# Perfectoid spaces-7

Now in this post we will use Witt vectors to show that the category of perfectoid $K$-algebras is equivalent to the category of perfectoid $K^{\flat}$-algebras(which are defined in this post).

Suppose that $A$ is a perfectoid $K$-algebra, which means that the Frobenius $F: A^o/p\rightarrow A^o/p$ is surjective. Then just like the process of passing $K^o$ to $K^{\flat o}$, we can construct $A^{\flat}=\varprojlim_F A^o/p[1/p']$ where $p'\in \varprojlim_F K^o/p$ is an element with the same valuation as $p$ in $K$.

Next we show how can we get back from $A^{\flat}$.

We first note that any element $x^{\flat}=(x_0,x_1,...)\in A^{\flat o}$ has a unique representative $(x^{(0)},x^{(1)},...)\in (A^o)^{\mathbb{N}}$ such that $x^{(n)}=x_n(\text{mod} p),(x^{(n)})^p=x^{(n-1)},\forall n$. Note that we can write $x^{(n)}$ explicitly as $x^{(n)}=\lim_m(x_{n+m}')^{p^m}$ where $x_{n+m}'$ is any lift of $x_{n+m}$ in $A^o$. In fact, this is just the Teichmüller representative of $x_n$ in $A^o$.

We then take $W(A^{\flat o})$ the Witt vector ring of $A^{\flat o}$, and set $T: W(A^{\flat o})\rightarrow A^o, a=(a_0,a_1,...)\mapsto \sum_n p^na_n^{(n)}$. We can show that:

Proposition-1:$T$ is a surjective ring homomorphism.

If this is true, then in some sense, after passing to localization, we get a functor from the category $K^{\flat}-Perf$ of perfectoid $K^{\flat}$-algebras to the category $K-Perf$ of perfectoid $K$-algebras. Combining with the first step from $K-Perf$ to $K^{\flat}-Perf$, we get an equivalence between these two categories.

Proof(of Proposition-1):We first show that $T$ is a ring homomorphism. We can show $T$ is a morphism modulo $p^n$ for all $n>0$, which implies that $T$ is a morphism. The definition of $T$ shows that we can consider just the elements of $W(A^{\flat o})$ of the form $(a_0,a_1,...,a_{n-1},0,0,...)$, that is, we can consider just $T_n:W_n(A^{\flat o})\rightarrow A^o/p^nA^o$. We can show that this $T_n$ can factorize through $W_n(A^{\flat o}/pA^{\flat o})$, that is, there are two maps $f:W_n(A^{\flat o})\rightarrow W_n(A^{\flat o}/p^nA^{\flat o})$ and $S_n:W_n(A^{\flat o}/p^nA^{\flat o})\rightarrow A^o$ such that $S_n\circ f=T_n$. Indeed, if we set $f(a_0,a_1,...,a_{n-1})=(a_0^{(n)},a_1^{(n)},...,a_{n-1}^{(n)})(\text{mod}p^n)$, $S_n(y_0,y_1,...,y_{n-1})=\sum_ip^i y_i^{p^{n-i}}$, then $S_n\circ f(a_0,...,a_{n-1})=\sum_ip^i (a_i^{(n)})^{p^{n-i}}=\sum_ip^i a_i^{(i)}=T_n(a_0,a_1,...,a_{n-1})$. Note that it is easy to see if there is a morphism of rings $s: R\rightarrow R'$, then there is an induced morphism of rings $s':W_n(R)\rightarrow W_n(R')$. So if we consider the projection $\pi:A^{\flat o}/p^n\rightarrow A^{\flat o}/pA^{\flat o}$, then we get the induced morphism $\pi':W_n(A^{\flat o}/p^nA^{\flat o})\rightarrow W_n(A^{\flat o}/pA^{\flat o})$, and the composed map of $f$ and $\pi'$ is just $\pi'\circ f: W_n(A^{\flat o})\rightarrow W_n(A^{\flat o}/pA^{\flat o}),(a_0,...,a_{n-1})\mapsto (a_0^{(n)},...,a_{n-1}^{(n)})(\text{mod}p)$. Note that this time $\pi'\circ f$ is a morphism of rings. Indeed, look at what the addition and multiplication means here: $\pi'\circ f$ just takes an element $a=(a_0,...,a_{n-1})$ to each of their $n$-th component $(a_{0,n},...,a_{n-1,n})$. And since the addition and multiplication of $W_n(A^{\flat o})$ is just component for component, so this shows that $\pi'\circ f$ is indeed a morphism of rings. On the other hand, we can show that $S_n$ depends only on elements modulo $p$. That is to say, $S_n((y_0+p z_0,...,y_{n-1}+p z_{n-1}))=\sum_i p^i (y_i+p z_i)^{p^{n-i}}$. Yet for $j>0$, we have that in $p^i{p^{n-i}\choose j}y_i^{p^{n-i}-j}(p z_i)^{j}$, the coefficient is $p^{i+j}{p^{n-i}\choose j}$. If $j, then $p^{n-i}|{p^{n-i}\choose j}$, thus $p^n|p^{i+j}{p^{n-i}\choose j}$. If $j=p^{n-i}$, then we still have that $p^n|p^{i+j}{p^{n-i}\choose j}$. So all in all, we showed that $S_n((y_0+p z_0,...,y_{n-1}+p z_{n-1}))\simeq S_n((y_0,...,y_{n-1}))(\text{mod}p)$. So, this means that $T_n$ can factorize through $W_n(A^{\flat o}/pA^{\flat o})$. Moreover, $\pi'\circ f$ is a morphism of rings, and $S_n$ is too, this shows that $T_n$, their composition is also a morphism of rings.

