Other Should I Become a Mathematician?

[eastside00_99]

Is the subject of non-commutative geometry just a continuation of algebraic geometry with emphasis on non-commutativity? If not, what are its prerequisites?

there is non-commutative algebraic geometry and non-commutative geometry, I believe the are somewhat different. Could be wrong about that.

 

[mathwonk]

and spend time reading and thinking. i am realizing i have spent my life collecting books and not reading them. do not worry about compiling lists of books, just find one good one and actually sit down and read it. i made the first half of my career just out of carefully reading one good paper, the classic on abelian integrals by andreotti and mayer, and working on the thoughts it inspired.

 

[Cincinnatus]

well that was in 1967. i too was young. indeed the world was...

paul monsky, now retired, was so young i thought he was a student.

I've heard Paul Monsky speak, he's retired but he still hangs around the department almost every day. He's a very good speaker as I recall.

 

[Gib Z]

Mathwonk, are mathematicians generally visual? I mean do most approach an idea visually whenever possible? I've been debating with myself: a strong mathematician does not need to resort to visual representations; to me it seems like "phony" mathematics.

I personally don't think it a bad thing. Being able to understand and interpret things in different ways is always a good thing. Terrence Tao backs me up on this point =]

 

[eastside00_99]

there is non-commutative algebraic geometry and non-commutative geometry, I believe the are somewhat different. Could be wrong about that.

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.rmi/1063050166"

I am sure it is not for the faint of heart...but if you want something even harder try being a Logician and an Algebraic Geometer--i.e., use algebraic geometry to tell you things about model theory and vise verse: http://search.barnesandnoble.com/booksearch/isbninquiry.asp?r=1&ean=9783540648635"

 

[pivoxa15]

bright students just impress you. you are always impressed if someone teaches you something you don't know, or notices something you do not.

Have you had students who lacked the prerequistes or were weak on them but still were able to impress you? If so how come?

even fields medalist rene thom felt that when seeing grothendieck,

rene thmo felt...? What do you mean?

How would you compare grothendieck with J.P.Serre? One 'better' then the other? Or different types? Who was more of a genuis?

I can't believe grothendieck got into politics at around 42. No stamina? Or just got too tired with the abstraction? Or pschological problems?

 

[Gib Z]

Everything is a fault with you, isn't it...Perhaps he just wanted a change? Perhaps he felt his true calling was in politics? Maybe he thought he could help other people that way? By the way, mathwonk meant "Even Fields medalist, Rene Thom, felt that when seeing Grothendieck...". And mathwonk already answered your question, in the quote you took. Example - I may be weak at a subject from lack of exposure to it, but If I took that required that subject as a prerequisite, and managed to do fine in it, it indeed might be enough to impress someone. The fact that you HAVEN'T learned that subject yet still managed to do fine is what could impress someone.

 

[mathwonk]

It is no more meaningful for me to compare people like Grothendieck and Serre, than for the fat guys on a sports show to compare athletes. Both were and are amazing.

Serre is still working, and I heard him give a fantastic lecture a few years ago, summarizing and organizing a large body of work in arithmetic and algebraic geometry, apparently almost off the top of his head.

Grothendieck was a very complex person, and you might enjoy reading one of the articles about his life in the Notices of the AMS. Even though he quit young, he accomplished far more than most people in a much longer period. From his own remarks, he may have overdone the hard work, and needed a rest.

Other reasons for his stopping work were his involvement in politics, and the reluctance of the scientific establishment to continue his mathematical support in the face of some evidence he was actually interested more in doing politics than mathematics.

An unfunded grant proposal he wrote at the end of his active mathematical life, has provided stimulus for years of work by others continuing to this day. See the book, geometric galois actions (2 volumes) by Leila Schneps.

http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521596428

 

[mathwonk]

I am impressed by an intelligent or insightful remark about material we are discussing, even allowing, or maybe more so because of, a student's lack of prior familiarity with it.

It is not a matter of how much knowledge a person has, but how much insight they generate when they do hear about something. We don't value just repeating information, but actively processing it.

