# Relation between homotopy groups of $S^n$ and homotopy groups of $SO(n)$, $O(n)$

What is the relation between the homotopy groups of spheres $S^n$ and the homotopy groups of the special orthogonal groups $SO(n)$ (resp. $O(n)$)?

This question occurred to me in the context of classifying real vector bundles over spheres via homotopy classes of maps. One can show that (see for example Hatcher):
For $k>1$, there is a bijection $$[S^{k-1},SO(n)]\longleftrightarrow Vect^n(S^k)$$ Here $[S^{k-1},SO(n)]$ denotes the set of homotopy classes of maps $S^{k-1}\to SO(n)$ and $Vect^n(S^k)$ denotes the set of isomorphism classes of real rank $n$ vector bundles over $S^k$. Furthermore one has that the map $$\pi_i(SO(n))\longrightarrow [S^i,SO(n)]$$ which ignores the basepoint data is a bijection, so one can essentialy classify real vector bundles over spheres via the homotopy groups of $SO(n)$.

I'm also interested in the following:
What is the relation between homotopy groups of spheres and the classification of real vector bundles over spheres?

Any references (as well as examples) would be much appreciated.

-
If I'm not mistaken, I think that maybe the whole point of using K-theory to calculate chromatic homotopy theory is somehow where the connection between vector bundles and homotopy groups of spheres may lie. – Jon Beardsley Nov 7 '11 at 16:53

$O(n)$ is diffeomorphic to 2 disjoint copies of $SO(n)$, so the homotopy groups of $O(n)$ are those of $SO(n)$.

The relationship between the homotopy groups of the spheres and $SO(n)$ come from the fiber bundle $SO(n)\rightarrow SO(n+1)\rightarrow S^n$ which takes, say, the first column of a matrix in $SO(n+1)$ and considers it as a unit vector in $\mathbb{R}^{n+1}$.

Any time you have a fiber bundle (or more generally, a fibration), you get a long exact sequence of homotopy groups. A portion of it is

$$...\rightarrow \pi_k(SO(n))\rightarrow\pi_k(SO(n+1))\rightarrow \pi_k(S^n)\rightarrow \pi_{k-1}(SO(n))\rightarrow ...$$

Then, e.g., since $SO(2) = S^1$, this tells you that $SO(3)$ has the same homotopy groups as $S^2$ except that $\pi_2(S^2) = \mathbb{Z}\neq \{0\}= \pi_2(SO(3))$.

-
Thank you for your answer. Do you happen to know any references? – Dave Nov 7 '11 at 17:04
Hatcher's Algebraic topology book talks about the the fact that a fiber bundle gives rise to a long exact homotopy sequence of homotopy groups. Lee's Smooth Manifolds proves that the natural action of $SO(n)$ on $SO(n+1)$ has quotient diffeomorphic to $S^n$. The fact that you get a fiber bundle comes from a more general fact: If a compact Lie group $G$ acts freely on a manifold $M$, then you get a fiber bundle $G\rightarrow M\rightarrow M/G$ (where $M/G$ is also a manifold in a canonical fashion). Unfortunately, I don't know of a reference for it. – Jason DeVito Nov 7 '11 at 17:57
Thank you, this is very helpful. – Dave Nov 7 '11 at 18:34

You also might be interested in the following connection: The J-homomorphism, it can be viewed as a morphism from homotopy groups of $SO(n)$ to the stable homotopy groups of spheres. Indeed its image is always a direct summand of $\pi_{*}^s$. Moreover this has a deep connection to the socalled surgery long exact sequence in surgery theory and the connection between stable homotopy theory and framed bordism.

I think the original papers of Adams ("On the groups $J(X)$") should be a reference and of course http://en.wikipedia.org/wiki/J-homomorphism .

-