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I am going to give a presentation about p-adic numbers. While studying what p-adic numbers even are, I got stuck at the geometric interpretation of the open balls. I have read many times that every point in those balls are considered to be the centrum of the balls. I do believe that one because I could easily follow the mathematical prove. I just cannot interpret it physically...

For example, let say we have the metric space Q_3 with the 3-adic metric. Then we can have a open ball of radius (1/3)^0. In that ball we have three identical disjoint open balls of radius (1/3). And in each of those balls we again have three open balls with radius (1/3)^2 etc. See the picture :D But then we can interpret those balls as a graph. The vertex in the middle is the ball with radius one. The vertices to the right and bottom left are the three balls with radius (1/3). AND NOW MY PROBLEM: I dont understand what the vertex on the top left is. Can someone please make this clear to me?

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  • $\begingroup$ I got it from this YouTube video: youtube.com/… At time 15:43, you can see the picture. This guy is also explaining why there is vertex on the left bottom. Unfortunately I still do not get him. @RobArthan $\endgroup$ – esmo Apr 9 '17 at 7:45
  • $\begingroup$ In the tree representation every vertex has degree 4 if it forks by 3 "going down". We need an edge to connect "up to the top" $\endgroup$ – Henno Brandsma Apr 9 '17 at 9:35
  • $\begingroup$ Aaaahhhhh now I do get it! Super. Thank you @HennoBrandsma $\endgroup$ – esmo Apr 9 '17 at 9:43
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In any ultrametric space (so a metric space $(X,d)$ with the strengthened triangle inequality $d(x,z) \le \max(d(x,y), d(y,z))$ for all $x,y,z$) any point $y$ for an open ball $B(x,r)$ is its "centre": $B(y,r) = B(x,r)$:

$z \in B(y,r)$, then $d(x,z) \le \max(d(y,z), d(x,y)) < r$, as $d(x,y) < r$ from $y \in B(x, r)$, and $d(y,z) < r$ from $z \in B(y,r)$. So $z \in B(x,r)$. The proof of the reverse: $z \in B(x,r)$ is in $B(y,r)$ is entirely similar. That makes it harder to interpret these balls geometrically, I think.

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  • $\begingroup$ Yes I did understand this prove when I saw it the first time. But I do not understand the graph in the picture. @Henno Brandsma $\endgroup$ – esmo Apr 9 '17 at 9:03
  • $\begingroup$ @esmo The key is seeing the graph isomorphism with the downward forking tree. Topologically we get a Cantor set really. $\endgroup$ – Henno Brandsma Apr 9 '17 at 9:34
  • $\begingroup$ Very similarly you get that every triangle in an ultrametric space is an equilateral triangle. From that you see directly (kind of geometrically) that every point of an open Ball must be its centre. $\endgroup$ – Rudi_Birnbaum Mar 8 '18 at 21:22

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