# de Rham-invariants of a Riemannian metric

$$\DeclareMathOperator{\Sym}{Sym}$$For $$N>0$$, consider the $$O_N$$-representations $$V = \mathbb R^N$$ and $$M_n = \ker (\Sym^n{V}\otimes\Sym^2 V\to \Sym^{n+1} V\otimes V)$$ (the irreducible $$GL_n$$-representation corresponding to the partition $$(n+2) = n + 2$$). There is an equivariant map $$\rho_n:M_{n}\to M_{n-1}\otimes V$$, induced from the "desymmetrization" $$\Sym^{n} V\to \Sym^{n-1} V\otimes V$$. From this, we can produce a derivation on the (graded) symmetric algebra $$\Sym^* (\bigoplus_{n\ge 2} M_n)\otimes\Lambda^* V$$ which vanishes on the degree $$1$$ generators coming from $$V$$, and on the degree $$0$$ generators coming from $$M_n$$ is given by $$\rho_n$$. Since $$M_{n+1}\xrightarrow{\rho_{n+1}} M_n\otimes V\xrightarrow{\rho_n\otimes\operatorname{id}_V}M_{n-1} V\otimes V^{\otimes 2}$$ factors through $$M_{n-1} V\otimes \Sym^2 V$$, this derivation squares to $$0$$, producing a cochain complex $$C_N$$. Moreover, there are obvious $$O(N)$$-equivariant inclusions $$C_N\to C_{N+1}$$.

## Question

What is $$\varinjlim_{N\to\infty}H^*(C_N)^{O(N)},$$ i.e. the $$O(N)$$-invariant part of the cohomology of this complex in the infinite-dimensional limit $$N\to \infty$$?

# Background/Motivation

The representations $$M_n$$ arise as the possible $$n$$-th partial derivatives of the metric tensor in normal coordinates. More invariantly, the symmetrization of the iterated covariant derivative $$\nabla^n R$$ of the Riemann curvature tensor takes values in the bundle associated to $$M_{n-2}$$. The individual terms of the complex are then diffeomorphism-invariant formulae for building a $$k$$-form out of the metric tensor (the existence of normal coordinates shows that any such formula must be given by contracting indices of products of covariant derivatives of the curvature tensor), and the differential on the complex is given by the de Rham differential.

In particular, one can identify the Chern-Weyl representatives of the Pontryagin classes explicitly: Using the identification $$M_2\cong \Sym^2(\Lambda^2 V)$$, there is an equivariant map $$\Sym^k M_2\to \Sym^k(\Lambda^2 V)\otimes\Sym^k(\Lambda^2 V)\to \Lambda^{2k} V\otimes \Sym^k(\Lambda^2 V)$$, and for $$k$$ even the subspace $$\Sym^k(\Lambda^2 V)^{O_n}$$ contains the $$k/2$$-th power of the invariant pairing on $$\Lambda^2 V$$; taking adjoints, one obtains an invariant element $$p_{k/2}\in \Lambda^{2k} V\otimes \Sym^k M_2$$, which (up to combinatorial factors) gives the trace of the $$k$$-th power of the curvature tensor. Using the degree grading of the complex $$C_N$$ (where $$V$$ has degree $$1$$ and $$M_n$$ has degree $$(n+2)$$) it's easy to see that this class defines a nontrivial cocycle in $$C_N$$. Thus a more precise question is

## Question*

Is the inclusion $$\mathbb R[p_1,p_2,\dots]\to \varinjlim_{n\to\infty}H^*(C_n)^{O(n)},$$ an isomorphism?

A class which is not in the image would give an interesting (i.e. not coming from a characteristic class of the tangent bundle) de Rham invariant of a Riemannian metric. Note that such invariants exist for each finite dimension $$N$$, at least when restricting to $$SO(N)$$-invariants (e.g. the product of the scalar curvature and the volume form), but this seems unlikely in the infinite-dimensional limit.

• You might want to have a look at the following paper by Peter Gilkey: Local invariants of real and complex Riemannian manifolds, Proc.Symp. in Pure Math. 30 (1977), 107–110. This and other papers of Gilkey from around that time solve the problem of computing the cohomology of the ring of the forms constructible canonically from a metric and its derivatives. He gives the expected answer that this cohomology ring is generated by the polynomials in the Pontrjagin forms. Jan 31 at 20:25

$$\DeclareMathOperator{\Sym}{Sym}\DeclareMathOperator{\Map}{Map}$$After Robert Bryant's helpful comment, I was able to find a positive answer to Question* as Theorem 1.2 in the article Gilkey, Peter B. Local invariants of an embedded Riemannian manifold. (English) Ann. Math. (2) 102, 187-203 (1975). Let me summarize the argument here:
Using the injection $$M_n\hookrightarrow V^{\otimes n+2}$$ and H. Weyl's determination of the $$O(n)$$-invariants of $$V^{\otimes k}$$, the equivariant maps $$\Sym^k \bigoplus_{n\ge 2} M_n\to \Lambda^p V$$ are spanned by the ones given by antisymmetrizing $$p$$ of the indices and pairing the remaining ones using the invariant bilinear form. There is a "weight grading" on the space of invariants, where the subspace $$\Map\left(\bigotimes_{k} M_{n_k},\Lambda^p V\right)^{O(n)}$$ has weight $$r -\sum_k n_k$$ (in terms of Riemannian metrics in normal coordinate form, this describes the behaviour under rescaling of the metric) which is preserved by the differential. Observe that the antisymmetrization of any three indices vanishes on $$M_n$$, so that we must have $$r\le 2k$$; it follows that there are no elements of positive weight and that for weight $$0$$ we must have $$n_i = 2$$, which is readily checked to correspond to the construction of (polynomials in) the Pontryagin classes I gave above.
It remains to see that the subcomplex of negative weight has vanishing cohomology. For this, Gilkey extends to the complex $$\Map(\bigoplus_{k_1,k_2}\Sym^{k_1} V\otimes\Sym^{k_2} \bigoplus_{n\ge 2} M_n,\Lambda^* V)^{O(n)}$$ and cites an argument from another paper that the cohomology of this complex vanishes in a suitable range depending on the dimension; as far as I can tell, this uses the Poincaré lemma on the $$\Sym^* V$$-component together with an inductive/filtration argument. Interpreting the original complex as diffeomorphism-invariant differential operators $$g\mapsto P(g)$$ from Riemannian metrics to differential forms, he then considers their behaviour under infinitesimal conformal transformations $$g\mapsto g+ \epsilon hg$$; if $$P$$ is a cocycle, i.e. takes values in closed forms, this defines a cocycle $$Q:h,g\mapsto \partial_\epsilon P(e^{\epsilon h}g)|_{\epsilon = 0}$$ of the above complex, so that the above argument gives another operator $$h,g\mapsto R(h,g)$$ such that $$Q =\mathrm dR$$. We obtain \begin{align*} P(e^{\epsilon h}g) &= P(g) + \epsilon Q(h,g)\\ &= P(g) + \epsilon \mathrm d R(h,g) \end{align*} Taking $$h = 1$$, the left-hand side is $$(1 + \epsilon\operatorname{wt}(P))P(g)$$; if this weight is nonzero, comparing coefficients shows that $$P = \mathrm d\frac{R(1,g)}{\operatorname{wt}(P)}$$.