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Before I formulate my question, let me remind P. B. Gilkey's characterization of Pontryagin forms,following the paper "On the heat equation and the index theorem" by Atiyah, Bott, Patodi.

By definition, a $q$-form valued invariant of Riemannian structures is a function $\omega$ which attaches to any smooth Riemannian manifold $(M,g)$ a $q$-form on it such that for any open imbedding of another smooth manifold $f\colon M'\to M$ one has $$\omega(M',f^*g)=f^*\omega(M,g).$$

Such an $\omega$ is called regular if the following conditions hold: in any local coordinate system the components of $\omega(M,g)$ are given by polynomials in:

1) the components $g_{ij}$ of the metric tensor;

2) a finite number of derivatives $\frac{\partial^\alpha}{\partial x^\alpha}g_{ij}$;

3) the inverse of $\det g=\det(g_{ij})$.

Furthermore such an $\omega$ is called homogeneous of weight $k$ if for any $\lambda>0$ $$\omega(M,\lambda^2g)=\lambda^k\omega(M,g).$$

Theorem (Gilkey). The only regular $q$-form valued invariants of structures of non-negative weights are polynomials in the Pontryagin forms (thus must have weight 0).

QUESTION. Is there anything known about $q$-form invariants of negative weights, or on other tensor-valued invariants (not forms)? I am particularly curious in the case of functions, i.e. 0-forms of negative weight.

Below are some examples of such regular $0$-form (function) valued invariants of negative weight.

Example 1. The scalar curvature of the metric. It has weight -2.

Example 2. More generally, let us denote by $R_{ijkl}$ the Riemann curvature tensor, and by $R^{ij}_{kl}$ this tensor with the first two indices lifted up using the metric. For any even number $e\in [0,n],\, n:=\dim M$ let us define the following functions on $M$: \begin{eqnarray*} H_e=\sum_{\alpha,\beta}sgn(\alpha,\beta)R^{\beta_1\beta_2}_{\alpha_1\alpha_2}\dots R^{\beta_{e-1}\beta_e}_{\alpha_{e-1}\alpha_e}, \end{eqnarray*} where the sum runs over all sequences of elements in the set $\{1,2,\dots,n\}$ such that all elements in $\alpha$ are distinct, all elements in $\beta$ are also distinct, and elements of $\beta$ are just permutation of $\alpha$. The sign of this permutation is denoted above by $sgn(\alpha,\beta)$.

Remark. If in Example 2 one takes $e=2$, one recovers the scalar curvature from EXample 1. Expressions from Example 2 have interesting geometric meaning: these are coefficients in the H. Weyl's formula for volume of an $\varepsilon$-neighborhood of a submanifold in a Eucliden space.

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You are really asking a question about invariant theory applied to the curvature tensor and its covariant derivatives. The case of scalar invariants (of any weight) was, in principle, worked out by Weyl a long time ago. I'd have to look in the literature to get the exact references (and I'm home this weekend, away from MathSciNet), but they shouldn't be hard to find. In any case, the negative weight scalars are generated by a systematic process using contractions of the curvature and its covariant derivatives with coefficients that depend on the metric $g$.

For example, in the case $n=2$, i.e., for surfaces, the story is as follows:

First, at differential order $0$, there is $g$, a nondegenerate section of $S^2(T^*)$, and its cometric $g^\sharp$, a (nondegenerate) section of $S^2(T)$. These have weight $+1$ and $-1$ respectively. Then you get all of the zeroth-order regular invariants by applying algebraic operations to these. For example, you have $\det(g)$, a nonvanishing section of $S^2\bigl(\Lambda^2(T^*)\bigr)$, which has weight $2$ and order $0$. (Note that the "area form", which would be a section of $\Lambda^2(T^*)$, is not a regular invariant, since it is not polynomial in the coefficients of $g$ in local coordinates.) Obviously, by algebra, one can generate a lot more, such as $g\otimes g$, a section of $S^2(T^*)\otimes S^2(T^*)$ and $g^{(2)}$, a section of $S^4(T^*)$, etc. However, none of these invariants (other than the trivial constant function) are regular scalar invariants of weight $0$ and order $0$.

Next, starting with the metric $g$, one can take one derivative to get the Levi-Civita connection and then another derivative to get the Gauss curvature $K = K(g)$, which has weight $-1$ (and order $2$). (You should check your conventions, because you seem to be claiming that $K$ has weight $-2$, but one obviously has $K(\lambda g) = \lambda^{-1} K(g)$.) Obviously, the $m$-th power of $K$ has weight $-m$ and order $2$, and any regular scalar invariant of weight $-m$ and order at most $2$ is a constant multiple of this $K^m$. Meanwhile an expression such as $K^2\,\det(g)$ is a regular invariant of weight $0$ and order $2$, it's just not scalar-valued.

Now, to generate higher order invariants, you do the following: Define $DK = D^1K = \mathrm{d}K$, which is a regular invariant of weight $-1$ and order $3$ that is a section of $T^* = S^1(T^*)$. Then, for $m>1$, inductively define $D^mK = \mathsf{S}(\nabla_g D^{m-1}K)$, the symmetrization of the covariant derivative $\nabla_g D^{m-1}K$, a section of $T^*\otimes S^{m-1}(T^*)$. Thus, $D^mK$ is a section of $S^m(T^*)$ that is a regular invariant of weight $-1$ and order $m{+}2$.

