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Does this mean that the first homotopy group in some sense contains more information than the higher homotopy groups? Is there another generalization of the fundamental group that can give rise to non-commutative groups in such a way that these groups contain more information than the higher homotopy groups?

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Interesting question. I don't think that there's necessarily less information in an abelian group: the commutativity of the higher groups is really a consequence of a geometrical fact about spheres, and not really a restriction. Might also be worth considering the relative homotopy groups: $\pi_1$ becomes a set and $\pi_2$ non-abelian. But these are just thoughts, not a real answer. –  Paul VanKoughnett Mar 4 '11 at 2:52

4 Answers 4

Thinking about the higher homotopy groups as just groups is in some sense missing the point. The higher homotopy groups are not just abelian groups: they are $\pi_1$-modules, for one thing.

More loftily, from the n-categorical point of view, the homotopy groups are really just a convenient stand-in for a more fundamental structure, the fundamental $\infty$-groupoid of a space. Roughly speaking, the fundamental $\infty$-groupoid is a gadget that incorporates information about the paths between points, homotopies between paths, homotopies between homotopies between paths, and so forth.

It is possible to truncate the fundamental $\infty$-groupoid into a collection of easier-to-understand objects, the fundamental $n$-groupoids $\Pi_n$:

  • The fundamental $0$-groupoid $\Pi_0$ is just the set of connected components.
  • The fundamental $1$-groupoid is the groupoid of homotopy classes of paths between points; it is a generalization of the fundamental group that is independent of basepoint. If the space is connected, the fundamental $1$-groupoid is equivalent to the category with a single object whose morphisms are the elements of the fundamental group $\pi_1$.
  • The fundamental $2$-groupoid is the $2$-groupoid of paths and homotopy classes of homotopies between them; it is a generalization of the action of $\pi_1$ on $\pi_2$ that is independent of basepoint.

And so forth: more generally the fundamental $n$-groupoid is a generalization of the relationship between the first $n$ homotopy groups. Unfortunately I can't think of a nice reference to these ideas off the top of my head; I've gleaned them from several sources. The references in Baez and Shulman's Lectures on n-categories and cohomology might be a good start.

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Although, if the space is simply connected the $\pi_1$-module structure doesn't give any information. –  Grumpy Parsnip Mar 5 '11 at 1:43
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A generalization of the $\pi_1$-action is the graded quasi-Lie algebra structure coming from the Whitehead product, which takes the form $\pi_m \otimes \pi_n \rightarrow \pi_{m+n-1}$. This structure (tensored with $\mathbb{Q}$) is what Sullivan exploits for his fundamental theorem of rational homotopy theory, that taking a simply-connected space (up to rational homotopy equivalence) to its commutative $\mathbb{Q}$-dga induces an equivalence of categories. –  Aaron Mazel-Gee Mar 5 '11 at 19:39
    
The notion of a fundamental 2-groupoid of a space is more delicate; see the paper Hardie, K. A.; Kamps, K. H.; Kieboom, R. W. A homotopy 2-groupoid of a Hausdorff space. Papers in honour of Bernhard Banaschewski (Cape Town, 1996). Appl. Categ. Structures 8 (2000), no. 1-2, 209–234. and the papers which follow from this. However one can define a fundamental 2-groupoid of a pair of spaces, or more generally a map of spaces. –  Ronnie Brown Apr 13 '12 at 14:21

Let $\mathcal{C}$ be a category with finite limits and a final object. In general, if $Y$ is an object in $\mathcal{C}$ such that $\hom(X, Y)$ is naturally a group for each $X \in \mathcal{C}$, then $Y$ is called a "group object" in $\mathcal{C}$; that is, there is a multiplication map $Y \times Y \to Y$ and an inversion $Y \to Y$ and an identity $\ast \to Y$ (for $\ast$ the final object) that satisfy a categorical version of the usual group axioms (stated arrow-theoretically). In the case of interest here, $\mathcal{C}$ is the homotopy category of pointed topological spaces, and the statement that the homotopy groups are groups is the statement that the spheres $S^n$ are group objects in the opposite category -- in other words, $S^n$ is a so-called "H cogroup." When one writes out the arrows, one ends up with a "comultiplication map" $S^n \to S^n \vee S^n$ and a map $S^n \to \ast$ ($\ast$ the point) that satisfy the dual of the usual group axioms, up to homotopy.

