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Homotopy groups of n-spheres are about embedding n-spheres into a manifold of dimension k.

I want to understand what does operationally mean $\pi_n (S^k)$ when $n >k$. The definition of embedding i'm familiar with requires that the embedded manifold (in this case n-spheres) are always lower or equally dimensional.

In short, i don't have any intuition whatsoever what does it mean to embed a 2-sphere inside, say, a 1-sphere (a loop). What kind of mapping would that be?

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I hate to be a pain, but embedding is the wrong word. It has technical meanings in both topology and smooth manifold theory which prevent any map $S^m \rightarrow S^n$ from ever being an embedding unless $m=n$(and I think the map has to be homotopic to a generater of the $m$-th cohomology class as well). That being said I don't know much about higher homotopy groups unfortunately. I definitely can't visualize them. Have you looked up the Hopf fibration for (the lowest dimensional non-trivial) example? – John Stalfos Dec 8 '12 at 15:08
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John has it right: the mistake is in the word "embedding".

Homotopy groups are defined as (pointed) homotopy classes of (pointed) maps from a sphere into a space (which does not have to be a manifold).

It turns out that there is no nontrivial way to map a $2$-sphere to a circle, but after that things get interesting...

  • Consider the map $S^3 \rightarrow \mathbb{C}P^1$ where we consider $S^3 \subset \mathbb{C}^2 - \{0\}$. This is not nullhomotopic, and so it is a nontrivial way to map a $3$-sphere to $\mathbb{C}P^1$. (This is called the Hopf fibration, after identifying the target with $S^2$). One way to prove this is to show that the fibers over two different points are linked nontrivially.
  • In general, the problem of calculating the homotopy groups $\pi_{n+k}S^{n}$ for $n$ large is equivalent to studying what is called "framed cobordism of $k$-manifolds". That is, we can say something about how complicated a map from a big sphere to a small sphere is by studying it's fiber which will (after a bit of wiggling) be some $k$-manifold. Saying that the map is nontrivial translates into saying that the manifold you get this way is nontrivial in some specific manner (called cobordism).
  • The above problem is hard and is, in some sense, the (unattainable) holy grail of algebraic topology: if we just cared about embeddings of spheres, our jobs would be over (or we'd become knot theorists...). Many nontrivial maps are known, and it is known that there a lots and lots.
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