2. Symmetric tensors

In this section we recall basic formulas for symmetric tensors as well as the definition and first properties of conformal Killing tensors. More details can be found in [ Reference Heil, Moroianu and Semmelmann21 ].

2·1. The symmetric algebra of a vector space

Let $({\mathrm{T}}, g)\,:\!={\mathrm{R}}^n$ be the standard Euclidean vector space of dimension n. We denote with ${\mathrm{Sym}}^k {\mathrm{T}} \subset {\mathrm{T}}^{\otimes k}$ the k-fold symmetric tensor product of ${\mathrm{T}}$ . Elements of ${\mathrm{Sym}}^k {\mathrm{T}}$ are symmetrised tensor products

(2·1) \begin{equation} v_1 \cdot \ldots \cdot v_k\,:\!= \sum_{\sigma \in S_k} \, v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(k)},\end{equation}

where $v_1, \ldots, v_k$ are vectors in ${\mathrm{T}}$ . In particular we have $v\cdot u = v\otimes u + u \otimes v$ for $u, v \in {\mathrm{T}}$ . Some authors (see [ Reference Paternain, Salo and Uhlmann33 , page 156]) use another convention for the symmetric product and divide by $k!$ in (2·1).

Using the metric g, one can identify ${\mathrm{T}}$ with ${\mathrm{T}}^*$ . Under this identification, $g\in{\mathrm{Sym}}^2 {\mathrm{T}}^*\simeq{\mathrm{Sym}}^2 {\mathrm{T}}$ can be written as $g=(1/2) \sum_i \mathbf{e}_i \cdot \mathbf{e}_i$ , for any orthonormal basis $\{ \mathbf{e}_i\}$ . The direct sum ${\mathrm{Sym}}\,{\mathrm{T}}\,:\!=\bigoplus_{k\ge 0} {\mathrm{Sym}}^k {\mathrm{T}}$ is endowed with a commutative product making ${\mathrm{Sym}}\, {\mathrm{T}}$ into a $\mathbb{Z}$ -graded commutative algebra. The scalar product g induces a scalar product on ${\mathrm{Sym}}^k {\mathrm{T}}$ , also denoted by g, defined by

\begin{align*}g(v_1 \cdot \ldots \cdot v_k,w_1 \cdot \ldots \cdot w_k) \;=\; \sum_{\sigma \in S_k}\, g(v_1, w_{\sigma(1)}) \cdot \ldots \cdot g(v_k, w_{\sigma(k)}).\end{align*}

With respect to this scalar product, every element K of ${\mathrm{Sym}}^k {\mathrm{T}}$ can be identified with a symmetric k-linear map (i.e. a polynomial of degree k) on ${\mathrm{T}}$ by the formula

\begin{align*}K(v_1,\ldots,v_k)=g(K,v_1\cdot\ldots\cdot v_k).\end{align*}

For every $v\in {\mathrm{T}}$ , the metric adjoint of the linear map $v\cdot\,:\, {\mathrm{Sym}}^k {\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k+1} {\mathrm{T}}, \; K \mapsto v \cdot K$ is the contraction $v\lrcorner\,:\, {\mathrm{Sym}}^{k+1} {\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k} {\mathrm{T}}, \; K \mapsto v \lrcorner \, K$ , defined by $(v \lrcorner \, K) (v_1, \ldots, v_{k-1}) = K(v, v_1, \ldots, v_{k-1})$ . In particular we have $v \lrcorner \, u^k = k g(v, u) u ^{k-1},$ for all $ v, u \in {\mathrm{T}}$ .

We introduce the linear map $ \deg \,:\, {\mathrm{Sym}}\, {\mathrm{T}} \rightarrow {\mathrm{Sym}}\, {\mathrm{T}}$ , defined by $\deg (K) = k K$ for $ K \in {\mathrm{Sym}}^k {\mathrm{T}}$ . Then we have

\begin{align*}\sum_i \mathbf{e}_i \cdot \mathbf{e}_i \lrcorner \, K = \deg(K), \quad \sum_i \mathbf{e}_i \lrcorner \mathbf{e}_i \cdot K = n K + \deg(K),\end{align*}

where $\{\mathbf{e}_i\}$ denotes an orthonormal frame of $({\mathrm{T}}, g)$ . Note that if $K\in {\mathrm{Sym}}^k {\mathrm{T}}$ is considered as a polynomial of degree k then $v \lrcorner K$ corresponds to the directional derivative $\partial_v K$ and the last formula is nothing else than the well-known Euler formula on homogeneous functions.

Contraction and multiplication with the symmetric tensor ${\mathrm{L}}\,:\!=\sum_i \mathbf{e}_i\cdot \mathbf{e}_i = 2g$ defines two additional linear maps:

\begin{align*}{\Lambda}\,:\, {\mathrm{Sym}}^k {\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k-2} {\mathrm{T}}, \quad K \mapsto \sum_i \mathbf{e}_i \lrcorner \, \mathbf{e}_i \lrcorner \, K\end{align*}

and

\begin{align*}{\mathrm{L}}\,:\, {\mathrm{Sym}}^{k-2} {\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k} {\mathrm{T}}, \quad K \mapsto {\mathrm{L}} \cdot K,\end{align*}

which are adjoint to each other. It is straightforward to check the following algebraic commutator relations

(2·2) \begin{equation}[\, \Lambda, \, {\mathrm{L}} \,] \;=\; 2n\, {\mathrm{id}} \;+\;4 \deg,\quad [\, \deg, {\mathrm{L}} \,] = 2\, {\mathrm{L}}, \quad [\,\deg, {\Lambda}\,] = -\,2 \,\Lambda,\end{equation}

and for every $v\in T$ :

(2·3) \begin{equation} [\, \Lambda, \, v \,\cdot \,] \;=\; 2\, v \, \lrcorner \,, \quad[\, v \lrcorner \,,\, {\mathrm{L}} \, \,] \;=\; 2\, v \cdot\,,\quad[\,{\Lambda}, \, v \lrcorner \, \,] \;=\; 0 \;=\; [\, {\mathrm{L}}, \, v \cdot \,].\end{equation}

For ${\mathrm{T}} = \mathbb{R}^n$ , the standard ${\mathrm{O}}(n)$ -representation induces a reducible ${\mathrm{O}}(n)$ -representation on ${\mathrm{Sym}}^k {\mathrm{T}}$ . We denote by ${\mathrm{Sym}}^k_0 {\mathrm{T}}\,:\!= \ker( {\Lambda}\,:\, {\mathrm{Sym}}^k {\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k-2} {\mathrm{T}})$ the space of trace-free symmetric k-tensors.