The reason that the higher homotopy groups are abelian and $\pi_1$ is not is that $S^n$ is an abelian H cogroup for $n \geq 2$ and not for $n=1$. This is basically a consequence of the Eckmann-Hilton argument (namely, there are two natural and mutually distributive ways of defining the H cogroup structure of $S^n$, depending on which coordinate one chooses; they must be equal and both commutative).

Now to your more general question. So one can define covariant functors from the pointed homotopy category to the category of groups: just pick any H cogroup object and to consider maps from it into the given space. An easy way of getting these is to take the reduced suspension of any space $X$, $\Sigma X$, and to note that $\Sigma X$ can be made into an H cogroup (in kind of the same way as $S^n$ is---actually, the $S^n$ is a special case of this).

One may object that considering suspensions is not really anything new, because homotopy classes of $\Sigma X = S^1 \wedge X$ into a space $Y$ is the same as considering homotopy classes of maps $S^1 \to Y^X$ when $X$ is reasonable (say locally compact and Hausdorff), so we really have a variant of the fundamental group.

Finally, there is the question of whether all functors from the pointed homotopy category to the category of groups can be expressed in this way, that is, whether it is representable. On the pointed homotopy category of CW complexes, there are fairly weak conditions that will ensure representability.

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For $hom(X,Y)$ to be a group, isn't it $Y$ that you need to be a group object (or $X$ to be a cogroup object)? By definition of a product, for $f,g\in hom(X,Y)$ you get $X\stackrel{f\times g}{\rightarrow} Y\times Y \stackrel{\mu}{\rightarrow} Y$. –  Aaron Mazel-Gee Mar 4 '11 at 18:23
    
@Aaron: Dear Aaron, thanks for the correction. –  Akhil Mathew Mar 5 '11 at 1:21

These problems puzzled the early topologists: in fact Cech's paper on higher homotopy groups was rejected for the 1932 Int. Cong. Math. at Zurich by Hopf and Alexandroff, who quickly proved they were abelian. We now know this is because group objects in groups are abelian groups. However group objects in the category of groupoids are NOT just abelian groups, but are equivalent to crossed modules, which occurred in the 1940s in relation to second relative homotopy groups, $\pi_2(X,A,x)$. It turns out that there is a nice double groupoid $\rho_2(X,A,x)$ consisting of homotopy classes of maps of a square $I^2$ to $X$ which map the edges to $A$ and the vertices to $x$. (The proof that the compositions are well defined is not quite trivial!). Using this Philip Higgins and I proved a 2-d van Kampen Theorem, published in Proc. LMS 1978, i.e. 34 years ago, from which one can deduce new results on the nonabelian second relative homotopy groups, as crossed modules over the fundamental group.

This is the start of using strict higher homotopy groupoids for obtaining nonabelian calculations in higher homotopy theory -- see the web page in my comment, and references there.

This idea came from examining in 1965 a proof of the 1-dim van Kampen theorem for the fundamental groupoid, and observing that it ought to generalise to higher dimensions if one had the right homotopical gadgets. It took years to get the idea that this could be done for pairs of spaces, filtered spaces, or $n$-cubes of spaces, but apparently not easily just for spaces, or spaces with base point.

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One geometrical fact from which the non-commutativity of the fundamental group is the following: two objects on a line can't switch relative position (i.e. left and right) through homotopy, as they are unable to "pass" each other. Two objects in a higher dimensional space can, however; so intuitively, it seems that a homotopy theory based on mappings of $I^n$ will naturally be abelian for $n \ge 2$.

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this seems off-topic. OP is not looking for an intuitive explanation for why $\pi_n$ is abelian for $n \ge 2$. –  Soarer Mar 4 '11 at 5:40
    
@Soarer This is not off topic: see my answer in preparation, on using higher homotopy groupoids of filtered spaces or $n$-cubes of spaces, and my page Higher dimensional group theory –  Ronnie Brown Apr 13 '12 at 11:57

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