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I was looking around on https://en.wikipedia.org/wiki/Homotopy_group, and saw that the definition of $\pi_n(X)$ is the set of homotopy classes of maps that map $S^n \to X$ (with fixed base points $a\in S^n$ and $b\in X$). Reading https://blogs.scientificamerican.com/roots-of-unity/higher-homotopy-groups-are-spooky, I see that even with $\pi_2(S^2)$, it begins to get difficult to understand (at least for me) what sort of "wrapping" we are doing; and of course $\pi_3(S^2)$ is even more confusing. But I had a question; what would happen if we change the definition to only consider homotopy classes of EMBEDDINGS $S^n \hookrightarrow X$? In this case, wouldn't the nastiness with higher homotopy groups go away, since $S^n$ can not embed in $S^i$ for $i<n$ (I don't know if this is true, but it is just a guess based on intuition)? I must admit I find this definition more intuitive regarding "detecting holes", or I guess in particular "spherical holes"; the official definition seems to me to be more about "interactions around holes" than the holes themselves.

As for group structure, maybe linear combinations, like what they have for cycles in homology groups (Intuition of the meaning of homology groups)? I have not studies these subjects in depth, but would appreciate any perspectives/references/pointers in a good direction. Basically, my question is this: it seems as if using an embedding $S^n \hookrightarrow X$ will make higher homotopy of spheres more in line with intuition (like what homology theory gives). Can this idea be made possible, with further details?

D.R.
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    It's somewhat unclear what you are asking. You want perspectives/references/pointers on what? On some kind of "other" definition than $\pi_n(X)$? – Lee Mosher Apr 30 '21 at 00:24
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    Ofc higher-dimensional spheres don't embed in lower-dimensional ones, look up Invariance of Domain Theorem. Anyway, so for you the torus will have trivial 'homotopy'" – John Samples Apr 30 '21 at 00:24
  • @LeeMosher yes pointers on this "new definition" I have written. – D.R. Apr 30 '21 at 01:21
  • @JohnSamples wouldn't $\pi_1$ of the torus still be non-trivial as it has loops around the two "holes" that aren't homotopic to 0? And for higher dimensions https://math.stackexchange.com/questions/915690/how-to-compute-homotopy-groups-of-torus I think the torus already has trivial homotopy. – D.R. Apr 30 '21 at 01:24
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    Well, then the short answer is that your "new definition" does not exist, as far as I know. Also the definition is not fleshed out, for example you haven't explained the group structure. – Lee Mosher Apr 30 '21 at 01:42
  • Since you are asking for pointers: Learn Algebraic Topology. Once you are familiar with the subject, feel free to edit your question (or ask a new question along these line) to make it more precise and answerable. – Moishe Kohan Apr 30 '21 at 02:15
  • @LeeMosher Although I haven't explained the group structure, one can still answer questions about what kind of holes this detects, right? For example, unlike the official definition of homotopy, higher homotopy for spheres with this "new definition" are trivial. In fact I think this "new definition" yields that $\pi_i (S^j)$ is non-trivial if and only if $i=j$. I think this is interesting, so I wonder what other things this definition says. – D.R. Apr 30 '21 at 03:27
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    Another problem with your definition is that being an embedding is not invariant under homotopy: Consider a straight line in $\Bbb R^n$ and deform it into any selfintersecting path (which you can do as $\Bbb R^n$ is contractible). It looks like you might want to use some notion of isotopy instead of homotopy... – Jonas Linssen Apr 30 '21 at 05:28
  • I think that for a more intuitive definition of holes, you may have a look at simplicial homology. The question whether a sphere embeds into some space might be nontrivial sometimes, but its a slightly different story – Peter Franek Apr 30 '21 at 05:50
  • Oh sorry, my comment got cut off for some weird reason! Yeah, I'm talking about two-spheres. Another issue will be tame vs wild embeddings; you might have problems getting excision axioms and local versions. Also what does 'tame' mean for general spaces? It makes sense for manifolds and some other sorts of stratified spaces, but it won't make sense for a lot of other important spaces. – John Samples Apr 30 '21 at 05:59
  • You might be interested in looking at the homotopy groupoids btw, they capture a bit of what you want, maybe you can get a more natural groupoid structure than group structure. – John Samples Apr 30 '21 at 06:02

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It's not a crazy thing, it's just that they are not called homotopy groups!

Let me say two words on why not to use embeddings only. A weak equivalence between spaces is a map that indices isomorphism on all the homotopy groups. A famous theorem of Whitehead says that a topological space is weakly equivalent to a CW-complex. In other words, with respect to the homotopy that the homotopy groups are able to capture, you can always suppose your space locally looks like a point, a line, a triangle, a tetrahedron... You would not have this theorem with your definitions, at least not with the classical proof (I am convinced that a counterexample van be found). For me, this is "why" we like homotopy groups: because we like CW complexes. There is a whole theory (quillen model category) that explains this stuff.

Also, they are the non commutative analog of the homology of a chain complex. The context in which the two concepts becomes the same thing is the one of stable (infinity) categories, or triangulate categories with a t-structure.

Now we come to your definition. The space of embeddings is indeed studied a lot, but usually when I speak of embedding I think about $X$ being a manifold and an embedding being a smooth embedding. In particular, for $n=1$ and $X= \mathbb{R}^3$ it becomes the space of knots, which is of great interest. But beware: the notion of homotopy must be carried out in the space of embeddings, so at any fixed time you must have an embedding. This translates formally the fact that it's not fair to unknot a knot by "going through" itself. From this picture, you can see that the first "embedding homotopy group" is highly non trivial for $X= \mathbb{R}^3$: for example it's impossible to unknot the treefoil knot. My entire PhD thesis is dedicated to understand a tiny brick of this enormous space. Despite its great importance, it does not have many properties that we expect from a homotopy group, the worst one being it's not homotopy invariant: for example $\mathbb{R}^3 \simeq *$ is contractible, but the space of embeddings of $S^1$ in a point is empty.

Keep having new ideas! That's what a mathematician does. And don't be sad if your idea does not work (it will happen thousands times).

Andrea Marino
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  • Re. counterexample: look up something like the pseudocircle. In fact McCord has shown that every finite complex is weakly equivalent to a finite topological space (and vice versa). Of course $S^n$ cannot embed in any finite space because of its cardinality. – Tyrone Apr 30 '21 at 11:21
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I don't think so, because there would go several of the most interesting aspects of the subject: no more Hopf fibration, no more stable and unstable homotopy. The Hopf-like fibrations, along with Van Kampen, allows for various exciting developments, and these would all be wiped clean off the map.

Thus we shouldn't just consider embeddings of $S^n $ into $X $, for higher homotopy. This would seem rather absurd, throwing away interesting mathematics.

Not to mention, then we would, in all likelihood, lose the Hurewicz theorem.

While Smale always said starting over with fresh definitions was desirable, it just seems that this would flush a whole gorgeous theory down the drain.

That being said, if you could show you have a group structure, somehow, maybe you could get something new.

Speaking of Smale, we went through a paper of his on immersions of spheres with Mladen Bestvina when I was a graduate student at Ucla. Sounds familiar, eh?