Something about the relationship between Plucker’s embedding and diff. forms

I hope everything is OK since I just start trying posting by mail.

The reason of being categorized into A.G. is that the idea is inspired by Plucker embedding in Algebraic geometry. Therefore I’d like to introduce this notion first.

There are several ways to give grassmanian a topology, manifold structure. Plucker embedding is one that in addition treat it as a variety.

denotes all the subspaces with dimension r in a linear space . One may embed it into where P means projectivization, i.e. quotient scalars.(who tells me how to insert wedge in tex). Choose a subspace in , choose any basis of , one may readily verify always refers to the same element when the subspace is fixed. This is the Plucker’s embedding.

When taking scalars into consideration, if V is endowed with a pos. definite quadratic form, one may view as a “cube” with some sort of “volume”. Therefore diff. forms can be viewed as continuingly varying cubes. Probably this is why we generalize the original concept of “differential” dated back to Newton’s age by introducing the anti-commutative tensors.

On a previous algebra problem

I asked the previous ring problem on Mathoverflow, titled "Can we deduce that two rings R1 and R2 are isomorphic if their polynomial ring is iso.?". I shall record the two answers here for future convenience (though I’m not able to understand them for now).

In one paper cited by them, the counterexample is given explicitly as below, but I’m not able to prove it.
are said to be the counterexample.

Moreover, It seems there is a common geometric background lying behind. My seniors told me these examples probably involving Algebraic K-Theory. There is still a long way in learning alg. geometry.

Below are answers.

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This was recently asked and answered at http://math.stackexchange.com/questions/13504/does-rx-cong-sx-imply-r-cong-s As answered there, it is possible to find two non-isomorphic commutative rings whose polynomial rings in one variable are isomorphic. An example is given in http://www.ams.org/journals/proc/1972-034-01/S0002-9939-1972-0294325-3/home.html

Tobias

Let X be an affine variety with two non-isomorphic vector bundles V and W that become isomorphic after adding a trivial line bundle to each. Then the coordinate rings of the total spaces of V and W should yield a counterexample. (Though you might have to do some extra work to verify that the total spaces of V and W are non-isomorphic as varieties.)
The counterexample cited by Tobias takes X to be the 2-sphere, V the tangent bundle to X, and W the trivial plane bundle over X.

Steven Landsburg

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A topology trick in algebraic geometry

Probably in strict terms, this cannot be called a trick. But for some people who is strongly inspired by point-set topology, this may evoke their interest in algebraic geometry, for eg. MCZeng.:)

I will put what I said yesterday afternoon in precise term.

DEFINITION: An n-dimensional projective space is a quotient space ~ where ~ iff there exists a , .

REMARK: Points in projective space can be represented by homogeneous coordinates where , , .

DEFINITION: can be endowed with Zariski topology where is closed iff there is a set of polynomial such that is the common zero locus of (called the defining polynomial)

As an easy excercise, one may check on his own that this satisfies the axioms of a topology. The relatively hard part is the finite union of closed set. One may use the fact that is Noetherian, i.e. for any ideal , is finitely generated.

DEFINITION: can be endowed with Zariski topology where is closed iff there is a set of homogeneous polynomial with complex coefficient such that is the common zero locus of (also called the defining polynomial), i.e., iff for all

Notice a homogeneous polynomial has such property: , where d is the degree of . Therefore, it is well-defined.

REMARK: can be covered by n+1 open subsets, where

THEOREM: is homeomorphic to under Zariski topology.

First check is open. The continuous map is easily given by dividing by . Notice that for a closed subset in , there is an intimate relationship between defining polynomial and where .

Our aim is to prove this theorem:

THEOREM: is defined by the zero locus of homogeneous polynomials if and only if is defined by the zero locus of polynomials when viewing as .

Actually this theorem does not involve any topological concept. This is difficult if one tackles directly, since it becomes even confusing when discussing points in . Actually I can’t think of an efficient way to do it elementarily.

Recall a simple theorem in point-set topology:

THEOREM: In a topological space , is covered by open sets . is closed iff is closed in for all .

Now, this theorem directly yields our result when Zariski topology is given. This is the power of point-set topology.

A hard problem

A sophomore asked me, given two rings , and is isomorphic to , could we deduce that is iso. to or give a counterexample?
It’s not at all easy! I have no idea for now.
PID, UFD, Noether/Artin, Local, these are invariants too specific to deal with such a general problem. Is there any versatile invariant of rings just like cohomology of topological spaces?

So the number theory is all about tricks and more tricks?

I was reading Jean-Pierre Serre’s A Course in Arithmetic right now. Thought group theory and Galois theory were not hard for me, but I had had almost no training in number theory. So I guessed it may take time to get familiar with it.

I still think number theory is awful because of the tricks so my study has been no obvious progression. Well, at least I was getting to know how the group theory is applied to specific objects such as . So I was to jot down what I learned today.

Considering Brother Zeng is constantly browsing this blog, I should explain the concepts a bit, although some of them is rather basic or trivial.

DEFINITIONS: For conveniency, we always write , and denote the group with the operation of multiplication as . Given a field , equals to the order of its identity under addition, but when the order is infinite, we simply write . For eg., .