The whole game is about not just calculating but thinking, something it is very hard to convince some students to attempt.

I am always sending out signals, probing for a response, for the sign that one of them has reached a receptive neuron in the student's brain. Imagine sending signals into outer space and waiting for a meaningful response. When you hear one it is exciting.

If all a student does is come to class, write down what is said, and memorize it, hoping to be asked the same things on a test, his brain is not even connected to the learning process. We are always looking for the one who also reflects on the consequences of what they hear, and generates questions about it, and maybe in a blue moon, answers.

For example, if you tell students all continuous functions are integrable, you might think that one in a thousand might ask you back, whether also some non continuous functions are integrable.

For most students, the difference between the statement "all continuous functions are integrable", and "all integrable functions are continuous" is not even visible. I spend much of my teaching life trying to think of new ways to provoke students to ask about and to see such differences.

I use such examples as "if you get all A's your dad promised to buy you a car", and ask whether that means that if he buys the car then you must have got all A's? Most student get this example, but do not always use the same thought process with mathematical statements.

 

[mathwonk]

By the way I still remember the day a calculus student did ask me exactly what conditions on a function were equivalent to (Riemann) integrability. When I quickly explained to him that the function needed to be bounded and have a set of discontinuities which could be covered by starting with an arbitrarily short interval, cutting it up into a sequence of subintervals, and laying those down on top of the discontinuities, he stared in amazement and delight.

He had apparently been waiting a long time for someone who knew the answer to his question, just as I had been waiting for someone who cared about the answer to the question.

 

[mathwonk]

you do not have to be a genius to ask good questions. that one question can be asked in a million ways. every theorem that proves A implies B, allows the question: "does B imply A"? If not, what other conditions must be added to B for it to imply A?

This question should be asked of yourself mentally, every time you see a theorem. The goal is not to impress your teacher by asking a question, but to educate yourself as to the meaning of the theorem you have seen.

To help answer the question for yourself, analyze the proof that A implies B. See if the same proof, or a very similar one, can prove B with weaker hypotheses on A. Or ask whether stronger hypotheses on A allow an easier proof.

In the example above, it is hard to prove that a continuous function is integrable, but easy to prove it for uniformly continuous ones, and also for monotone, possibly discontinuous, ones.

in my mind it is stupid for books and teachers always to state the theorem as "continuity implies integrability", but not prove it, rather than to state and prove one of the easier versions.

At least after stating the usual version, they might point out that since one can integrate on separate intervals separately and add, that it follows that a bounded function with a finite number of discontinuities is integrable. Even this is seldom seen. The standard calc book authors are mostly just as uncurious as the weakest students.

If you begin to ask "why?" when you hear a statement, you are already way above the student who just asks what was said. That question "why?" is the beginning of understanding. The next step is trying to answer it.

 

[tronter]

and spend time reading and thinking. i am realizing i have spent my life collecting books and not reading them. do not worry about compiling lists of books, just find one good one and actually sit down and read it. i made the first half of my career just out of carefully reading one good paper, the classic on abelian integrals by andreotti and mayer, and working on the thoughts it inspired.

When you read the books, mathwonk, do you mean also completing the exercises? Do you read books in a linear order?

 

[mathwonk]

well i read few books nowadays, but as a student i recall noticing that if i only read and did no exercises, that i learned basically nothing.

so yes that is part of it, i recall now that s a grad student reading herstein or some such book, i tried every single exercise.

indeed some authors say explicitly one should at least attempt every problem.

as a researcher reading papers, zariski recommends just reading the statement of the theorems and then proving them oneself. i used this method as a student reading mumford's red book of algebraic geometry. it is very useful. even if you only get half the proof before getting stuck, you then see clearly what the key idea was that you missed.

yes do the exercises and write up the proofs. e.g. someone earlier who said he was only beginning the study of algebraic geometry mentioned he had already read george kempf's book on algebraic varieties. i found myself wondering how thoroughly he meant he had read it, since fully grasping the contents of kempf's book would place one rather far along in sophistication.

i still have not mastered everything in kempf, e.g. his proof of Serre duality and riemann roch, and the details of the proofs, especially in chapter 9 on families of cohomology groups gave me plenty of problems, some still not resolved.

