Mirislee's Pages

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In this post, we explore some properties of the ring and analytic germs in .

First we can show that the local ring is a noetherian UFD. Note that we have the results in algebra that

  1. if is a UFD, then is a UFD;
  2. if is a UFD and are relatively prime, then there exist and nonzero such that .

Theorem. The local ring is a UFD.

Proof. Prove by induction on . When , is a field and clearly a UFD. Suppose is a UFD. Then is a UFD. By Weierstrass preparation theorem, it remains to show that any irreducible factor of a Weierstrass polynomial in is also irreducible in . However, this follows from the facts that any Weierstrass polynomial can be written as a product of irreducible Weierstrass polynomials, and that any irreducible Weierstrass polynomial is irreducible in . These facts can be proved by direct applications of Weierstrass preparation theorem.

Recall that a commutative ring is called noetherian if every ideal in is finitely generated. Hilbert’s basis theorem tells us that if is noetherian then is also noetherian.

Theorem. The local UFD is noetherian.

Proof. Prove by induction on . The result is trivial when . Now suppose that is noetherian. Then is noetherian. Consider any non-trivial ideal and take a nonzero . Up to a change of coordinates, we may assume with a Weierstrass polynomial and a unit. Then . Since is noetherian, the ideal is generated by finitely many elements . For any , Weierstrass division theorem gives with , and then . It follows that . We conclude that is finite generated.

We can consider the zero set of a germ . In fact, what we consider is a germ of sets. Define an equivalence relation on the collection of subsets of by that if and only if there exists an open neighborhood of such that . An equivalence class of with respect to this equivalence relation is called a germ of sets in . Now we can define to be the germ given by the zero set of . Moreover, we can define for a subset its zero set to be the germ given by the common zero set of functions in . Note that the inclusion, union and intersection of germs are all well-defined. We can see that

Proposition. Suppose is irreducible. If vanishes on , i.e., , then divides .

Proof: We may assume is a Weierstrass polynomial with degree . Since is irreducible, and are relatively prime in , and hence there exists a nonzero such that It follows that if has a multiple root in , then is a zero of . Meanwhile, from the identity theorem we see that has an empty interior.

Weierstrass division theorem yields with a Weierstrass polynomial of degree less than . We must have as . Therefore for any outside , has at least distinct roots, implying that for the degree reason. It follows that is identically zero and divides .

Proposition. If is a holomorphic function on a neighborhood of such that is irreducible in , then is irreducible in for in a sufficiently small neighborhood of .

Proof: We may assume is a Weierstrass polynomial. If is reducible in , then with , and hence This shows that the germ of the set of points such that is reducible in is contained in . It suffices to show that is a proper subset of . This is direct from the preceding proposition.

Proposition. If and are holomorphic functions in a neighborhood of such that they are relatively prime in , then and are relatively prime in for in a sufficiently small neighborhood of .

Proof: Still we may assume are both Weierstrass polynomials. Then they are relatively prime as elements in and hence there exist and nonzero such that . This holds on a sufficiently small neighborhood around .

If have a non-unit common factor in , then divides and hence belongs to . It follows that is zero, and hence and are both zero, a contradiction.

Now we consider the relations between ideals of and germs in . A germ in is called analytic if there exist such that as germs. For a germ in , we let to be the set of elements satisfying .

Using the definitions and the fact that is noetherian, we can see that is an ideal of for any germ , and is an analytic germ in for any , where is the ideal generated by . Moreover, the assignments and are both inclusion-reserving. We also have for any analytic germ in and for any ideal .

An analytic germ is called irreducible if any decompostion such that and are analytic germs implies either or .

Proposition. An analytic germ in is irreducible if and only if is a prime ideal in .

Proof: First suppose that is irreducible and . Then we have and hence . Since is irreducible, we must have either or , which implies either or .

Now suppose that is a prime ideal and with analytic germs. If , then , which suggests that is a proper subset of . If we further have , then we can take and . Since , , contradicting to that is a prime ideal.

