Christol’s theorem and the Cartier operator February 11, 2010Posted by David Speyer in Algebraic Geometry, characteristic p, Number theory.
Let’s suppose that we want to compute , and we have already been given written out in base as . Here is a small prime and we want to do this conversation repeatedly for many ‘s.
Remember that and thus . So, start with , multiply it by , then by , then by and so forth. When you get to the end, read off .
We can precompute the effect on of multiplying by , for . Then we can compute just by scanning across the base representation of and applying these precomputed maps to the finite set .
The precise way to say that this is a simple, one-pass, process is that it is a computation which can be done by a finite-state automaton. Here is the definition: let , and be finite sets (input, states and output), and (the start). For each , let be a map . We also have a map (readout). Given a string in , we compute . So our input is a string of characters from , and our output is in .
We can say that our example above shows that is computable by a finite-state automaton. (In our example, the sets , and all have cardinality , but I do not want to identify them.)
This is a special case of an amazing result of Christol et al: Let be a sequence of elements of . Then can be computed by a finite-state automaton if and only if the generating function is algebraic over !
We have just explained the case . The reader might enjoy working out the cases (the Fibonacci numbers) and (the Catalans).
In this post, I will use Christol’s theorem as an excuse to promote the Cartier operator, an amazing tool for working with differential forms in characteristic .
Reformulating the problem
Let be a power series in . Define the power series , , …, by . So . Notice that for . It will be convenient for future generalization to write the defining equation of the ‘s as
Let be the set of all power-series which can be obtained from by applying the operators finitely many times. So consists of , , …, , , , …, , and so forth.
Please stare at the following claim until it is obvious to you: The function can be computed by an FSA if and only if the set is finite. Indeed, in that case, we can take .
Thus, Christol’s theorem is that is algebraic if and only if is finite. It is convenient to generalize from to an arbitrary perfect field of characteristic . We continue to define the operators by equation . So, for example, if is some element of , then .
The right generalization of Christol’s theorem is then the following: is algebraic over if and only if the -vector subspace of spanned by is finite dimensional. Obviously, this reduces to Christol’s theorem when .
If is finite dimensional, it is fairly easy to see that , and in fact all of , must be algebraic. Taking , …, a basis for . we have equations of the form
for constants in . These are polynomial equations in variables with coefficients in . One can show that the solution is isolated and is therefore algebraic over .
We will now focus on the other direction. Let be a finite extension of , contained in , and let . We need to show that is finite dimensional.
Surprising fact: takes to itself
Nothing like this is true in characteristic . Working over , if we extract the terms with exponent divisible by from the power series , we get , where is a primitive -th root of unity. Although is algebraic over , it is not in .
In orer to show , it is enough to do the case , as .
We have the formula
Since , the extension is separable (exercise!). We can use this to show that takes to itself. The proof is simply implicit differentiation: Any satisfies a polynomial relation . So , where the subscripts mean differentiation with respect to the first and second inputs of . In particular, is in . We used that was separable to ensure that we didn’t divide by zero.
This shows that is in . Note also that . I leave it as an exercise for the reader to check that, if satisfies , then . Hint: Remember that is perfect!
Putting the above together, the right hand side of is in . So the carry to itself.
Making the ‘s coordinate free
But is infinite dimensional over . Why is finite dimensional? Thinking geometrically, is the field of merormorphic functions on some complete curve over . To prove a finite dimensionality theorem, we will need to work globally on . But the definition of the is very dependent on the choice of the coordinate , which is a local choice. In this section, I will show how to get away from that choice.
Define a map by . Here is the miracle: The map does not depend on the choice of coordinate .
Let’s see an example. Clearly, . Let . So . Now, . So and !
This miracle deserves a proof, but I won’t give it. I’ll just make an analogy: In characteristic zero, there is a residue map, which takes the differential form to the coefficient of . A differential form can be locally written as if and only if its residue is zero. And, working purely in formal power series, it is somewhat challenging to show that the residue map is well defined. plays the same role in characteristic : We have if and only if can be written as , and it is true but subtle that does not depend on the choice of coordinates. But the residue is a mere number, while is a differential form, so is far more powerful! The operator is merely the simplest example of a Cartier operator, an operation on differential forms which exists only in characteristic .
The fact that is independent of the choice of coordinate should be a hint that it sheafifies. Indeed, let be the Kahler-differentials of over . The elements of are precisely for , but the formulation of does not depend on the choice of . Then is a well-defined map from to itself. Moreover, if does not have a pole at some point , then neither does .
Finishing the proof
We now restate our goal, changing all of our functions to differentials.
Let be an element of . Define the operations , , …., from to by . Fix a differential form . Let be the -vector space spanned by those differential forms we get by applying the to . We want to show that is finite dimensional.
Let be a point of . As we have already mentioned, if and do not have poles at , then all the forms in are likewise pole-free.
Suppose that is one of the finitely many points where or has a pole. If has a pole of order , and has a pole of order at some point, then an easy computation shows that has a pole of order . Feeding this bound into itself, has a pole of order at most . Continuing in this manner, no element of will have a pole of order more than at .
The exact numbers don’t matter. The moral is that there are only finitely many points where the forms in can have poles; and the orders of the poles at those points are bounded. So the vector space is finite dimensional. QED.