Talk:Chern class

Latest comment: 23 days ago by TakuyaMurata in topic Typos?

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The hairy ball theorem is a very bad example to give not so much because it concerns the real differentiable case (after all, the tangent bundle to the Riemann sphere is indeed a complex bundle) but because the bundle is stably trivial (in particular, its Stiefel-Whitney classes vanish) and stably trivial bundles are precisely the sort of thing Chern classes cannot detect. --Gro-Tsen 03:41, 17 July 2005 (UTC)Reply


"where H is Poincaré-dual to the hyperplane  ". How can a class H be dual to the hyperplane  ?

The hyperplane H represents a homology class in  , which is Poincare-dual to  , which is where the cohomology   lives. That is,   of the bundle equals  , where   stands for Poincare-dual and   is the hyperplane. Is your complaint simply that the text does not say "... dual to the class of the hyperplane..." (which is technically correct but pedantic, I think)? Joshuardavis 16:03, 26 September 2005 (UTC)Reply
[I don't know why this section is not a section.] But the phrase "Poincaré dual to the class of the hyperplane" is exactly what the article should say. It is 100% irrelevant if that seems "pedantic" to a professional who is well familiar with the subject matter. Many people reading this article will be students just learning the material, and if the addition of only a few words makes the article clearer, then that's what is important here.
This ought to be completely obvious to anyone editing Wikipedia.Daqu (talk) 15:22, 4 October 2015 (UTC)Reply

Revert anonymous edit

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I reverted this anon edit:

The field strength F may be viewed as a matrix of 4-forms (or 2 2-forms), and thus the eigenvalues of F are 4-forms as well. Each Chern class is then a 4k (2x2k)form.

back to the original

The field strength F may be viewed as a matrix of 2-forms, and thus the eigenvalues of F are 2-forms as well. Each Chern class is then a 2k form.

I'm not sure what the anonymous editor was thinking ... perhaps they were counting the number of spinor indecies instead of vector indecies? Or were they counting two real components instead of one complex component? linas 02:26, 18 October 2005 (UTC)Reply

Grothendieck or Atiyah?

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I'm not sure that it is correct to attribute the splitting principal to Grothendieck. I thought it was due to Atiyah, although I could be wrong. Can anyone provide a reference? 151.204.6.171

Implications of Chern Classes

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I know that the Chern classes are obstructions to finding non-vanishing vector fields. What I don't know, and hence i think it would be nice to add to this entry, is what the converse implies. I.e. if I know that the first Chern class vanishes, then what can I conclude? for exemple, if the first Stiefel-Whitney class vanishes then the bundle is orientable - but it doesn't imply that the bundle has a nowhere vanishing section (eg.: the tangent space of the 2-sphere. All of its SW classes vanishes, as it does for any n-sphere, but it doesn't have a nowhere zero section).

Different ways to define Chern Classes

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There are a lot of different ways to define Chern Classes. Some of them differ by signs. I think the article should point this out. A good reference is Borel/Hirzebruch, Characteristic Classes and Homogeneous Spaces II, Amer. J. Math 81, 1959. They really figure out the different definitions in the Appendix. Spaetzle (talk) 11:39, 27 October 2011 (UTC)Reply

Assessment comment

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The comment(s) below were originally left at Talk:Chern class/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Can this be made more accessible? A longer lead would help. Maybe more motivation and some applications? Geometry guy 20:38, 9 June 2007 (UTC)Reply


The lines: For example, the first few terms are ch(V) = dim(V) + c1(V) + c1(V)2/2 − c2(V) +...

would benefit by extension to a further term by clarifying the general form of the terms.