The final conclusion of the relations between ideals and germs is given by the analytic version of Nullstellensatz.

Theorem (Nullstellensatz). If is an ideal, then where is the radical of , i.e., the ideal of all elements such that for some .

The complete proof is too involved to be presented, so we will just prove the pricipal ideal case of Nullstellensatz.

Proof (of principal case): Suppose is nonzero and . We may assume with a unit and irreducible. Then it suffices to show that each divides . However, this is clear since vanishes on and hence .

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For a holomorphic function of one variable, we can consider the decomposition where is an -order zero of and is holomorphic. We want to obtain a similar description of holomorphic functions of several variables near a zero.

To find such decomposition, we need to figure out a special kind of holomorphic functions which share the same zeros with the original holomorphic functions. Consider the Weierstrass polynomials (in , of degree ), which have the form where are holomorphic functions in on a neighborhood of such that

We see that for any fixed such that the above Weierstrass polynomial is defined, has exactly zeros (with multiplicities) as a holomorphic function in .

Lemma. Suppose is a holomorphic function on a neighborhood of , and is a -order zero of the function in . Then there exists a polydisc and such that for each , has exactly zeros (with multiplicities) on as a holomorphic function in .

Proof: There exists such that is the only zero of on . Then there exists such that for any and .

Let be the number of zeros (with multipicities) of on . The argument principle implies that and hence is holomorphic in on by a lemma from the previous post. Thus must be constant, i.e.,

Theorem (Weierstrass preparation theorem). Suppose is a holomorphic function on a neighborhood of , such that and is a nonzero holomorphic function in . Then there exists a polydisc around , a unique Weierstrass polynomial in and a holomorphic function on such that

Proof: Take the same polydisc and the same as the proof of the lemma. Suppose is a -order zero of . By the lemma, we can let be the zeros of on for each .

By residue theorem we have and hence the left hand side is holomorphic in . It follows that the function defined by is a Weierstrass polynomial in a neighborhood of , such that and have exactly the same zeros on this neighborhood.

For each , there is a unique such that is holomorphic in on and that . We can see from the Cauchy integral formula in and a lemma from the previous post that is also holomorphic in , and hence holomorphic in .

The uniqueness of is clear from the fact that and must have the same zeros. The uniqueness of then follows from the identity theorem.

For each holomorphic function on an open subset , we denote the zero set of by . The Weierstrass preparation actually gives a description of in a neighborhood of a zero. The construction of the holomorphic in the theorem can be modified to a proof of the following Riemann extension theorem.

Theorem (Riemann extension theorem). Suppose is a nonzero holomorphic function on a connected open set and is holomorphic and bounded. Then can be extended to a holomorphic function .

Another Weierstrass theorem is concerned with the Euclid algorithm of holomorphic functions.

Theorem (Weierstrass division theorem). Suppose is a holomorphic function on a neighborhood of and is a Weierstrass polynomial in of degree defined around . Then there exists a sufficiently small neighborhood of , a uniquely determined holomorphic function on and a uniquely determined Weierstrass polynomial in of degree less than , such that

Proof: The unique statement can be proved in a similar way to the unique statement of an ordinary Euclid algorithm. Now we show the existence. Taking a sufficient small neighborhood we may define such that is holomorphic for and . It remains to show that is a Weierstrass polynomial with degree less than the degree of .

Suppose Then we can write where are determined by and holomorphic in and . It follows that where are holomorphic in .

For the preceding theorems, we often need to take smaller and smaller neighborhoods. In convenience, we introduce the concept of germs of functions.

Suppose is a fixed point. Let be the set of holomorphic functions defined in some neighborhood of . Define an equivalence relation on by that if and only if and agree with each other on a sufficiently small neighborhood of . An equivalence class of with respect to this equivalence relation is called a germ of holomorphic functions near , and the collection of all such germs is denoted by .

