Timeline for Cycles covering the edges of the graph corresponding to the Van Kampen diagram of a presentation of a group
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jul 22, 2015 at 7:57 | comment | added | Sh.M1972 | No, any Van Kampen diagram over a given presentation is in fact a geometric figure for deducing a "result" of relators. So, as Derek Holt mentioned above, if we consider the diagram $\Gamma$ which has just two cells with boundary labels $R_1$ and $R_2$, then this diagram will not have any information about $R_3$. | |
Jul 15, 2015 at 9:23 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
deleted 87 characters in body
|
Jul 15, 2015 at 9:07 | comment | added | Alireza Abdollahi | @DerekHolt: Yes. This answers negatively my question: You take the graph $\Gamma$ as follows $1 \overset{a}{\longrightarrow} 2$, $1 \overset{b}{\longrightarrow} 2$, $1 \overset{c}{\longrightarrow} 2$, $1 \overset{d}{\longrightarrow} 2$. A graph with two vertices $1,2$ and four directed edges between these two vertices. Are there examples in which we have no multiple edges between vertices? It depends to how one choose the graph $\Gamma$. If one consider the graph $\Gamma$ with three disjoint cylces, then the answer to my question is positive. Anyway, thanks for your try to answer my question | |
Jul 15, 2015 at 8:17 | comment | added | Derek Holt | I am still unsure whether I understand. Let $G = \langle a,b,c,d \mid ab^{-1},bc^{-1},cd^{-1} \rangle$ and conisder a van Kampen diagram for $ad^{-1}$ with three interior regions. Can we not take $\mathcal{C}$ to be the cycles labelled $R_1=ab^{-1}$ and $R_2=cd^{-1}$? | |
Jul 15, 2015 at 7:56 | comment | added | Alireza Abdollahi | @HJRW: An example of the graph $\Gamma$: upload.wikimedia.org/wikipedia/commons/thumb/8/81/Abelian.jpg/… | |
Jul 15, 2015 at 7:49 | comment | added | Alireza Abdollahi | @DerekHolt: What I mean by the graph $\Gamma$: it is a directed labelled graph in which we have at least one cycle corresponding to each relation $r_i$. The graph $\Gamma$ is not unique: e.g., one may consider $m$ disjoint cycles each of them corresponds to a relation $r_i$; or one may glue properly the cycles to have a graph with less number of connected components. May be referring to the Van Kampen diagrams is not necessary and confusing; I am trying to draw such a graph $\Gamma$ for an specific example but at the moment I do not know how to do it in MathJax... | |
Jul 15, 2015 at 7:39 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
[Edit removed during grace period]
|
Jul 15, 2015 at 7:24 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
added 688 characters in body
|
Jul 14, 2015 at 21:13 | comment | added | HJRW | I too am confused. A van Kampen diagram is a geometric proof that one element $\gamma$ of the free group on the generators $x_1,\ldots,x_n$ maps to the trivial element in $G$. It does not tell you anything about the whole of $G$. Perhaps you meant 'presentation complex'? | |
Jul 14, 2015 at 18:10 | comment | added | Derek Holt | I am confused. $\Gamma$ might not involve all of the relations $r_1,\ldots,r_m$. For example $\Gamma$ could just consist of a single loop labelled $r_1$, in which case $k=1$ and $R_1=r_1$. | |
Jul 14, 2015 at 14:09 | history | edited | Alireza Abdollahi | CC BY-SA 3.0 |
added 90 characters in body
|
Jul 14, 2015 at 13:45 | history | asked | Alireza Abdollahi | CC BY-SA 3.0 |