A Explicit examples of calculations of Chern classes

nrqed

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Hello,

I would like to find explicit calculations of Chern classes on various bundles. I mean examples where the steps are worked out explicitly, in terms of explicit connections. I have only found abstract and non explicit "examples". I only really understand things when I can work out several explicit examples.

Thanks!

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mathwonk

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i was going to suggest looking at the book by kodaira and morrow but i seem to have given it away. i guess you know this but few books do calculations directly from connexions, rather they mostly use axioms, in particular the whitney product formula. indeed once you know the axioms the classes are uniquely determined by them so they are all that is needed to make calculations at least in theory. in fact they are quite useful in practice, and one easily computes eg the chern classes of any smooth hypersurface in projective space by easy algebra. The point is that the class plus that of its known normal bundle equals that of projective space which is known. i have myself benefited from the discussions and calculations in bott-tu, and hirzebruch. as i mentioned also i once slogged through some more explicit calculations in kodaira and morrow involving curvature, but that was mostly to prove kodaira vanishing. the only explicit calculation i ever saw anyone do was the first chern class of a line bundle on a one dimensional manifold, showing it to be the oriented number of zeroes of a section, and that was not at all enlightening.

i have a short discussion of computations of chern classes, derived from these sources, in my notes on the riemann roch theorem starting on page 48:
http://alpha.math.uga.edu/~roy/rrt.pdf

lavinia

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The simplest case is perhaps the Chern class of an oriented 2 plane bundle with a Riemannian metric. For a specific example take any surface with a Levi-Civita connection for instance the standard connection on the 2 sphere.

An oriented 2 plane bundle with a Riemannian metric has a complex structure in which multiplication by $i$ is counter clockwise rotation by 90 degrees. By this is meant that if $e_1$ and $e_2$ is an oriented orthonormal basis then $ie_1=e_2$ and $ie_2=-e_1$.

If $∇$ is the connection on the 2 plane bundle with connection 1 form $ω_{12}$ with respect to $e_{1}$ $e_{2}$, then the connection on the complex line bundle is $∇e_1=ω_{12}⊗e_2= ω_{12}⊗ie_1 = iω_{12}⊗e_1$.

In this case the Chern form is just $(1/2π)i$ times the trace of the 1 by 1 complex matrix {$iΩ_{12}$} where $Ω_{12}$ is the curvature 2 form. For the tangent bundle of a surface, $Ω_{12}$ is the Gauss curvature times the area element of the surface.

- More generally, the top Chern class of a complex n-plane bundle is equal to the Euler class of the underlying oriented 2n-plane bundle. If the 2n-plane bundle has a Riemannian metric then its Euler class is cohomologous to the Pffaffian applied to the matrix of curvature 2 forms of any metric compatible connection. (Multiplication of matrix elements is replaced by the wedge product.)

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mathwonk

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I am a novice but I am pretty sure lavinia's first 4 paragraphs, augmented by the Gauss Bonnet theorem, imply my remark that for a (complex) one manifold, the chern class of a (complex) line bundle is the weighted number of zeroes (-poles) of a meromorphic section. The euler class of course is the cohomology class dual to the homology class of the zeroes of a regular section. I try to keep in mind that roughly characteristic classes are obstructions to the existence of families of sections that are everywhere independent, i.e. they are dual to homology classes of loci where a family of section is "degenerate".

Ok the most computational looking discussion I can find in my own library is that in Griffiths and Harris, Principles of algebraic geometry, starting on page 400, with a computation of the classes of projective space on page 408, using Whitney's product formula of course, and an obstruction theoretic interpretation on page 413.

mathwonk

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The following discussion is well intended but admittedly that of a novice:

It is perhaps worth keeping in my mind that apparently characteristic classes originally had nothing to do with connexions. They were topological invariants defined to measure the size of the degeneracy locus of a family of sections. It was then discovered that they could be defined via differential forms and curvature forms. In effect this is the content of the gauss bonnet theorem and its modern generalizations. I.e. the fact that different generic sections of the same n plane bundle on an n manifold have the same number of zeroes, allows one to define the euler class as the number of those zeroes. Then (on a surface) the classical gauss bonnet theorem says one can recapture that euler number as an integral of the curvature form.

