# Pontryagin Forms and Special Holonomy

Let $(M,g)$ be a Riemannian manifold. Recall that the $k^{th}$-Pontryagin class is a topological invariant which, by classical Chern-Weil theory, can be represented using the so-called Pontryagin forms. These forms are obtained by appropriate constant linear combinations and wedge multiplications of the closed $4p$-forms $\operatorname{Tr}(R^{p})$, where

$$\operatorname{Tr}(R^{p})(X_1,....,X_{4p}):=\sum_{\sigma}(-1)^{|\sigma|}\operatorname{Tr}(R(X_{\sigma(1)},X_{\sigma(2)})\circ\cdots\circ R(X_{\sigma(4k-1)},X_{\sigma(4p)}))$$

and $R$ is the curvature tensor of $(M,g)$. It is known that the Pontryagin forms depend only on the Weyl part of the curvature tensor (in particular they vanish for conformally flat manifolds). In concrete cases, they can in principle be explicitly computed.

However I am interested in the case where $M$ has special holonomy (contained in) $SU(5)$, i.e. it is a not necessarily compact Calabi-Yau $5$-fold. My question is:

1. Are there any conditions under which the Pontryagin forms (or, equivalently, the above trace forms) are proportional to the appropriate powers of the Kahler form $\omega$?

2. More generally, are there any conditions under which the Pontryagin forms are parallel?

Note that the $k^{th}$-Pontryagin form of a Kahler manifold is always of type $(2k,2k)$ so that, when the value $h^{2k,2k}$ of the Hodge diamond is one, this form is necessarily a multiple of $\omega^{2k}$ up to exact terms.

Any example using algebraic geometry or any argument which might enlighten some properties (if any) of Pontryagin forms for manifolds with special holonomy in the sense of Berger's classification are appreciated.

Here are a few remarks to show that your Questions 1 and 2 are almost equivalent. (What the answers are is another matter, and I expect that to be somewhat difficult, as I'll explain below.) Of course, Question 1 is a special case of Question 2 in the OP's situation because, if a closed form is a scalar multiple of $\omega^2$ or $\omega^4$ on $M^{10}$, then that scalar multiple must be constant, and hence the form is parallel. What's interesting is that, in spite of appearances, having the Chern-Weil representatives of the Pontryagin forms be parallel when the holonomy is a subgroup of $\mathrm{SU}(5)$ is nearly the same as having these forms be scalar multiples of $\omega^{2k}$.

First, for Question 2, you can reduce to the case in which the holonomy acts irreducibly: If $(M,g)$ is a (simply-connected) Riemannian manifold whose holonomy preserves a parallel splitting $TM = V_1\oplus V_2$, then, locally, $g$ is a product metric and the formula for the total Pontryagin class $p(TM) = p(V_1)p(V_2)$ and that fact that the holonomy is the product of the holonomies acting on the two subbundles $V_1$ and $V_2$ shows that the Chern-Weil representatives of the classes in $p(TM)$ are parallel with respect to the holonomy of the product if and only if each of the two factor metrics has the property that its Chern-Weil representatives of the classes $p(V_i)$ are also parallel. By reduction, then, it suffices to classify the metrics $(M,g)$ for which the Pontryagin forms are parallel and the holonomy acts irreducibly.

Second, although you say you are mainly interested in the case of an $(M^{10},g)$ whose holonomy is (contained in) $\mathrm{SU}(5)$, you might as well generalize your question to considering the case of $(M^{2n},g)$ with holonomy (contained in) $\mathrm{SU}(n)$. Since, by the first remark, we can assume the holonomy acts irreducibly, that means, by the Berger classification, that the holonomy has to be either $\mathrm{SU}(n)$ or, if $n$ is even, $\mathrm{Sp}(n/2)$.

In the case the holonomy is $\mathrm{SU}(n)$, the only parallel forms of type $(2k,2k)$ are the (even) powers of the Kähler form $\omega$, so, in this case, having the Pontryagin forms be parallel implies having them be (constant) multiples of the powers of $\omega$. Meanwhile, for $k<n$ the only multiples of $\omega^k$ that are closed are the constant multiples. In particular, when $n=5$, if the Chern-Weil representative of $p_1(M)$ (respectively,$p_2(M)$) is a multiple of $\omega^2$ (respectively, $\omega^4$), then it is a constant multiple, and hence parallel.

In the case the holonomy is $\mathrm{Sp}(n/2)$, things are a little more complicated: In this case, the algebra of parallel forms is generated by $\omega$, $\phi$, and $\bar\phi$, where $\phi$ is a parallel holomorphic $(2,0)$-form. Thus, the parallel $(2k,2k)$-forms are constant linear combinations of the forms $\omega^{2(k-j)}\wedge\phi^j\wedge\bar\phi^j$ for $0\le j\le k\le n/2$. Since we only care about the cases $n\le 5$ for the OP's problem, we really only have to deal with $n=2$ and $n=4$ in this situation. When $n=2$, since $\mathrm{Sp}(1)=\mathrm{SU}(2)$, this case is already handled. When $n=4$, the only case to deal with is $k=1$, where there are two parallel $(2,2)$-forms, $\omega^2$ and $\phi\wedge\bar\phi$. However, in this case, because there is a $2$-sphere of parallel complex structures and because the Pontryagin form $p_1(M,g)$ will have to be of type $(2,2)$ with respect to all the complex structures of this family, one finds that, if the Chern-Weil Pontryagin form $p_1(M,g)$ is to be parallel, it must be a constant multiple of $\omega^2 + \phi\wedge\bar\phi$. (Here, I am assuming that $\phi$ and $\omega$ have been suitably normalized.) This multiple can't be zero because it is essentially the squared norm of the curvature; if it were zero, the metric would be flat, and the holonomy wouldn't be $\mathrm{Sp}(2)$.

Thus, the only case that you care about in which the two Questions essentially differ is the case of an $(M^{10},g)$ that is a product $(N^8,h)\times (\mathbb{C},g_0)$ where $g_0$ is the flat metric and $(N^8,h)$ has holonomy $\mathrm{Sp}(2)$ with $p_1(N,h)$ being a (constant) multiple of $\omega^2 + \phi\wedge\bar\phi$ and $p_2(N,h)$ being a constant multiple of the volume form.

Finally, there is the question of whether any of these irreducible cases can actually occur. Here, I don't have much to say. I do not know of a single example of even a local, non-flat Calabi-Yau metric for which the square norm of the curvature tensor is constant, even in complex dimension $2$, which is the very first nontrivial case. The problem gets more and more overdetermined as the dimension goes up, so it seems unlikely that such metrics exist, even locally, but I think that actually writing down a proof might not be easy.

• @R. Bryant. Thanks a lot for your answer and the comments on the possibility of the existence of such metrics. Jun 8 '15 at 10:11