It is well known that ${\mathrm{Sym}}^k_0 {\mathrm{T}}$ is an irreducible ${\mathrm{O}}(n)$ -representation and we have the following decomposition into irreducible summands

\begin{align*}{\mathrm{Sym}}^k {\mathrm{T}} \;\cong\; {\mathrm{Sym}}^k_0 {\mathrm{T}}\;\oplus\; {\mathrm{Sym}}^{k-2}_0 {\mathrm{T}} \;\oplus\; \ldots \,\end{align*}

where the last summand in the decomposition is $\mathbb{R}$ for k even and ${\mathrm{T}}$ for k odd. The summands ${\mathrm{Sym}}^{k- 2i}_0 {\mathrm{T}}$ are embedded into ${\mathrm{Sym}}^k {\mathrm{T}}$ via the map ${\mathrm{L}} ^i$ . Corresponding to the decomposition above any $K \in {\mathrm{Sym}}^k {\mathrm{T}}$ can be uniquely decomposed as

\begin{align*}K \;=\; K_0 \;+\; {\mathrm{L}} K_1 \;+\; {\mathrm{L}} ^2K_2 \;+ \; \ldots\end{align*}

with $K_i \in {\mathrm{Sym}}^{k-2i}_0 {\mathrm{T}}$ , i.e. $\Lambda K_i = 0$ . We will call this decomposition the standard decomposition of K. In the following, the subscript 0 always denotes the projection of an element from ${\mathrm{Sym}}^k {\mathrm{T}}$ onto its component in ${\mathrm{Sym}}^k_0 {\mathrm{T}}$ . Note that for any $v \in {\mathrm{T}}$ and $K \in {\mathrm{Sym}}^k_0 {\mathrm{T}}$ we have the following projection formula

(2·4) \begin{equation} (v \cdot K)_0 \;=\; v \cdot K \;-\; \tfrac{1}{n + 2 k-2}\, {\mathrm{L}} \, (v \lrcorner \, K).\end{equation}

Indeed, using the commutator relations (2·2) we have ${\Lambda} ( {\mathrm{L}} \, (v \lrcorner \, K)) = (2n + 4(k-1)) \, (v \lrcorner \, K)$ , since $\Lambda $ commutes with $v \lrcorner \, $ and $\Lambda K=0$ . Moreover $\Lambda (v \cdot K) = 2\, v \lrcorner \, K$ . Thus the right-hand side of (2·4) is in the kernel of ${\Lambda}$ , i.e. it computes the projection $(v \cdot K)_0$ .

2·2. Conformal Killing tensors

Let $(M^n, g)$ be a Riemannian manifold with Levi-Civita connection $\nabla$ . All the algebraic considerations above extend to vector bundles over M, e.g. the ${\mathrm{O}}(n)$ -representation ${\mathrm{Sym}}^k {\mathrm{T}}$ defines the real vector bundle ${\mathrm{Sym}}^k {\mathrm{T}} M$ . The ${\mathrm{O}}(n)$ -equivariant maps ${\mathrm{L}} $ and ${\Lambda}$ define bundle maps between the corresponding bundles. The same is true for the symmetric product $\cdot$ and the contraction $\lrcorner$ . We will use the same notation for the bundle maps on M.

Next we will define first order differential operators on sections of ${\mathrm{Sym}}^p {\mathrm{T}} M$ . We have

\begin{align*}{\mathrm{d}}\,:\, C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M) \rightarrow C^\infty(M,{\mathrm{Sym}}^{k+1}{\mathrm{T}} M), \quad K \mapsto \sum_i \mathbf{e}_i \cdot \nabla_{\mathbf{e}_i}K,\end{align*}

where $\{\mathbf{e}_i\}$ denotes from now on a local orthonormal frame. The symmetric tensor ${\mathrm{d}} K$ is the complete symmetrisation of $\nabla K$ , in the sense that

(2·5) \begin{equation}\begin{split}g({\mathrm{d}} K,X^{k+1})&=\sum_i g (\nabla_{\mathbf{e}_i} K,\mathbf{e}_i\lrcorner X^{k+1}) = (k+1)\sum_i g (\nabla_{\mathbf{e}_i} K, g(\mathbf{e}_i,X)X^{k})\\ &=(k+1) g (\nabla_{X} K, X^{k})\end{split}\end{equation}

for every $X\in {\mathrm{T}} M$ . The formal adjoint of ${\mathrm{d}}$ is the divergence operator ${\mathrm{d}}^*$ defined by

\begin{align*}{\mathrm{d}}^*\,:\, C^\infty(M,{\mathrm{Sym}}^{k+1} {\mathrm{T}} M) \rightarrow C^\infty(M,{\mathrm{Sym}}^{k}{\mathrm{T}} M), \quad K \mapsto - \sum_i \mathbf{e}_i \lrcorner\, \nabla_{\mathbf{e}_i}K.\end{align*}

As an immediate consequence of the definition we have that the operator ${\mathrm{d}}$ acts as a derivation on the algebra of symmetric tensors, i.e. for any $K_1 \in C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M)$ and $K_2 \in C^\infty(M,{\mathrm{Sym}}^l {\mathrm{T}} M)$ the following equation holds

\begin{align*}{\mathrm{d}} ( K_1 \cdot K_2) \;=\; {\mathrm{d}} K_1 \cdot K_2 \,+\,K_1 \cdot {\mathrm{d}} K_2.\end{align*}

Moreover, an easy calculation proves that the operators ${\mathrm{d}}$ and ${\mathrm{d}}^*$ satisfy the commutator relations:

(2·6) \begin{equation}[ \, \Lambda, \,{\mathrm{d}}^*\,] \;=\; 0 \;=\; [\,{\mathrm{L}},\, {\mathrm{d}}\,], \quad [\, \Lambda, \,{\mathrm{d}}\,] \; =\; -2 {\mathrm{d}}^*, \quad [\, {\mathrm{L}},\, {\mathrm{d}}^*\,] \;=\; 2 \,{\mathrm{d}}.\end{equation}

We also consider the operator

\begin{align*}{\mathrm{d}}_0\,:\, C^\infty(M,{\mathrm{Sym}}^k_0 {\mathrm{T}} M) \rightarrow C^\infty(M, {\mathrm{Sym}}^{k+1}_0{\mathrm{T}} M), \quad K \mapsto ({\mathrm{d}} K)_0.\end{align*}

According to (2·4), we have ${\mathrm{d}}_0 K={\mathrm{d}} K + ({1}/({n + 2k-2}))\, {\mathrm{L}} \, {\mathrm{d}}^* K$ for every $K\in C^\infty(M,{\mathrm{Sym}}^k_0 {\mathrm{T}} M)$ . The formal adjoint ${\mathrm{d}}_0^*\,:\, C^\infty(M,{\mathrm{Sym}}^{k+1}_0 {\mathrm{T}} M) \rightarrow C^\infty(M, {\mathrm{Sym}}^{k}_0{\mathrm{T}} M)$ is clearly equal to the restriction of ${\mathrm{d}}^*$ to $C^\infty(M,{\mathrm{Sym}}^{k+1}_0 {\mathrm{T}} M)$ .