Of course it takes time for a good book to sink in, and there is no harm in reading one several times. and no there is no real requirement to read linearly. Indeed that can be very discouraging and even boring. Trying to read the interesting later parts can also provide incentive to read the earlier duller parts.

I used to read linearly, but i seldom got very far, and usually missed the most important parts. Often the later part is the part that was discovered and used first, and the earlier part is only the background someone filled in years later to nail down all the details. That stuff is much duller than the results, but may be needed for understanding them.

Since it is so hard to read a whole book, I try to isolate one good piece of it, one interesting theorem, and understand that. I once spent a lot of time trying to understand the implicit function theorem. Unfortunately I started with very high powered abstract, infinite dimensional versions, so even though I knew very powerful points of view, I did not grasp what the theorem meant in the simplest cases until much later.

E.g. i did not realize it just says how to recognize a point on a plane curve, near which the curve looks like the graph of a smooth function, or equivalently near which the curve has just one tangent line. i.e. it identifies non singular points of a curve. analytically it tells when you can solve an equation like f(x,y) = 0, for y = g(x). i.e. when you can get all the variables x on one side, and y on the other, at least locally.

I thought it meant something about factoring smooth maps locally through closed projections. i.e. one of the most powerful versions of the theorem is a necessary and sufficient criterion for writing a map locally, and after smooth change of coordinates, simply as a linear projection. But what good is this?, what does it mean? what does it do for you?

 

[mathwonk]

someone asked how a student impresses me. when i was teaching out of kempf, i found some misprints that rendered his proof unintelligible that cech cohomology agrees with derived functor cohomology in case i guess of coherent sheaves.

i corrected the proof of his lemma 9.2.1. page 115, in the special case we needed and presented that, but a student filled it in more generally. that is what i meant about a student teaching you something. that impressed me, and it gave me something explicit to say in a letter of recommendation for him. there were at least 2 students in that course who I recall improved my presentation, and i still remember their names. both are now good mathematicians in their own right.

 

[eastside00_99]

I have read thoroughly the first half (chapt. 1-4) and peeked at the more advanced stuff on coherent sheaves and sheaves of differentials. As I am trying to get a solid understanding of schemes through reading Quing Liu's book, I like to read ahead and get a feel for what one can do with schemes. I do this quite often actually; for instance, I picked up Abelian Varieties by Mumford and did not have a prayer for reading that book, but I worked at it and took a theorem of his book and brought it down to a more specific situation--i.e., a complex torus. I think Algebraic Varieties is a natural prerequisite for Abelian Varieties...but I am not sure of that.

But, yeah, I agree with you Mathwonk, I personally spend greats amount of time not reading anything but writing out my thoughts on specific material. This is by far one of the most useful strategies for me as when I write my knowledge forms a coherent whole instead of just sporadic problems and facts. For instance, you mentioned that EGA says that the actual most general framework for sheaves is posets. Well, I thought about this all yesterday to try and come up with the most succinct definition of a pre-sheaf.

 

[eastside00_99]

A Prelude to Presheaves:

In Functors, we gave a precise definition of a functor and illustrated it with examples. In this post, we will continue this line of thought by discussing a very important example of a functor known as a pre-sheaf. Pre-sheaves are preludes to Discontinuous Sheaves which are in turn preludes to Sheaves. This is in fact a prelude to a prelude to a prelude since we will work through our definition of a pre-sheaf given at the end of this post in the next post. As they are used in topology, differential geometry, analysis, and, most importantly to us, algebraic geometry, we will spend some time on developing an understanding of pre-sheaves, discontinuous sheaves, and sheaves.

Definition 1. Let X and Y be sets. The Cartesian product of X and Y is the set {(x,y)|x ∈ X, y ∈ Y }. This set is denoted by X × Y, and ΠX_α denotes the Cartesian product of family of sets {X_α} indexed by the indexing set Α ={i | X_i ∈ {X_α}}.

Definition 2. A relation on a family of sets {X_α} is a set L ⊂ ΠX_α.