John McKay24.200.155.110 (talk) 10:40, 16 December 2008 (UTC)Reply

Last edited at 10:40, 16 December 2008 (UTC). Substituted at 01:52, 5 May 2016 (UTC)

various dots

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A TeX oddity: Of course \dots between commas gets rendered as \ldots and \dots between plus signs and some other binary operation symbols and binary relation symbols gets rendered as \cdots, but \dots between two instances of \oplus gets rendered as \ldots, not as \cdots. (Whether it works that way in genuine TeX and LaTeX as opposed to the thing we use here I haven't checked yet. 19:10, 5 May 2016 (UTC)

Chern classes box product

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This page should discuss chern classes of box products of vector bundles over a product space. — Preceding unsigned comment added by 71.212.185.82 (talk) 04:30, 24 August 2017 (UTC)Reply

Buried basics

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The material in "Properties of Chern classes" and "classical axiomatic definition" seems like it ought to be much higher up. I don't even think the article says that the kth Chern class lives in H^{2k} before section 6! Cyrapas (talk) 15:42, 21 September 2017 (UTC)Reply

Todo

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Chern Classes in Algebraic Geometry

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This page should have a section for chern classes in algebraic geometry. This should start with basic computations but also give non-trivial examples from enumerative geometry. Some places to look are

Todo:

Great. Please keep in mind that on Wikipedia it is even more important that material be reliably sourced than that it be correct. So please work on the citations as you add new material. Mgnbar (talk) 02:13, 30 September 2017 (UTC)Reply

Some serious problems

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At one point the article reads:

"The observation that a Chern class is essentially an elementary symmetric polynomial can be used to "define" Chern classes. Let Gn be the infinite Grassmannian of n-dimensional complex vector spaces. It is a classifying space in the sense that, given a complex vector bundle E of rank n over X, there is a continuous map

 

unique up to homotopy. "

This is ridiculous. It is necessary to say that the map pulls back the universal bundle to get the given complex vector bundle (or some equivalent qualifying phrase) for this map to be "unique up to homotopy".

Later the article states:

"For example, consider the quintic threefold in  ."

The article does not bother to link this to any definition of "quintic threefold" — as though every reader knows exactly what this phrase means.

Yet there is no such thing as "the quintic threefold" because there are many different quintic threefolds.

These types of problems are the result of sloppy writing. That does not help Wikipedia.50.205.142.50 (talk) 22:42, 3 May 2020 (UTC)Reply

I attempted to fix these problems. It's almost as quick to fix them as it is to complain about them. John Baez (talk) 00:49, 16 October 2024 (UTC)Reply

The Chern classes of V can therefore be defined as the pullback of the Chern classes of the universal bundle

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"The Chern classes of V can therefore be defined as the pullback of the Chern classes of the universal bundle." Such recursive statements are unhelpful to the uninitiated. 24.177.184.68

One just has to read the next sentence, to see how the "recursion" is terminated. But I'll think about the wording. Mgnbar (talk) 11:56, 5 April 2021 (UTC)Reply

Serious problems in the "In algebraic geometry" section

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There are a number of problems with the "In algebraic geometry" section. None of it is sourced. Axiom 5 of the axiomatic description is wrong (the total Chern class is not a ring homomorphism). There are also many grammar issues "In particular, we can find X is a spin 4-manifold if 4 - d is even..."

I'm not an expert, so there may be other mathematical errors I didn't notice.

Hasire (talk) 16:20, 23 May 2021 (UTC)Reply

Typos?

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The article says:

  • For example,[1] for  ,
    when  ,  
    when  ,  
(cf. Segre class#Example 2.)

It doesn't say what   is supposed to be here, and it's giving two different formulas for the same thing. I bet the second   is supposed to read  , and we can skip talking about the mysterious  . But that's just my guess! John Baez (talk) 00:39, 16 October 2024 (UTC)Reply

Here, I think the r is the rank of E. So the formulae are different since they depend on r. —- Taku (talk) 18:30, 16 October 2024 (UTC)Reply
  1. ^ Use, for example, WolframAlpha to expand the polynomial and then use the fact   are elementary symmetric polynomials in  's.