A symmetric tensor $K \in C^\infty(M, {\mathrm{Sym}}^k {\mathrm{T}} M)$ is called conformal Killing tensor if there exists some symmetric tensor $ k \in C^\infty(M, {\mathrm{Sym}}^{k-1} {\mathrm{T}} M)$ with ${\mathrm{d}} K = {\mathrm{L}} \, k$ . Note that K is conformal Killing if and only if its trace-free part is conformal Killing. Indeed, since ${\mathrm{d}}$ and ${\mathrm{L}}$ commute, if $K = \sum_{i\ge 0} {\mathrm{L}}^i K_i$ , with $K_i \in C^\infty(M,{\mathrm{Sym}}^{k-2i}_0 {\mathrm{T}} M)$ is the standard decomposition of K, then ${\mathrm{d}} K = \sum_{i \ge 0} {\mathrm{L}}^i {\mathrm{d}} K_i$ , so ${\mathrm{d}} K$ is in the image of ${\mathrm{L}}$ if and only if ${\mathrm{d}} K_0$ is in the image of ${\mathrm{L}}$ . More precisely we have the following characterisation (see also [Reference Heil, Moroianu and Semmelmann21, lemma 3·3]): a symmetric tensor $ K \in C^\infty(M, {\mathrm{Sym}}^k {\mathrm{T}} M)$ is a conformal Killing tensor if and only if

(2·7) \begin{equation}{\mathrm{d}} K_0 = - \tfrac{1}{n + 2k-2} \, {\mathrm{L}} {\mathrm{d}}^* K_0.\end{equation}

or, equivalently, if and only if the symmetric tensor K satisfies the condition $ {\mathrm{d}}_0 K_0 = 0$ .

Let E be a real vector bundle over M with connection $\nabla^{E}$ . We extend ${\mathrm{d}}$ and ${\mathrm{d}}_0$ to twisted operators

\begin{align*}{\mathrm{d}}\,:\, C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M \otimes E) \to C^\infty(M,{\mathrm{Sym}}^{k+1} {\mathrm{T}} M \otimes E),\end{align*}

\begin{align*}{\mathrm{d}}_0\,:\, C^\infty(M,{\mathrm{Sym}}^k_0 {\mathrm{T}} M \otimes E) \to C^\infty(M,{\mathrm{Sym}}^{k+1}_0 {\mathrm{T}} M \otimes E),\end{align*}

defined on decomposable elements by

\begin{align*}{\mathrm{d}} (K \otimes \xi) = {\mathrm{d}} K \otimes \xi + \sum_i(\mathbf{e}_i \cdot K) \otimes \nabla_{\mathbf{e}_i}^{E}\xi,\quad {\mathrm{d}}_0 (K \otimes \xi) ={\mathrm{d}}_0 K \otimes \xi + \sum_i(\mathbf{e}_i \cdot K)_0 \otimes \nabla_{\mathbf{e}_i}^{E}\xi,\end{align*}

obtained from the tensor product of Levi–Civita and $\nabla^{E}$ connections. In this case, sections in $\ker {\mathrm{d}} $ are called twisted Killing tensors and sections in $\ker {\mathrm{d}}_0$ are called twisted conformal Killing tensors.

3. Weitzenböck formulas

Let $(M^n, g)$ be an oriented Riemannian manifold with Riemannian curvature tensor R. Let ${R}\,:\, \Lambda^2 {\mathrm{T}} M \rightarrow \Lambda^2 {\mathrm{T}} M$ be the curvature operator defined by $g({R} (X \wedge Y), Z\wedge U) = R(X,Y,Z, U)$ . With this convention we have ${R} = - \, {\mathrm{id}}$ on the standard sphere.

Let $P=P_{{\mathrm{SO}}(n)}M$ be the frame bundle of M and let VM be the vector bundle associated to P via a ${\mathrm{SO}}(n)$ -representation $\rho\,:\, {\mathrm{SO}}(n) \rightarrow {\mathrm{Aut }} (V)$ , where ${\mathrm{Aut }}(V)$ denotes the isometries of a Euclidean vector space $(V, g_V)$ . Then the curvature endomorphism $q(R) \in {\mathrm{End}} \, VM$ is defined as

(3·8) \begin{equation}q(R) \;\,:\!=\; \tfrac12 \, \sum_{i, j} (\mathbf{e}_i \wedge \mathbf{e}_j)_\ast {R}(\mathbf{e}_i \wedge \mathbf{e}_j)_\ast.\end{equation}

Here $\{\mathbf{e}_i\}, i = 1, \ldots n$ , is a local orthonormal frame of ${\mathrm{T}} M$ and for $X\wedge Y \in \Lambda^2 {\mathrm{T}} M $ we define $(X \wedge Y)_\ast = \rho_\ast (X \wedge Y)$ , where $\rho_\ast\,:\, {\mathfrak{so}}(n) \rightarrow {\mathrm{End}}(V)$ is the differential of $\rho$ . In particular, the standard action of $\Lambda^2{\mathrm{T}} M$ on ${\mathrm{T}} M $ is written as $(X\wedge Y)_\ast\, Z \;=\; g( X,\,Z)\,Y \;-\; g(Y,\, Z) \, X = (Y \cdot X \,\lrcorner - X \cdot Y \, \lrcorner \,) Z$ . This is compatible with

\begin{align*}g( (X \wedge Y)_\ast Z, U) \;=\; g( X \wedge Y, Z \wedge U) \;=\; g(X,Z) \, g(Y, U) \;-\; g( X, U)\, g( Y, Z).\end{align*}

Let ${\mathrm{T}} =\mathbb{R}^n$ be the standard representation of ${\mathrm{SO}}(n)$ defining the tangent bundle ${\mathrm{T}} M$ . Then any ${\mathrm{SO}}(n)$ -equivariant endomorphism ${\mathrm{p}} \in {\mathrm{End}}_{{\mathrm{SO}}(n)} ({\mathrm{T}} \otimes V)$ induces an ${\mathrm{SO}}(n)$ -equivariant element $\tilde {\mathrm{p}}\in{\mathrm{Hom}}_{{\mathrm{SO}}(n)}({\mathrm{T}} \otimes {\mathrm{T}} \otimes V, V)$ defined by

\begin{align*}\tilde {\mathrm{p}}(a\otimes b \otimes v)\,:\!= (a \, \lrcorner\otimes \, {\mathrm{id}}) \, {\mathrm{p}}(b \otimes v),\qquad \forall\ a, b \in {\mathrm{T}},\ v \in V.\end{align*}

We note at this point that equivariant objects give rise to global parallel sections which we will denote by the same letter; for instance p defines a parallel section $p \in C^\infty(M, {\mathrm{End}}({\mathrm{T}} M \otimes V M))$ . Important examples of such endomorphisms are the orthogonal projections ${\mathrm{p}}_i,\, i =1, \ldots, N$ , onto the summands in an ${\mathrm{SO}}(n)$ -invariant decomposition ${\mathrm{T}} \otimes V = V_1 \oplus \ldots \oplus V_N$ . Another example is the so-called conformal weight operator $B \in {\mathrm{End}} ({\mathrm{T}} \otimes V)$ introduced in [ Reference Gauduchon15 ] (see also [ Reference Calderbank, Gauduchon and Herzlich8 ]) and defined as

\begin{align*}B(b \otimes v)\,:\!= \sum_i \mathbf{e}_i \otimes (\mathbf{e}_i \wedge b)_\ast v.\end{align*}

The corresponding element $\tilde B \in {\mathrm{Hom}} ({\mathrm{T}} \otimes {\mathrm{T}} \otimes V, V)$ is given by

\begin{align*}\tilde B(a \otimes b \otimes v) = (a \wedge b)_* v.\end{align*}

For every equivariant orthogonal projector ${\mathrm{p}} \in {\mathrm{End}}_{{\mathrm{SO}}(n)} ({\mathrm{T}} \otimes V)$ we define a first order differential operator $P\,:\!= {\mathrm{p}} \nabla $ .