Definition 3. A binary relation on a set X is a subset L of X × X.

Definition 4. Let X be a set and L be a binary relation. We say that L is a partial order if and only if
a. (reflexive) ∀ x ∈ X, (x,x) ∈ L,
b. (antisymmetric) (x,y),(y,x) ∈ L ⇒ x=y ∀ x,y ∈ X,
c. (transitive) (x,y),(y,z) ∈ L ⇒ (x,z) ∈ L ∀ x,y,z ∈ X.

Definition 5. A set X together with a partial order L is called a partially ordered set (or, for short poset).

Example 1. Consider the set of integers Z={...,-3,-2,-1,0,1,2,3,...}. Let a,b ∈ Z. We write a≤b if b is a successor of a.

Example 2. Again, consider the set of integers Z and two elements a,b ∈ Z. We write a=b if b is not a successor of a and a is not a successor of b. This is also an example of an Equivalence Relation. Every equivalence relation is a partial order.Example 3. Let X be a set. Let Y ⊂ ℘(X)="the power set of X"="the set of all subsets of X" such that there exists an element Z∈Y where Z⊂A ∀ A ∈Y. We call Z the infimum of X. Now, the relation ⊂ is a partial order on Y. Let A,B ∈ Y be such that A ⊂ B. Then we may define a function (known as the inclusion map) ι: A → B by ι(x) = x. This function is trivially injective and monotone. Y is a trivial example of a category whose objects are elements of Y and whose collection of morphisms between two sets A and B, are just the singleton sets containing the inclusion map defined above (c.f. Categories, More on Cats). As a final note to this example, the fact that Y is a poset with an infimum makes it a semilattice. More generally, a semilattice is a poset with an infimum. We will use these terms interchangeably from now on.

Definition 6. Let Y and Y' be posets as described in Example 3 and Z and Z' be their respective infimums. A Presheaf is a contravariant functor from the category Y to the category Y' such that F(Z)=Z'.

 

[eastside00_99]

Presheaves:

In this post we will investigate the definition of a pre-sheaf which was given at the end of A Prelude to Presheaves in more detail. As we have already seen many times, a covariant functor (resp. contravariant functor) can be defined by an explicit pushfoward (resp. pullback). For example, the transpose of a linear transformation defines a pullback as well does * in the post Categories. What is therefore not explicit in our definition of a pre-sheaf is this pullback. We will make this clear now. As in the previous post, let Y be a collection of subsets of a set X which contains an infinum (i.e., Y has a semilattice structure). Then ⊂ defines a partial order and thereby a morphism ι between two objects in the category Y. Suppose there exists a pre-sheaf structure on Y defined by a contravariant functor F from Y to a category Y' where Y' is also a semilattice defined over a set X'. Let A be an object in Y or in other words we have A ∈ Y. Then a functor must map this object to an object in Y'. We denote this object F(A). It also must map a morphism from A to B where A and B are objects of Y to a morphism from F(B) to F(A) (it is from F(B) to F(A) instead of F(A) to F(B) since F must be a contravariant functor). We denote this morphism F(ι_A,B) by ρ_B,A where ι_A,B: A → B is the inclusion map defined in the previous post. So, F(ι_A,B)=ρ_B,A:F(B) → F(A) is a morphism in the category Y'. In other words, we have the following diagram

 

[eastside00_99]

Presheaves Continued:

Now, we already have that F(Z)=Z' and this will be our first property that explicitly describes a pre-sheaf. We also have ι_A,A=id_A by definition of the inclusion map. Since F is a contravariant functor, we have id_F(A)=F(id_A)=F(ι_A,A)=ρ_A,A. This will be the second property that eplicity describes pre-sheaves. Finally, In our definition of a contravariant functor (c.f., Functors), we must have the following:
if ι_A,B:A → B and ι_B,C:B → C, then ρ_C,A=F(ι_A,C)=F(ι_B,Cι_A,B)= F(ι_A,B)F(ι_B,C)=ρ_B,Aρ_C,B. Again, this is the pullback action characteristic of contravariant functors. This is the third and finally property that will explicitly describe a presheaf.

