
Definition
The space is Hausdorff.
If is compact, then is finite.
has no limit at .
There exists a nonempty open proper subset of .
Example
Finite fields with the discrete topology.
Linearly ordered fields with the order topology.
If is any field, then the field of formal Laurent series with the valuation topology (in fact any valued field). Note that is closed, and that the induced topology on is the discrete topology.
Let be a topological field, and consider an algebraic extension . Then as a vector space over . The product topology then induces a topological field.
We fix a topological field .
Definition
The number is unique, and we write .
This is preserved by sums, products, quotients, composition and so on. Moreover differentiability implies continuity. We also have the CauchyRiemann equations.
We will next consider more specific contexts: minimality in particular.
Let be minimal and write for its algebraic closure . Any rational function is differentiable.
Consider the case of . If converges in a neighborhood of zome , then there is some open such that is definable in .
On , the function is not definable in any minimal expansion of . Fact: let be open such that is definable in some minimal expansion of . Then the imaginary part of must be bounded. Indeed, defining on , and the set (project onto the axis if you must) must be bounded by minimality. Conversely, in , each such is definable whenever is definable. In particular is defined and injective on the strip of with . So is definable.
Question
Answer
Consider the field of formal Puiseux series over , where is a positive infinitesimal. The field is realclosed. Let denote its algebraic closure. For all (formal power series), we have a function . Write for the structure . Robinson and Lipshitz showed that this is minimal. Any such function can be dedinably extended to .
We fix an minimal expansion of a realclosed field , and write for the algebraic closure of .
Everything will be definable.
Definition
Example
Given any closed curve and a definable continuous map . We have maps
choosing so that . Write . For simplicity, assume that is not locally constant anywhere. The only possible discontinuities lie in , which by minimality, is finite. Write be the discontinuities. We add as a point, in order to define the winding number of as follows. For , write
We define the winding number of as
xample
Example
Remark
Let us give a few properties of those winding numbers.
If is not surjective, then .
Let be definably holomorphic. Assume that and are homotopyequivalent, i.e. there is a continuous with and . Then . This relies on the facts that is dedinably connected in , and that the function is locally constant.
For continuous definable , we have .
If one reverses the parametrization of , then .
Definition
Lemma
Proof. Write . Since , there is a with for all . So we can consider . Also set
Note that for all circles around . Since , for sufficiently close to , the element is close to . In particular, picking a sufficiently small circle around , the function is not surjective. So . We conclude that .
Write for the closed unit disk in , and for the unit circle, parametrized counterclockwise.
Main Lemma. Let be definable and differentiable on . Let . We have
Example
Proof of the Main Lemma. We first prove i. We can use an homotopy to shrink continuously, and as the radius of tends to , the curve is close to , so will not be surjective (it will only cover a small angle). So the winding number of is , hence the result.
Let us now prove ii. Fix a definably connected and open component of which contains . Since is definably connected (because is and is continuous and definable), the point is not isolated in , so is infinite. Consider the open set . We claim that the set
has dimention . Indeed assume for contradiction that has dimension . Then in particular has codimension . But has differential everywhere, so partitioning , we see that is locally constant, so takes only finitely many values. So is finite: contradicting the previous argument.
The statements of the main lemma are firstorder, so we can move to a sufficiently saturated elementary extension, and consider a generic point in over . Then is at as an function. Moreover is invertible (i.e. ), then the inverse function theorem for minimal structures gives that contains an open set, whence in particular contains an open set.
Pick be generic, so . We claim that is finite and that is nonzero on this set. Assume for contradiction that is infinite. Let . Then since it contains the generic point . So is surjectve for all such infinite fibers, which is impossible since the dimentin of is . Assume for contradiction that there is with , and write for the set of such 's. Then again has dimension . By definable choice, there is a such that for all , there is a with and this yields a similar contradiction.
0302: Lecture 5
Removal of singularities à la Riemann. Let be open and nonempty, let and let be definable, differentiable and bounded. Then there is a unique such that the extension of to with is differentiable.
Proof. Set for and . Since is bounded, we have , so is differentiable on , so by the previous theorem, the function is differentiable at , with . We then extend by continuity and see that is differentiable at .
Using the previous result and the maximum principle, one can prove the following:
Theorem
We start with an minimal fact:
Proof. Assume for contradiction that this is not the case. So for all there is a definable path converging to such that tends to . So tends to . So . But this is the frontier of a set of dimension , whereas has dimension : a contradiction.
We fix an open set , a point and a definable and differentiable . Assume that is not constant around . We define the order of at as follows. Recall that for sufficiently large , the number is constant (where is the circle around of radius ). We then define to be that integer.
