# Huber spaces-5

In this post we will say something about the completion of $f-$adic rings.

Suppose $A$ is a topological ring, and $I_0$ is an ideal of definition for $A$, then we set $\hat{A}=\varprojlim A/I_0^n$. We call $\hat{A}$ the Hausdorff completion of $A$.

Note that the original definition for Hausdorff completion is to take the topological completion using Cauchy sequences and then module the closure of the zero element. We can show that for a topological ring, these two definitions are the same. Thus $\hat{A}$ is a Hausdorff space(this can also be seen from the first definition, since each $A/I_0^n$ is a discret space). Moreover, this shows that the above definition is independent of the choice of the ideal of definition.

As is expected, the Hausdorff completion is characterized by some universal property. That is:

$A$ is a topological ring, and $\hat{A}$ its Hausdorff completion with the morphism of topological rings $i:A\rightarrow \hat{A}$, then for any $B$ Hausdorff complete topological ring and any morphism of topological rings $f:A\rightarrow B$, there exists a unique morphism of topological rings $f':\hat{A}\rightarrow B$ such that $f'\circ i=f$.

Moreover, we can see that $ker(i)=\bar{\{0\}}$.

As for $f-$adic rings, we have that:

If $A$ is an $f-$adic ring, then $\hat{A}$ is also an $f-$adic ring. Suppose $B$ is a ring of definition for $A$ and $I$ is an ideal of definition associated to $B$ for $A$, then $\hat{B}=\varprojlim B/I^n$ is a ring of definition for $\hat{A}$, and $\hat{I}=\varprojlim I/I^n$ is an ideal of definition. Besides, there is an isomorphism of topological rings between $A\bigotimes_B\hat{B}$ and $\hat{A}$.

It might be interesting to see what the morphism between these two rings is. For any element $(a_n)_n\in \hat{A}$(viewed as a Cauchy sequence in $A$), then according to the definition of Cauchy sequence, there is an $N$ such that for any $n\geq N$ we have that $a_n-a_N\in B$ since $B$ is open. In this way, the element $(a_n-a_N)_n$ actually lies in $\hat{B}$(we disregard of the first finite terms), so we get a decomposition $(a_n)_n=(a_n-a_N)_n+(a_N)_n$ with the first factor in $\hat{B}$ and the second factor in $A$. So we can define $j: \hat{A}\rightarrow \hat{B}\bigotimes_BA, (a_n)_n\mapsto (a_n-a_N)_n\bigotimes_B1+1\bigotimes_Ba_N$. It is not hard to show that $j$ is well defined, that is to say, if there is another representative $(a_n')_n$ equivalent to $(a_n)_n$, and another decomposition of $(a_n')_n$, then there image are the same. This is a simple verification.

# Huber spaces-4

In the previous posts, we have just talked about one single $f-$adic ring. In this post, we will say something about the morphisms between $f-$adic rings.

Suppose that $A,B$ two $f-$adic rings with $A_0,B_0$ their rings of definition respectively, and $I_0,J_0$ their ideals of definition respectively. A ring morphism $F:A\rightarrow B$ is called an adic morphism if $F(A_0)\subset B_0$ and $F(I_0)B_0$ is an ideal of definition associated to $B_0$ for $B$.

One of the most basic problem concerning this definition is whether the definition depends on the choice of rings of definition. Before all, we have a small lemma which turns out to be very useful later:

In an $f-$adic ring $A$, $U\subset A$ is a subset, and $U\cdot A$ is open in $A$, then for any $V$ open neighborhood of $0$ in $A$, we have that $U^n\cdot V$ is also open in $A$ for any positive integer $n$.

This is a rather surprising result. Suppose that $A_0$ is a ring of definition for $A$ and $I_0$ is an ideal of definition for $A$ associated to $A_0$. Up to changing $I_0$ for its some power $I_0^n$, we can assume that $I_0\subset V\bigcap (U\cdot A)$. Write $G$ the finite set of generators of the ideal $I_0$. Then since $G\subset I_0\subset U\cdot A$, we can find a finite set $G'$ of $A$ such that $G\subset U\cdot G'$. Since $G'$ is finite, thus bounded, so $G'\cdot I_0^N\subset V$ for some $N$. So we have that $I_0^{N+1}=G\cdot I_0^N\subset U\cdot G'\cdot I_0^N\subset U\cdot V$, thus showing that $V\cdot U$ is open. Then $U\cdot U\cdot V$ is also open, and we can deduce that $U^n\cdot V$ is open for all $n$.