The miraculous fact that cohomology can be completely captured by classes of differential forms leads to the question of actually finding differential forms that will represent certain geometrically given cohomology classes. In the case of the degeneracy loci originally defining characteristic classes, chern's theorem seems to do this. Then in later times his general gauss bonnet theorem becomes the, now quite unmotivated, definition of chern classes. I.e. one applies an abstract invariant polynomial to the curvature form and proves the resulting form is independent of the connexion. This reverse historical development is given in Griffiths Harris, but with historical comments which are enlightening. The historical version occurs in the classic book by Steenrod, Topology of fiber bundles.

The upshot is that understanding the connexion / curvature version of chern classes is perhaps best achieved by studying the proof of the general gauss bonnet theorem, e.g. in Griffiths Harris. In general while I sometimes find that source challenging in terms of precise technicalities, it is often highly valuable for intuitive insight.

Ok here is a reference to an author who is very meticulous about details, Michael Spivak, Comprehensive introduction to differential geometry. In volume V, as virtually the last sequence of topics, he discusses the definition of the characteristic classes using comnnexions and the gauss bonnet theorem. That should be worth a look.

another terse and challenging, but obviously highly authoritative source, is the little book "Complex manifolds without potential theory", by Chern himself, pages 46-49, and the section 8 on the grassmann manifold, pages 64-79. He gives one of my favorite definitions of chern classes: namely every bundle is classified by a map to the grassman which is unique up to homotopy. since chern classes pull back to chern classes by such a map, you only need to know the chern classes of the grassman, and then can define those for your bundle by pullback. Thus computing the chern classes of the "universal bundle" on the grassman is essentially the only case, and by far the most important one. this book seems to be available:

https://www.springer.com/us/book/9780387904221

if it were me, i would start by computing the chern class of the tautological line bundle on P^1, then on P^n, then move up to the classes of the universal bundle on the grassman. For a reference in the case of line bundles on riemann surfaces, R.C.Gunning computes the chern class of a line bundle explicitly in the proof of lemma 14, section 7, page 100, Princeton mathematical notes, 1966. i.e. he computes a differential form that represents it.

here is a free online version of his updated book, which contains that computation in the first 10 pages:

https://web.math.princeton.edu/~gunning/book.ps

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lavinia

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Another good example is the Chern class of a magnetic monopole. For each magnetic charge value one gets a connection on a different $U(1)$ bundle over the 2 sphere.
The Chern form is easily calculated as the field strength of a gauge field.

You may like to review the following thread.

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lavinia

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I am a novice but I am pretty sure lavinia's first 4 paragraphs, augmented by the Gauss Bonnet theorem, imply my remark that for a (complex) one manifold, the chern class of a (complex) line bundle is the weighted number of zeroes (-poles) of a meromorphic section. .
x
To elaborate Mathwonk's point: I think this connects the meromorphic section to the Gauss Bonnet Theorem - not completely sure if this is right though.

Give the Riemann surface a Riemannian metric. Since a Riemann surface is canonically oriented, the unit circle bundle $E$ is a principal $U(1)$bundle with $U(1)$ acting by rotation along the fibers. Choose a connection 1 form $ω$ on $E$ with curvature $Ω=dω$.

The meromorphic section is a vector field with isolated singularities so dividing it by its length away from the singularities gives a section $s$ of the unit circle bundle. The pull back $s^{*}(Ω)$ equals the Euler class except at finitely many points so its integral over the surface minus the singular points equals the integral of the Euler class over the entire surface.

Since $s^{*}(Ω)$ is exact ($s^{*}(Ω)=s^{*}(dω)=ds^{*}(ω)$) its integral can be computed by summing the integrals of $s^{*}(ω)$ over the boundary circles $C_{i}$, of a set of small disks $D_{i}$ centered at the singularities $p_{i}$ then taking the limit of these boundary integrals as the radii of the disks shrink to zero.