Next we show that $T$ is surjective. This is simple as we can proceed as follows: if we can find for each $a'\in A^{\flat}$ an element $a\in W(A^{\flat o})$ such that $T(a)=a'(\text{mod}p)$, then we are done. Indeed, if this is so, then consider $(a'-T(a))/p\in A^{\flat}$, we can again find a $b\in W(A^{\flat})$ such that $T(b)=(a'-T(a))/p(\text{mod}p)$. This means that $a'=T(a+p' b)(\text{mod}p^2)$ (here we can choose $p'\in W(A^{\flat o})$ such that $T(p')=p(\text{mod} p^2)$) since $T$ is a morphism of rings. So we can iterate this process to infinity and get a sequence $a,b,c,...\in W(A^{\flat o})$. It is not hard to see that using the topology of $W(A^{\flat o})$, we can show that $a+p' b+p'^2 c+...$ converges in $W(A^{\flat o})$. So this means that there is an element in $W(A^{\flat o})$ which is sent to $a'$ for any $a'\in A^{\flat}$, thus showing that $T$ is surjective, which finishes the whole proof.

Now as we have remarked, we first construct a functor $S$ from $K-Perf$ to $K^{\flat}-Perf$, and then another functor $T:K^{\flat}-Perf\rightarrow K-Perf$, which sends each perfectoid $K^{\flat}$-algebra $A^{\flat}$ to $T(W(A^{\flat o}))[1/p]=A$. The problem with this construction is that, we should know that $A^{\flat}$ corresponds to some $A$ in $K-Perf$. To get rid of this restriction, we show the following:

Proposition-2: $Ker(T)$ is a principal ideal of $W(A^{\flat o})$.