If K is a section of VM, then $\nabla^2K = \sum_i \mathbf{e}_i \otimes \mathbf{e}_j \otimes \nabla^2_{\mathbf{e}_i, \mathbf{e}_j}K$ is a section of the bundle ${\mathrm{T}} M \otimes {\mathrm{T}} M \otimes VM$ . Here for vector fields X, Y on M we denote $\nabla^2_{X, Y} K\,:\!= \nabla_X \nabla_Y K - \nabla_{\nabla_X Y} K$ ; then the curvature endomorphism is given by $R_{X, Y} = \nabla^2_{X, Y} - \nabla^2_{Y, X}$ . We can thus obtain natural second order operators by applying elements of the bundle ${\mathrm{Hom}} ({\mathrm{T}} M\otimes {\mathrm{T}} M\otimes V M, V M)$ to $\nabla^2 K$ .

Lemma 3·1 ([Reference Semmelmann34, proposition 3·1 and lemma 3·6]). The following relations hold:

\begin{align*}\tilde B \nabla^2 = q(R), \qquad \tilde {\mathrm{p}} \nabla^2 =- P^* P,\end{align*}

where $P^*$ is the formal adjoint of P.

Proof. Let $(\mathbf{e}_i)$ be a local orthonormal frame of ${\mathrm{T}} M$ , parallel at the point where the computations are done (i.e. satisfying $\nabla_{\mathbf{e}_i} \mathbf{e}_j = 0$ for all i, j). The first formula is immediate:

\begin{align*}\tilde B \nabla^2=\sum_{i,j}(\mathbf{e}_i\wedge \mathbf{e}_j)_*\nabla^2_{\mathbf{e}_i,\mathbf{e}_j}=\tfrac12\sum_{i,j}(\mathbf{e}_i\wedge \mathbf{e}_j)_*R_{\mathbf{e}_i,\mathbf{e}_j}=q(R).\end{align*}

In order to prove the second one, we first compute the formal adjoint of $\nabla$ . For all sections $\varphi$ of VM and $\psi$ of ${\mathrm{T}} M\otimes VM$ we have

\begin{align*} g(\nabla\varphi,\psi)&=&g\left(\sum_i \mathbf{e}_i\otimes\nabla_{\mathbf{e}_i}\varphi,\psi\right)=\sum_ig(\nabla_{\mathbf{e}_i}\varphi,(\mathbf{e}_i\lrcorner\otimes{\mathrm{id}})\psi)\\&=&\sum_i \mathbf{e}_i(g(\varphi,(\mathbf{e}_i\lrcorner\otimes{\mathrm{id}})\psi))-\sum_ig(\varphi,(\mathbf{e}_i\lrcorner\otimes{\mathrm{id}})\nabla_{\mathbf{e}_i}\psi).\end{align*}

Since the first term in the last equation is the codifferential of the 1-form $X\mapsto -g(\varphi,(X\lrcorner\otimes{\mathrm{id}})\psi)$ , we obtain $\nabla^*=-\sum_i(\mathbf{e}_i\lrcorner\otimes{\mathrm{id}}) \nabla_{\mathbf{e}_i}$ . Using this formula, together with the fact that $\nabla {\mathrm{p}}=0$ , ${\mathrm{p}}^2={\mathrm{p}}$ and ${\mathrm{p}}^*={\mathrm{p}}$ , we then compute:

\begin{align*} \tilde {\mathrm{p}} \nabla^2&=\tilde {\mathrm{p}} \left(\sum_{i,j}\mathbf{e}_i\otimes \mathbf{e}_j\otimes\nabla^2_{\mathbf{e}_i,\mathbf{e}_j}\right)=\sum_{i,j}(\mathbf{e}_i\lrcorner\otimes {\mathrm{id}}) {\mathrm{p}} (\mathbf{e}_j\otimes\nabla^2_{\mathbf{e}_i,\mathbf{e}_j})\\&=\sum_{i,j}(\mathbf{e}_i\lrcorner\otimes {\mathrm{id}}) \nabla_{\mathbf{e}_i} \left({\mathrm{p}} (\mathbf{e}_j\otimes\nabla_{\mathbf{e}_j})\right) = \sum_{i}(\mathbf{e}_i\lrcorner\otimes {\mathrm{id}}) \nabla_{\mathbf{e}_i} ({\mathrm{p}} \nabla)\\&=-\nabla^* {\mathrm{p}} \nabla=-\nabla^* {\mathrm{p}}^* {\mathrm{p}} \nabla=-P^* P.\end{align*}

Let us now consider the orthogonal projections ${\mathrm{p}}_s,\, s =1, \ldots, N$ , onto the summands in an ${\mathrm{SO}}(n)$ -invariant decomposition ${\mathrm{T}} \otimes V = V_1 \oplus \ldots \oplus V_N$ . The above result shows that whenever the conformal weight operator B can be expressed as a linear combination of the projections ${\mathrm{p}}_s$ , i.e. $B=\sum_s a_s{\mathrm{p}}_s$ for $a_s \in \mathbb{R}$ , we obtain a corresponding Weitzenböck formula:

(3·9) \begin{equation} q(R)=-\sum_s a_s \,P_s^* P_s\end{equation}

on sections of VM, where $P_s$ are the first order differential operators defined by $P_s(K)\,:\!={\mathrm{p}}_s(\nabla K)$ for every section K of VM, giving a section of ${\mathrm{T}} M \otimes V M$ .

This universal Weitzenböck formula was considered for the first time in [ Reference Gauduchon15 ] and later extended and generalised for other holonomy groups in [ Reference Semmelmann and Weingart36 ]. In fact, the irreducible summands $V_s$ appearing in the decomposition of $\mathrm T \otimes V$ are all pairwise non-isomorphic as ${\mathrm{SO}}(n)$ representations. Thus the projections ${\mathrm{p}}_s$ form a basis of ${\mathrm{End}}_{{\mathrm{SO}}(n)}(\mathrm T \otimes V)$ and there is an explicit formula for expressing the coefficients $a_s$ in terms of the highest weights of V and $V_s$ (see [Reference Semmelmann and Weingart36, corollary 3·4]).

We consider now another ${\mathrm{SO}}(n)$ -representation E with an invariant scalar product and the corresponding vector bundle EM over M, together with the induced metric. Let $\nabla^E$ be any metric connection on E, with curvature tensor denoted by $R^E$ . For simplicity, we still denote by $\nabla^E$ the tensor product connection $\nabla\otimes{\mathrm{id}}_{EM}+{\mathrm{id}}_{VM}\otimes \nabla^E$ on $VM\otimes EM$ . The projections ${\mathrm{p}}_s:{\mathrm{T}} \otimes V\to {\mathrm{T}} \otimes V$ define projections ${\mathrm{p}}_s\otimes {\mathrm{id}}:({\mathrm{T}} \otimes V)\otimes E\to ({\mathrm{T}} \otimes V)\otimes E$ and, correspondingly, differential operators $P_s^E\,:\!=({\mathrm{p}}_s\otimes {\mathrm{id}}) \nabla^E$ , acting on $VM\otimes EM$ .

