Timeline for Deciding when an infinite graph is connected
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Apr 5, 2010 at 1:33 | comment | added | Igor Belegradek | @damiano, I do not see how the case of arrangements in the ball is any easier, but yes, it would be very helpful if by some miracle there is a algorithm checking connectedness in this case. | |
| Apr 5, 2010 at 1:23 | comment | added | damiano | @IB: thanks for the clarification to the comment I erased! Indeed, if the intersections have to happen within a unit ball, then things can be very intricate. Btw, would a reasonable answer to this question be enough for you? (Not that I know how to solve this one, just curiosity.) | |
| Apr 5, 2010 at 1:19 | comment | added | Igor Belegradek | @damiano: actually you are right, this is what happens in $\mathbb C^n$, but things become different once we only pay attentions to intersections inside the unit ball, so in my comments above substitute $\mathbb C^n$ by the unit ball in $\mathbb C^n$ with the complex hyperbolic metric. | |
| Apr 5, 2010 at 1:05 | comment | added | damiano | @IB: sorry, i had erased my comment, since it did not seem relevant! It simply seemed to me that two hyperplanes either intersect or they are parallel and as soon as two intersect, then any other hyperplane must intersect one of the two, since it cannot be parallel to both. | |
| Apr 5, 2010 at 1:00 | comment | added | Igor Belegradek | @damiano: complex hyperplanes have codimension $2$, so I do not see why connectedness comes for free. In truth I have thought more of the case when $\mathbb C^n$ is replaced with the unit ball in $\mathbb C^n$ with the complex hyperbolic metric, and in this case I have tons of examples of arrangements whose correspondning graphs aren't connected. | |
| Apr 5, 2010 at 0:37 | comment | added | Igor Belegradek | Thanks! I really do not know much about the graph beyond what I stated. There is a closely related problem that is easier to explain. There the graph arises from a locally finite arrangement of hyperplanes in $\mathbb C^n$, where vertices correspond to hyperplanes, and two vertices are adjacent if and only if the hyperplanes intersect. This graph need not not locally finite though. | |
| Apr 4, 2010 at 19:07 | comment | added | François G. Dorais | I guess a much simpler example has vertex -17 and also vertex s when T halts in exactly s steps; no edges. However, this feels even more like cheating than the above example. Regardless, I think the question would have a much better answer if the the relevant graphs were more thoroughly explained. | |
| Apr 4, 2010 at 18:47 | history | edited | François G. Dorais | CC BY-SA 2.5 |
fixed typos
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| Apr 4, 2010 at 18:30 | history | edited | François G. Dorais | CC BY-SA 2.5 |
new example, now locally finite
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| Apr 4, 2010 at 17:30 | history | answered | François G. Dorais | CC BY-SA 2.5 |