We therefore have the following more explicit definition of a pre-sheaf: Let X and X' be sets. Let Y ⊂ ℘(X) and Y' ⊂ ℘(X') both be semilattices under the partial order ⊂. Then a pre-sheaf is a assignment F which sends subsets of X contained in Y to subset F(X) of X' contained in Y' together with a restriction mapping ρ_B,A whenever B ⊂ A in Y' such that

1. F(Z)=Z' where Z (resp. Z') is the infimum of Y (resp. Y'),

2. ρ_A,A = id_A for all elements A in Y', and

3. ρ_B,Aρ_C,B=ρ_C,A whenever C⊂B⊂A in Y'.

This is the definition one would usually see in a textbook on Modern Algebraic Geometry. But, what has been done here is offer a window into how one could study sheaves in themselves as objects instead of as pullbacks or pushforwards as would be done in a majority of those books. Category theory and lattice theory are then the answer to how to do this. These tools are what are used to define Cohomology of sheaves and thereby prove the Weil Conjectures. Much later, we will attempt to illuminate this work a little. One step in this direction, will be to define morphisms between pre-sheaves. We will need this idea in order to obtain this illumination, but more imperatively we will need this idea in order to have a solid understanding of sheaves.

 

[eastside00_99]

Heres my notes on functors also:

Functors

The previous two post have really been on the foundations on mathematics and specifically the foundation of Algebraic Geometry. There is one more detail that we need which is a precise definition of a functor. Fix two categories X and Y. Let F be a function from the objects of X, Ob_X, to the objects of Y, Ob_Y, and a function from the morphisms of X, Mor_X, to the morphisms of Y, Mor_Y. We call F a funtor if

1. F([A,B]_X) is a subset of [F(A),F(B)]_Y ,

2. F(e_A) = e_F(B) for every object in Ob_X, and

3. for a:A-->B and b:B-->C, we have F(ba) = F(b)F(a).

Stricly speaking F is called a covariant functor as X and Y are fixed categories (more on this later in the post). In addition we have a contravariant functor given by

1*. F([A,B]_X) is a subset of [F(B),F(A)]_Y

2*. Same as 2

3*. for a:A-->B and b:B-->C,F(ba)=F(a)F(b)

We have already alluded to one example of a contravariant functor in the post Categories which we will make clearer in the next post. We offer two other examples:

Example 1. First note that Vector spaces of finite dimension over a field k (which you can think of as real or complex numbers) form a category, denoted Vec, where the objects are vector spaces, the morphism are linear transformations (or matrices), and the multiplication is given, again, by composition of functions (or multiplication of the matrices representing the linear transformations). Let A and B be vector spaces and T:A-->B a linear transformation between them. Then we can form what is known as the transpose of the linear transformation T, denoted T^t. We do this as follows:

(1) Hom(A,B)=[A,B]_Vec is the set of all linear transformations from A to B.
(2) Hom(A,k) where k is our field (substitute the real numbers for k if you wish) is the set of all linear functionals on A--i.e., an element f in Hom(A,k) is a linear transformation from A to k.
(3) Hom(A,k) forms a vector space of finite dimension (in fact dim(Hom(A,k))=dim(A)) an so is an object in Ob_Vec. In fact, the set of all linear functions (here they are all assumed to be bounded as dimension is finite) will form a category as they form a commutative ring with unity (c.f., More on Cats). What we want is a function from the category finite dimensional vector spaeces to the category of linear functions on finite dimensional vector spaces over k which is a functor.

Now, the transpose of a linear transformation will satisfy such a definition as is shown in 1-3 if we can define it correctly. So, letting the category of linear functions on finite dimensional vector spaces over k be denoted by X, we have a function from one category to another, t: Vec --> X, defined by sending T to T^t. We now must define T^t.