Write for the continuation on . We have , whence
by a previous theorem.
Again write for the continuation on . If , then we claim that
Indeed shrinking (hence ) sufficiently, we can obtain that lie outside of .
In particular must be unbounded near . We claim that
(3.1) 
Let be a neighborhood of , let and such that
(3.2) 
for all . Set
for all . The function is differentiable, and bounded by (3.2). So is a removable singularity for . So . We conclude since is unbounded near that , whenc .
Let us show that
Set
on a [épointé] neighborhood of . By (3.1), we can extend to by setting . Now we know that , whence for all sufficiently small (i.e. whenever ).
Theorem
for all .
Proof. Let be sufficiently small, so . Note that , so setting on , we have . By the previous trichotomy, the function can be extended to with ; hence the result.
Corollary
Corollary
Let be a nonempty open definable set, and let .
From the proof of Theorem 3.1, we deduce:
Theorem
If is a pole, then
for a fixed sequence .
Corollary
from germs at of differentiable and definable functions to power series in is injective.
Theorem
Theorem
Proposition
Question
Example
This is definable in , and for all , the function is a polynomial of degree . But for the function is not a polynomial, so this doesn't give a negative answer to the previous question.
Proposition
Proposition
be the Laurent series associated to , where is as in the previous proposition. Then the function
is definable.
Let us now go back to the classical setting . Let be a simple closed curve and let have finitely many residues in . Recall that in the previous notations. We have .
So if is as above and is a definable family of simple closed curves, and is finite and uniformly definable, then the function
is also definable.
Theorem
Proof. For all , the function is polynomial by the one variable corresponding result. By minimality, the degrees of corresponding polynomials when ranges in are bounded by some . So . Loooking at , we can conclude by induction.
In fact, we have a result from
Assume that is definable and differentiable on a nonempty open set , and let such that is simply connected for some neighborhood of . Then exists in . Indeed recall that for definable, and nonconstant and differentiable on , then is finite on . Indeed assume that as infinitely many limit points around . Then sufficiently close to , one can also arrange that is injective and that it have nonzero derivative. So the inverse map of the restriction will be differentiable on an open set. Then sends an infinite subset of to , so must be constant: a contradiction.
Question
We now work with , so . Apparently the results should still be valid in the more general context.
Definition
a definable ,
a finite cover by definable subsets ,
For all , a definable bijection into an open subset of such that the transition maps are holomorphic.
Note that the transition maps are definable.
Example
7.1. this speaks to Lou's remark being relevant: can we not define this abstractly rather than always having to find a embedded representation?
Fact: Every compact analytic manifold is definably biholomorphic (in the sense of manifolds) to a definable manifold.
Definition
Definition
The only compact definale submanifolds of are the finite sets.
Theorem
Definition
It can be showed that in fact can be covered by finitely many such sets and functions.
Example
(By Chow's theorem, every analytic subset of is an algebraic variety.)
The basic problem: is a definable manifold, we have a definable open , and a definable analytic subset of as per Definition 7.6. When is the closure of in an analytic subset of ? So did we “add singularities” by taking the closure?
Example
We recall a classical result of removal of singularities:
RemmertStein theorem. Let be a manifold, let be a analytic subset, and set . If is irreducible analytic subset, and , then is an analytic subset of .
A bunch of minimal ROS results:
Assume that for all nonempty definable open . Then is an analytic susbet of .
Let be a definable analytic subset with . Then is an analytic susbet of .
A corollary of is that
Corollary
Proof. Set for each . Each is a locally analytic definable subset of . The set is an analytic subset of if and only if is closed, if is dense in , and if for all nonempty definable open .
It follows that we have a:
Definable Chow theorem. If is a definable analytic subset, then is algebraic.
Corollary
Exercise
Assume that we have a lattice where is an basis of . Write with its definable structure of manifold. Then we saw that is isomorphic to an elliptic curve .
We may assume that where . Recall that and are isomorphic if and only if there is a with (where is the standard action of on ). Moreover, there is a holomorphic and transcendental surjective map with
where the isomorphism is as abelian varieties, analytic manifolds. The function is called the invariant.
Set
Then each orbit of has excatly one representative in . It follows that is still surjective (and injective).
Theorem
Proof. Consider the function . Then is definable in by previous results. Now on any bounded part of , the function is definable on in . For , we have , and we see that is the punctured disk centered on . Recall that in particular for all , and as well. So we can factor by and get an analytic map with
Fact: , so , i.e. is a pole of , and we can write where are analytic, definable in . So is definable in .