Since $\sum_sa_s({\mathrm{p}}_s\otimes{\mathrm{id}})=B\otimes {\mathrm{id}}$ on ${\mathrm{T}}\otimes V\otimes E$ , Lemma 3·1 yields at once

(3·10) \begin{equation}\widetilde{B\otimes {\mathrm{id}}} (\nabla^E)^2=-\sum_s a_s \, (P_s^E)^* P_s^E,\end{equation}

acting on sections of $VM \otimes EM$ . It remains to compute the action of the left-hand side operator. If $K\otimes \xi\in C^\infty(M,VM\otimes EM)$ is a decomposable section and $(\mathbf{e}_i)$ is an orthonormal frame parallel at the point of interest, we have

\begin{equation*}\begin{split}& (\widetilde{B\otimes {\mathrm{id}}} (\nabla^E)^2)(K\otimes \xi) \\&=\widetilde{B\otimes {\mathrm{id}}}\left(\sum_{i,j}\mathbf{e}_i\otimes \mathbf{e}_j\otimes(\nabla^E)^2_{\mathbf{e}_i,\mathbf{e}_j}(K\otimes \xi)\right)\\&=\widetilde{B\otimes {\mathrm{id}}}\left(\sum_{i,j}\mathbf{e}_i\otimes \mathbf{e}_j\otimes\left(\nabla^2_{\mathbf{e}_i,\mathbf{e}_j}K\otimes \xi+\nabla_{\mathbf{e}_i}K\otimes\nabla^E_{\mathbf{e}_j}\xi\right.\left.+\nabla_{\mathbf{e}_j}K\otimes\nabla^E_{\mathbf{e}_i}\xi+K\otimes(\nabla^E)^2_{\mathbf{e}_i,\mathbf{e}_j}\xi \right)\right)\\&=\sum_{i,j}\left(((\mathbf{e}_i\wedge \mathbf{e}_j)_*\nabla^2_{\mathbf{e}_i,\mathbf{e}_j}K)\otimes \xi+((\mathbf{e}_i\wedge \mathbf{e}_j)_*\nabla_{\mathbf{e}_i}K)\otimes\nabla^E_{\mathbf{e}_j}\xi\right.\\&\left.\qquad\qquad\qquad\qquad\qquad+((\mathbf{e}_i\wedge \mathbf{e}_j)_*\nabla_{\mathbf{e}_j}K)\otimes\nabla^E_{\mathbf{e}_i}\xi+ (\mathbf{e}_i\wedge \mathbf{e}_j)_*K \otimes(\nabla^E)^2_{\mathbf{e}_i,\mathbf{e}_j}\xi \right)\\&=(q(R)K)\otimes \xi+\tfrac12\sum_{i,j} (\mathbf{e}_i\wedge \mathbf{e}_j)_*K \otimes R^E_{\mathbf{e}_i,\mathbf{e}_j}\xi,\end{split}\end{equation*}

where the two middle terms cancel each other due to the skew-symmetry in i,j. Denoting by $q(R)^E$ the linear operator acting on (decomposable) sections of $VM\otimes EM$ by

(3·11) \begin{equation} q(R)^E(K\otimes \xi)\,:\!=(q(R)K)\otimes \xi+\tfrac12 \sum_{i,j}(\mathbf{e}_i\wedge \mathbf{e}_j)_*K\otimes R^E_{\mathbf{e}_i,\mathbf{e}_j}\xi,\end{equation}

the previous relation (3·10) implies the twisted Weitzenböck-type formula

(3·12) \begin{equation}q(R)^E=-\sum_s a_s(P_s^E)^* P_s^E\qquad \hbox{on } C^\infty(M,VM\otimes EM).\end{equation}

We now consider the case of interest for us, namely $V={\mathrm{Sym}}^k_0 {\mathrm{T}}$ , where ${\mathrm{T}}\,:\!=\mathbb{R}^n$ is the standard ${\mathrm{O}}(n)$ representation of highest weight $(1,0,\ldots, 0)$ . Recall the classical decomposition into irreducible ${\mathrm{O}}(n)$ representations (e.g. see [ Reference Semmelmann and Weingart36 , p. 511-512]):

(3·13) \begin{equation}{\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}} \;\cong\; {\mathrm{Sym}}^{k+1}_0{\mathrm{T}} \;\oplus\; {\mathrm{Sym}}^{k-1}_0{\mathrm{T}} \;\oplus\; {\mathrm{Sym}}^{k,1} {\mathrm{T}},\end{equation}

where ${\mathrm{Sym}}^k_0 {\mathrm{T}}$ is the irreducible representation of highest weight $(k, 0, \ldots, 0)$ and ${\mathrm{Sym}}^{k,1} {\mathrm{T}}$ is the irreducible representation of highest weight $(k,1,0, \ldots, 0)$ . We note that ${\mathrm{Sym}}^{k+1}_0{\mathrm{T}}$ is the so-called Cartan summand. Its highest weight is the sum of the highest weights of ${\mathrm{T}}$ and ${\mathrm{Sym}}^k_0{\mathrm{T}}$ .

For later use, let us first express the operator q(R) on symmetric tensors in a more convenient way.

Lemma 3·2. For every $K\in {\mathrm{Sym}}^k{\mathrm{T}} M$ , the following relation holds:

\begin{align*}q(R)(K)=-\sum_{i,j,k} R_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_k\lrcorner(\mathbf{e}_j\cdot \mathbf{e}_k\cdot (\mathbf{e}_i\lrcorner K)).\end{align*}

Proof. For every skew-symmetric endomorphism A of ${\mathrm{T}} M$ (identified with a section of $\Lambda^2 {\mathrm{T}} M$ ) we have $A_*K = \sum A\mathbf{e}_i\cdot (\mathbf{e}_i\lrcorner K).$ In particular, for $A=X\wedge Y$ we get $(X\wedge Y)_*K=Y\cdot (X\lrcorner K)-X\cdot(Y\lrcorner K)$ . We then compute using the symmetries of the Riemannian curvature tensor:

\begin{eqnarray*}q(R)(K)&=&\tfrac12\sum_{k,l}(\mathbf{e}_l\wedge \mathbf{e}_k)_*(R_{\mathbf{e}_l,\mathbf{e}_k})_*K=\tfrac12\sum_{i,k,l}(\mathbf{e}_l\wedge \mathbf{e}_k)_*(R_{\mathbf{e}_l,\mathbf{e}_k}\mathbf{e}_i\cdot (\mathbf{e}_i\lrcorner K))\\&=&\sum_{i,k,l}\mathbf{e}_k\cdot \mathbf{e}_l\lrcorner(R_{\mathbf{e}_l,\mathbf{e}_k}\mathbf{e}_i\cdot \mathbf{e}_i\lrcorner K)=\sum_{i,k,l}\mathbf{e}_l\lrcorner(\mathbf{e}_k\cdot R_{\mathbf{e}_l,\mathbf{e}_k}\mathbf{e}_i\cdot (\mathbf{e}_i\lrcorner K))\\&=&\sum_{i,j,k,l}\mathbf{e}_l\lrcorner(\mathbf{e}_k\cdot \mathbf{e}_j\cdot \mathbf{e}_i\lrcorner K)g(R_{\mathbf{e}_l,\mathbf{e}_k}\mathbf{e}_i,\mathbf{e}_j)=-\sum_{i,j,k}R_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_k\lrcorner(\mathbf{e}_k\cdot \mathbf{e}_j\cdot (\mathbf{e}_i\lrcorner K)).\end{eqnarray*}