T^t needs to be a morphism (i.e. a linear transformation since we have (by 3) that X consists of finite dimensional vector spaces) either from Hom(A,k) to Hom(B,k) or from Hom(B,k) to Hom(A,k). It is in fact a function T^t: Hom(B,k)-->Hom(A,k) given by if f is in Hom(B,k)--i.e., a linear functional on B, then T^t(f)(x) = f(T(x)) for all x in A. We have just defined a contravariant functor from Vec to X (or we can also view this as a contravariant functor from Vec to Vec). As you will notice, the way T^t is defined is the same way f^* was defined in the post Categories. Again, we will speak about the contravariant functor * in an upcomming post.

Example 2. This example will be much more trivial than example 1. Perhaps it is best to read this example first an then go back to 1. The identity function on a Category C sending an object A in Ob_C to A and a morphism a in [A,B]_C to a also defines a functor which is obviously a covariant functor.

Now let's see if we can recharacterize the notion of contravariant functor from a category X to a category Y in terms of the dual of a category and use Example 1 as a guide. A contravariant functor F is given by F(ba)=F(a)F(b). The claim is that this is equivalent to saying F is a covariant functor from X to Y^* (the dual category of Y). Let's check this:

(1) F([A,B]_X) is a subset of [F(B),F(A)]_Y= [F(A),F(B)]_Y^* which is part 1 of the definition of a covariant functor from the category X to the category Y^* (c.f., More on Cats for the definition of the dual of a category).

(2) there is nothing to check if we switch from considering F as a function from X to Y to F as a function form X to Y^* (the objects of Y and Y^* are the same so F(A) can be considered as an element of Ob_Y or Ob_Y^*).

(3) F(ba) = F(a)F(b) is exactly what the element (F(b),F(a)) goes to under the multiplicative map of the dual of Y induced by the multiplicative map of Y. More explicity, if a is in [A,B] and b is in [B,C]. Then F(a) is in [F(B),F(A)]_Y and F(b) is in [F(C),F(B)]_Y and multiplying we have F(ba)= F(a)F(b) in [F(C),F(A)]_Y.

Therefore, F is a covariant funtor from X to Y^*, and so from now on when we speak generally about functors, without aid of specific examples, we will be thinking in terms of either all functors being covariant or contravariant (lets just say covariant). The application of this to Example 1 is therefore obvious if one can say what is the dual of linear functions on finite dimensional vector spaces to k. It is all just notation and should be checked as an mental exercise, but as a hint the morphism should be Hom(k,A)--i.e. linear transformations from k to A.

Now, as another example of a Category, the class of all Categories, denoted Cat, is also a category, where its objects are categories, its morphisms being covariant functors, and multiplication is contained a priori in the definition of a covariant functor. Further, as an example of the dual of a Category, the dual of Cat, Cat^*, is just again the category whose objects are categories, but this time whose morphisms are contravariant functors, and again where multiplication is given a priori. We now can give a final example of a functor:

Definition Let F and G be functors from a Category A to a Category B. A natural transformation T from F to G is a function taking objects of A to morphisms of B such that

(i) T(X): F(X) --> G(X) for all objects X of A.
(ii) if a:X-->Y, then we have G(a)T(X) = T(Y)F(a).

(i) says that each object X of A yields a morphism T(X) from the objects F(X) to G(X) which are contained in B.
(ii) gives a commutative diagram between the functors and our function of categories T.

Example 3. As already noted, the class of categories is itself a category denoted Cat. Also, the class of functors is also a category whose morphisms are given by evaluating a functor at categories. The natural transformation gives a map from category of categories to the category of functors.

 

[pivoxa15]

well i read few books nowadays,

May I ask why?

 

[eastside00_99]

Actually my definition of a presheaf is wrong as it does not make the restriction maps morphisms in some other category other than that defined by inclusion maps (i.e. in my definition the restriction maps are inclusion maps). Here is the most general form I could get which will be consistent with the way I have seen the definition in other books

Definition 3. (Presheaf) Let X be a set and Y be a collection of subsets of X which form a semilattice under ⊂ and Z its infimum. Let C be a category which also has a binary relation ~ on its objects such that Z' is the infimum of Ob_C under ~. A presheaf on X is a contravariant functor F from the category defined by Y to the category C such that F(Z)=Z'

 

