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Suppose we have a family of compact oriented even dimensional spin manifolds $\{Y_x\}$ parameterized by a compact even dimensional manifold $X$. The $Y_x$'s are all diffeomorphic to some $Y$, of dimension $n$, and fit together to form a fiber bundle $\pi : Z \rightarrow X$ with fiber $Y_x=\pi ^{-1}(x)$. $TZ$ has the subbundle $V:=\text{ker }\pi_*$ which is tangent to the fibers. There may be a family of coefficient bundles also and we obtain a family of twisted Dirac operators $D_x:\Gamma(S^+_x\otimes E_x)\rightarrow \Gamma (S^-_x\otimes E_x)$. The index of the family gives rise to an element $\text{ind} D \in K(X)$, which is the virtual vector bundle $[\text{ker } D_x]-[\text{coker }D_x]$ when the dimension of both spaces are constant. Finally, there is a map $\text{H} ^{\*}(Z,\mathbb{R})\rightarrow \text{H} ^{\*-n}(X,\mathbb{R})$ known as the Gysin homomorphism or integration over the fibers map. We'll use the latter terminology writing the map $\int_Y$ and regarding cohomology classes as living in de Rham cohomology. The Atiyah-Singer index theorem gives

$$\text{ch }(\text{ind } D)= \int _Y \hat A (V) \text{ch}(E)$$


What general results exist regarding the components of the Chern character of the index bundle, or equivalently the results of the integration over the fibers map, for twisted Dirac operators?

To illustrate, an immediate answer is that the zero cohomology (virtual rank) is the index of the Dirac operator on $Y$. A more interesting answer is that in some cases that might be all one obtains: it is a result of Borel-Hirzebruch that the signature is strictly multiplicative in all bundles where $\pi_1$ of the base acts trivially on the rational cohomology of the fibers. The signature is the index of a certain twisted Dirac operator. If we have a family of these operators such that $Z\rightarrow X$ satisfies the condition involving the fundamental group, then the strict multiplicativity gives $\text{ch}(\text{ind }D)=\int_Y \hat A (V)ch(E)=\text{sign }(Y)$. A priori one could expect higher degree cohomology classes. It seems interesting that these vanish.

If the question is too vague or broad, I would be happy knowing

Are there any instances in which there are known relations between the Chern character of the index bundle and the Chern classes of $X$?

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The answer to your second question is no. The smooth structure on the base space $X$ is completely irrelevant for index theory on fibre bundles on $X$. See Atiyah-Singer "Index of elliptic operators IV". –  Johannes Ebert Apr 1 '11 at 9:04
    
That's good to know. Thank you for your response! I will read up on it some more. –  charris Apr 1 '11 at 14:33
    
Pontryagin classes only depend on the topological structure of the manifold. Is it ever possible that there could be relations between the Chern character of the index bundle and the Pontryagin classes on $X$, or am I still missing something? –  charris Apr 2 '11 at 1:15
    
Yes, you are missing something. The Atiyah-Singer family index theorem holds for any compact Hausdorff base $X$. The definition of a smooth bundle and a fibrewise elliptic operator needs to be adjusted a bit. Only the homotopy type of $X$ and isomorphism class of the smooth fibre bundle matters. Of course, if $X$ is a closed manifold, then so is $Z$. An elliptic operator $D$ on $Z$ restricts to an elliptic operator $D_{fib}$ on the fibres. Of course $ind(D)$ and the index bundle of $D_{fib}$ are related. –  Johannes Ebert Apr 2 '11 at 9:19

1 Answer 1

Your question is not so clearly stated; so I take the opportunity to interpret it and write a little essay.

As far as I understand your question, you found out that the multiplicativity result by "Borel-Hirzebruch" (I think it is actually due to Hirzebruch-Chern-Serre) implies that the higher Chern character of the index bundle of the signature operator are zero and now you want to see a more conceptual explanation. You should have a look into Atiyah's "The signature of fibre bundles". The following is definitly in the spirit of that paper, though I did not find the statement there.

