These pages are a brief written report on the construction of -adic integration. Our final goal is to give sense of integrating continuous function defined from , or at least some particular subsets like , with value in . We develop a theory similar to Riemann integration for real valued functions: we define the integral as the limit of Riemann sums. The main result is Theorem 4.2. which states that for continuous functions the sequence of Riemann sums actually converges to a limit in . We will give particular attention to Bernoulli distribution (in chapter 3) and measures (in chapter 4), because they are the most common and useful. For example they can be used to prove Kummer’s congruences. This work is mostly based on "Koblitz" .
Some topology on Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\textstyle \\}
We will consider the topology induced by the metric on . The standard basis of this topology consists of all sets of the form Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\displaystyle \begin{split} a+(p^N)&=a+p^N\mathbb{Z}_p\\&=\{x\in\mathbb{Q}_p|~x=a+tp^N, t\in\mathbb{Z}_p \}\\&=\{x\in\mathbb{Q}_p|~|x-a|_p\leq p^{-n}\}\\&=B_{p^{-n-1}}(a). \end{split}}
These sets are called open balls. They are, open, but also closed.
There is also a really unexpected property:
Proposition 1. 'Any (open) ball is compact in .'
Proof. Consider the open ball for some and . We will prove that is sequentially compact and therefore (since it is a metric space) compact. We first observe that for any , . In fact, if , then . We can fix such that , so if we expand we get Errore del parser (funzione sconosciuta '\label'): {\displaystyle \label{starequation} \tag{$\star$} x=\sum_{i=-\alpha}^{\infty}{a_ip^i}}
Now we want to prove that any sequence in has a convergent subsequence. To do so let to be a sequence, any of the can be written in the form [starequation]. We observe that for all , is between and , so it can take only possible values. Thus there is a subsequence of elements with for some . Iterating the process, we build a subsequence with , and so on until . Then is clearly a convergent subsequence of the original sequence . Thus is sequencially compact. ◻
We say that a subset of is compact open if it is compact and it is open.
Proof. Suppose
and
. We have:
thus
.
Errore del parser (errore di sintassi): {\textstyle \\}
If
we can apply this result in both directions, thus
. ◻
Proposition 2. 'An open subset of is compact if and only if it is a finite union of disjoint open balls.'
Proof. Finite union of disjoint open balls are both open and compact (because balls are compact!). Conversely, if a subset is open, it is the union of open balls, and because of the compactness we can take just a finite number of them. Now Lemma Lemma 1 is sufficient to conclude. ◻
Recall 1.1. For all , , for some and .
Recall 1.2.For all p, . Therefore, if , then .
Definition 1. Let X and Y be two topological spaces. A map is called locally constant if it is constant in the neighborhood of each point.
Observe that in a connected space, locally constant function are not that interesting, because they have to be globally constant.
Lemma 2. 'Let be a ball, .'
Proof. Suppose
, then
for some
. Thus
for
. We have proven
Conversely if
, then
Thus
and we have the desired equality. ◻
Lemma Lemma 2 means that we can divide a ball of length into (disjoint) balls of length . Repeating this process, Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} a+(p^N)&=\bigcup \limits_{b_0=0}^{p-1}\bigcup \limits _{b_1=0}^{p-1} a+b_0p^N+b_1p^{N+1}+(p^{N+2})\\&=\bigcup \limits _{b_0,...,b_k=0}^{p-1} a+b_0p^N+...+b_kp^{N+k}+(p^{N+k+1}). \end{split}}
So we can split any ball into smaller balls of arbitrary length. This construction is also unique: suppose is an ball and , with each and of length . If , for some and by Lemma Lemma 1 , without loss of generality we can set . Thus if we proceed we get: and for all . It is easy to see that the same property still holds if is a finite union of balls. Therefore we have the following:
Corollary 1. If is compact open, it can be written in a unique way as the disjoint union of a finite number of balls of length , for any N big enough.
We notice that this corollary implies that any ball is totally bounded. Since balls are closed they are also complete, because is complete. Therefore this is another proof of Proposition Proposition 1.