Proof(of Proposition-2): Now for any $a\in W(A^{\flat o})$ such that $T(a)=\sum_n p^n a_n^{(n)}=0$, this means that $a_0^{(0)}=0(\text{mod}p)$, so $a_0^{(0)}=p^{(0)} b_0$ for some $b_0\in A^o$. For any $n>0$, since $(a_0^{(n)})^{p^n}=a_0^{(0)}$, this means that $b_n=a_0^{(n)}/p'^{(n)}\in A^o$ since $A^o$ is integrally closed in $A$. So, in this way, we find an element $b'=(b_0,b_1,...)$ which is in $A^{\flat o}$ such that $a_0=p' b'$. If we define two elements $b=(b',0,0,...),\tilde{p}=(p',0,0,...)\in W(A^{\flat o})$, then for $c=a-\tilde{p}b$, we have that $c_0=a_0-p' b'=0$, and for $n>0$, $c_n=a_0^{p^n}-(p'b')^{p^n}+p(...)=0(\text{mod}p)$. So, this means that $c=a-\tilde{p}b\in pW(A^{\flat o})$. We can take a factor from $W(A^{\flat o})$ such that $a-(\tilde{p}-p)b\in pW(A^{\flat o})$. And note that this time $T(a-(\tilde{p}-p)b)=T(a)-T(\tilde{p}-p)T(b)=-T(b)(T(\tilde{p})-T(p))=-T(b)(p^{(0)}-p)=0$, so we can redo the above process with $(a-(\tilde{p}-p)b)/p\in W(A^{\flat o})$, and we shall get some $c\in W(A^{\flat o})$ such that $(a-(\tilde{p}-p)b)/p-(\tilde{p}-p)c\in pW(A^{\flat o})$, then again we proceed to infinity and finally we get, using the completeness of $W(A^{\flat o})$, that $a=(\tilde{p}-p)(b+p c+p^2 d+...)\in (\tilde{p}-p)W(A^{\flat o})$, this shows that $Ker(T)=(\tilde{p}-p)$, an ideal generated by $\tilde{p}-p$.

Note that this description of $Ker(T)$ gives an intrinsic definition of $T(W(A^{\flat o}))$ using the kernel of $T$, that is $A=W(A^{\flat o})/(\tilde{p}-p)$.

So in this way we have these two functors $S,T$ the definitions of which depend only on the tilting $K\rightarrow K^{\flat}$.

One word about the correspondence of $A$, a perfectoid $K$-algebra and $A^{o}$(the same for $K^{\flat}$-algebras). Note that, since $A$ is a Banach $K$-algebra, so for any $a\in A$,there is some $k\in K^*$ such that $||k a||<1$, so $k a\in A^o$. Note that for any $k\in K-K^o$, we have that $K=K^o[k]$, so this shows that $A=A^o[1/(1/k)]=A^o[k]$. So, in we see that there is a one-to-one correspondence between the category of perfectoid $K$-algebras and the category of $K^o$-algebras.

Combining all the above results, we arrive at proving that the category of perfectoid $K$-algebras and the category of perfectoid $K^{\flat}$-algebras are equivalent.

In this post I will say something about an interesting mechanic, a linkage.

I will not define what a linkage is. We can consult this page of wikipedia on linkage. From mathematics’ point of view, a linkage is, roughly speaking, a finite set of segments of finite length each as a subset of $\mathbb{R}^3$(most of the time, they are straight segments) such that their union is a connected subspace of $\mathbb{R}^3$. What this mathematical definition does not say is that, these segments can move(rigid motion in Euclidean space), and their intersection points(if they have any) do not change their position on each segment to which the intersection point belongs. These two last points are the essential characteristics of a linkage.

Perhaps the simplest example of a linkage is just one straight segment with one end fixed. If the other end move in a plane, then its locus will be a circle, as the following graph shows(the straight segment is $l$ and the point end fixed is $A$):

A circle

Another famous example is the lever:

A lever

Of course we can again use a lever to draw circles, yet the most important use of levers do not perhaps lie here, we use it to carry heavy things. Of course the mechanism of these two uses are the same, that is, convert a rotary move into another rotary move yet their radii may be different. Using the law of conservation of energy, we can use a little force to move a heavy stone.

