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Classicly, for a spin Riemannian manifold $M$, the $\hat{A}(M)$ genus will be $0$, if the scalar curvature is positive.

The proof is to use the Lichnerowicz formula. we have the index of the Dirac operator will be $0$, i.e., $$Ind(D_+)=0.$$ On the other hand, by the index theorem of Atiyah and Singer, we have $$Ind(D_+)=\hat{A}(M).$$ So we get $$\hat{A}(M)=0.$$

My question is "Can we have a different method to proof this result? without using the Lichnerowicz formula or without using the index theorem" Maybe a formal proof or explanation.

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Do you mean without use of the Lichnerowicz formula and without use of the index theorem? Or do you mean without use of the Lichnerowicz formula but with use of the index theorem? –  Kelly Davis Sep 28 '11 at 12:05
Schoen and Yau dealt with positive scalar curvature obstructions using variational methods, but I don't think they identified the $\hat{A}$-genus explicitly. The problem is, as far as I know, the $\hat{A}$-genus doesn't have a geometric interpretation other than that it is the index class for the Dirac operator. –  Paul Siegel Sep 28 '11 at 13:34
Hi, Kelly Davis. I want to say "without using of the Lichnerowicz formula or without using of the index theorem". –  shu Sep 28 '11 at 14:07
I'm agree with Paul Siegel. One of the motivation to pose this question is that I want to understand the $\hat{A}$ genus by some geometric methods –  shu Sep 28 '11 at 14:11
I would guess the answer would go something like: write the $\hat{A}$ genus in terms of the curvature two-form, then use a diagonalization of the curvature two-form, positivity of the scalar curvature, and the fact that $\hat{A}$ is an integer, i.e. $M$ is spin, to prove that the only integer $\hat{A}(M)$ can be is $0$. Admittedly, I have not thought about the last steps in detail, but with all the factors of 1/24 and 1/5760 and...in the $\hat{A}$ genus I'd guess it'll work. –  Kelly Davis Sep 28 '11 at 15:26

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up vote 16 down vote accepted

After thinking about this question for two hours, my belief is strengthened that there cannot be a proof of the vanishing of the A-genus for spin manifolds with positive scalar curvature without using both the index theorem for the Dirac operator and the Lichnerowicz formula (or other analytic techniques). In the comments to the question, a proof using the integrality of the $\hat{A}$-genus and Chern-Weil theory has been suggested. I wish to argue that there is absolute no reason to believe such a proof might work out.

\textbf{Objections to the use of the integrality theorem.} The manifold $CP^2$ has positive scalar curvature and $\hat{A}$-genus $1 /16$. It is of course not spin. Take the connected sum of 16 copies of $CP^2$ and call it $M$. The $\hat{A}$-genus of $M$ is $1$; $M$ is still not spin. The famous surgery lemma by Gromov and Lawson implies that $M$ has a metric of positive scalar curvature, because $M$ is obtained from the manifold $\coprod_{16} CP^2$ by a sequence of surgeries of codimension $4 \geq 3$. So there are connected pscm manifolds with integral but nonzero $\hat{A}$-genus. More generally, it can be shown that any simply connected, high-dimensional, oriented and non-spin manifold admits a metric of positive scalar curvature. This also goes back to Gromov and Lawson, and Sven F\"uhring worked out the details in his thesis:


This result shows that the integrality of the $\hat{A}$-genus has nothing to do with positive scalar curvature and that the spin hypothesis has to be used in a more subtle way.

\textbf{Objections to the use of Chern-Weil theory:} At a first glance, it seems tempting to use Chern-Weil theory for such questions, because it expresses characteristic classes in terms of curvature. Moreover, a conceptual proof of the Gauss-Bonnet theorem can be given with help of this theory. But an examination of the argument quickly shows that this is a lucky accident. The first accident is that the Euler class has a direct geometric-topological meaning, which allows to relate it to another topological invariant (the Euler number), while the complicated rational polynomial in the Pontrjagin classes has no such meaning (Paul Siegel alluded to this point in his comment).

The second accident is that the integrand of the Gauss-Bonnet-Chern theorem only depends on the curvature tensor and not on the metric itself (though to define the integrand, one needs to know that the connection preserves \emph{some} metric).