Next we want to describe projections and embeddings of the three summands. By (2·4), the map ${\mathrm{q}}_1\,:\, {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0{\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k+1}_0 {\mathrm{T}}$ onto the first summand is defined as

(3·14) \begin{equation}{\mathrm{q}}_1(v \otimes K) \;\,:\!=\; (v \cdot K)_0 \;=\; v \cdot K \;-\; \tfrac{1}{n + 2k - 2}\, {\mathrm{L}} \, (v \lrcorner \, K).\end{equation}

The adjoint map ${\mathrm{q}}_1^*\,:\, {\mathrm{Sym}}^{k+1}_0 {\mathrm{T}} \rightarrow {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}}$ is easily computed to be

(3·15) \begin{equation}{\mathrm{q}}^*_1(K) = \sum_i \mathbf{e}_i \otimes (\mathbf{e}_i \lrcorner \, K).\end{equation}

Note that for any vector $v\in {\mathrm{T}}$ , the symmetric tensor $v\lrcorner \, K$ is again trace-free, because $\, v \lrcorner \,$ commutes with ${\Lambda}$ . Since ${\mathrm{q}}_1 \, {\mathrm{q}}^*_1 = (k+1) \, {\mathrm{id}}$ on ${\mathrm{Sym}}^{k+1}_0{\mathrm{T}}$ , we conclude that

(3·16) \begin{equation}{\mathrm{p}}_1 \;\,:\!=\; \tfrac{1}{k+1} \, {\mathrm{q}}^*_1 \, {\mathrm{q}}_1\,:\, {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}} \;\rightarrow \;{\mathrm{Sym}}^{k+1}_0 {\mathrm{T}} \;\subset \; {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}}\end{equation}

is the orthogonal projection onto the irreducible summand of ${\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}}$ isomorphic to ${\mathrm{Sym}}^{k+1}_0{\mathrm{T}} $ .

Similarly the map ${\mathrm{q}}_2\,:\, {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}} \rightarrow {\mathrm{Sym}}^{k-1}_0 {\mathrm{T}}$ onto the second summand in the decomposition (3·13) is given by the contraction map

(3·17) \begin{equation}{\mathrm{q}}_2 ( v \otimes K)\,:\!= v \lrcorner \, K.\end{equation}

In this case the adjoint map ${\mathrm{q}}_2^*\,:\, {\mathrm{Sym}}^{k-1}_0 {\mathrm{T}} \rightarrow {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}}$ is computed to be

(3·18) \begin{equation}{\mathrm{q}}^*_2(K) \;=\; \sum_i \mathbf{e}_i \otimes (\mathbf{e}_i \cdot K)_0\;=\;\sum_i \mathbf{e}_i \otimes \left(\mathbf{e}_i \cdot K - \tfrac{1}{n+2k-4} \, {\mathrm{L}} \, (\mathbf{e}_i \lrcorner \, K)\right).\end{equation}

It follows that

\begin{align*}\;{\mathrm{q}}_2 \, {\mathrm{q}}^*_2 = (n+k-1) \, {\mathrm{id}} \;-\; \tfrac{2k - 2}{n + 2k - 4} \, {\mathrm{id}}=\tfrac{(n+2k-2)(n+k-3)}{n+2k-4}\,{\mathrm{id}},\end{align*}

so the projection onto the irreducible summand in ${\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}}$ isomorphic to ${\mathrm{Sym}}^{k-1}_0 {\mathrm{T}}$ is given by

(3·19) \begin{equation}{\mathrm{p}}_2 \;\,:\!=\; \tfrac{n+2k-4}{(n+2k-2)(n+k-3)} \, {\mathrm{q}}_2^* \, {\mathrm{q}}_2\,:\, {\mathrm{T}}\otimes {\mathrm{Sym}}^k_0 {\mathrm{T}} \;\rightarrow \; {\mathrm{Sym}}^{k-1}_0 {\mathrm{T}} \;\subset \; {\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}},\end{equation}

valid for $n \geq 3$ and $k \geq 1$ . The projection ${\mathrm{p}}_3$ onto the third irreducible summand in ${\mathrm{T}} \otimes {\mathrm{Sym}}^k_0 {\mathrm{T}}$ is obviously given by ${\mathrm{p}}_3 = {\mathrm{id}} - {\mathrm{p}}_1 - {\mathrm{p}}_2$ .

The algebraic considerations above extend to vector bundles over M. In particular, the operators ${\mathrm{d}}_0\,:\, C^\infty(M, {\mathrm{Sym}}^k_0 {\mathrm{T}} M)\to C^\infty(M, {\mathrm{Sym}}^{k+1}_0 {\mathrm{T}} M)$ and ${\mathrm{d}}_0^*\,:\, C^\infty(M, {\mathrm{Sym}}^k_0 {\mathrm{T}} M)\to C^\infty(M, {\mathrm{Sym}}^{k-1}_0 {\mathrm{T}} M)$ introduced above can be described as

(3·20) \begin{equation} {\mathrm{d}}_0 K = {\mathrm{q}}_1\nabla K,\qquad{\mathrm{d}}_0^* K = - {\mathrm{q}}_2\nabla K,\end{equation}

for every section $K \in C^\infty(M, {\mathrm{Sym}}^k_0 {\mathrm{T}} M)$ . By (2·7) together with (3·14) and (3·16) we see that the kernel of $P_1= {\mathrm{p}}_1 \nabla$ consists exactly of trace-free conformal Killing tensors. The kernel of $P_2= {\mathrm{p}}_2 \nabla$ are the divergence free tensors, i.e. tensors in $\ker {\mathrm{d}}_0^*$ .

An easy calculation using the explicit formulas for ${\mathrm{q}}_1$ and ${\mathrm{q}}_2$ proves the following relation on ${\mathrm{T}} \otimes {\mathrm{Sym}}^k_0{\mathrm{T}}$ (see [Reference Heil, Moroianu and Semmelmann21, proposition 6·1]):

\begin{align*}B = k\, {\mathrm{p}}_1 \; - \; (n+k-2) \, {\mathrm{p}}_2 \;-\, {\mathrm{p}}_3.\end{align*}

As explained above, this yields the Weitzenböck-type formula.