So, in some sense, the abstraction of linkages in mechanics is to create a linkage such that it can convert a particular type of movement into another type. Rotary motion into rotary motion, straight-line motion into straight line motion are easy. So the next problem is to convert a rotary motion into a straight-line motion, and vice versa.

The first such linkage was perhaps due to Pierre Frédéric Sarrus, a French mathematician. The linkage is called a Sarrus linkage. This linkage is not well known(on youtube, I find some demonstrations like this and this), mainly because of another much simpler linkage, yet with wider applications. That is Peaucellier linkage. It was also invented by a French(together with a Lithuanian Jews).

Note that Sarrus linkage do not use any guideways to convert rotary motion into straight-line motion. The main reason for this is that there are two hinges not parallel, as the following graph shows:

Note that the two planes $L_1,L_2$(with red segments) can simply be replaced by two segments $b_1,b_2$. The hinge $h_1$ forces the motion of $L_2$ to be in a plane perpendicular to $h_1$, similar for $h_2$, thus the motion of $L_2$ has to be the intersection of these two planes, that is a straight line since $h_1,h_2$ are not parallel. Note that this linkage is inevitably $3$-dimensional, since the mechanism of it is so.

As to Peaucellier linkage, it is a plane linkage, which means that all its segments are in one plane, that is:

Here the segments $OA=OC$(having the same length) and $AB=BC=CD=DA$. The point of this linkage is that, if the point $B$ moves in a circle passing through $O$, then the point $D$ will move on a straight line. If $B$ moves on a straight line not passing through $O$, then $D$ will move on a circle passing through $O$.

This linkage is a beautiful demonstration of the inversion of circles. In some sense, this is just elementary geometry. Look at the following graph:

Here is a simple proof. Suppose that $B$ moves on a circle $l_1$, with center $R$, then by symmetry it is clear to see that $O,B,D$ are co-line. Suppose that $AC\bigcap BD=E$, then since $ABCD$ is a rhombus, thus $AC\perp BD$. So, using Pythagorean theorem, we see that $OB*OD=OE^2-BE^2=OE^2-(AB^2-AE^2)=OA^2-AB^2$ which is a constant. Suppose that the diameter passing through $O$ of $l_1$ is $T$(which is not in the graph) and its intersection with a straight line $l_2$ passing through $D$ is $S$. Now consider the triangles $\triangle OBT\simeq \triangle OSD$. Then we have that $OT*OS=OB*OD$. Since $OT$ is the diameter, a constant, thus $OS$ is also a constant. This means that the projection of $D$ onto the diameter $OT$ is always the same point $S$, thus showing the $D$ always moves on the line $l_2$. For the second part, it is totally similar. In the wikipedia’s article, there is a vivid demonstration of this linkage, I copy it here:

We have said that Peaucellier linkage is a simple application of the inversion of circles. Why? In fact, we have shown that $OB*OD=OA^2-AB^2$, we can set $r^2=OA^2-AB^2$ with $r>0$. Then we can see easily that $B,D$ are inversion points one to the other about the circle, centered at $O$, of radius $r$. Then why the locus of $D$ should be a straight segment when $B$ moves on a circle passing through $O$? This is a very good question, and the answer to this point uses the essential characteristic of circle inversion: it is a conformal mapping. This means that circle inversion conserves angles, sending an angle $\pi/3$ to an angle $\pi/3$, etc. Note that, since the above inversion is about the circle centered at $O$, so this inversion maps $O$ to infinity $\infty$. Since a circle passing through $O$ is sent to a circle passing through $\infty$, so this image must be a straight line(any circle will not pass through a point at infinity), so this explains why the locus of $D$ is a straight segment. And also we require that the locus of $B$ must pass through $O$. This is shown in the following graph:
As for the second case, where $D$ moves on a circle passing through $O$, the mechanism is similar to the previous one, as the following graph shows:
Note that in this case, the straight line $l_1$ of the locus of $B$ must pass through the circle $l_2$ of the locus of $D$.