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I was searching for online sources on chern classes.

One version is that given a vector bundle $E$ over space $M$, $i^{th}$ chern class is an element of cohomology $H^{2i}(M,\mathbb{Z})$.

Another version is that given a vector bundle $E$ over space $M$, $i^{th}$ chern class is an element of cohomology $H^{2i}(M,\mathbb{R})$.

I got confused sufficiently and came to a conclusion that in first case $M$ was just a topological space(manifold) in which case there is only one obvious notation of cohomology that is singular cohomology with standard choice of coefficients, integers. In second case they are considering smooth manifolds. So, there is a notion of differential forms and deRham cohomology and they are considering deRham cohomology.

But then this Wikipedia article https://en.m.wikipedia.org/wiki/Chern_Weil_homomorphism in subsection chern classes and chern characters says chern class is an element of $H^{2i}(M;\mathbb{Z})$. Moreover it says it is an element in image of chern Weil homomorphism where as chern Weil homomorphism has its codimain as deRham cohomology ring.

Can some one help me to clarify this confusion.

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    $\begingroup$ Cohomology with real coefficients is a graded vector space over $\mathbb{R}$, and it contains cohomology with integer coefficients as a graded lattice. It is a fact (either part of the definition or a theorem, depending on how you set things up) that the Chern classes lie in this lattice, so there is no inconsistency between the various statements in your question. $\endgroup$ Commented Jan 24, 2018 at 20:36

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Chern classes can be defined in multiple different but roughly equivalent ways. As you have seen, they can be defined to be certain classes in $H^{2i}(X;\mathbb{Z})$, or they can be defined to be certain classes in $H^{2i}(X;\mathbb{R})$ (in the latter case, usually because they arise naturally via de Rham cohomology). However, these two definitions are very closely related and there is usually no confusion in calling both of them "Chern classes".

To relate the definitions, recall that any space $X$, there is a natural homomorphism $H^*(X;\mathbb{Z})\to H^*(X;\mathbb{R})$, induced by the inclusion map $\mathbb{Z}\to\mathbb{R}$. The two types of Chern classes mentioned above are related by this natural homomorphism: if you take the Chern class in $H^{2i}(X;\mathbb{Z})$ and apply the natural homomorphism $H^{2i}(X;\mathbb{Z})\to H^{2i}(X;\mathbb{R})$, you get the Chern class in $H^{2i}(X;\mathbb{R})$. When people talk about classes in de Rham cohomology as actually being in $H^{*}(X;\mathbb{Z})$, they really mean they are in the image of this homomorphism.

In general, beware that the homomorphism $H^{2i}(X;\mathbb{Z})\to H^{2i}(X;\mathbb{R})$ may not be injective (the kernel is the torsion elements of $H^{2i}(X;\mathbb{Z})$), and so the integral Chern classes are "stronger" than the real Chern classes. For instance, a complex line bundle is determined up to isomorphism by its first integral Chern class, but it is not determined by the first real Chern class (if $H^2(X;\mathbb{Z})$ has torsion).

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  • $\begingroup$ Can you confirm what cohomology we are talking about when we write $H^*(X;\mathbb{Z}$. It is the singular cohomology right? $\endgroup$
    – user312648
    Commented Jan 25, 2018 at 2:39
  • $\begingroup$ Yes, $H^*(X;\mathbb{Z})$ is singular cohomology. $H^*(X;\mathbb{R})$ is either singular or de Rham cohomology (it doesn't make a difference as long as $X$ is a manifold so they are both defined, since they are naturally isomorphic by the de Rham cohomology). $\endgroup$ Commented Jan 25, 2018 at 2:48
  • $\begingroup$ Ok ok. Thank you,... You said, When people talk about classes in de Rham cohomology as actually being in $H^*(x;\mathbb{Z}$ they really mean they are in the image of this homomorphism... in fourth axiom in giving axiomatic definition of chern classes (in Kobayashi Nomizu, Vol $2$, page 307) they say that $-c_1(E_1)$ is the generator of $H^2(P_1(\mathbb{C};\mathbb{Z})$ where $E_1$ is the complex line bundle over $p_1(\mathbb{C})$ that they have constructed before.. That was also a confusion, now it is clear... $\endgroup$
    – user312648
    Commented Jan 25, 2018 at 3:06
  • $\begingroup$ I was confused that they were giving axioms saying chern classes are elements of $H^{2i}(X;\mathbb{R})$ and suddenly they said for complex line bundle $-c_1(E_1)$ is generator of $H^*(P_1\mathbb{C})$. So, you are saying they mean $-c_1(E_1)$ as an element of $H^2(P_1\mathbb{C};\mathbb{R}) $ is just the image of generator of $H^2(P_1\mathbb{C};\mathbb{Z})$ under the homomorphism you have defined. Is it correct. $\endgroup$
    – user312648
    Commented Jan 25, 2018 at 3:13
  • $\begingroup$ That's correct. $\endgroup$ Commented Jan 25, 2018 at 3:22

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