1
$\begingroup$

This question asked here claimed that the fundamental group of $L(n):= U(n)/O(n)$ was $\cong \mathbb{Z}$ using a long exact sequence from the fibration $U(n) \to U(n)/O(n)$:

$$\cdots \to \pi_1(O(n)) \to \pi_1(U(n)) \to \pi_1(L(n)) \to \pi_0(O(n))\to \pi_0(U(n)) $$

However, they wrote that $\pi_0(O(n)) = \{1\}$, which is false. The sequence of groups is actually (for $n>2$):

$$ \mathbb{Z}/2 \to \mathbb{Z} \to \pi_1(L(n)) \to \{\pm 1\} \to \{1\}$$

I don't see how this implies $\pi_1(L(n))\cong \mathbb{Z}$. The leftmost map is zero and $\pi_1(L(n)) \to \{\pm 1\}$ is a surjection, but that doesn't mean $\pi_1(L(n)) \cong \mathbb{Z}$ (for instance, it could be $\mathbb{Z}\oplus \mathbb{Z}/2$).

  • 1
    The last part of Bebep's answer in the linked question gives another way to compute $\pi_1(L_n)$ using a different fibration. – 2017-01-29
  • 1
    Yes; perhaps I should have been more clear. I just wanted to know if this particular fibration could work (and perhaps I just wasn't seeing something). – 2017-01-29

0 Answers 0