(3·21) \begin{equation} q(R) K = - k \, P_1^*P_1 K \,+\, (n+k-2)\, P_2^*P_2 K \,+ \, P_3^*P_3\, K,\end{equation}

for any section K of ${\mathrm{Sym}}^k_0 \,{\mathrm{T}} M$ . In the present situation it is easy to get the coefficients for B by a direct calculation. Alternatively one can use the general formula in terms of highest weights mentioned above.

Now, if EM is a Euclidean vector bundle associated to a representation E of ${\mathrm{SO}}(n)$ with metric connection $\nabla^E$ , we denote by ${\mathrm{p}}_i^E\,:\!={\mathrm{p}}_i\otimes{\mathrm{id}}_{E}$ , by ${\mathrm{q}}_i^E\,:\!={\mathrm{q}}_i\otimes{\mathrm{id}}_{E}$ and by

(3·22) \begin{equation}P^E_i\,:\!={\mathrm{p}}_i^E \nabla^E,\quad i = 1, 2, 3,\end{equation}

and obtain as before the twisted counterpart of (3·21)

(3·23) \begin{equation} q(R)^E = -k \, (P_1^E)^*P_1^E \,+ \,(n+k-2) \, (P_2^E)^*P_2^E \,+ \,(P_3^E)^*P_3^E,\end{equation}

acting on sections of ${\mathrm{Sym}}^k_0 \,{\mathrm{T}} M\otimes EM$ .

Since ${\mathrm{p}}_i^E$ are orthogonal projectors, we have $({\mathrm{p}}_i^E)^* {\mathrm{p}}_i^E={\mathrm{p}}_i^E$ , so using (3·16) and recalling that ${\mathrm{d}}_0 = {\mathrm{q}}_1^E \nabla^E$ (similarly to (3·20)), we obtain

\begin{align*}(P_1^E)^*P_1^E=(\nabla^E)^* ({\mathrm{p}}_1^E)^* ({\mathrm{p}}_1^E) \nabla^E=(\nabla^E)^* ({\mathrm{p}}_1^E) \nabla^E=\tfrac1{k+1}(\nabla^E)^* ({\mathrm{q}}_1^E)^* {\mathrm{q}}_1^E \nabla^E=\tfrac1{k+1}{\mathrm{d}}_0^* {\mathrm{d}}_0,\end{align*}

and similarly using (3·19), yields

\begin{align*}(P_2^E)^*P_2^E= \tfrac{n+2k-4}{(n+2k-2)(n+k-3)}{\mathrm{d}}_0 {\mathrm{d}}_0^*.\end{align*}

From these last two equations, together with (3·23) we obtain the following:

Proposition 3·3. (Twisted Weitzenböck formula). The following formula holds for sections of ${\mathrm{Sym}}^k_0 \,{\mathrm{T}} M\otimes EM$ :

(3·24) \begin{equation} q(R)^E = -\tfrac{k}{k+1} {\mathrm{d}}_0^* {\mathrm{d}}_0 + \tfrac{(n+k-2)(n+2k-4)}{(n+2k-2)(n+k-3)} {\mathrm{d}}_0 {\mathrm{d}}_0^* + (P_3^E)^*P_3^E.\end{equation}

4. Fourier analysis in the fibers of the unit tangent bundle

Further details on this section can be found in [ Reference Paternain27 ], [ Reference Paternain, Salo and Uhlmann32 , section 2].

4·1. Functions on the unit tangent bundle

We denote by SM the unit tangent bundle of (M,g) and by $\pi\,:\, SM \rightarrow M$ the projection on the base. There is a canonical splitting of the tangent bundle to SM as:

\begin{align*}T(SM) = \mathbb{V} \oplus \mathbb{H} \oplus \mathbb{R} X,\end{align*}

where X is the geodesic vector field, $\mathbb{V}\,:\!= \ker d \pi$ is the vertical space and $\mathbb{H}$ is the horizontal space defined in the following way. Define the connection map $\mathcal{K}\,:\, T(SM) \rightarrow {\mathrm{T}} M$ as follows: let $v \in SM, w \in T_{v}(SM)$ and a curve $(-\varepsilon,\varepsilon) \ni t \mapsto v(t) \in SM$ such that $v(0)=v, \dot{v}(0)=w$ . Denoting $x(t)\,:\!=\pi(v(t))$ , we have $\mathcal{K}_{v}(w)\,:\!= \nabla_{\dot{x}(t)} v(t)|_{t=0}$ . We denote by $g_{\mathrm{Sas}}$ the Sasaki metric on SM, which is the canonical metric on the unit tangent bundle, defined by:

\begin{align*}g_{\mathrm{Sas}}(w,w')\,:\!= g(d \pi(w), d\pi(w')) + g(\mathcal{K}(w),\mathcal{K}(w')).\end{align*}

Then the horizontal bundle $\mathbb{H}$ is defined as the orthogonal complement of X inside $\ker \mathcal{K}$ .

We define the normal bundle $\mathcal{N} \rightarrow SM$ whose fiber at $v \in SM$ is given by $\mathcal{N}_v\,:\!= v^\bot \subset T_{\pi(v)}M$ . Then $d \pi\,:\, \mathbb{H} \rightarrow \mathcal{N}, \mathcal{K}\,:\, \mathbb{V} \rightarrow \mathcal{N}$ are both isometries and all these bundles over SM are isomorphic. We will freely identify them in the following. In particular, we will think of the normal bundle $\mathcal{N}$ as the tangent bundle to the spheres.

For $x \in M$ , the unit sphere

\begin{align*}S_xM = \left\{ v \in T_xM \;|\; |v|^2_x = 1\right\} \subset SM\end{align*}

(endowed with the Sasaki metric) is isometric to the canonical sphere $(\mathbb{S}^{n-1},g_{\mathrm{can}})$ . We denote its Laplace operator by $\Delta_x$ . Let $\Delta_{\mathbb{V}}$ be the vertical Laplacian acting on $f \in C^\infty(SM)$ as $\Delta_{\mathbb{V}}f(v)\,:\!= \Delta_{\pi(v)}(f|_{S_{\pi(v)}M})(v)$ , for every $v\in SM$ . For $k \geq 0$ and $x\in M$ , we introduce

\begin{align*}\Omega_k(x) = \ker\left(\Delta_x- k(n+k-2){\mathrm{id}}\right),\end{align*}

the spherical harmonics of degree k. Observe that $\Omega_k \rightarrow M$ defines a vector bundle over M, and that $C^\infty(M, \Omega_k)$ is naturally identified with a subspace of $C^\infty(SM)$ . Given $f \in C^\infty(SM)$ , it can be decomposed as $f = \sum_{k \geq 0} {f}_k$ where ${f}_k \in C^\infty(M,\Omega_k)$ is the projection of f onto spherical harmonics of degree k. We call Fourier degree of f, denoted by $\mathrm{deg}(f)$ , the maximal integer $k_0 \in \mathbb{Z}_{\geq 0}$ (if it exists) such that ${f}_{k_0} \neq 0$ ; otherwise we set $\deg(f) = \infty$ . We will also say that f has finite Fourier content if its degree is finite, that it is odd (resp. even) if it only contains odd (resp. even) spherical harmonics.

