I am going to ask a question, which could be a stupid one.

I am reading a paper "an index theorem in differential K-theory". The first paragraph of section 8.28 recalls a filtration of K-theory $K_{(i)}(X)$, let say we work with $K^0$, which consists of elements $x\in K(X)$ with the property that pulling back via any continuous map $f:Y\to X$, where $Y$ is any finite simplicial complex of dimension less than $i$, is 0, i.e., $f^*x=0$. I am ok with this definition.

In the next paragraph it says if $[E]\in K_{(i)}(X)$, then the degree $i$ component $\textrm{ch}^{(i)}(E)$ of the Chern character $\textrm{ch}(E)$, which lives in $H^{2i}(X; \mathbb{Q})$, can be refined to an integral cohomology class $H^{2i}(X; \mathbb{Z})$. I don't understand, but this fact seems to be trivial to others. I think I missed something, and please let me know why this is true.

Thanks a lot.


You seem to misunderstand the grading a bit. Look at $X=CP^2$ and the $K$-theory class $H-1$, where $H$ is the Hopf bundle. Then $ch(H-1)= 1+z+\frac{z^2}{2}-1$, with $z$ a generator of $H^2 (CP^2;Z)$. Let $i=2$. Each map $Y \to CP^2$ from an $(i-1)$-dimensional $Y$ is nullhomotopic, since $CP^2$ is simply connected. Therefore, $H-1$ belongs to $K^{0}_{(2)} (CP^2)$. On the other hand, the component of $ch(H-1)$ in $H^4(CP^2)$ does not refine to an integral class.

What I can offer you is an explanation of the fact that the $i$th component of the Chern character of an element $x \in K^{0}_{(2i)} (X)$ is integral. Assume, for simplicity, that $X$ is a finite CW complex. Let $f: X \to BU$ be a classifying map for the $K$-theory class $x$. The assumption says that $f|_{X^{(2i-1)}}$ is nullhomotopic. By the homotopy extension property, $f$ factors through the $(2i-1)$-connected complex $Y=X/X^{(2i-2)}$. By the universal coefficient theorem in cohomology, it is enough to prove that $ch_i$ evaluates to an integer on each integral homology class $a \in H_{2i}(Y, Z)$. By Hurewicz, it is enough to verify that for each $g: S^{2i} \to Y$, $\langle g^{\ast} ch_i, [S^{2i}]\rangle$ is integral.

The arguments so far reduce the question to the following result:

Theorem: let $V \to S^{2n}$ be a complex vector bundle. Then $\langle ch(V), [S^{2n}] \rangle$ is integral.

This is a highly nontrivial result and it is not really elucidated by taking appropriate definitions of the Chern classes. In fact, you need either Bott periodicity or the index theorem for Dirac operators. Here are the two alternative arguments.

Bott periodicity theorem (this is done in, say, somewhere in the book ''Topology of Lie groups'' by Mimura and Toda). You have to prove that the Chern character $K^{0}_{cpt} (C^n) \to H_{cpt}^{ev} (C^n;Q)$ takes integral values. Since the Chern character is multiplicative, this amounts to proving that the Bott class $b \in K^{0}_{cpt} (C)$ has $ch(b)$ integral (this reduction uses Bott periodicity!). But $ch(b)\in H^{2}_{cpt} (C)$ is the generator, by direct computation!

Index theorem: The Hirzebruch class of $S^{2n}$ is $\mathcal{L}=1$. Take the signature operator $D_V$, twisted by the bundle $V$. The index is

$$ ind (D_V)= \langle \mathcal{L}(TS^{2n}) ch (V); [S^{2n}]\rangle= \langle ch (V); [S^{2n}]\rangle $$

and the left-hand side is integral. Note that the version of the index theorem that can be proved using the heat kernel is enough.

  • $\begingroup$ In your example of Hopf bundle, you mean $H-1\in K_{(2)}^0(\mathbb{C}P^2)$, right? $\endgroup$ – GRR Oct 23 '13 at 7:19
  • $\begingroup$ Yes, and I fixed it. $\endgroup$ – Johannes Ebert Oct 23 '13 at 15:58

This seems to follow from the fact that the Chern classes are integral, an observation which is more mysterious using some definitions than others. For instance, I know of no direct way to prove that the Chern classes defined using Chern-Weil theory are integral; I would be very pleased if someone knows a way.

Using other definitions it is at least more standard; see the answers to this question, for instance: Cohomology ring of BG.

  • $\begingroup$ So the filtration doesn't play a role anyway, right? $\endgroup$ – ThiKu Oct 22 '13 at 13:56
  • $\begingroup$ The filtration is important for the statement. @Paul: the leading term of the Chern character is $1/n! c_n +$ stuff that involves nontrivial products of Chern classes, so there is no easy reason why it is integral, see my answer. $\endgroup$ – Johannes Ebert Oct 22 '13 at 21:18
  • $\begingroup$ To prove that the Chern classes from the Chern-Weil formalism are integral is much simpler, even if you never heard about integral homology. I figured it out yesterday, preparing a lecture course. Here is the argument. Let $V \to M$ be a complex vector bundle on a closed $2n$-manifold and let $p$ be any monomial in the Chern classes of $V$. Then $\int_M p$ is integral. Proof by induction on $rk(V)$. If $V$ is a line bundle, you have to show that $\int_M c_{1}(V)^n$ is integral. Let $f: M \to CP^n$ be a classifying map for $V$ (cellular approximation or general position creates such a map). $\endgroup$ – Johannes Ebert Oct 22 '13 at 21:24
  • $\begingroup$ Then $\int_M c_{1} (V)^n= deg (f) \int_{CP^n} c_{1}^{n}$. The latter integral can be evaluated explicitly and the value is $deg (f)$ (slightly lengthy, but direct). For the inductive step, consider $q:PV \to M$, the projectivization of the rank $r$ bundle $V$. Then $q^{\ast} V$ splits as $L \oplus W$, $L$ a line bundle. Let $z$ be the first Chern class of $L$. Then $\int_M x = \int_{PV} q^{\ast} x z^{r-1}$ for each class $x$ on $M$. But $q^{\ast} p$ is an integral linear combination of terms of the form $z^i p_i$, where $p_i$ is a monomial in the Chern classes of $W$. $\endgroup$ – Johannes Ebert Oct 22 '13 at 21:29
  • $\begingroup$ Now let $L \to N$ is a line bundle on a manifold and $p$ a monomial in the Chern classes of a rank $r-1$ vector bundle $V \to N$. Let $S \subset N$ be the zero locus of a regular section of $L^{\oplus s}$. Then $\int_N c_{1}^{s} p= \int_S p$. This formula provides the inductive step. $\endgroup$ – Johannes Ebert Oct 22 '13 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.