So, in the proof we used in an essential way the fact that $I_0$ is finitely generated.

$F:A\rightarrow B$ is an adic morphism, then (1) $F$ is bounded(sending bounded sets to bounded sets);(2) if $A_0$ is a ring of definition for $A$ and $B'$ is an open subset of $B$ such that $F(A_0)\subset B'$, then there exists a ring of definition $B_0$ for $B$ such that $F(A_0)\subset B_0\subset B'$;(3)if $A_1$ is a ring of definition for $A$ and $I_1$ the associated ideal of definition, and $B_1$ a ring of definition for $B$ such that $F(A_1)\subset B_1$, then $F(I_1)B_1$ is an ideal of definition for $B$ associated to $B_1$.

So the last point shows that the definition of adic morphism does not depend on the choice of rings of definition.

The first point is rather straightforward, since in the definition, $F(A_0)\subset B_0$, so any subset of $A_0$ is sent to a bounded subset of $B$. For any bounded subset $S\subset A$(not necessarily contained in $A_0$), $S\cdot I_0^N\subset A_0$ for some $N$, so $F(S\cdot I_0^N)\subset (F(I_0)\cdot B_0)^N\cdot F(S)$ is bounded. Yet $F(I_0)\cdot B_0$ is an ideal of definition, thus showing that $F(S)$ is bounded. The second point is easy. For the third point, we need a lemma:

Lemma: two rings of definition $A_0\subset A_1\subset A$, and $I_0\subset A_0$ an open ideal of $A_0$, then $I_0$ is an ideal of definition associated to $A_0$ if and only if $I_0\cdot A_1=I_0A_1$ is an ideal of definition associated to $A_1$.

Proof: If $I_0$ is an ideal of definition, then for any open neighborhood $U$ of $0$ in $A$, $U\bigcap A_1$ is again open in $A_1$, thus $I_0^N\subset U\bigcap A_1\subset U$. Moreover, $A_1$ is bounded, thus $(I_0\cdot A_1)^M=I_0^M\cdot A_1\subset U$ for some $M>N$. On the other hand, $I_0$ is open, and $A_1$ is a subring, thus $((I_0\cdot A_1)^n)$ defines a fundamental system of neighborhoods of $0$ in $A$, so $I_0\cdot A_1$ is an ideal of definition associated to $A_1$. Conversely, if $I_0A_1$ is an ideal of definition, then $I_0\subset I_0A_1$  and that $I_0$ is open show that $(I_0^n)$ defines a fundamental system of neighborhoods of $0$ in $A$, thus $I_0$ is an ideal of definition associated to $A_0$.

Now return to our proof to the third point. We take the original notations in the definition of adic morphism. And we consider $A_2=A_0\cdot A_1$, it is also a ring of definition. If we write $B_2=B_0\cdot B_1$, then $F(A_2)\subset B_2$. Since $I_1$ is an ideal of definition associated to $A_1$, so $I_0^M\subset I_1^N\subset I_0$ for some $M,N$, so $F(I_0^M)B_0\subset F(I_1^N)B_0\subset F(I_1)B_0$, so $F(I_1^N)B_0$ is open in $B$. And thus $F(I_1^N)B$ is open in $B$, so according to one proposition above, we have that $F(I_1^N)B_1$ is open in $B$. Moreover, for any open set $U\subset B_1$ containing $0$, then $F(I_1^{LN})B_0\subset F(I_0)^LB_0\subset U$. Thus $F(I_1^{LN})B_1\subset F(I_1^{LN})B_0B_1\subset F(I_0^L)B_0B_1$. We can choose $L$ large enough such that $F(I_0^L)B_0B_1\subset U$, thus we have that $F(I_1^{LN})B_1\subset U$, showing that $(F(I_1^{LN})B_1)$ defines a fundamental system of neighborhoods of $0$ in $B$, and thus $F(I_1^{LN})B_1$ is an ideal of definition associated to $B_1$ for $B$, and so is $F(I_1)B_1$ which completes the proof.

