Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been looking at intersection forms and the Arf invariant recently and I got a comment to one of my previous questions related to this.

So I looked at framed manifolds. There seems to be quite a bit of knowledge required that I don't have yet and I struggled to get the big picture.

Question 1: What's the easiest example of a framed manifold? Or say, two manifolds since I will want to compute their Arf invariant to show that they're not homeomorphic.

Question 2: While I read about framing I also came across framed links. (Unfortunately also without example). In the linked article, what exactly is "an extension of a tame knot to an embedding of the solid torus ... to $S^3$"? I'll assume I can always think of $S^3$ as the Euclidean space $\mathbb R^3$.

Question 3: Just to be sure: a framed knot or link is not also a manifold, or is it? I mean, I can see that $S^1$ embedded into $\mathbb R^3$ is a manifold but I also assume that there must be more structure available in both cases, otherwise they wouldn't be treated as separate theories, right?

Thanks for your help.

share|cite|improve this question
Wouldn't the easiest example of a framed manifold be $\mathbb{R}^n$? Its tangent bundle is just $T\mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$. More generally, any Lie group $G$ has trivial tangent bundle, since you can "spread" the tangent vectors at $e \in G$ by left translations. So $\mathbb{S}^1$ and $\mathbb{S}^3$ would also be examples... or did I misunderstand something? – student Sep 6 '12 at 21:28
up vote 8 down vote accepted

Matt -

Sorry to send you on a wild goose-chase! (Or is it a wild-goose chase?) Anyways, I learned about the Arf invariant because of the Kervaire Invariant One problem (which I'll describe briefly below), and so I assumed (wrongly, obviously) that this was also the reason you had come across it too.

As for your Question 1, @student is correct that a Lie group is always frameable -- you can choose any basis of its Lie algebra, and then transport it via (say) left-multiplication to the entire manifold (i.e., extend to a left-invariant framing). There are of course framed manifolds that can't be made into Lie groups, but this is definitely the easiest class of examples. In fact, the cobordism ring of (stably) framed manifolds is precisely the ring of stable homotopy groups of spheres! This is easy to prove via something called the "Pontrjagin-Thom construction", but this ring is, on the whole, extremely mysterious. However, much about it is known; some people like to say that our understanding of $\pi_*^S$ is a yardstick for our progress in algebraic topology.

As for your Question 2, an extension of a knot to an embedded torus is just a closed tubular neighborhood of the submanifold $S^1$ sitting inside $S^3$. And you're right that for the purposes of knot theory, you can often just think of this as $\mathbb{R}^3$; knot theorists like to compactify the ambient space for various reasons (e.g. so that they can talk about the "hyperbolic volume" of the knot's complement, which believe it or not is an invariant of the knot (i.e. it's invariant under isotopy)), but certainly any knot will miss some point, which you can take to be the point at infinity (so that your knot sits in $\mathbb{R}^3 \subset S^3 = \mathbb{R}^3 \cup \{\infty\}$).

As for your Question 3, a knot or link is a different collection of data from a manifold, but that data involves manifolds. People like to be very careful about what structure they take to be part of the data they care about, so for instance a manifold might be defined as a topological space which admits local charts such that etc etc etc; that is, they don't take the local charts as part of the data.

And now, on the to Kervaire Invariant One problem. The Kervaire invariant of an even-dimensional manifold is the Arf invariant of the intersection form on its middle-dimensional mod-2 homology. Incidentally, this was first used (by Kervaire, of course) to construct a piecewise-linear but non-smoothable 10-dimensional manifold, the first of its kind. (One can show that the Kervaire invariant vanishes on all smooth 10-manifolds.)

Now, the Kervaire Invariant One problem asks: Which framed manifolds have nontrivial Kervaire invariant? As I've indicated above, this is secretly a question about stable homotopy groups of spheres. This problem is around half a century old, and people chipped away at it bit by bit, but for the most part it withstood anything remotely resembling a complete resolution. You can read the blow-by-blow yourself at the wikipedia article, but the most striking result known up through 2009 was that it could only be nonzero in dimensions of the form $2^k-2$ (due to Browder in the '60s).

But then in 2009, Hill, Hopkins, and Ravenel (the "HHR" I referred to in my comment on your other question) stunned the algebro-topological world with a proof that the Kervaire invariant vanishes above dimension 126. Since there are known examples in dimensions 2, 6, 14, 30, and 62, this leaves only the case of dimension 126 still open. Nutty, right?

share|cite|improve this answer
This might be a bit much, but I'll again plug Snaith's article: (also on the arXiv). Section 3 in particular might be useful – Juan S Sep 6 '12 at 22:34
That looks like a nice survey. Why might it be a bit much, have you plugged it before? – Aaron Mazel-Gee Sep 6 '12 at 23:08
On this post. By 'might be a bit much' I meant could be at too high a level – Juan S Sep 7 '12 at 4:08
This answer is awesome, still reading, hang in there before I accept. – Rudy the Reindeer Sep 7 '12 at 9:28
@JuanS Thanks loads, I somehow overlooked your comment to my other post! : ( – Rudy the Reindeer Sep 7 '12 at 9:29

Your Answer


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.