By change of variables, $∫_{C_{i}}s^{*}(ω) = ∫_{s(C_{i})}ω$ so the boundary integrals can be done in the unit circle bundle on the images of the $C_{i}$.

Since $ω$ is a connection form, its restriction to any fiber circle is just the length element $dθ$ so its integral over any parametrized closed curve on a fiber is $2π$ times the winding number of the curve around the fiber.

One think of the unit circle bundle above any of the the disks $D_{i}$ as a solid torus $D_{i}×S^{1}$ with the fiber above the singularity as the equatorial circle running through the middle. $s(C_{i})$ is a loop that winds around the equator as it loops around the boundary of the solid torus. $ω$ calculates the winding number plus an error term that depends on the projection of the loop onto the disk. This second term goes to zero as the radii of the disk shrinks to zero.

Notes:
- The connection does not need to be a Levi-Civita connection so this proof is a little more general than the Gauss-Bonnet Theorem.
- In general the Euler class is defined for sphere bundles and these may not come from vector bundles. The structure group of the bundle may not be reducible to the general linear group.
- The Euler class of a sphere bundle always pulls back to an exact form $dω$ on the total space of the bundle. $ω$ always has the property that its integral over any fiber sphere is equal to 1.
-

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nrqed

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Another good example is the Chern class of a magnetic monopole. For each magnetic charge value one gets a connection on a different $U(1)$ bundle over the 2 sphere.
The Chern form is easily calculated as the field strength of a gauge field.

You may like to review the following thread.

Thank you so much lavinia for all the helpful comments. I will also go through that thread you recommended, as it seems to be quite explicit.
I still have to absorb everything you and mathwonk wrote. I learn mostly by working out very explicit calculations before moving on to general statements.

Meanwhile, I am trying to do the calculation over CP1 using the Levi-Civita connection to do an explicit calculation. I calculated the connection $A$ using the usual induced metric on the sphere and then calculated $F = dA + A \wedge A$ but what I found was a that F is traceless, which would mean a vanishing Chern class. Am I doing something completely wrong by following this approach? If not, I will give more details of the steps of my calculation to find where my mistake is.

Thank you again!

lavinia

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@nrqed

I would guess that you are treating the connection as a real matrix rather than as a complex matrix. The real matrix will always have trace zero since $ω_{11}=ω_{22}=0$.

This is a good point since it emphasizes why Chern classes are only defined for complex vector bundles.

- $A∧A$ is zero..

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nrqed

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@nrqed

I would guess that you are treating the connection as a real matrix rather than as a complex matrix. The real matrix will always have trace zero since $ω_{11}=ω_{22}=0$.

This is a good point since it emphasizes why Chern classes are only defined for complex vector bundles.

- $A∧A$ is zero..

Thank you so much for replying. That's an important point that I was missing, indeed. I am trying to do the calculation using the usual spherical coordinates $r, \theta, \phi$, I am curious about how the complexification is done in that approach. I will post more as soon as I have a minute.

Thank you!!

lavinia

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- Maybe this will clarify this point.

A connection on a complex vector bundle may be thought of as a complex linear map from the space of sections of the bundle $E$ into the the smooth sections of the tensor bundle $TM^{*}⊗E$ that satisfies the Leibniz rule

$∇fs= df⊗s + f∇s$. Here $TM^{*}$ is the space of smooth real linear complex valued 1 forms on the manifold.

A connection on a real vector bundle is defined in the same way except everything is real linear.

If $s_{i}$ is a basis of sections of the bundle over an open neighborhood then

$∇s_{i} = Σ_{j}ω_{ij}⊗s_{j}$ for some one forms $ω_{ij}$. This is a matrix of complex valued $R$ linear 1 forms.