The following are some simple consequences of the above results:

$A,B,C$ three $f-$adic rings, and $F:A\rightarrow B,G:B\rightarrow C$ ring morphisms, then (0)if $F$ is adic morphism, then it is continuous; (1)if $F,G$ are adic morphism, then so is $G\circ F$;(2)$F,G$ are both continuous, and $G\circ F$ is adic, then so is $G$;(3) $A'\subset A,B'\subset B$ are open subrings such that $F(A')\subset B'$, then $F$ is adic if and only if $F|_{A'}: A'\rightarrow B'$ is adic.

The zeroth and the first points are easy. For the second point, since $A_0$ ring of definition for $A$ is bounded, so is $F(A_0)$. Since $C_0$ ring of definition for $C$ is open, so is $G^{-1}(C_0)$, thus there exists a ring of definition $B_0$ for $B$ such that $F(A_0)\subset B_0\subset G^{-1}(C_0)$. And it is not difficult to find an ideal of definition $J_0$ associated to $B_0$ such that $G(J_0)C_0$ is an ideal of definition for $C_0$. The third point is not difficult, either.

The last proposition concerns with adic morphisms between Tate rings:

$A,B$ two $f-$adic rings and $F:A\rightarrow B$ is a continuous ring morphism, if $A$ is Tate ring , then $B$ is Tate ring, too, and $F$ is adic morphism.

The proof is not difficult. Since $A$ has a unit element $a$ that is topologically nilpotent, then its image in $B$ is also a unit, and also topologically nilpotent, thus showing that $B$ is Tate ring. What is more, if $B_0$ is a ring of definition containing $a$, then $A'=F^{-1}(B_0)$ is an open subring, so it is $f-$adic ring. Since $A'$ contains $a$, a unit topologically nilpotent element, so $A'$ is Tate ring, so there is a ring of definition $A'_0$ for $A'$ such that $A'=A_0'[a^{-1}]$. Since $A'$ is open in $A$, this shows that $A'_0$ is a ring of definition for $A$. Then it is easy to show that $F_{A'}:A'\rightarrow B_0$ is adic, and so is $F$.

In the next post, I will say something on the completion of $f-$adic rings.

# Huber spaces-3

in this post we will say something on Tate rings.

A ring $A$ is a Tate ring if it is an $f-$adic ring and there is a unit element $a\in A^*$ such that $a\in A^{oo}$.

Note that for any ring of definition $A_0$ for a Tate ring $A$, $A_0\cdot A=A_0$ is open, thus $A^{oo}\subset \sqrt{A_0}$, yet according to the definition, $A^{oo}\bigcap A^*\neq\emptyset$, so $A_0$ contains a unit element.

Suppose that $a_0\in A_0\bigcap A^*\bigcap A^{oo}$, then the morphism $f: A\rightarrow A, a\mapsto a a_0$ is an isomorphism. So, it sends the open set $A_0$ to an open set $a_0A_0$. Thus $a_0A_0$ is open in $A_0$, too. Suppose one ideal of definition associated to $A_0$ for $A$ is $I_0$, then $I_0^n\subset a_0A_0$ for some $n$. Yet on the other hand, $a_0^k\rightarrow 0$, so $a_0^k\in I_0$ for some $k$, thus $a_0^kA_0\subset I_0A_0=I_0$, this shows that the $I_0-$adic topology on $A_0$ and the $a_0A_0-$adic topology on $A_0$ are the same. So, the ideals $(a_0^mA_0)$ defines a fundamental system of neighborhoods of $0$ for $A$. What is more, for any open neighborhood $U$ of $0$ in $A$, and any element $a\in A$, there is another neighborhood $V$ of $0$ in $A$ such that $aV\subset U$. Yet there is some positive integer $k$ such that $a_0k\in V$, thus $a_0^k a\in U$, this shows that $a_0^n a\rightarrow 0$. So, $a_0^N a\in A_0$ for some $N$. This means that $A=A_0[a_0^{-1}]$. It is not difficult to show that

A ring $A$ is Tate ring if and only if there is a ring of definition $A_0$ with an ideal of definition $a_0A_0$ such that $A=A_0[a_0]$.