We can see that has a natural ring structure with identity given by the constant function . It can be verified that the units in are the germs given by the functions such that . This imples that is a local ring, whose unique maximal ideal consisting of the germs given by the functions vanishing at . We may use the same notation for a function and the germ it determines for short.

Now we can formulate Weierstrass theorems using the language of germs.

Theorem (Weierstrass preparation theorem). If satisfies that and that is nontrivial, then can be uniquely decomposed as , where is a Weierstrass polynomial and is a unit in .

Theorem (Weierstrass division theorem). Let and let be a Weierstrass polynomial of degree . Then there exist uniquely determined and a Weierstrass polynomial such that .

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The theory of holomorphic functions of several variables is the foundation of complex geometry, which uses as its local model. We try to generalize some results from the one variable case, and explore some different phenomena in the several variables case.

Suppose is a point in and satisfies for each . Then we define the polydisc to be the set consisting of the points such that

For a continuous function defined on a connected open subset , we say is holomorphic on if holds on . Meanwhile, we can define for each , called the partial derivative of with respect to .

Applying the result of the one variable case, we obtain the following formula.

Theorem (Cauchy integral formula). Suppoose is a continuous function such that is holomorphic with respect to each single component at each point in . Then for each the following holds

The above formula can be used to obtain the the power series expansion of a holomorphic function. Suppose is a connected open subset and is holomorphic. Then for each , there is a polydisc such that where

The preceding results enable us to generalize the identity theorem, the maximal module principle and the Schwarz lemma to the several variables case.

Theorem (Identity theorem). Suppose is a connected open subset in and is holomorphic. If vanishes on a neighborhood of some , then equals zero on the entire .

Proof: Let be the subset of consisting of the points on a neighborhood of which vanishes. Then is nonempty and open. By the local power series expansion of holomorphic functions, we see that if and only if all derivatives of vanish at , implying that is also closed in . It follows that is exactly the entire .

Theorem (Maximal module principle). Suppose is a connected open subset in and is holomorphic. If attains a maximal value at , then is constant on .

Theorem (Schwarz lemma). Suppose and let . Suppose is holomorphic with , and for each . Then for each , we have where

Proof: For a fixed , let Applying the one-variable Schwarz lemma to the function given by we obtain that

An interesting phenomenon which is not expected for the one variable case is the following theorem.

Theorem (Hartogs’ theorem). Suppose and and satisfy that for each . Then any holomorphic function can be uniquely extended to a holomorphic map .

We refer to the proof of this theorem in Huybrechts’ book.

Lemma. Suppose is an open subset of , is a neighborhood of for some , and is a holomorphic function. Then the function given by is holomorphic.

Proof: Since is compact, we can cover it with finitely many neighborhoods with such that on each neighborhood the power series expansion of converges uniformly, and hence commutes with the integral. This yields a power series expansion of locally.

Now we turn to the proof of Hartogs’ theorem.

Proof of Hartogs’ theorem: For any such that for each , gives a holomorphic function on the annulus .

Consider the Laurent expansion

Then since the lemma implies that is holomorphic for .

Meanwhile, the function is holomorphic for any when , which suggests that when and . It follows from the identity theorem that holds whenever . (This is where we need )

Define This power series converges uniformly on the disc , as the maximal module of each can only be attained on the boundary, and that the power series converges on the annulus.

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In this post, we are going to prove the Riemann mapping theorem, which characterizes the simply connected open subsets in .

Theorem (Riemann mapping theorem). Suppose is a simply connected nonempty proper subset in and . Then there exists a unique bijective and holomorphic with nonvanishing derivative such that and .

Remark: A bijective and holomorphic function with nonvanishing derivative is usually called a conformal function. This is equivalent to that the function has a holomorphic inverse.

First let us consider the uniqueness statement.

Theorem (Schwarz lemma). Suppose is holomorphic and satisfies . Then for each and . If either there is a nonzero such that , or , then for some .