Proof. Suppose , with compact open. Notice that , this is a disjoint union of open sets and on each of these sets is constant. If we take any , it is in one and only of these sets, thus it has a neighborhood where is constant. Conversely, suppose is locally constant, which means for all , there exists a neighborhood of (we call it ), where is constant. Since , there exist such that . By Lemma Lemma 2 we can split each into balls of length , for some N big enough, and this partition is unique. Thus we have split into a finite number of disjointed balls, and on each of these balls is constant. Therefore, we can write as a linear combination of the characteristic functions associated with these balls. ◻
In these notes we will often omit the "-adic" in front of "distributions".
We can see -adic distributions in another interesting and useful way. If is a -adic distibution, it restricts to a map from the set of compact open (we will call this set ) to : which is (finitely) additive. Conversely if we have such an additive map we can extends it to a -adic distribution and the extension is clearly unique. From now on, we will identify and in the sense that we will see a -adic distribution on also as a map from to .
Proof. By proposition Proposition 2, any compact-open is the finite disjoint union of open balls. So if , , thus the only possible definition to be additive is . We now have to prove that this is a good definition, i.e. it is independent of the partition of U. First we notice that iterating we have: Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} \mu(a+(p^N))&=\sum_{b_0=0}^{p-1}\mu(a+b_0p^N+(p^{N+1}))\\&=\sum_{b_0=0}^{p-1}\sum_{b_1=0}^{p-1}\mu(a+b_0p^N+b_1p^{N+1}+(p^{N+2}))\\&=\sum \limits _{b_0,...,b_k=0}^{p-1}\mu(a+b_0p^N+...+b_kp^{N+k}+(p^{N+k+1})). \end{split}}
Suppose . Now we set , and we split each into smaller balls of length . By (for ) we have . This means that in the end the value of depends only on his value on smaller intervals. By the previous observation we know the decomposition is unique, thus the definition of is independent of the partition. At this point is a map from to which is clearly additive. Thus extends to a p-adic distribution. ◻
Examples 1.
- : Suppose X is any compact open in and , we can define the Dirac distribution concentrated at as Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\textstyle \mu_{\alpha}(U)=\begin{cases} 1, & \mbox{if }\alpha\in U \\ 0, & \mbox{otherwise }\ \end{cases}}
for any compact-open
- : We define the Haar distribution on the compact open (it is a compact open since it is the ball of radius 1 centered at 0) as for all (i.e and ). We observe that we have defined the distribution only on balls, but it is sufficient to get a distribution by Prop Proposition 4. In fact .
- : We define the Mazur distribution on the compact open as for all (i.e and ), where is a positive integer such that . It is easy to show that in any ball exists one and only with that property, thus by Lemma Lemma 1 we have that , so this is a good definition. As before we can verify that , thus it extends to a distribution. For example , this may be apparently strange, because in real analysis we usually have distributions with positive value, but here , thus it does not make sense to think it as a "negative number". Someone could ask why this appears in the definition. There are actually two answer to this question. The first one is the fact that if we don’t do it, we don’t get a distribution, since the condition of Proposition Proposition 4 is not true anymore. The second and more interesting answer is related to Bernoulli numbers and it is the purpose of the next paragraph.
- Integrating locally constant functions: Let be the function which associate to any his first digit in p-adic expansion (i.e the coefficient of in its expansion). This is clearly locally constant and naturally splits into p subsets: . is constant on each and it has value b, thus . If we want to integrate respect to a distribution we get . For example if we have , thus . Instead, if , thus .
We will see that distribution are not enough to extends integration to continuous functions, and then we need a more particular tool.
By definition we can write .Errore del parser (errore di sintassi): {\textstyle \\}
We write the first few Bernoulli numbers: Errore del parser (errore di sintassi): {\displaystyle B_0=0,\ B_1=-\frac{1}{2},\ B_3=0,\ B_4=-\frac{1}{30}\}
It is not difficult to show that for every odd integer greater than 1, .Errore del parser (errore di sintassi): {\textstyle \\}
Now we define in a similar way the Bernoulli polynomials.