The Levi-Civita connection on a surface is a real(as opposed to complex) connection and with respect to an oriented orthonormal basis $e_{1}$ $e_{2}$ is $∇e_1 = ω_{12}⊗e_2$ and $∇e_{2} = ω_{21}⊗e_2$ with $ω_{21}=-ω_{12}$ and $ω_{11}=ω_{22}=0$. The $ω_{ij}$'s are real valued 1 forms. So the matrix of the connection viewed as an $R$ linear map is skew symmetric with zeros on the main diagonal. Similarly the curvature is a 2×2 skew symmetric matrix of real valued 2 forms. Its trace is zero.

Since this bundle is oriented one can turn it into a complex vector bundle by defining multiplication by $i$ as $ie_{1} = e_{2}$ and $ie_{2} = -e_{1}$ on any oriented orthonormal basis.

The Levi-Civita connection becomes a connection on this complex vector bundle since $∇e_{1} = ω_{12}⊗e_{2} = ω_{12}⊗ie_{1}= iω_{12}⊗e_{1}$ and $∇e_{2} = -ω_{12}⊗e_{1} = ω_{12}⊗ie_{2}= iω_{12}⊗e_{2}$.

The connection 1 form matrix is now 1×1 with single entry $iω_{12}$ and the curvature 2 form matrix is also 1×1 with single entry $iΩ_{12}$. Its trace is $iΩ_{12}$.

One knows from the axioms for Chern classes that $iΩ_{12}$ is a constant multiple of the Chern class. This constant is $2π$ The constant can be determined from the Chern class of any complex line bundle for instance the tangent bundle of the 2 sphere.

- This description is covered in Milnor's book Characteristic Classes

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mathwonk

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@nrqed: have you looked at theorem 1.2 on pages 8-9 of the last link in post #5? does it seem relevant/useful?

nrqed

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@nrqed: have you looked at theorem 1.2 on pages 8-9 of the last link in post #5? does it seem relevant/useful?
Hello mathwonk,

Thank you for all your comments, I have read them and I want to understand them. I have looked at that theorem and I want to understand this but after having done an explicit calculation using the Levi-Civita symbols. I want to see how all the steps work in a very simple situation, in the most naive and simple minded manner before learning and using more powerful techniques. So I will try to complete the calculation for the sphere using the Levi-Civita connection and then will come back with more question on your posts.

Thanks again!

nrqed

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- Maybe this will clarify this point.

A connection on a complex vector bundle may be thought of as a complex linear map from the space of sections of the bundle $E$ into the the smooth sections of the tensor bundle $TM^{*}⊗E$ that satisfies the Leibniz rule

$∇fs= df⊗s + f∇s$. Here $TM^{*}$ is the space of smooth real linear complex valued 1 forms on the manifold.

A connection on a real vector bundle is defined in the same way except everything is real linear.

If $s_{i}$ is a basis of sections of the bundle over an open neighborhood then

$∇s_{i} = Σ_{j}ω_{ij}⊗s_{j}$ for some one forms $ω_{ij}$. This is a matrix of complex valued $R$ linear 1 forms.

The Levi-Civita connection on a surface is a real(as opposed to complex) connection and with respect to an oriented orthonormal basis $e_{1}$ $e_{2}$ is $∇e_1 = ω_{12}⊗e_2$ and $∇e_{2} = ω_{21}⊗e_2$ with $ω_{21}=-ω_{12}$ and $ω_{11}=ω_{22}=0$. The $ω_{ij}$'s are real valued 1 forms. So the matrix of the connection viewed as an $R$ linear map is skew symmetric with zeros on the main diagonal. Similarly the curvature is a 2×2 skew symmetric matrix of real valued 2 forms. Its trace is zero.

Since this bundle is oriented one can turn it into a complex vector bundle by defining multiplication by $i$ as $ie_{1} = e_{2}$ and $ie_{2} = -e_{1}$ on any oriented orthonormal basis.