Proof: Since , there is a holomorphic function such that for each . For each , the maximal value of on can only be attained on , implying that It follows that on . Therefore for any and . If either of the equality holds, then must be constant on by the maximal module principle. Thus for some and .

The uniqueness statement of the Riemann mapping theorem then follows from the Schwarz lemma. In fact, suppose and both satisfies the conditions, then we can consider the holomorphic functions and . The Schwarz lemma then implies that and are only differ by a rotation. Since their derivatives are both positive real at , they are actually the same.

Before the proof of the existence statement, we introduce the concept of univalent functions. A function on a connected open set is called univalent if is injective and holomorphic.

Theorem. Suppose is a connected open set in and is univalent. Then has no zero on .

Proof: Assume for some . Let . Then is a zero of with order . Take a neighborhood of on which is the only zero of . By a similar proof to the open mapping theorem, for each sufficiently near , the total order of zeros of in the neighborhood is . Since is nonzero at these zeros, they must be different zeros of order , contradicting the injectivity of . We conclude that cannot have any zero on .

Since a bijective and holomorphic function is naturally univalent, it must have nonvanishing derivative, and hence become conformal.

Theorem (Montel theorem). Suppose is a connected open set in , and is a nonempty family of holomorphic functions on such that is uniformly bounded on each compact subset. Then is equicontinuous on each compact subset, and each sequence in has a subsequence that is uniformly convergent on each compact subset.

Proof: First we show that for each , there is such that and is equicontinuous on . Since is open, we can take . Let , and we see that . There is such that for each and each , we have . Consider any . Take . For each such that , and each , we have

Then we show that is equicontinuous on each compact subset. Consider any compact subset and . For each , there is a neighborhood such that is equicontinuous on . Since form an open cover of and is compact, there are such that By the Lebesgue number lemma, we have a such that for , implies that they are contained in the same for some . As shown in the previous part, for each there is such that Let . Then we can see that for each such that , and each , we have .

Now we show the final conclusion. Suppose is a sequence in . Take a sequence of compact substes of such that that for ech , and that for each compact , there is such that . By the previous part and the Arzela-Ascoli theorem, there is a subsequence of such that converges uniformly on . Then there is a subsequence of such that converges uniformly on . Doing this repeatedly, we have a list of sequence such that converges uniformly on for each . Let . For a compact , take such that contains . We see that is a subsequence of , implying that converges uniformly on .

Now we step into the proof of the existence statement of the Riemann mapping theorem.

STEP 1. Each simply connected nonempty proper subset of can be conformally mapped into a subset of , with the given maps to .

Take . Since is simply connected, we have take a simple-valued holomorphic branch of We can directly see that is injective. Fix a . We claim that has a positive lower bound on . Otherwise, there is a sequence in such that . Then implying that , a contradiction. Thus is a bounded univalent mapping on . By a scaling and a translation, we get a conformal mapping from to a subset of that maps to .

Therefore we can assume that is contained in and that .

Lemma. Suppose is a connected open set in and is a sequence of univalent functions on such that converges uniformly to on each compact subset. Then either is constant or is univalent.

Proof: The Cauchy integral formula and the uniform convergence on each compact subset implies that is holomorphic. Assume that is nonconstant and not injective. Then there are in such that . Since is nonconstant, there is such that is the only zero of in . It follows from the argument principle that Noting that each is injective, and that we have the uniform convergence we obtain a contradiction. The conclusion follows.

STEP 2. The “maximal” univalent function from to preserving the origin is the desired conformal mapping.

Let be the family of all univalent functions from to that maps to . Since the identity function belongs to , is nonempty. It is clear that is uniformly bounded. Take such that . By the Cauchy integral formula, we have Thus is finite. Meanwhile, as the identity function belongs to , we have . Suppose is a sequence in such that . By the Montel theorem, there is a subsequence of converging uniformly to a function on each compact subsets on . We can then see that is univalent and from the lemma and the fact that . Thus with .

