Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers 2

up vote 4 down vote accepted

$O(n)$ 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 $\pi_2(S^2) = \mathbb{Z}\neq \{0\}= \pi_2(SO(3))$.

share|improve this answer
    
Thank you for your answer. Do you happen to know any references? –  Dave Hartman 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 Hartman Nov 7 '11 at 18:34
add comment

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 .

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.