PROPOSITIONS: Given a field , is either a prime or 0. When , we can always imbed into . When , we have , which indicates that the mapping: is a ring homomorphism. Besides, when is finite, for some n because can be seen as a vector space over the field .

THEOREM: All with the same n and p are isomorphic

This thm. is usually proved by the Isomorphism Extension Theorem, but JPS gives us another. Denote the algebraic closure of as . By the theory of polynomials over fields, the number of roots in of above polynomial is at most . Because the elements of the subfield in with order also satisfies the above polynomial, we get to know that if it exist, the subfield is unique. To construct it, consider the endomorphism:, and the set . Of course is a field with elements. Because we can trivially imbed a field with order into , we get to know it is isomorphic with .

THEOREM: is cyclic. (when , we denote as the unique field of order q)

After decompositing the abelian group, the conclusion is trivially followed. But JPS gives another proof. Let’s consider the Euler-fai function. denotes the number of integers from 1 to x that are coprime with x. We noted .

LEMMA: Given a finite group of order n. , Suppose the number of elements with order d is at most d. Then has to be cyclic.

, there is no second cyclic group with the same order as (h), and there are generating elements of (h). Let’s summed up the numbers of generating elem. of different cyclic subgroups. Since every elem. generates a cyclic group, the sum has to be bigger than n. But we know exactly. Since n|n, there has to be an elem. with order n. Thus H is a cyclic group generated by this one.

Now apply the lemma to .

======Above is the basic introduction. Now we came to the nonbasic part======

LEMMA 1: Field . Denote where f is a function on . When and , . Otherwise .

It’s easy to verify the trivial cases. And note the fact that .

THEOREM 1(Chevalley-Warning): let . The number of the common zeros of can be divided by p

Consider . Prove that when x is the common zero, P(x)=0. Otherwise P(x)=1. So the number of common zeros is where S is the same as the former lemma. Since P is a polynomial, prove that . Because , and the fact of the former lemma, the thm. holds.

(IT'S TRICKY! IT SEEMS TO ME WITHOUT TRICKS, THERE IS NO DEEP RESULT.)

REMARK: It’s an important case when and are the quadratic forms. Since the 0 is a common zero, the theorem indicates there are nontrivial zeros.

I didn’t know why JPS remark this. He said a quadratic form is a conic hyper surface(?). I admitted I knew nothing about it.

=======Quadratic reciprocity law======

THEOREM 2: . When , every unit is a square. Otherwise, all the square unit in forms a subgroup of , which is the kernel of the mapping:.

Note that the roots of is 1 and -1.

This theorem is an example of how to use the language of group theory to intepret the concepts in number theory, such as the square units. And this kind of translation offers a new way to look into it.(Well, we were just trivially considering the kernel of a mapping from an ABELIAN group to itself…Yes, it’s not exciting enough)

The remaining part of quadratic reciprocity law is not exciting except for another trick:

DEFINITION:

PROBLEM: Calculate

Consider the primitive root of . Let . Because , . . Simple calculation implies that if , otherwise .

(OH TRICK…)

When we looked back we may figure out we had only learned one deep theorem, i.e., Chevalley theorem. The rest of it were translations of trivial facts to the modern language. They were both enlightening.

But why Chivalley or Warning had tried to contruct such a P(x)?…

What is metric?

We call satisfying three ptc. properties a metric.
What about ? Or more broadly, what about where is a set with ordered, .

I proved that for any metric d, it can be transformed into a Q-valued metric with its topology unchanged. I attach the details below.

For any metric space , we assume d<=1 without loss of generality. we define a metric where . Obviously, the n is unique. After some annoying verification, we got to know is the metric we were expecting.

This told us the properties of a metric topology has nothing to do with R. Heuristically we noticed that the metric topology is only related with a structure of infinitum in R or Q or some other set. But what is it?

Notes about topological group

A topological group(I should use top. group in abbr. below) is a group with a topology satisfying T1 and that the mappings of and are continuous. The latter condition is rather strong for a topological space, but it’s natural, since we need some universal properties for our topological groups. First of all, we have to derive some essential properties.

Thm.1 For a top. group with an open subset , , , , and are open.

Thm.2 Any top. group is regular(T1+T3).

Thm.3 For a subgroup lying in a top. group , is also a top. group when given a subspace topology.

Thm.4 For a top. group with a closed regular subgroup , we have that is also a top. group.

To decide whether a group is a top. group, we have the following theorem, which resembles a similar theorem in group theory.

Thm.5 For a group with a topology, is a top. group iff. is continuous.

The proofs are trivial when one is well aware of the latter part of the definition. Important examples:


when given the topology induced by natural norm. To verify it, we may adopt a trick, a transformation in basic calculus (at least I used it).

+++++

A notable and relatively nontrivial result of a topological group:

let be a topological group. If is a subgroup of , then is also a subgroup.

The proof is simple, once realizing the fact that a continuous map maps limit points into the closure of the image.

[to be continued]

A group with Krull topology might be a top. group. I had noticed something subtle in Krull topology when learning Infinite Galois Extensions. Since we had the concept of top. group here, I may have a try to look into it.