Today I’m going to be studying this paper by Theodor Christian Herwig to learn about $p$-adic analysis.

An absolute value on a field $\Bbb{K}$ is a map $|.|: \Bbb{K}\to \Bbb{R}_+$ which satisfies the usual absolute value conditions; namely $|x|\geq 0$ and $|x|=0\iff x=0$; $|xy|=|x||y|$, and $|x+y|\leq |x|+|y|$. An absolute value which is non-archimedian also satisfies the following additional property: $|x+y|\leq\max\{|x|,|y|\}$.

Why’re we doing all this? Why are we trying to define a function, that serves the purpose of a norm in most settings, on an algebraic object that might not have any such structure? This is because we want to do topology on such algebraic objects. We want to be able to study a particular object from as many angles and perspectives as we want. An analogy would be representation theory: trying to study groups using properties from Linear Algebra.

Why “non-archimedian” though? Where does this term even come from? Archimedian means that for any $\epsilon>0$, we can construct an $n\in\Bbb{B}$ such that $\frac{1}{n}<\epsilon$. In other words, we can construct arbitrarily large integers. When we have a non-Archimediam valuation, what we’re saying is that we cannot build arbitrarily large integers. For example, let $N_1$ and $N_2$ be positive integers. Then we should have $N_1+N_2>\max\{N_1,N_2\}$. However, this is not the case. Numbers seem to behave “funny” here. This is all that non-Archimedian implies.

A $p$-adic valuation $v_p:\Bbb{Z}-\{0\}\to \Bbb{R}$ is defined in the following way: $v_p(a)$ is the largest power of prime $p$ that divides $a$. This map can be extended to $\Bbb{Q}^\times$ in the natural way: $v_p(a/b)=v_p(a)-v_p(b)$.

We can see that $v_p$ has the following two properties: $V_p(xy)=v_p(x)+v_p(y)$ and $v_p(x+y)\geq \min\{v_p(x),v_p(y)\}$. Clearly, these properties are not suggestive of the kind of non-archimedian absolute value properties we’re looking for. We shall rectify that now.

Let $|x|_p=p^{-v_p(x)}$. Hence, if $p^n|x$, where $n$ is the highest power of $p$ that divides $x$, then we map $x\to \frac{1}{p^n}$. Is this the non-archimedian absolute value that we were looking for? Yes. Just checking for the non-archimedian property, we see that $|x+y|=\max\{|x|,|y|\}$ when at least one of $x$ and $y$ is a multiple of $p$, and is less than $\max\{|x|,|y|\}$ if neither are multiples of $p$ but their sum $x+y$ is.

Let us now solve an exercise that illustrates how this absolute value metric works really well. Find a sequence of real numbers that converges to $32$, $7$-adically. Answer- Consider the sequence $\{a_n\}_{n\in\Bbb{N}}=\{32+7^n\}_{n\in\Bbb{N}}$. This is clearly a divergent sequence under the Euclidean metric- a metric that we’re most used to. However, here we see that $|a_n-32|_p=|7^n|_p=\frac{1}{7^n}$. Clearly, $|a_n-32|\to 0$. Hence, we see that this sequence does indeed converge to $32$ under the $7$-adic norm.

Wait. Norm? How do we know that the $p$-adic absolute value function is a norm? We’ll check for triangle inequality. Clearly $|x-y|\leq \max\{|x-z|,|z-y|\}$. Hence, as both $|x-z|$ and $|z-y|$ are positive numbers, it follows that $|x-y|\leq |x-z|+|z-y|$. Hence, the $p$-adic absolute value function is indeed a norm.

From this point on, $|x|$ will be the $p$-adic norm on $x$, and not the Euclidean norm. Now we shall prove that if $|x|\neq |y|$, then $|x+y|=\max\{|x|,|y|\}$. There are two major steps in this proof. The first is to prove that $|-x|=|x|$. How do we see that? If $p^n|x$ then $p^n|-x$ too. Hence, $v_p(x)=v_p(-x)$. Therefore, $|-x|=p^{v_p(-x)}=p^{v_p(x)}=|x|$.

The second “trick” in this proof is the following: let $|y|>|x|$. Now $|y|=|y+x-x|\leq \max\{|y+x|,|-x|\}=\max\{|y+x|,|x|\}$. As $|y|>|x|$, we must have $|y|\leq |x+y|$. Now remember that by the non-archimedian property of this norm, we have $|x+y|\leq |y|$ (note that $\max\{|x|,|y|\}=|y|$). We’re therefore done.

A corollary from the above theorem says that in such a $p$-adic space, every triangle is isoceles. What does this mean? Say we have three points $x,y$ and $z$. Why do at least two of $|x-y|,|x-z|$ and $|y-z|$ have to be the same? Assume that there are two sides that are not the same. Take the third side; which is the sum of the two. Now use the theorem proved above. In fact, we can conclude something stronger: the sides of the isoceles triangle that are of the same length are of a length greater than or equal to the third side.

Now we will begin a study of some non-trivial properties of the $p$-adic metric space.

•  If $a\in B(x,r)$, then $B(a,r)=B(x,r)$. This goes completely against any intuition that the Euclidean metric may have given us. We will now prove this non-intuitive fact. Let $p\in B(x,r)$. Then $|x-p|. Let us now calculate $|a-p|$. We know that $|a-p|\leq\max\{|a-x|,|x-p|\}. Hence, we have $B(a,r)\subset B(x,r)$. Similarly, we can prove that $B(x,r)\subset B(a,r)$, which would then imply that $B(x,r)=B(a,r)$.
• Any open set (ball) $B(x,r)$ is closed. How do we see that? Take any boundary point $b$ of $B(x,r)$. Then for any $s>0$, we have $B(b,s)\cap B(x,r)\neq\emptyset$. Now let $p\in B(b,s)\cap B(x,r)$. Then $|x-b|\leq\max\{|x-p|,|p-b|\}=\max\{r,s\}$. We can make $s. Hence, we have $|x-b|, which would prove that $b\in B(x,r)$.
• If $a,b\in\Bbb{K}$, and for $r,s>0$ we have $B(a,r)\cap B(b,s)\neq\emptyset$, then $B(a,r)\subset B(b,s)$ or $B(b,s)\subset B(a,r)$. In fact, even the converse is true. How do we see this? For any $p\in B(a,r)\cap B(b,s)$, we have $B(p,r)=B(a,r)$ and $B(p,s)=B(b,s)$. If $s, then $B(p,s)\subset B(p,r)$, which implies $B(b,s)\subset B(a,r)$. Similarly, if $r, we have $B(a,r)\subset B(b,s)$. What if $s=r$? Then we have $B(a,r)=B(b,s)$. The converse is obvious. If one set is a subset of another, then their intersection is trivially non-emptyset, assuming the sets are non-empty too.

We will discuss the completion of $\Bbb{Q}$ to $\Bbb{Q}_p$ in the next post.