 \implies \pi_1=0$ in this specific case?

Aloizio Macedo
  • 34,292
  • As the deleted answers indicate, you should be using the long exact sequence in homotopy for a fiber bundle instead. – Qiaochu Yuan Oct 05 '15 at 22:59
  • You would be able to deduce $\pi_1 = 0$ from $H_1 = 0$ if you knew that $\pi_1$ was abelian by the Hurewicz theorem. Aside from the case of H-spaces it's unclear how you would come to know this without knowing that $\pi_1 = 0$. – Qiaochu Yuan Oct 06 '15 at 03:12

1 Answers1

4

These are all sphere bundles over the respective manifolds, so you have a fiber bundle $S^k \to T^1(M) \to M$, $k+1$ the real dimension of $M$. Passing to the long exact sequence of homotopy groups you have the exactness of $$\pi_1(S^k) \to \pi_1(T^1(M)) \to \pi_1(M).$$

For $n \geq 1$ and your manifold $\neq S^1$, the last term is zero, so you have a surjection $\pi_1(S^k)\to \pi_1(T^1(M))$. In any case, this means that $\pi_1(T^1(M))$ is a quotient of $\Bbb Z$ or $0$, so is abelian, and hence $\pi_1(T^1(M)) = H_1(T^1(M))$.

In fact, other than the case $k = 1$ for $M = S^2 = \Bbb{CP}^1$, $k > 1$, so $\pi_1(T^1(M))$ is automatically trivial.