[mathwonk]

well that is very impressive eastside. my own bent is to be more interested in examples than in definitions. so my observation about posets being abstract referred to the fact that interesting examples of sheaves have to do with sheaves of regular functions on algebraic varieties or complex manifolds, and throwing away the topology obscures that for me. I have kind of gotten away from such very abstract stuff. But if you are attracted by it, go for it. Some of the smartest and most productive mathematicians I have known at georgia are of a similar mind.

i don't read books much because i don't have a lot of time, and books usually are not at an advanced research level, papers are moreso. but i read some in archimedes and euclid and hartshorne's geometry books last semester while teaching. (teaching is why i have so little time, so i try to combine teaching with some reading.)

 

[mathwonk]

eastside, since you have a knack for abstraction, maybe you could help me understand kempf's argument in the general cohomology section i had trouble with. i recall he computes the cohomology of the sheaves O(n) on projective space by adding them all together, making a ring, pulling back to affine space and comouting the cohomology of the whole business, then pushing it back down to P^n. But then he claims he gets the desired result one degree at a atime, and I do not see why the push down process he described should be degree preserving. I.e. I do not see why he can separate out the various O(n) one at a time in his conclusion. This is a standard argument, but he seems to have omitted a key step in justifying it.

 

[eastside00_99]

I guess the statement is

when n>1, the (n-1)th cohomology module H^(n-1)(O_A^n) is the graded module k(X_1)^P_1 *** (X_n)^P_n where the sum is over all p in Z^n such that p_i <=-1 and the module structure is implied.

I'm actuall confused by the statement a little as the direct sum is over all p in Z^n such p_i<=-1 and I guess that is not completely clear. I think he means that the sum is over all p in Z^n such P_i<=-1 for some i=1,...,n.

 

[eastside00_99]

I have been looking at it and think it would take quite a while for me to understand how to prove this theorem. I don't think I even remotely understand the way cohomology of sheaves is defined much less how to compute the cohomology groups. I will make it a goal of mine to understand this though as soon as possible.

 

[waht]

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.rmi/1063050166"

I am sure it is not for the faint of heart...but if you want something even harder try being a Logician and an Algebraic Geometer--i.e., use algebraic geometry to tell you things about model theory and vise verse: http://search.barnesandnoble.com/booksearch/isbninquiry.asp?r=1&ean=9783540648635"

Thanks, didn't know there was an algebraic non-commutative geometry. Brilliant. I am particularly intrigued by non-commutative geometry led by Alain Connes, as Mathwonk said.

I found a free ebook on this subject by Alain Connes. It seems there is some K-theory in there.

 

[mathwonk]

say eastside, maybe i recall being in your situation as a student. sheaves seem very exotic and there are many ways to look at them, and it is fun to play with the functor point of view.

but it turns out they are not very important in themselves. what a sheaf is, is just an important type of coefficients for cohomology. so one should not obsess over different ways to view sheaves, just choose one and get on with the main business of learning sheaf cohomology.

i.e. cohomology is the main thing. in fact shemes are also to my mind much less important than cohomology. so it is preferable to know cohomology of varieties, than to know schemes without cohomology.

at least that's my point of view after all these years of studying the subject, and it is why i chose kempf's book for my class.

of course there are other points of view on everything, but i highly recommend beginning to study cohomology.

this is all part of my philosophy that theorems are more interesting than definitions.

 

[eastside00_99]

I will take your advice to heart. Kempf's explanation of cohomology is also just really hard to follow for me. I feel like I have to start from the beginning and learn every fact and see every detail. But, maybe I can start to understand this cohomology today without that.

 

[eastside00_99]

But, thanks, for it is often hard to distinguish, as a student, what is essential and what is not, so I end up having to treat everything as essential.

 

[mathwonk]

take a look at miles reids webpage for a free book called "chapters on algebraic surfaces", where he gives a quick intro to cohomology.

the point is to start using it, and not bog down in the definition and construction of it.

or read serre's paper FAC, or i could send you some notes i wrote using baby cech cohomology to begin, mostly just H^1. that's very intuitive.