Higher Order π Groups, The 1 Complex

π Groups of the 1 Complex

Let S be a connected 1 dimensional cw complex. Let f map Sⁿ (n > 1) into S. Thus f represents a homotopy class of the n sphere in S.

Call the image T, a subspace of S. Since the domain is compact and the range is hausdorff, T is closed. Its intersection with every edge is closed. Suppose an intersection includes p, but does not include points on either side of p. This is a disconnected region of T. However, the image of T is connected, so this can't happen at all; or else all of T fits in one closed portion of the interval. In any case, we only need introduce at most 2 new endpoints to indicate where T starts and/or stops along the interval. Make this refinement for every interval, and T becomes a subcomplex of S. In fact we can consider T on its own, as a 1 complex.

Since T is the image of a compact set it is compact, and requires finitely many cells. Thus T is a finite cw complex.

Suppose T contains a cycle, in the graph theoretic sense. Pull the vertices back, through f, to the sphere. (Choose any point in the preimage of each vertex; it doesn't matter.) Let a chain of arcs in the sphere pass through these points, and define a cycle, as a graph, and in homology. Since the sphere is simply connected, this is 0 in h₁. Apply f, and the induced homology homomorphism must take 0 to 0. Thus the chain of arcs in T becomes 0 in h₁(T). Compute h₁(T) using the cellular boundary operator. Our cycle is indeed a cycle, and there is no boundary from above, hence this is a nontrivial element of h₁. This is a contradiction, hence T has no cycles, and is a tree.

Select any vertex as the root, and retract all of T onto its root. (This is where we need T to be a finite cw complex.) The resulting function, which is in the same homotopy class as f, maps the entire sphere onto a point. Finally, since S is connected, any point can slide along to any other point. There is but one homotopy class of the sphere, and it is trivial.

The same result holds for pointed spaces. Make sure the image of the base point of the sphere becomes a vertex in T, and make it the root of the tree.

Universal Cover

Here is a similar proof. Turn S into a simplicial complex, and build a simply connected universal cover U, which is also a 1 dimensional simplicial complex. Rewrite f as g into U, followed by the projection p into S. Reason as above to show the image of g is contractable to a point. Project this homotopy down to S, and f is contractable to a point. From there the proof is the same.

Circle

Since the circle S¹ is a 1 complex, its π groups are known. π₁ = Z, and all higher order π groups are trivial.

π₁ and h₁

Since the higher order π groups of S are trivial, we only need find π₁(S) to characterize the homotopy groups of S. While we're at it, let's derive the homology of S; it's not hard.

Let S be a 1 complex. Let p be a point in S. If p is part of an edge, the edge containing p is an open path connected set. If p is a vertex, bring in all the open edges incident to p and find an open path connected set. Therefore S is locally path connected. The components of S are the path components of S, and we may as well analyze them one component at a time. Therefore, assume S is connected.

We know that h₀ = Z. Since there are no 2 cells, h₂ and beyond is 0. We only need evaluate h₁.

Let T be a spanning tree for S, and build the cycle basis. This is generated by the chords of S, i.e. the edges of S that are not in T.

Let r be any vertex of T, which will act as the root of the tree. Let a path start at r and connect to any other vertex in T. This path is compact, and touches finitely many cells. Every point of T is finitely many steps away from r. Use this to contract T onto r. For the first half second, all the edges adjacent to r contract linearly onto r. Vertices that were one unit away from r now coincide with r. Repeat this step, moving twice as fast. Now everything two steps away hass been pulled onto r. Contract again and again, geometrically faster, so that after one second all of T shrinks down to r. As the spanning tree shrinks to a point, Each chord becomes its own loop. This retract of S has the same homology, and the same homotopy classes. Therefore h₁ is a free Z module, with each loop acting as a generator. Similarly, π₁(S) is the free group with these loops as generators.

Remember that the retraction is an isomorphism, in homology or in homotopy. So in the original complex S, the cycle basis, running through the spanning tree, generates π₁ and h₁.

A Different Cycle Basis

Consider another set of cycles that acts as a cycle basis relative to T. Contract T to a point and each generator is a set of loops. It isn't one generator one loop any more. One generator could be the loop A, and another could be A+B. In terms of homology, each generator is a sum of chords, where the chords are known to be a basis for h₁. assume the rank is n, whence are alternate cycle basis must also have rank n. In other words, n generators are involved. These span the chords mod 2, but they need not span the entire Z module. You have to go to four dimensions, i.e. four chords, to see this. Let the generators form a matrix of all ones, except for zeros down the main diagonal. This has a determinant of -3, which is 1 mod 2, but is not a unit in the integers. Certain elements of h₁ are not spanned. Best to stay with the standard cycle basis; one cycle for each chord.

π₁(S) fairs no better. Label the chords of S as A B and C. Build a cycle basis: A+B, B+C, and A+B+C. This time the algebra is noncommutative. The first generator becomes AB in the free group of S, not BA. Let the generators be AB, BC, and BAC. After some trial and error, you can show that A is not accessible from these three generators. (You can find a more formal proof here.) Our alternate cycle basis spans a proper subgroup of the fundamental group of S.

k Connected

A space S is k connected if π_i(S) = 0 for i ≤ k. Thus 1 connected is path connected and simply connected.