The Levi-Civita connection becomes a connection on this complex vector bundle since $∇e_{1} = ω_{12}⊗e_{2} = ω_{12}⊗ie_{1}= iω_{12}⊗e_{1}$ and $∇e_{2} = -ω_{12}⊗e_{1} = ω_{12}⊗ie_{2}= iω_{12}⊗e_{2}$.

The connection 1 form matrix is now 1×1 with single entry $iω_{12}$ and the curvature 2 form matrix is also 1×1 with single entry $iΩ_{12}$. Its trace is $iΩ_{12}$.

One knows from the axioms for Chern classes that $iΩ_{12}$ is a constant multiple of the Chern class. This constant is $2π$ The constant can be determined from the Chern class of any complex line bundle for instance the tangent bundle of the 2 sphere.

- This description is covered in Milnor's book Characteristic Classes
Thank you again, lavinia, I realize that you had already explained all this in post number 3 and I apologize for not having absorbed it back then.
Now I can get the correct result using tetrads and forms, so that's a start. I still want to recover the result using the more "naive" approach of working with the Levi-Civita symbols, but it seems that there is a nontrivial change of basis between the two approaches and I have to sort this out. I will probably be back with very simple questions within a day or two (and then I will be ready to tackle the more advanced concepts that you and mathwonk mentioned).

Thank you both, mathwonk and lavinia.

nrqed

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-

The Levi-Civita connection becomes a connection on this complex vector bundle since $∇e_{1} = ω_{12}⊗e_{2} = ω_{12}⊗ie_{1}= iω_{12}⊗e_{1}$ and $∇e_{2} = -ω_{12}⊗e_{1} = ω_{12}⊗ie_{2}= iω_{12}⊗e_{2}$.
Just a short question: why are the diagonal elements $ω_{11},ω_{22}$ zero? I mean it must be imposed from the get go that $\omega$ must be skew symmetric? What is the reason for this condition? Or is this something that follows from a calculation? I am asking because I am trying to do the calculation the following way (which is what I have learned in a GR course a very long time ago so I might be forgetting some points that would answer my question):

I use for tetrad basis (I am working on the sphere) $e^\theta = r d \theta$ and $e^\phi=r \sin \theta d\phi$. Then I use, for example

$$\nabla e^\theta = d e^\theta + \omega^\theta_j \wedge e^j = \omega^\theta_\theta \wedge e^\theta + \omega^\theta_\phi \wedge e^\phi =0$$

and so on. Now, if I assume $\omega^\theta_\theta= \omega^\phi_\phi=0$ I recover the correct result for everything. But it seems that I have to take these conditions as imposed from the outside, they do not follow from the approach I am following. Am I missing something?

lavinia

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@nrqed: have you looked at theorem 1.2 on pages 8-9 of the last link in post #5? does it seem relevant/useful?
I think one can connect Theorem 1.2 to the Gauss-Bonnet Theorem directly. Choose a Riemannian metric that is compatible with the complex structure on the Riemann surface. Then cover the surface with isothermal coordinates. Isothermal means that $ds^{2}= r_{α}(dx^{2}+dy^2) = r_{α}|dz_{α}|^2$ for some positive function $r_{α}$ on each coordinate domain $U_{α}$.

If $φ$ is a meromorphic 1-form then $φ= f_{α}dz_{α}$ on $U_{α}$ for some meromorphic function $f_{α}$.

One calculates $|f_{α}|^2|dz_{α}|^2 = |f_{α}|^2ds^{2}/r_{α}$. On overlapping domains $φ$ can be written in two ways, $φ=f_{α}dz_{α} = f_{β}dz_{β}$ so

$|f_{α}|^{2}ds^{2}/r_{α} = |f_{β}|^{2}ds^{2}/r_{β}$ or $|f_{α}|^2/r_{α} = |f_{β}|^2/r_{β}$. Rewriting this equation gives $|f_{β}/f_{α}|^2 = r_{β}/r_{α}$ or in the notation of Theorem 1.2 $|λ_{αβ}|^2 r_{α}= r_{β}$ so $r_{α}/|f_{α}|^{2}$ gives a globally defined function $γ$ on the surface as in the statement of Theorem 1.2. By Theorem 1.2 the integral of $d(∂ln(γ)$ is an invariant of the tangent bundle.