So, in this sense, Tate rings are very special $f-$adic rings: the ideal of definition is a principal ideal, and in some sense, we can define its all integer powers, $I_0^n(n\in \mathbb{Z})$ such that they form a fundamental system of neighborhoods of $0$ and $A=\bigcup_{n\in\mathbb{Z}} I_0^n$.

In the next post, we will define morphisms between $f-$adic rings.

# Huber spaces-2

In the previous post we have introduced the adic rings, in this post we will introduce a type of ring which includes adic rings as open subrings.

A topological ring $A$ is called $f-$adic ring if it contains an open subring $A_0$ for which the induced subspace topology is an $I_0-$adic topology where $I_0$ is a finitely-generated ideal of $A_0$.

The $A_0$ in the definition is called a ring of definition for $A$ and the ideal $I_0$ is called an ideal of definition for $A$. We have three equivalent characterization for the ring of definition. That is

If $A$ is an $f-$adic ring, then the following three statements are equivalent: (1)$A_0\subset A$ is a ring of definition; (2)$A_0$ is open and an adic ring; (3)$A_0$ is open and bounded;

Proof: Before we show (1) implies (2), we have a little remark: $(I^n)$ defines a fundamental system of neighborhoods of $0$ of $A$(note here, not of $A_0$). This is rather trivial once we notice that each element  of $A$ is bounded and $I^n$ are all open in $A$. The implication of (1) to (2) is trivial. That (2) implies (3) is also trivial. So it remains to show the implication of (3) to (1). Suppose a ring of definition for $A$ is $B_0$ and an ideal of definition for $A$ in $B_0$ is $J$, since $A_0$ is open, we have that $J^n\subset A_0$ for some $n$. So, if $G_J$ is the finite set of generators of $J$, then $G_J^n\subset A_0$, too. So, we define $I_0$ to be the ideal genertated by $G_J^n$($I_0=G_J^nA_0$), which is finitely generated. Moreover, $I_0\subset J^n$, and $I_0=G_J^nA_0\supset G_J^nJ^n\supset J^{2n}$(here we use a rather simple yet important fact, that is, $G_JJ=J^2$, which garantes that $G_JJ$ is open, otherwise it will be very difficult to show this point), thus proving that the $I_0$-adic topology and the induced subspace topology on $A_0$ are the same. What is more, $I_0$ is finitely generated, thus $A_0$ is a ring of definition.

The followings are some simple properties of $f-$adic rings:

$A$ is an $f-$adic ring, (1)if $A_0,A_1$ are two rings of definitions for $A$, then $A_0\bigcap A_1$, $A_0\cdot A_1$ are again rings of definition;(2) Subset $B$ of $A$ is bounded and contains $1$, and subset $C$ of $A$ is an open subring, what is more, $B\subset C$, then there exists a ring of definition $A_0$ for $A$ such that $B\subset A_0\subset C$;(3) $A^o$ is the union of rings of definition for $A$.

The proofs are not very difficult. For the first point, it is sufficient to show that $A_0\bigcap A_1,A_0\cdot A_1$ are open and bounded(to show that $A_0\cdot A_1$ is bounded, note that for any $U$ an open neighborhood of $0$ in $A$, there exists $V$ another open neighborhood such that $A_0\cdot V\subset U$. Note that, $A_0\cdot V$ is an open set! Thus there exists a third open neighborhood $W$ of $0$ in $A$ such that $A_1\cdot W\subset A_0\cdot U\subset V$, thus we have that $A_0\cdot A_1\cdot W\subset A_0\cdot A_0\cdot U=A_0\cdot U\subset V$, thus showing the boundedness of $A_0\cdot A_1$). For the second point, we can take a ring of definition, say $A_0$ for $A$, then noting that since $C$ is an open subring, $A_0\bigcap C$ is a ring of definition for $C$, making $C$ into an $f-$adic ring. It is easy to verify that $B\cdot (A_0\bigcap C)$ is also a ring of definition for $C$, thus also for $A$(since $C$ is open in $A$); For the third point, since any $a\in A^o$ is power-bounded, thus there exists a ring of definition $A_0$ containing $(a^n)$. So $A^o$ is contained in the union of rings of definition. Conversely, since any ring of definition is bounded, so is any element in it, thus showing that the union of rings of definition is contained in $A^o$, and this finishes the proof for the third point.

