2
$\begingroup$

Let $X$ be a path connected topological space and $x_0\in X$. It is known that taking homology class of a loop $f \in \pi_1(X,x_0)$ gives us a homomorphism from the fundamental group to the first homology group. This homomorphism is surjective and its kernel is the commutator of the fundamental group: $N = [\pi_1(X,x_0), \pi_1(X,x_0)]$.
now the question is about the loops inside $N$.

$\mathbf {Theorem}$: a loop $f:(S^1,s_0) \to (X,x_0)$ is in $N$ iff it can be extended to a map $F: \Sigma \to X$ where $\Sigma$ is the sphere with $m\geq 0$ handles and one hole, and $F|_{\partial \Sigma} = f$. moreover, when it is the case, the number of handles, $m$ equals the number of commutators in which $f$ factors.

in order to prove I tried to use a technique found in Hatcher's book, page 167: I presented $f$ as a boundary of some chain of singular simplices of dimension 2. I built a $\Delta$ - complex with a 2 - simplex for every one in the chain, and glued them by the edges which represent a canceling pair in the chain. As Hatcher noted, every edge is in one canceling pair besides one edge, which represents $f$. Then our simplex is a a collection of 2-simplices, such that the all the edges are glued by pairs besides one edge, which represent the loop $f$. of course we can extend $f$ to this simplex by the singular maps in the chain.
so the question becomes why this complex is the sphere with handles and one hole, and why the number of handles is the number of commutators which factor $f$.

I would appreciate any kind of help, and of course a solution.

Thank you very much.

$\endgroup$
1
$\begingroup$

The proof that $\Sigma$ is a sphere with $m \ge 0$ handles and $1$ hole is an application of the classification theorem for surfaces with boundary: for every compact, connected, oriented surface $\Sigma$ there exists $m \ge 0$ and $k \ge 0$ such that $\Sigma$ is a sphere with $m$ handles and $k$ holes, where $k$ equals the number of components of the boundary of $\Sigma$; and your surface $\Sigma$ is a compact, connected surface whose boundary is connected, corresponding to the loop $f$.

The reason that $m$ corresponds to a product of $m$ commutators is because if $\Sigma$ is a sphere with $m$ handles and $1$ hole then $\pi_1(\Sigma)$ is a free group having a free basis consisting of $2m$ elements $a_1,b_1,...,a_m,b_m$, and the boundary of the surface, which corresponds to $f$, is represnted by the following product of $m$ commutators: $$(*) \quad [a_1,b_1] [a_2,b_2] ... [a_m,b_m] $$ To visualize that last point is, again, a tool in the classification of surfaces. Take a polygon with $4m+1$ sides and label the sides as follows: leave one side blank; then label the others in order as $$a_1, b_1, a_1^{-1}, b_1^{-1},...,a_m, b_m, a_m^{-1},b_m^{-1} $$ Put arrows on the labelled sides to indicate direction: a clockwise arrow for a $-1$ exponent; a counterclockwise arrow otherwise. Now identify sides with matching letters, making sure to match up the arrows. The resulting quotient space is the sphere with $m$ handles and $1$ hole. The labelled sides give the free basis elements of the fundamental group of the quotient surface. The unlabelled side gives you the unique boundary component $f$ of the labelled surface. The polygon itself gives you the path homotopy between $f$ and the commutator $(*)$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thank you, can you elaborate on the last statement? why the boundary of the surface is represented by the product of the commutators? $\endgroup$ – D. Hershko Mar 31 at 6:35
  • $\begingroup$ I added the requested elaboration. $\endgroup$ – Lee Mosher Mar 31 at 12:44
  • $\begingroup$ thank you very much $\endgroup$ – D. Hershko Mar 31 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.