In any isothermal chart the curvature form is $(-1/2)Δln(r_{α})dx∧dy$ where $Δ$ is the Laplacian.

Going back to the global function $γ=r_{α}/|f_{α}|^{2}$, $|f_{α}|^{2}$ is harmonic so $(-1/2)Δln(γ)dx∧dy$ is the same as the curvature form away from the zeros and poles of the 1 form.

- In the notation of Theorem 1.2 $(-1/2)Δln(γ)dx∧dy = id(∂ln(γ)$. This formula says that the curvature form is exact away from the zeros and poles of the meromorphic 1 form. In general the Euler class is exact over the domain of a non-zero section of an oriented vector bundle.

- If one did not know the Gauss Bonnet theorem, then Theorem 1.2 would imply that the integral of the curvature form is an invariant of the tangent bundle equal to $2π$ times the Chern class.

- If one knows by the Gauss- Bonnet Theorem then Theorem 1.2 implies that the Chern class of the tangent bundle is $2π$ times the Euler characteristic of the surface.

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romsofia

and so on. Now, if I assume $\omega^\theta_\theta= \omega^\phi_\phi=0$ I recover the correct result for everything. But it seems that I have to take these conditions as imposed from the outside, they do not follow from the approach I am following. Am I missing something?
They follow from metric compatibility! $\omega_{ij} = -\omega_{ji}$ if $i=j$ it follows that $\omega_{ii}=-\omega_{ii} = 0$ That is how I understand it anyway, there might be some more rigorous underlying condition.

lavinia

Gold Member
They follow from metric compatibility! $\omega_{ij} = -\omega_{ji}$ if $i=j$ it follows that $\omega_{ii}=-\omega_{ii} = 0$ That is how I understand it anyway, there might be some more rigorous underlying condition.
From metric compatibility one has for the orthonormal basis $e_1$ $e_2$

$0 = e_{1}⋅<e_1,e_{1}> = 2 <∇_{e_1}e_1,e_1> = 2 <ω_{11}(e_1)e_1 + ω_{12}(e_1)e_2,e_1>= ω_{11}(e_1)$ and
$0 = e_{2}⋅<e_1,e_{1}> = 2 <ω_{11}(e_{2})e_{1} + ω_{12}(e_{2})e_{2},e_{1}>= ω_{11}(e_{2})$

So $ω_{11}=0$ since it is zero on a vector basis.
Similarly $ω_{22}=0$

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nrqed

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From metric compatibility one has for the orthonormal basis $e_1$ $e_2$

$0 = e_{1}⋅<e_1,e_{1}> = 2 <∇_{e_1}e_1,e_1> = 2 <ω_{11}(e_1)e_1 + ω_{12}(e_1)e_2,e_1>= ω_{11}(e_1)$ and
$0 = e_{2}⋅<e_1,e_{1}> = 2 <ω_{11}(e_{2})e_{1} + ω_{12}(e_{2})e_{2},e_{1}>= ω_{11}(e_{2})$

So $ω_{11}=0$ since it is zero on a vector basis.
Similarly $ω_{22}=0$
Ah, yes, of course! Thank you so much. I cannot express how much your help has been useful! Especially this issue of the complexification, which was driving me crazy since I kept getting a vanishing Chern class. Thank you for all the time you took to explain things, it is very generous of you.

I have a couple of (hopefully) quick related questions.

First, am I right then in concluding that one cannot talk about the Chern class of the Mobius band since it is not orientable?