The following result says something on $A^o$ and $A^{oo}$. Combining the first and the third points, we get that $A^o$ is actually an open subring of $A$.

$A$ is $f-$adic ring, then $A^o$ is integrally closed; for any subset $S\subset A$, $S\cdot A$ is open if and only if $A^{oo}\subset \sqrt{S\cdot A}$.

For the first point, we observe that if $a\in A$ satisfies $a^n+b_{n-1}a^{n-1}+...+b_1a+b_0=0$ with $b_0,b_1,...,b_{n-1}\in A^o$, we choose $A_0$ a ring of definition containing $b_0,...,b_{n-1}$, then $A_0[a]=A_0\cdot\{1,a,a^2,...,a^{n-1}\}$. Yet the second set $\{1,a,a^2,...,a^{n-1}\}$ is a finite set, thus bounded, and so $A_0[a]$ is bounded. It is clearly open, thus it is a ring of definition, and this shows that $a$ is power-bounded, thus $a\in A^o$, so $A^o$ is integrally closed. For the second point, if $S\cdot A$ is open, then for any $a\in A^{oo}$, $a^n\rightarrow 0$, so $a^N\in S\cdot A$, thus $a\in \sqrt{S\cdot A}$, and so $A^{oo}\subset \sqrt{S\cdot A}$. If $A^{oo}\subset \sqrt{S\cdot A}$, then note that since $(I_0^n)$ defines a fundamental system of neighborhoods of $0$, so each element in $I_0$ is topologically nilpotent, thus $I_0\subset A^{oo}$. That is to say, $I_0\subset \sqrt{S\cdot T}$. Yet $I_0$ is finitely generated, so there exists an integer $n$ such that $I_0^n\subset S\cdot T$, thus showing that $S\cdot T$ is open.

In the next post, I will say something about Tate-rings, which are a special kind of $f-$adic rings.

# Huber spaces-1

One of the motivating examples for Huber spaces comes from some application of valuation theory in Riemann surfaces.

Suppose that $k$ is a algebraically closed field and $k\subset K$ is a field extension of transcendental degree $1$. Then there is a very famous result in the theory of Riemann surfaces which says that each such extension $K$ corresponds (in some sense uniquely) to a Riemann surface over $k$. One fascinating proof to this result is to utilize valuation theory.

We say a function $v:K\rightarrow \mathbb{Z}\bigcup \{\infty\}$ is a discret valuation if $v(0)=\infty,v(1)=1$, and $v|_{K^{*}}$ is a group homomorphism to the group $\mathbb{Z}$. What is more, $v(a+b)\geq min(v(a),v(b))$(here the infinity should be understood as the positive infinity). Then we write $V(K/k)$ as the set of all such discret valuations. And for any $k-$algebra contained in $K$, we define $U(B)=\{v\in V(K)|v(B)\subset \mathbb{N}\bigcup \{\infty\}\}$. Then we let $B$ run over all the $k-$algebras contained in $K$ and take all these $U(B)$ as a basis of topology, so that the topology thus generated is verified to give a Riemann surface structure to $V(K/k)$. This is the essential part of the proof.

So here we want to generalize the pair $(K,B)$, and that is where Huber spaces arise.

First we say something on topological rings. A commutative ring $(A,+,*)$ (with unit) with some topology said to be a topological ring if the $A\times A\rightarrow A, (a,a')\mapsto a-a'; A\times A\rightarrow A,(a,a')\mapsto a*a'$ are both continuous. Some typical examples are the $p-$adic integers $\mathbb{Z}_p$.