It remains to show that is surjective onto . Assume that is not surjective. Then there is . Let Then is a conformal mapping from to itself that switches and . It follows that does not contain , and hence there is a simple-valued holomorphic branch of Consider It can be seen that is univalent with , i.e., . Let and Since is not injective, it cannot be a rotation, and then the Schwarz lemma suggests that . Noting that , we have If is positive, then , contracting to that is the supremum. Otherwise , a contradiction as well. In conclusion, must be surjective.

The proof of the Riemann mapping theorem is now completed.

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We will focus on complex geometry for quite a long time from now on. Before we step into the world of complex manifolds and related structure, a review of holomorphic functions of one variable is of great help. The first part will focus on some conclusions which are useful in the theory of holomorphic functions of several variables, and the second part will devote to the Riemann mapping theorem.

Suppose is a connected open set in and is a function. Then is called holomorphic if for each , the limit exists. Viewing as a subset of , we have the partial derivatives of with respect to and . It can be seen that is holomorphic on if and only if the Cauchy-Riemann equation holds on , with the derivative Let then the above conditions become

An important part of the complex analysis is concerned with the integral of a function along a contour (piecewise differentiable path).

Theorem (Cauchy-Goursat). Suppose is a connected open set in and is a holomorphic function. Then for each contour that is contractible in ,

Using Cauchy-Goursat theorem, we can consider contours of simpler shapes without changing the value of the integral. An important fact about the holomorphic functions is that they are actually smooth and analytic, shown by the following theorems.

Theorem (Cauchy integral formula). Suppose is a connected open set in and is holomorphic. Then exists and is holomorphic on for each , with where is a simple contour that can be contracted to in .

Theorem (Taylor series). Suppose is a connected open set in and is holomorphic. Then for each , the Taylor expasion holds in a neighborhood .

Corollary. Suppose is a connected open set in and is holomorphic. If is a zero of with order , i.e., then there is a holomorphic function such that and

By this corollary, we may consider a contour contracted to such that there are no other zeros of inside , since is nozero in a neighborhood of . Noting that we have implying that where the last equality holds from the Cauchy integral formula and the Cauchy-Goursat theorem as is holomorphic inside . This extends to the argument principle.

Theorem (Argument principle for zeros). Suppose is a connected open set in and is holomorphic. If are the zeros of inside a simple contour , with orders , then

Theorem (Extension theorem). Suppose is a connected open set in , and is bounded and holomorphic. Then can be extended to a holomorphic function .

Proof: Define by for and where is a sufficiently small contour aorund contained in . The bounded of can imply the holomorphic property of at .

Another important corollary of the Cauchy integral formula is the maximal module principle.

Theorem (Maximal module principle). Suppose is a connected open set in and is holomorphic. If attains a maximal value , then is constant on .

Proof: Consider the inequality which holds for sufficiently small and implies that is constant aroud . The Cauchy-Riemann equations then implies that is constant near , and an application of the Taylor series and the path-connectedness of shows that is constant on the whole .

Next we come across the theorem which decribes the topological property of a holomorphic function.

Theorem (Open mapping theorem). Suppose is a connected open set in and is a non-constant holomorphic function. Then is an open mapping.

Proof: It suffices to show that is an open set. Take any and let . There is such that is the unique zero of in . Let and take . We claim that . Consider any and let . We have and . Thus where the last equality holds for the argument principle. We then see that has a zero in .

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The following are my notes of some topics in mathematics: (Updating)

  1. Differential geometry
    Notes of the book Lectures on Differential Geometry by S. S. Chern.
    LaTeX source
    PDF file

  2. Category theory
    Notes of the seminar on category theory.
    LaTeX source
    PDF file

  3. Sheaf theory
    Notes of the book Topologie Algebrique et Theorie des Faisceaux by Roger Godement.
    LaTeX source
    PDF file

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