The problem with scalar curvature is that it depends on both, the connection and the metric, while the Chern-Weil forms only depend on the connection and not on the metric.

Of course there is a universal formula for the $\hat{A}$-form of any Riemann manifold $(M,g)$ of the form $\hat{A} = f(jet^2 g) \cdot vol_g$, where $f$ is an algebraic function of the $2$-jet of the metric. I have not worked out the concrete formula for $f$; for $4$-dimensional manifolds, this should not be too complicated and might be a worthy exercise. I predict that you won't find a formula for $f$ that depends on the scalar curvature in a controllable way. Another remark is that if such a formula gives a result for positive scalar curvature, it should give a result for negative scalar curvature as well. On the other hand, any manifold of large dimension has a metric of negative scalar curvature (Lohkamp).

The conclusion I draw is that Chern-Weil theory is insensitive to the sign of scalar curvature. The next problem is that Chern-Weil theory is insensitive to the spin condition as well: the map $Sym^{\ast} (\mathfrak{so}_n)^{SO(n)}\to Sym^{\ast} (\mathfrak{spin}_n)^{Spin(n)}$ on the space of invariant polynomials is an isomorphism. In other words, the spin condition does not yield any relation between Chern-Weil classes. The integrality of the $\hat{A}$-genus cannot be obtained by a local consideration; this would equate it to another manifestly integral class, which is not the case.

The view that positivity should enter the proof in a more global way is furthermore supported by Gromov's h-principle, a special case of which states that any open manifold (no compact component) has a metric of positive scalar curvature. This implies that any obstruction to positive scalar curvature has to arise in an essential global way.

My conclusion of the above discussion is that the integrality result forgets the relevant information from the spin condition and that Chern-Weil theory is not the correct recipient for the information coming from the positivity of the curvature. Philosophically, integrality is too global to capture the spin condition and Chern-Weil theory is too local to capture the sign of the curvature.

It is a pity that Misha Gromov is not active on this site.....

EDIT: here is an example which shows that any hope to use Chern-Weil theory to obtain information on scalar curvature is in vain.

Fact 1: Let $\nabla_g$ be the Levi-civita connection of the Riemann manifold $(M,g)$, then for all $c>0$, $\nabla_{cg} = \nabla_g$. Fact 2: the scalar curvature of $(M,cg)$ is $c^{-1} scal_g$. Fact 3: the scalar curvature of a product is the sum of the scalar curvature.

Now take $S^2$, $g_0$ the standard metric with $scal_{g_0}=1$; $F$ a genus $\geq 2$ surface with metric $g_1$ and $scal_{g_1}=-1$. Then the product $M=S^2 \times F$ with product metric $(c_0 g_0 )\times (c_1 g_1)$ has scalar curvature $c_{0}^{-1} - c_{1}^{-1}$. Suitable choices of $c_i$ yield any real value for the scalar curvature. But by fact 1, the Levi-Civitta connection and hence the Chern-Weil invariants do not depend on $c_i$.

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There are even more dramatic statements which demonstrate that the vanishing of the $\hat{A}$ genus due to positive scalar curvature could not possibly be explained locally. Suppose that $M$ is a manifold which is quasi-isometric to $\mathbb{R}^n$ and that $M$ has an open set $U$ quasi-isometric to a half-space on which the scalar curvature is uniformly positive. Then the $\hat{A}$ genus still vanishes. (In fact $U$ can be quite a bit smaller, though not too small.) –  Paul Siegel Sep 29 '11 at 1:48
Do you mean that the universal cover of $M$ is quasiisometric to euclidean space? –  Johannes Ebert Sep 29 '11 at 7:04
@Johannes Cool. In the 5min I thought about the problem I didn't try to think at all about counter examples, just wrote the first thing that popped into my head in the way of a proof sketch. The problem is more difficult than I expected –  Kelly Davis Sep 29 '11 at 8:30
Thank you Johannes Ebert! I like this answer! –  shu Sep 29 '11 at 18:58
Oops - I did indeed mean the universal cover of $M$. –  Paul Siegel Oct 3 '11 at 14:41

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