Geometric picture of $\mathbb{Q}_p$

Question

So when we complete $\mathbb{Q}$ wrt to the Euclidean abs. value, we have a nice geometric picture of filling a line (the real line) with the missing points that are almost everywhere. Is there an nice geometric analogue so that we can picture the completion $\mathbb{Q}_p$ wrt to the $p$-adic abs. value?

Answer 1

Here is a picture, described in words.

First, a general element of $\mathbb Q_p$ can be written as a bi-infinite sequence of the form $$\ldots a_i \ldots a_1 a_0 . a_{-1} a_{-1} \ldots a_{-k} \ldots $$ where $a_i \in \{0,\ldots,p-1\}$ for each $i$, and $a_i$ is eventually zero as one goes far enough to the right (i.e. there exists $j$ such that $a_i = 0$ if $i<j$). What that notation represents, of course, is the standard infinite series representation of an element of $\mathbb Q_p$: $$\sum_{i \ge j} a_i p^i $$ Think of that notation like decimal notation where $a_i \in \{0,\ldots,9\}$ for each $i$, except that in decimal notation one instead requires that that the $a_i$'s are all zero if one goes far enough to the left, and instead of a "decimal point" we have something called a "$p$-adic point".

Now let's turn that sequence notation into a picture, which I invite you to draw. It will be a drawing of an infinite tree. The "ends" of this tree will form the field $\mathbb Q_p$

I'll start by describing the $\mathbb Z_p$ tree whose ends form the ring of integers $$\mathbb Z_p = \left\{\sum_{i \ge 0} a_i p^i\right\} $$ First draw a single vertex labelled $.0$

Then draw $p$ segments extending upwards from the vertex $.0$ whose upper vertices are labelled by appending $0,1,...,p-1$ to the label $.0$ giving labels $$0.0 \qquad 1.0 \qquad 2.0 \qquad\ldots\qquad (p-1).0 $$ From each of those vertices draw $p$ segments extending upwards, appending $0,1,...,p-1$. For example, extending upwards from $0.0$ are $p$ segments whose upper vertices are labelled $$00.0 \qquad 10.0 \qquad 20.0 \qquad\ldots\qquad (p-1)0.0 $$ and extending upwards from $1.0$ are $p$ segments whose upper vertices are labelled $$01.0 \qquad 11.0 \qquad 21.0 \qquad\ldots\qquad (p-1)1.0 $$ After doing this you will now have a total of $p^2$ segments of length $2$ extending upwards from the vertex $.0$

Continue this recursively: from a vertex labelled $a_i....a_0.0$, there are $p$ segments extending upwards labelled $$0a_i....a_0.0 \qquad 1a_i....a_0.0 \qquad 2a_i....a_0.0 \qquad \cdots \qquad (p-1)a_i....a_0.0 $$ Taking the union over $i=0,\ldots,\infty$ one obtains the $\mathbb Z_p$ tree, an infinite tree extending upward from the vertex $0.$

Elements of $\mathbb Z_p$ are in one-to-one correspondence with infinite rays in the $\mathbb Z_p$ tree that extend upwards from their base vertex $0.$.

The actual elements of $\mathbb Z_p$ can themselves can be visualized geometrically as the "space of ends" of the tree. In the case $n=2$ one pretty much gets the standard picture of the middle thirds Cantor set (known to represent infinite trinity numbers of the form $.b_1 b_2 b_3 ...$ where $b_i \in \{0,2\}$, so one obtains $\mathbb Z_2$ from that representation by replacing each $2$ with at $1$ and reflecting the representation across $.$)

One thing that pops out of this picture is the embedding $\mathbb Z \subset \mathbb Z_p$. The additive identity is just the leftmost ray starting from $0.$, namely $\ldots000000.$ More generally the non-negative integers $\mathbb Z_{\ge 0} = \{0,1,2,\ldots\}$ consist of all rays starting at $0.$ that, above some point, always takes the leftmost segment heading upwards, i.e. the segment corresponding to appending another $0$. The negative integers $\mathbb Z_{<0} = \{-1,-2,-3,\ldots\}$ are similarly described, namely all rays starting at $0.$ which, above some point, always take the rightmost segment heading upwards, corresponding to appending $p-1$ (thanks to @TorstenSchoeneberg for this point).

With this description of $\mathbb Z$, it's fun to try to visualize three operations: adding $+1$; additive inversion; and multiplication by $p$.

Using this geometric description of $\mathbb Z_p$, the geometric description of the whole of $\mathbb Q_p$ now proceeds by another recursion extending in the downwards direction.

For the first step of the downward recursion, off to the right of the vertex $.0$ place $p-1$ more vertices in a row, altogether labelled $$.0 \qquad .1 \qquad .2 \qquad\cdots\qquad .(p-1) $$ Extending downwards from each vertex in this row draw $p$ edges ending at a common vertex labelled $.00$ and then atop each of the vertices in this row attach separate copies of the $\mathbb Z_p$ tree extending upwards. One now the $p^{-1} \cdot \mathbb Z_p$ tree, representing $$p^{-1} \cdot \mathbb Z_p = \left\{\sum_{i \ge -1} a_i p^i\right\} $$ And now continue the recursion, representing the increasing sequence of inclusions $$\mathbb Z_p \subset p^{-1} \cdot \mathbb Z_p \subset p^{-2} \cdot \mathbb Z_p \subset $$ as an increasing sequence of inclusions of trees and of their ends.

I'll say the first few words of the 2nd step of the downward recursion: off to the right of the vertex $.00$, place $p-1$ additional vertices, altogether labelled $$.00 \qquad .01 \qquad .02 \qquad \ldots \qquad .0(p-1) $$ and then, atop each of those vertices, attach separate copies of the $p^{-1} \cdots \mathbb Z_p$ tree .........

Answer 2

Of course $\mathbb Z_p$ is a self-similar fractal. The whole is made up of $p$ parts, each with radius shrunk by factor $1/p$, and so on.

Image by Heiko Knospe

The $3$-adic integers are in the gray circle. That is decomposed into three yellow circles (called $0$, $1$, and $2$ mod $3$, if you like). Each of those is decomposed into three green circles (according to mod $9$). Each of those is decompolsed into three blue circles (according to mod $27$). Further subdivisions are not shown in the picture. You have to imagine them.

Topologically, $\mathbb Z_p$ is a Cantor set.

What about the $p$-adic numbers $\mathbb Q_p$? For that we have to go upward as well as downward. The gray circle, together with two more just like it make up a larger circle; $3$-adic numbers with a terms up to a $3^{-1}$ term. Then three of those make a still larger circle; the $3$-adic numbers with terms up to a $3^{-2}$ term. And so on upward. Of course a $3$-adic number has $3^{-N}$ terms for only finitely many $N$.