Then, all of the regular invariants of order at most $m{+}2$ and some fixed weight are obtained as invariant weighted homogeneous polynomial contractions starting with the following tensors $$ g, g^\sharp, K, D^1K, D^2K,\ldots, D^mK, $$ with the individual weights being $1$ for $g$ and $-1$ for all the rest.

Thus, for example, $|\nabla K|^2$ is the contraction of $g^\sharp$ with two copies of $D^1K$ using the canonical contraction mapping $S^2(T)\otimes T^*\otimes T^*\to\mathbb{R}$. It has weight $-3$ and (differential) order $3$. Meanwhile $\Delta K$, the Laplacian of $K$ is the natural contraction of $g^\sharp$ with $D^2K$, so it has weight $-2$ and order $4$.

Past this point, it becomes an algebra problem to write down generators and relations for the ring $I_{m+2}$ of regular scalar invariants of order at most $m{+}2$ on a surface. By a theorem of Hilbert, $I_{m+2}$ is a finitely generated polynomial ring that is graded by weight. For example, $I_0 = I_1 = \mathbb{R}$, while $I_2 = \mathbb{R}[K]$ and $I_3 = \mathbb{R}\bigl[K, |\nabla K|^2\bigr]$. $I_4$ is not a free polynomial ring, but it can be written as one modulo a single relation, if I remember correctly.

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  • $\begingroup$ Thanks for the interesting information. I would appreciate a reference. Also the case which might be of particular interest for me is expressions in $g$ which involve at most 2 derivatives. $\endgroup$
    – makt
    Apr 4 '15 at 17:22
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    $\begingroup$ For surfaces, expression in $g$ that involve at most $2$ derivatives have to belong to $I_2 = \mathbb{R}[K]$, so they are only the powers of $K$ (if they have a fixed weight). This is completely classical. In higher dimensions, if it only involves derivatives of order at most $2$ in $g$, it must be a polynomial in the Riemann curvature tensor (with coefficients that could depend on $g$). Again, this is a consequence of Weyl's theorem and invariant theory. $\endgroup$ Apr 4 '15 at 19:47
  • $\begingroup$ Thank you. The result which I would like to have in the case of 2 derivatives is that any such function is a polynomial in $H_e$'s (in the 2 dimensional case this would be equivalent to what you said). Also probably I do not know which Weyl's theorem you referred to. $\endgroup$
    – makt
    Apr 5 '15 at 3:59
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    $\begingroup$ @MKO: This isn't true already in dimension $n=3$: There, you would only get $H_0=1$ and $H_2$, the scalar curvature. However, there are two more that are algebraically independent: When $n=3$, the Riemann curvature tensor is essentially the same as the Ricci curvature, which is a quadratic form, so each of the three symmetric functions of the eigenvalues of the Ricci curvature (with respect to $g$) gives an example, with $H_2$ being the trace. When $n>2$, the number $N$ of algebraically independent rational invariants of order at most $2$ is $$N=\frac{(n{+}3)(n)(n{-}1)(n{-}2)}{12}\ .$$ $\endgroup$ Apr 5 '15 at 10:56
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The objects you are after are sometimes called invariant tensors or curvature tensors. The first observation that simplifies their classification is that, by the invariance (or covariance, if you wish) criterion, the dependence on $\frac{\partial^\alpha}{\partial x^\alpha}g_{ij}$ always factors through the Riemann curvature tensor and its covariant derivatives. This result is sometimes known as the Thomas replacement theorem (see arXiv:gr-qc/9404030 by Anderson & Torre for a proof).

In certain branches of mathematical physics, computing and classifying invariant tensors once constituted a minor industry. See for instance Fulling et al. CQG 9 1151, as well as citations therein and thereof. There, the classification is not exactly by weight, but by rank (number of tensor indices), order (of the highest derivative of the metric) and degree (number of factors of the Riemann tensor or one of its derivatives). But I think you should be able to reconstruct the weight from that information.

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  • $\begingroup$ Many thanks! It looks very interesting, I will have a look. Actually I do not care about the weight. Perhaps I am particularly interested in rank 0 tensors (functions) or order 2 (at most 2 derivatives). $\endgroup$
    – makt
    Apr 4 '15 at 17:23
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There are several of those appearing in the definition of the so-called Quermassintegrals. On a manifold of dimension $m$ there are $\lfloor m/2\rfloor+1$ such integrals

$$Q_m(M) =\int_M |dV_g|,\;\;Q_{m-2}(M)=\int_M s(x)\;|dV_g(x)|,\dotsc, Q_{m-2k}(M)=\int_M P_{k,g}(x) |dV_g(x)|,\dotsc $$

Where $P_k$ is an invariant of order $-k$ in the sense of your definition. $P_k$ is a universal polynomial of degree $k$ in the curvature of $g$. For its precise definition and more properties I refer to Section 9.3 of these lectures (see especially Thm. 9.3.11).

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  • $\begingroup$ You are right: the $H_e$'s are precisely (up to a constant) the integrands in the quermassintegrals of $M$. $\endgroup$
    – makt
    Apr 4 '15 at 11:14
  • $\begingroup$ Yes. And given Hadwiger's classification theorem, I do not believe that there is anything more. $\endgroup$ Apr 4 '15 at 11:16
  • $\begingroup$ Certainly there are more: any polynomial in $H_e$'s. They are not covered by the Hadwiger's theorem. Thus the question is if there are other invariants independent of $H_e$'s. $\endgroup$
    – makt
    Apr 4 '15 at 11:19

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