In non-archimedean analysis, we often require that the topology on $A$ is given by some ‘nice’ sets, in other words, one fundamental system of open sets of $0$ can be a system of subgroups of $(A,+)$. Note that, this is not at all a trivial requirement since that the most common spaces like the real numbers $\mathbb{R}$ do not have this property(and that is why there is non-archimedean analysis). Yet this requirement is on the other hand rather natural for the non-archimedean analysis.

Here we consider when such a system of subgroups of $(A,+)$ can give a topology to $A$ such that it becomes a topological ring. In the following, we will write $S\cdot T$ as the subgroup generated by elements of the form $st(s\in S,t\in T)$ in $A$.

Suppose that $\mathfrak{G}$ is a set consisting of subgroups of $(A,+)$, then $\mathfrak{G}$ defines a fundamental system of neighborhood of $0$ such that $A$ becomes a topological ring if and only if $\mathfrak{G}$ satisfies the following three points:(1)for any $G,G'\in\mathfrak{G}$, there is a $H\in\mathfrak{G}$ such that $H\subset G\bigcap G'$;(2)for any $x\in A$ and any $G\in\mathfrak{G}$, there exists a $H\in\mathfrak{G}$ such that $xH\subset G$;(3)for any $G\in\mathfrak{G}$, there is a $H\in\mathfrak{G}$ such that $H\cdot H\subset G$.

There is an important remark to make. The notion $S\cdot T$ of two sets is in some sense the same as $ST$ in the setting of non-archimedean analysis, since we can always find a system of subgroups as a fundamental system of neighborhood of $0$. In the following we shall use frequently this remark.

One important class of subgroups of a ring is the ideals. And this introduces the adic rings.

Definition: A ring $A$ is called an adic ring if there is an ideal $I\subset A$ such that the system $(I^n)_{n\in\mathbb{N}}$ defines a fundamental system of neighborhood of $0$.

Sometimes for precision, we call a ring an $I-$adic ring if the ideal in the definition is $I$. Note that an ideal $J\subset A$ is open in the $I-$adic topology if there is a positive integer $n$ such that $I^n\subset J$. So the $I-$adic topology and $J-$adic topology are the same if and only if there exist two positive integers $n,m$ such that $I^n\subset J,J^m\subset I$. This amounts to say that these two ideals are comparable. In particular, $I^i-$adic topology and $I^j-$adic topology are always the same($i,j>0$).

Suppose that $A'\subset A$ is an open subring, then the $I-$adic topology on $A$ gives an induced topology on $A'$. Since $A'$ is open, then $I^N\subset A'$ for some integer $N$. Then it is easy to see that the induced subspace topology on $A'$ and the $I^N$-adic topology on $A'$ are the same. We will also often use this fact.

Next is a very important notion in analysis, the boundedness. For a metric space, this is a rather clear concept, yet if the space has no metric, things becomes not so easy. Here we adopt the definition of boundedness from functional analysis.

$A$ is a topological ring, and $B\subset A$ is a subset. We say that $B$ is bounded in $A$ if for any open neighborhood $U$ of $0$ in $A$, there exists another open neighborhood of $0$ in $A$ such that $B\cdot V\subset U$.

Here if $A$ is a non-archimedean topological ring, then we can replace $B\cdot V$ by $BV$, and the definition remains unchanged. And we call an element $a\in A$ is power-bounded if $\{a^n\}$ is a bounded set. $a$ is called topologically nilpotent if $a^n\rightarrow 0$. And we write $A^o$ as the set of power-bounded elements of $A$ and $A^{oo}$ as the set of topologically nilpotent elements of $A$.

A little remark is that any element $a\in A$ as a single-element set $\{a\}$ in a topological ring is bounded due to the very simple fact that $A\times A\rightarrow A,(a,a')\mapsto aa'$ is continuous.

In the next post we will enlarge the notion of adic rings to $f-$adic rings.

# positive definite functions

I have encountered positive definite functions somewhere long before, and I think it should be during a probability course. These days when I attend a seminar on Brownian motions, this concept, positive definite function, recurs many times, and until then I realized that I did not know anything about this kind of function except perhaps its definition.