Definition 5. Consider
where
We call
-th Bernoulli polynomial:
In analogy with Bernoulli numbers, we notice that for a fixed , the -th Bernoulli polynomial evalued at is ! the -th coefficient of the Taylor expansion of .
Definition 6. We define the
-th Bernoulli distribution as
with
and
.
It is a distribution in the sense of Proposition Proposition 4, by the next:
Proposition 5. Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\displaystyle \label{1} \tag{1} \mu_{B,k}(a+(p^N))=\sum_{b=0}^{p-1}\mu_{B,k}(a+bp^N+(p^{N+1})).}
Thus extends to a p-adic distribution.
Examples 2.
- : , since we simply have .
- : , since we simply have Errore del parser (errore di sintassi): {\textstyle \\}
. This is the interesting reason why we need the in the definition of : we want it to be a Bernoulli distribution.
In these notes, we will often omit the "-adic" in front of "measures".
Proof. It is a straightforward application of the definitions, so we omit the tedious details. ◻
Definition 9. Let
such that
and
, we define the
-th Bernoulli measure regularized by
as
By Proposition Proposition 6 we know that is a distribution, we will prove that it is also a measure. To do so we are going to focus on which is the basic case and then we will extend to .
Examples 4.
- :
since is invariant for translation and .
- : Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &\mu_{1,\alpha}(a+(p^N))=\mu_{B,1}(a+(p^N))-\alpha^{-1}\mu_{B,1}(\alpha(a+(p^N)))\\=&\frac{a}{p^N}-\frac{1}{2} -\alpha^{-1}(\frac{<\alpha a>_N}{p^N}-\frac{1}{2})=\frac{a}{p^N}+\frac{1-\alpha}{2\alpha}-\frac{1}{\alpha}(\frac{\alpha a}{p^N}-\left[\frac{\alpha a}{p^N}\right])\\=&\frac{1-\alpha}{2\alpha}+\frac{1}{\alpha}\left[\frac{\alpha a}{p^N}\right], \end{split}}
where is the only integer between and which is congruent to modulo and is the greatest integer function.
Up to now we have proven is a measure, we want to extend this result to for all k.
Proof. For simplicity we omit to write
, any congruence in this proof should be thought in this sense. First we notice that
We can consider just these two first terms because the others have
in the denominator for
, thus when we multiply by
at least a factor
survives (and
removes every denominator of the polynomial). Thus,
Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &d_k\mu_{k,\alpha}(a+(p^N))\\\equiv &d_k\left(\mu_{B,k}(a+(p^N))-\alpha^{-k}\mu_{B,k}(\alpha a+(p^N))\right)\\ \equiv & d_kp^{N(k-1)}\left(B_k(\frac{a}{p^N})-\alpha^{-k}B_k(\frac{<\alpha a>_N)}{p^N}\right)\\\equiv & d_kp^{N(k-1)}\left( \frac{a^k}{p^{Nk}}-\frac{k}{2}\frac{a^{k-1}}{p^{N(k-1)}}-\alpha^{-k}\left((\frac{<\alpha a>_N}{p^N})^k-\frac{k}{2}(\frac{<\alpha a>}{p^N})^{k-1}\right)\right), \end{split}}
As in the previous example , in fact
also we notice that
, thus
. So the extreme right side of the previous congruence is congruent to
Now, using the binomial expansion, and omitting every term with
in the denominator for
, this is congruent to
Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &d_k p^{N(k-1)}\left(\left( \frac{a^k}{p^{Nk}}-\alpha^{-k}(\frac{\alpha^k a^k}{p^{Nk}}-\frac{k\alpha^{k-1}a^{k-1}}{p^{N(k-1)}}\left[\frac{\alpha a}{p^N}\right])\right) -\frac{k}{2}\left(\frac{a^{k-1}}{p^{N(k-1)}}-\alpha^{-k}(\frac{\alpha^{k-1} a^{k-1}}{p^{N(k-1)}}\right)\right)\\ \equiv &d_kka^{k-1}\left(\frac{1}{\alpha}\left[\frac{\alpha a}{p^N}\right]+\frac{1-\alpha}{2\alpha}\right)\equiv d_kka^{k-1}\mu_{1,\alpha}(a+(p^N)). \end{split}}
◻
Corollary 2. 'for all k and (for which we have defined it) is a measure.'