Theorem: "Let $Z \to X$ be an oriented smooth fibre bundle with fibre $Y$, closed, of dimension $4k$. Pick a fibrewise Riemann metric and let $D$ be the fibrewise signature operator. Assume that $\pi_1 (X)$ acts trivially on $H^{2k}(Y; \mathbb{R})$. Then the index bundle of the signature operator $ind(D)$ is trivial of virtual rank $sign(Y)$. Moreover, if the image of $\pi_1 (X)$ in $Gl(H^{2k} (Y); \mathbb{R})$ is finite, then the components of $ch(ind(D))$ in positive degree are zero."

Proof:

  1. A linear algebra fact. Given a finite-dimensional real vector space $V$ with a nondegenerate symmetric bilinear form $b$; of signature $(p,q)$. Consider the space $Q(b)$ of all pairs $(V_{+},V_{-})$, such that $\pm b|_{V_{\pm}} $ is positive definite and $W_{-} \oplus W_{+}=V$. $Q$ is a subspace of the product $Gr_p (V) \times Gr_q(V)$ of two Grassmannians. $Q(b)$ is diffeomorphic to the homogeneous space $O(p,q)/(O(p)\times O(q))$, and $O(p) \times O(q) \subset O(p,q)$ is a maximal compact subgroup. Therefore $Q$ is contractible (there should be a more direct proof, though).

  2. Assume that the $\pi_1$-action on $V:=H^{2k} (Z_x; \mathbb{R})$ is trivial. There is the intersection form $b$ on $V$; pick a decomposition $V=W_{+} \oplus W_{-}$ as before. Now consider the bundle $E:=H^{2k} (Z/X) \to X$; the fibre over $x$ is the cohomology $H^{2k} (Y_x; \mathbb{R})$. Because the $\pi_1$-action is trivial, the bundle is trivial and the splitting above gives a splitting $E=F_{+} \oplus F_{-}$ into two trivial subbundles.

  3. Now pick a Riemann metric on the fibres. The Hodge theorem identifies $E$ with the bundle of harmonic $2k$-forms and the Hodge star $\ast$ is an involution and it splits $E= E_{+} \oplus E_{-}$ into the sum of eigenspaces. More or less by definition, $ind (D) = [E_{+}]-[E_{-}]$. Now I claim that $E_{\pm}\cong F_{\pm}$. This is because both decompositions can be viewed as sections to the bundle $X \times Q(b)$ and since $Q(b)$ is contractible, they are homotopic.

  4. If the image of the monodromy is a finite group, say $\mu: \pi_1 (X) \to G$, you pass to the finite cover $\tilde{X} \to X$ corresponsing to the kernel of $\mu$. The pullback of $Z$ to $\tilde{X}$ has trivial monodromy and so the signature index bundle is trivial. But the induced homomorphism $H^{\ast} (X; \mathbb{Q}) \to H^{\ast}(\tilde{X}, \mathbb{Q})$ is injective, which allows to reduce the argument to the case of a trivial action. QED

After I have written all this, I realize that the main point is that the bundle $E$ has structural group $O(p,q)$ and it is flat as such a bundle. It allows a reduction of the structural group to $O(p) \times O(q)$ (by the contractibility of $Q(b)$), but not as a flat bundle. If you find a flat reduction, then $ch(Ind(D))=0$ (in positive degrees) by Chern-Weil theory. The index theorem is not really relevant here; it identifies $ch(ind(D))$ with the fibre integral of the $L$-class of the vertical tangent bundle.

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Thanks for the great response. Sorry my question wasn't completely clear. I was wondering if there were other situations like we're discussing where something "interesting" happens like the index bundle is actually trivial. But I was very interested in a conceptual understanding in this case, so your answer should really help me out a lot. I could be wrong, but I don't think the triviality of the index bundle (immediately) follow from the H-C-S result. Multiplicativity of the signature doesn't require strict multiplicativity (that the fiber integral ends up only in degree 0). –  charris Apr 2 '11 at 0:53

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