The most natural way to introduce positive definite function is perhaps to start from positive definite matrices. We all know what a positive definite matrix is. Here I assume always that the matrices are with real valued entries. Now consider a probability space, $(\mathbb{R}^n,\mathfrak{B},d\mu)$ where the $\sigma$-algebra on the space $\mathbb{R}^n$ is just generated by the Borel sets, and $d\mu$ is a finite measure. Then for $i,j=1,2,...,n$, we can define

$a_i=\int_{\mathbb{R}^n}x_id\mu,cov(i,j)=\int_{\mathbb{R}^n}(x_i-a_i)(x_j-a_j)d\mu$

Of course, we should assume that these quantities all exist. Then, we can easily show that the covariance matrix $Cov=(cov(i,j))_{i,j}$ is semi-definite positive.

One can show that, conversely, for any semi-positive definite matrix $Q$, there is a measure $d\mu$ on $\mathbb{R}^n$ such that the above quantities exist and its covariance matrix is just $Q$.

Then, one can ask, what if these quantities, $a_i,cov(i,j)$ do not exist? Then there appears some generalization. Note that the essential obstacle for the existence of these quantities is that these functions $x_i,(x_i-a_i)(x_j-a_j)$ are not bounded. Are there some important bounded functions related to a random variable or simply a finite measure? Yes, there are. And one of them is the characteristic function of a random variable or in other words, the Fourier transform of a finite measure:

$\hat{\mu}(x)=\int_{\mathbb{R}^n}e^{i}d\mu(y)$

where $$ is the inner product on $\mathbb{R}^n$.

Then how could we translate the semi-positive-definiteness of the covariance matrix into the language of characteristic functions? It is not so obvious. I will give it directly. For any positive integer $k$ and for any real numbers $b_1,b_2,...,b_k$ and any vectors $v_1,v_2,...,v_k$, we have that

$\sum_{i,j=1,...,k}\int_{\mathbb{R}^n}e^{i}d\mu(y)\geq0$

This is the definition for a positive definite function: we call a complex valued function $f:\mathbb{R}^n\rightarrow \mathbb{C}$ is semi-positive definite if for any positive integer $k$ and for any real numbers $b_1,b_2,...,b_k$ and any vectors $v_1,v_2,...,v_k$, we have that

$\sum_{i,j}f(v_i-v_j)b_i\bar{b_j}\geq0$

(a little remark is in order: there is another way around in Bayesian analysis which gives some more motivation for the introduction of (semi)positive definite functions, cf the wikipedia article). We drop the prefix ‘semi’ if the equality holds only for the case that $b_1=b_2=...=b_k=0$. So it is clear that the function $f(x)=\int e^{id\mu(y)}$ is a semi-positive definite function. If the measure $d\mu$ admits the existence of the above quantities, then we can show that $f$ is positive definite if and only if $Cov$ is a positive definite matrix. So, in this sense, we say that positive definite functions are generalizations of positive definite matrices.

There is a rather remarkable result which inverses the above process, that is the Bochner’s theorem.

Theorem(Bochner) If $f$ is a positive definite function on $\mathbb{R}^n$, then there is a finite measure $d\mu$ on $\mathbb{R}^n$ such that $f$ is the Fourier transform of $d\mu$.

Note that, according to this theorem, if $f$ is positive definite, then $f(x)$ tends to zero when $x$ tends to infinity. We can not see this point directly from the definition of positive definite functions.

Note that in the definition of positive definite functions, we use only the additive group structure of $\mathbb{R}^n$, this inspires us to again generalize the domain of definition to a general group. This will be the content of the next post.

# Brownian motion, Feynman integral-1

Brownian motion first appeared in biology. Later Einstein, Smoluchowsk and other people used heat equation to derive some equations for Brownian motions. They viewed Brownian motion as some diffusion process. Later Wienner gave a rigorous foundation of Brownian motions, he constructed the Wienner measure on the space of the paths of the Brownian motions(in general situation is the space $\Gamma=C_{(0)}([0,T],\mathbb{R}^d)$, consisting of continuous functions with value $0$ at time $t=0$). In economics, some economists proposed some models which turned out to be Brownian motion, too.