Second, what happens if one tries to use complex coordinates from the beginning? I am still considering CP1 and the tangent bundle. Let's say I use for coordinate z (and w=1/z on the second patch). Naively it seems that I will necessarily get zero for the curvature two form, since it seems to me that the connection has to be of the form $f(z) dz$, so both $d A$ and $A \wedge A$ are trivially zero. What am I missing? (maybe the connection will be of the form $f(z) dz + g(\bar{z}) d \bar{z}$? But I would not expect that).

lavinia

Gold Member
First, am I right then in concluding that one cannot talk about the Chern class of the Mobius band since it is not orientable?
That is correct.

Every complex vector bundle is orientable (proof?). So the tangent bundle of the Mobius band can not have a complex structure.

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lavinia

Gold Member
Ah, yes, of course! Thank you so much. I cannot express how much your help has been useful! Especially this issue of the complexification, which was driving me crazy since I kept getting a vanishing Chern class. Thank you for all the time you took to explain things, it is very generous of you.

I have a couple of (hopefully) quick related questions.

Second, what happens if one tries to use complex coordinates from the beginning? I am still considering CP1 and the tangent bundle. Let's say I use for coordinate z (and w=1/z on the second patch). Naively it seems that I will necessarily get zero for the curvature two form, since it seems to me that the connection has to be of the form $f(z) dz$, so both $d A$ and $A \wedge A$ are trivially zero. What am I missing? (maybe the connection will be of the form $f(z) dz + g(\bar{z}) d \bar{z}$? But I would not expect that).
Try the metric $ds^2= 4|dz|^2(1+|z|^2)^{-2}$ . I think this is the standard metric on the sphere.

lavinia

Gold Member
T

The upshot is that understanding the connexion / curvature version of chern classes is perhaps best achieved by studying the proof of the general gauss bonnet theorem, e.g. in Griffiths Harris. In general while I sometimes find that source challenging in terms of precise technicalities, it is often highly valuable for intuitive insight.
Another way to get comfortable with the Weil homomorphism is to see that the curvature forms satisfy the axioms for Chern classes.

Once one knows that the cohomology class of the Chern form is independent of the connection then one has well defined cohomology classes associated to each smooth complex vector bundle. One then shows that these classes satisfy the axioms for Chern classes.

The axioms are

Naturality: The Chern classes of an induced complex vector bundle bundle are the pull backs of the Chern classes of the target bundle.

There is a pull back connection on the induced bundle and its Chern forms are the pull backs of the Chern forms on the target bundle.

Product Formula: Given two complex vector bundles with connections then there is a connection on their Whitney sum whose Chern classes are the products of the Chern classes of the two bundles separately.

- The Chern form of a trivial bundle is cohomologous to zero.

- The Chern form of the canonical line bundle over projective space is cohomologous to its Chern class.

It is instructive to see that there is a Whitney sum formula for connections and that connections are natural. To me anyway, this is not obvious and it makes the Weil homomorphism seem less "miraculous' as you have said.

Example: The Gauss mapping of a closed surface in 3 space into the unit sphere, is covered by a morphism of tangent bundles which parallel translates tangent vectors at a point on the surface to the point on the sphere determined by the unit normal. These translated vectors are tangent to the sphere because they are perpendicular to the unit normal.

This bundle morphism shows that the tangent bundle of the surface is the pull back of the tangent bundle of the sphere. It follows that the pull back of the Chern form - which is the volume form of the sphere divided by 2π - is the Chern form of the induced connection. Since the determinant of the Gauss map is the Gauss curvature - classically this was the definition of Gauss curvature - the Chern form of the surface is the (1/2π) Gauss curvature times the area element of the surface.

So Naturality of the Chern classes immediately leads to forms in the curvature of a connection.

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lavinia

Gold Member
@mathwonk

- What source do you recommend to follow the theory of meromorphic forms?

- If I understand the link correctly, a divisor determines a line bundle and the ratios $f_{α}/f_{α}$ of local meromorphic function on intersections $U_{α}∩U_{β}$ are coordinate charts and equivalent divisors determine the same bundle.

- So equivalent divisors are Poincare dual to the Chern class of the line bundle. As a cohomology class in $H^2(M : Z)$ Is there a way to define the Chern class directly?

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