Proof. By Theorem
Theorem 1 Thus,
Errore del parser (funzione sconosciuta '\begin{split}'): {\displaystyle \begin{split} &|\mu_{k,\alpha}(a+(p^N))|_p \\ &\leq max(|\mu_{k,\alpha}(a+(p^N))-ka^{k-1}\mu_{1,\alpha}(a+(p^N))|_p,~ |ka^{k-1}\mu_{1,\alpha}(a+(p^N))|_p)\\& \leq max(|\frac{1}{d_k}|_p,~ 1), \end{split}}
where we used the fact that
and Proposition
Proposition 7. Since
is fixed when we choose k, we have proven that
is bounded on balls, and therefore, by Lemma
Lemma 3, on every compact-open. ◻
As we mentioned at the end of the third paragraph distributions are not sufficient to introduce integrals for continuous functions, but we can use measures to give a good and powerful definition of "-adic integrals". We will first give the basic idea of the construction of integrals for real valued function. The definition for is the same as we will see.
Definition 10. Let
be a continuous function defined on
, a compact subset of
. We define the Riemann sums as:
where
Here we are basically splitting the interval into smaller intervals Errore del parser (errore di sintassi): {\textstyle \\}
. Then, we take a "random" point in each subinterval and we multiply the value of the function in this point by the length of the subinterval. We want this sums to converge to something as grows to , so we can define the integral as the limit. This is actually true for real valued continuous functions. We want to do the same thing but with valued functions instead.
Definition 11. Let
be a measure and X a compact open,
. Suppose
Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\textstyle \\f:X\rightarrow\mathbb{Q}_{p}}
is a continuous function. We define the Riemann sums as:
where
is an element in
.
Proof. Since
is a measure we can fix a constant C such that
for any ball
. Since
is continuous on a compact set, it is also uniformly continuous, i.e. for all
there exist N s.t. for all
We have to prove
is a Cauchy sequence. To do so we will consider only positive integers N such that any
is either contained or outside
, and we try to estimate
for some
. Notice that:
where
is the residue of a
(in this way the
are the same as the
). Thus,
Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\displaystyle \begin{split} |S_{N,\{x_{a,N}\}}(f)-S_{M,\{x_{a,M}\}}(f)|_p&=\left |\sum\limits_{\underset{a+(p^N)\subseteq X}{0\leq a<p^N} }(f(x_{\overline{a},N})-f(x_{a,M}))\mu(a+(p^M))\right|_p \\& \leq Cmax(|f(x_{\overline{a},N})-f(x_{a,M})|_p). \end{split}}
But
. Therefore if we take N big enough, by the uniform continuity
. So we have proven the convergence of the Riemann sums.
Errore del parser (errore di sintassi): {\textstyle \\}
Now we only need to show the limit is independent of the choice of
. Suppose for example to fix another sequence of elements
such that
and
. Then
Errore del parser (SVG (MathML può essere abilitato tramite plug-in del browser): risposta non valida ("Math extension cannot connect to Restbase.") dal server "http://localhost:6011/it.wikipedia.org/v1/":): {\displaystyle \begin{split} |S_{N,\{x_{a,N}\}}(f)-S_{N,\{y_{a,N}\}}(f)|_p&=\left |\sum\limits_{\underset{a+(p^N)\subseteq X}{0\leq a<p^N}}(f(x_{a,N})-f(y_{a,N}))\mu(a+(p^N))\right|_p\\&\leq Cmax(|f(x_{a,N})-f(y_{a,N}|_p). \end{split}}
Since
and
are elements of
,
. Thus if we take N big enough
, by the uniform continuity of
. Therefore the sequences
and
converge to the same limit. ◻