The story did not stop here. Another aspect came from quantum mechanics. Feynman developed his own formulation of quantum mechanics using the Feynman integral. Later Kac found that removing the factor $\sqrt{-1}$ in the Feynman integral gives a solution to the heat equations.

So Feynman integrals, Brownian motions are related by the heat equations, or diffusion processes.

In this post I will try to give several formulations of the Brownian motions from different aspects.

The first approche relies on Gaussian measures on Hilbert spaces. So, we need first say something on this.

On the real line $\mathbb{R}$, we have the standard Lebesgue measure $dx$ with $dx([0,1])=1$. Then we define the Gaussian measure to be $d\mu=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{|x-\alpha|^2}{2\sigma^2}}dx$ where $\alpha, \sigma$ are two constants with $\alpha\in\mathbb{R},\sigma\in\mathbb{R}_{>0}$.

For superior dimensions, we say that a probability measure $d\mu$ on $\mathbb{R}^d$ is Gaussian if there is a linear map $h:\mathbb{R}^d\rightarrow \mathbb{R}$ such that the induced measure on $\mathbb{R}$, $d\mu\circ h^{-1}$ is Gaussian. We can show that the Fourier transform $\mu(y)=\int_{\mathbb{R}^d}e^{i}d\mu(x)$ of a Gaussian measure $d\mu$ is $\hat{\mu}(y)=e^{i-\frac{1}{2}}$ where $\alpha$ is a vector in $\mathbb{R}^d$ and $Q$ is a positive operator on $\mathbb{R}^d$. In fact, there is a one-to-one correspondence between the set of Gaussian measures and the set of the pairs $(\alpha,Q)$, so we can write $d\mu=N(\alpha,Q)$.

We can continue this process to any dimension, even infinite dimension. That is to say, if $(H,<,>)$ is a real Hilbert space, then a probability measure $d\mu$ on $H$ is said to be Gaussian if there is a continuous linear map, $h:H\rightarrow \mathbb{R}$ such that $d\mu\circ h^{-1}$ is a Gaussian measure on the real line. Just like the above statements, we can show that there exists a mean vector $m_{\mu}$ and a covariance operator $Q_{\mu}$(a vector $m\in H$ is a mean vector if $=\int_Hd\mu(z)$ for all $x\in H$, assuming that given $x$, the function $H\rightarrow \mathbb{R},z\mapsto $ is integrable. And a covariance operator is a symmetric, positive definite operator $Q:H\rightarrow H$ such that $=\int_Hd\mu(z)$, assuming again some integrability). We can show that $Q$ is a nuclear operator, which, roughly speaking, is so operator for which the trace can be defined and is finite, and is independent of the choice of basis. Conversely, we can show that for any such pair $(m,Q)$ with $Q$ a nuclear opeartor, there is a unique probability measure on $H$ which is Gaussian and $m_{\mu}=m,Q_{\mu}=Q$. In the same way, we write $d\mu=N(m,Q)$.

To define Brownian motion, we set $H=L^2([0,\infty))$(equipped with the usual Lebesgue measure) and $\mu=N(0,Q)$ where $Q$ is a symmetric, positive definite nuclear operator on $H$. Thus we can define the square root of $Q$, which we choose to be positive, too($\sqrt{Q}>0$). For any $t\geq0$, we write $z=\chi([0,t])$ the characteristic function of the interval, $[0,t]$. We can show that $\sqrt{Q}(H)$ is a dense subspace of $H$(what is more, we have that, $d\mu(\sqrt{Q}(H))=0$, this is a remarkable fact). So, there is a sequence $(z_n)\in (\sqrt{Q}(H))^{\mathbb{N}}$ such that $z_n\rightarrow z$. Then we define $B_t(x)=\lim_n$.

Then I say that $B_t$ is a Brownian motion. This means that $B_t$ is a random variable on $H$, and for $s, $B_t-B_s,B_v-B_u$ are independent and $\int_H|B_t(x)-B_s(x)|^2d\mu(x)=|t-s|$. These identities are not difficult to prove.

In the next post, I will try to establish Brownian motion from the point of view of diffusion process.