Understanding $\pi_3O(n)$

Question

I'm reading Hatcher and trying to understand his statement about the Bott periodicity for the orthogonal group $O(n)$. Using the fiber bundle structure $O(n-1)\hookrightarrow O(n) \to \mathbb{S}^{n-1}$, and assuming a large enough $n$, we know that $\pi_k\left (\mathbb{S}^{n-1}\right ) = 0$ for $k<n-1$, hence in lower dimensions this sets up isomorphisms $\pi_kO(n-1) \cong \pi_kO(n)$ for $k<n-2$ via the long exact sequence

$$ \ldots \to \;\; \pi_kO(n-1) \;\; \to \;\; \pi_kO(n)\;\; \to \;\; \pi_k\mathbb{S}^{n-1} \to \pi_{k-1}O(n-1) \;\; \to \ldots $$

The thing I don't understand is his computation of $\pi_3O(n) = \mathbb{Z}$. The way I originally argued this to myself is that there is essentially a chain of isomorphisms $\pi_kO(p) \cong \pi_kO(q)$ for $p \leq q$, hence we could reduce this to $$ \pi_3O(n) \;\; \cong \;\; \pi_3O(3) \;\; \cong \;\; \pi_3 \left (\mathbb{RP}^3\right ) \;\; \cong \;\; \pi_3\left (\mathbb{S}^3\right ) \;\; =\;\; \mathbb{Z} $$

where the congruence $\pi_3 \left (\mathbb{RP}^3\right ) \cong \pi_3\left (\mathbb{S}^3\right )$ comes from the covering space isomorphism. However, if I'm understanding the chain of isomorphisms correctly (which I'm probably not), then couldn't we argue that

$$ \pi_3O(n) \;\; \cong \;\; \pi_3O(2) \;\; \cong \;\; \pi_3\left (\mathbb{S}^1\right ) \;\; = \;\; 0? $$

I'm self-studying topology and don't have too many people to discuss this with. I'd like to understand the root of my misunderstanding. For reference, the statement as claimed in Hatcher is the following table:

$$ \begin{array}{ l | cccccccc} i \; \text{mod} \; 8 & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \pi_iO(n) & \mathbb{Z}_2 & \mathbb{Z}_2 & 0 & \mathbb{Z} & 0 & 0& 0 & \mathbb{Z} \\ \end{array} $$

Answer 1

So the issue lies with you assumption that $$\pi_k(O(n))=\pi_k(O(m))$$ for all $m\leq n.$ This is not true.

In general you have a fibration $O(n-1)\to O(n)\to S^{n-1}.$ Now $S^{n-1}$ has no homotopy in degrees less than $n-1$ and $\pi_{n-1}S^{n-1}=\mathbf{Z}.$

So in the long exact sequence you get

$$\ldots\to\pi_{n}S^{n-1}\to \pi_{n-1}O(n-1)\to \pi_{n-1}O(n)\to \pi_{n-1}S^{n-1}\to \pi_{n-2}O(n)\to \pi_{n-2}O(n-1)\to 0 $$ so in particular $$\pi_{n-1}O(n)\neq \pi_{n-1}O(n-1),$$ and also $$\pi_{n-2}O(n)\neq \pi_{n-2}O(n-1).$$

However from the same construction you can note that

$$\pi_{n-i}O(n-1)\cong \pi_{n-i}O(n)$$ for all $i>2.$

In general you can conclude from the same arguments that $$\pi_k(O(n))\cong \pi_{k}(O(n-1))$$ whenever $n>k+2$

Note that the computation in Hatcher is for the stable range i.e. large enough $n$ so that the homotopy groups start stabilising. It is not for all $n$, specifially lower dimension $n$.

In particular, the full periodicity shows up for the group $$O=\varinjlim O(n)$$

Answer 2

Left translation on the Lie group $O(n)$ induces homeomorphisms between path components, so there is a homeomorphism (diffeomorphism) $$O_n\cong SO_n\times\mathbb{Z}_2,\qquad \forall n.$$ In particular $$\pi_kSO_n\cong\pi_kO_n,\qquad \forall k\geq1.$$ Using the fibration sequences $$SO_{n-1}\rightarrow SO_n\rightarrow S^{n-1}$$ we see that what we really need to compute is $\pi_3SO_5$. That is, $$\pi_3O_n\cong\pi_3SO_5,\qquad\forall n\geq5.$$ Here we have used nothing more than $\pi_kS^n=0$ for $k<n$, $\pi_nS^n\cong\mathbb{Z}$, and exactness of the homotopy sequence of a fibration. In particular we have not calculated any of the homomorphisms in any sequence, save for when they are isomorphisms or trivial.

The easiest way to compute $\pi_3SO_5$ is to know something about its universal cover. By definition, the universal cover of $SO_n$ is the spinor group $Spin_n$. Since $\pi_1SO_n\cong\mathbb{Z}_2$ for $n\geq 3$ ($\pi_1SO_1=0$ and $\pi_1SO_2\cong\mathbb{Z}$), we have a 2-sheeted universal cover and a fibration $$\mathbb{Z}_2\rightarrow Spin_n\rightarrow SO_n.$$

In the special case $n=5$, as it happens, there is an exceptional isomorphism $$Spin_5\cong Sp_2$$ (i.e. a smooth group isomorphism). Now $Sp_n$ is the group of orthogonal transformations of the quaternionic space $\mathbb{H}^n$. This is defined in exactly the same way as the $SO_n$, with $\mathbb{R}^n$ replaced by $\mathbb{H}^n$, and the Euclidean inner product with the quaternionic form $(q_1,q_2)\mapsto q_1\overline q_2$. For the same reaons we have the fibration sequences for the $SO_n$, we have fibration sequences $$Sp_{n-1}\rightarrow Sp_n\rightarrow S^{4n-1}.$$ The factor of $4$ which appears can be traced to the fact that $\mathbb{H}$ is 4-dimensional over the reals, so the unit sphere in $\mathbb{H}^n$ is $S^{4n-1}$.

Now $Sp_1\cong SU_2\cong Spin_3\cong S^3$, so for $n=2$ the fibration sequence above is $$S^3\rightarrow Sp_2\rightarrow S^7.$$ In particular $$\pi_3Sp_2\cong \pi_3S^3\cong \mathbb{Z}$$ and hence $$\pi_3SO_5\cong \pi_3Spin_5\cong\mathbb{Z}.$$

Some comments on $\pi_3SO_n$ for $n<5$. We have $\pi_3SO_1=0$, since $SO_1$ is trivial. We have $\pi_3SO_2=0$, since $SO_2\cong S^1$ as the unit circle in the complex plane. We have $\pi_3SO_3\cong \mathbb{Z}$ as pointed out in the comments. As a space $SO_3$ is homeomorphic (diffeomorphic) to $\mathbb{R}P^3$ and to the unit sphere bundle in the tangent bundle of $S^2$ (using $SO_2\cong S^1$, the fibration sequence introduced above actually is the fibring of the unit tangent bundle over $S^2$).

The group $SO_n$ is simple for $n\neq 4$, and the final special case is that $SO_4$ is not simple. The fibration sequence $SO_3\rightarrow SO_4\rightarrow S^3$ splits (i.e. has a section). See here for a nice discussion. In any case this implies that there is a homeomorphism (diffeomorphism) $$SO_4\cong SO_3\times S^3$$ and in particular $$\pi_3SO_4\cong\pi_3SO_3\oplus\pi_3S^3$$ (this group is $\mathbb{Z}^2$).