Classnotes for January 10

(Difference between revisions)
 Revision as of 22:04, 22 Feb 2006Branimir (Talk | contribs)Group-Theoretical Defintions - corrected error← Go to previous diff Revision as of 22:04, 22 Feb 2006Branimir (Talk | contribs) Group-Theoretical DefintionsGo to next diff → Line 18: Line 18: - '''Defintion:''' ''Let $G_{1}$, $G_{2}$, and $H$ be groups such that there exist homomorphisms $\phi_{1} : H \rightarrow G_{1}$ and $\phi_{2} : H \rightarrow G_{2}$. Then the '''free product of $G_{1}$ and $G_{2}$ with amalgamated subgroup $H$''', denoted by $G_{1} \star_{H} G_{2}$, is defined to be $G_{1} \star G_{2}$ together with the additional relation that for every $h \in H$, $\phi_{1}(h)=\phi_{2}(h)$.'' + '''Defintion:''' ''Let $G_{1}$, $G_{2}$, and $H$ be groups together with homomorphisms $\phi_{1} : H \rightarrow G_{1}$ and $\phi_{2} : H \rightarrow G_{2}$. Then the '''free product of $G_{1}$ and $G_{2}$ with amalgamated subgroup $H$''', denoted by $G_{1} \star_{H} G_{2}$, is defined to be $G_{1} \star G_{2}$ together with the additional relation that for every $h \in H$, $\phi_{1}(h)=\phi_{2}(h)$.''

Group-Theoretical Defintions

The goal of this lecture is to state van Kampen's theorem (also known as the Seifert-van Kampen theorem), give a brief explanation of its plausibility, and supply several examples of its use. In order to do this, we need to first define the group-theoretical notion of the free product of two groups.

Definition: Let (G1, * 1) and (G1, * 2) be groups. Then the free product of (G1, * 1) and (G2, * 2), denoted by $G_{1} \star G_{2}$, is defined to be the group whose elements are words with letters in $G_{1} \sqcup G_{2}$ (i.e. the disjoint union of G1 and G2) and whose multiplication is the concatenation of words, with the empty word being the identity e, together with the relations:

1. $e_{G_{1}} = e = e_{G_{2}}$, where $e_{G_{i}}$ is the identity of (Gi, * i),i = 1,2;
2. for i = 1,2, if $a,b \in G_{i}$ then a * b = a * ib, where * without subscript denotes the multiplication of $G_{1} \star G_{2}$.

This definition can be extended readily to arbitrary finite free products by induction. One can alternatively define free products by means of a universal property (with uniqueness up to group isomorphism), and provide a rigorous construction to prove their existence. For this approach, see here (http://planetmath.org/encyclopedia/FreeProduct.html).

To see what a free product looks like, let us consider $\mathbb{Z} \star \mathbb{Z}$. If we call the generator of the first copy of $\mathbb{Z}$ by a, and the generator of the second copy by b, an example of an element of $\mathbb{Z} \star \mathbb{Z}$ would be a2ba − 3a, and an example of a calculation following the relations outlined in the definitions of a free product is a2ba − 3a * a2ba = a2ba − 1ba.

As shall soon be explained, it turns out that the concept of the free product is not quite sufficient for the statement of van Kampen's theorem. What we shall need instead is a slight modification, called the free product with amalgamated subgroup:

Defintion: Let G1, G2, and H be groups together with homomorphisms $\phi_{1} : H \rightarrow G_{1}$ and $\phi_{2} : H \rightarrow G_{2}$. Then the free product of G1 and G2 with amalgamated subgroup H, denoted by $G_{1} \star_{H} G_{2}$, is defined to be $G_{1} \star G_{2}$ together with the additional relation that for every $h \in H$, φ1(h) = φ2(h).

Once again, one can define free products with amalgamated subgroup by means of a universal property (with uniqueness up to group isomorphism), and provide a rigorous construction to prove their existence. For this approach, see here (http://planetmath.org/encyclopedia/FreeProductWithAmalgamatedSubgroup.html).

Van Kampen's Theorem

We are now in a position to finally state van Kampen's theorem:

Theorem (van Kampen): Let X be a topological space, let $U_{1},U_{2} \subset X$ be open, and suppose that $X = U_{1} \cup U_{2}$ and that $U_{1} \cap U_{2}$ is path-connected. Let $b \in U_{1} \cap U_{2}$, and for i = 1,2 let $\iota_{i} : U_{1} \cap U_{2} \rightarrow U_{i}$ be the inclusion map. Then $\pi_{1}(X,b) \approx \pi_{1}(U_{1},b) \star_{\pi_{1}(U_{1} \cap U_{2},b)} \pi_{1}(U_{2},b)$, where, for i = 1,2, the monomorphism from $\pi_{1}(U_{1} \cap U_{2},b)$ to π1(Ui,b) is taken to be i) * .

Let us see how this is plausible on an intuitive level. If we are given a loop in X with basepoint b in $U_{1} \cap U_{2}$, then, since $U_{1} \cap U_{2}$ is path-connected, one can apply a homotopy to the portion of the loop within $U_{1} \cap U_{2}$ to decompose the loop, as it were, into the concatenation of several loops, each lying entirely within U1 or U2. This motivates π1(X,b) being some sort of free product of π1(U1,b) and π1(U2,b). If we are given a loop which lies entirely within $U_{1} \cap U_{2}$, then we would like for it to correspond to a single element of π1(X,b), whether we view the loop as residing in U1 or in U2. In this manner, we see that the additional relations which distinguish the free product with amalgamated subgroup from the free product are entirely natural and, in fact, forced upon us, if we wish to avoid double counting of homotopy classes of loops residing entirely in the intersection. Thus, van Kampen's theorem is seen to be an entirely plausible result, but, as we shall see, this does not mean that the proof will be all that easy.

Examples

To provide further motivation for study of van Kampen's theorem, we now examine a number of examples of its application.

The Figure Eight $X = S^{1} \vee S^{1}$

Our open subsets U1 and U2 are provided by taking the one or the other copy of S1 together with a small neighbourhood of the join point b. Thus U1 and U2 are both readily homotopic to S1, the homotopy consisting of the retraction of the "antennae", whilst $U_{1} \cup U_{2}$, which is path-connected and readily homeomorphic to $\times$, and thus homotopic to a point by the obvious retraction. Thus, for i = 1,2, $\pi_{1}(U_{i},b) \approx \pi_{1}(S^{1}) \approx \mathbb{Z}$, whilst $\pi_{1}(U_{1} \cap U_{2},b)$ is trivial. Hence $\pi_{1}(U_{1} \cap U_{2},b)$ cannot impose any new relations on the free product $\pi_{1}(U_{1},b) \star \pi_{1}(U_{2},b)$, so, by van Kampen's theorem, $\pi_{1}(X,b) \approx \pi_{1}(U_{1},b) \star_{\pi_{1}(U_{1} \cap U_{2},b)} \pi_{1}(U_{2},b) = \pi_{1}(U_{1},b) \star \pi_{1}(U_{2},b) \approx \mathbb{Z}\star\mathbb{Z}$

Record of the Lecture

Scanned lecture notes

Available in PDF format.

Audio recording

1st hour lecture recording

2nd hour lecture recording