It can be proved that the operator X has the following mapping properties (see [Reference Paternain, Salo and Uhlmann32, section 3]):

\begin{align*}X\,:\, C^\infty(M,\Omega_k) \rightarrow C^\infty(M,\Omega_{k+1}) \oplus C^\infty(M,\Omega_{k-1}).\end{align*}

This is understood in the following sense: a section ${f}_k \in C^\infty(M,\Omega_k)$ defines in particular a smooth function in $C^\infty(SM)$ which we can differentiate in the X-direction and this only contains spherical harmonics of degree $k-1$ and $k+1$ . Taking the projection on higher degree (resp. lower degree), we obtain an operator $X_+\,:\, C^\infty(M,\Omega_k) \rightarrow C^\infty(M,\Omega_{k+1})$ of gradient type i.e. with injective principal symbol (resp. $X_-\,:\, C^\infty(M,\Omega_k) \rightarrow C^\infty(M,\Omega_{k-1})$ of divergence type) such that $X=X_+ + X_-$ and $X_+^*=-X_-$ (the latter being a mere consequence of the fact that $X^*=-X$ as X preserves the Sasaki volume (also known as the Liouville measure) on SM). As $X_+$ acting on spherical harmonics of degree k has injective principal symbol, its kernel is finite dimensional by elliptic theory. As a consequence of Lemma 5·3 we will later see that elements in the kernel of $X_+$ correspond to conformal Killing tensors, i.e. elements in the kernel of ${\mathrm{d}}_0$ as defined in Section 2·2.

4·2. Twist by a vector bundle

Let ${E} \rightarrow M$ be a real vector bundle over M equipped with a metric connection $\nabla^{E}$ . Consider the pullback bundle $\mathcal{E}\,:\!=\pi^*E \rightarrow SM$ equipped with the pullback connection $\nabla^{\mathcal{E}}\,:\!=\pi^*\nabla^E$ and introduce the first order differential operator

\begin{align*}\mathbf{X}\,:\!= \nabla^{\mathcal{E}}_X\,:\, C^\infty(SM,\mathcal{E}) \rightarrow C^\infty(SM,\mathcal{E}).\end{align*}

The connection $\nabla^{\mathcal{E}}$ also gives rise to differential operators:

\begin{align*}\nabla^{\mathcal{E}}_{\mathbb{H}},\ \nabla^{\mathcal{E}}_{\mathbb{V}}\,:\, C^\infty(SM,\mathcal{E}) \rightarrow C^\infty(SM,\mathcal{N} \otimes \mathcal{E}),\end{align*}

defined in the following way: for every section $f \in C^\infty(SM,\mathcal{E})$ , the covariant derivative $\nabla^{\mathcal{E}} f \in C^\infty(SM, T^*(SM) \otimes \mathcal{E})$ can be identified with an element of $C^\infty(SM,T(SM) \otimes \mathcal{E})$ by applying the musical isomorphism $T^*(SM) \rightarrow T(SM)$ induced by the Sasaki metric. Using the orthogonal projections $\bullet_{\mathbb{H}}$ and $\bullet_{\mathbb{V}}$ of T(SM) onto $\mathbb{H}$ and $\mathbb{V}$ , respectively, one can then define the operators:

\begin{align*}\nabla^{\mathcal{E}}_{\mathbb{H}} f\,:\!= d\pi( (\nabla^{\mathcal{E}} f)_\mathbb{H}), \qquad \nabla^{\mathcal{E}}_{\mathbb{V}}f\,:\!= \mathcal{K} ((\nabla^{\mathcal{E}} f)_\mathbb{V}),\end{align*}

which take values in the bundle $\mathcal{N} \otimes \mathcal{E} \rightarrow SM$ . In local coordinates, these operators have explicit expressions in terms of the connection 1-form and we refer to [Reference Guillarmou, Paternain, Salo and Uhlmann19, lemma 3·2] for further details.

If $(\xi_1,\ldots, \xi_r)$ is a local orthonormal frame of E, then smooth local sections f of $\mathcal{E}$ can be written as:

\begin{align*}f(v) = \sum_{j=1}^r f^{(j)}(v) \xi_j(x) \in \mathcal{E}_x,\qquad\forall v\in S_xM,\end{align*}

where $f^{(j)} \in C^\infty(SM)$ are locally defined functions. As before, each $f^{(j)}$ can be in turn decomposed into spherical harmonics. In other words, we can write $f = \sum_{k \geq 0}{f}_k$ , where ${f}_k \in C^\infty(M, \Omega_k \otimes E)$ .

As before, we can define the degree of $f \in C^\infty(SM,\mathcal{E})$ and we say that f has finite Fourier content if its expansion in spherical harmonics only contains a finite number of terms. The operator $\mathbf{X}$ maps

(4·25) \begin{equation}\mathbf{X}\,:\, C^\infty(M, \Omega_k \otimes E) \rightarrow C^\infty(M, \Omega_{k-1} \otimes E) \oplus C^\infty(M, \Omega_{k+1} \otimes E)\end{equation}

and can be decomposed as $\mathbf{X} = \mathbf{X}_+ + \mathbf{X}_-$ , where, if $u \in C^\infty(M,\Omega_k \otimes E)$ , $\mathbf{X}_\pm u \in C^\infty(M,\Omega_{k \pm 1} \otimes E)$ denote the orthogonal projections on the twisted spherical harmonics of degree $k \pm 1$ . The operator $\mathbf{X}_+$ is elliptic and thus has finite-dimensional kernel whereas $\mathbf{X}_-$ is of divergence type. Moreover, $\mathbf{X}_+^* = -\mathbf{X}_-$ , where the adjoint is computed with respect to the canonical $L^2$ scalar product on SM induced by the Sasaki metric and the metric on E. We also refer to the original articles of Guillemin–Kazhdan [ Reference Guillemin and Kazhdan16, Reference Guillemin and Kazhdan17 ] for a description of these facts and to [ Reference Guillarmou, Paternain, Salo and Uhlmann19 ] for a more modern exposition. It was shown in [ Reference Guillarmou, Paternain, Salo and Uhlmann19 , theorem 4·1] (see also [Reference Cekić, Lefeuvre, Moroianu and Semmelmann10, corollary 4·2] for a short argument) that flow-invariant sections, i.e. smooth sections in $\ker \mathbf{X}$ have finite Fourier content.

5. Symmetric tensors versus polynomial functions

Considering symmetric tensors in ${\mathrm{Sym}}^k {\mathrm{T}} M$ as (pointwise) homogeneous polynomials of degree k on ${\mathrm{T}} M$ , gives linear maps

(5·26) \begin{equation}\pi^*_k\,:\, C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M) \to C^\infty(SM), \qquad (\pi^*_k K)(v)\,:\!= \tfrac1{k!}g(K,v^k).\end{equation}

Note here that $(1/{k!}) v^k = v \otimes \dotsm \otimes v$ , where the tensor product is repeated k times.

Lemma 5·1. The linear map

\begin{align*}\pi^*\,:\!=\bigoplus_{k\ge 0}\pi_k^*:C^\infty\big(M,{\mathrm{Sym}} {\mathrm{T}} M\big)\to C^\infty(SM)\end{align*}

is an algebra homomorphism.

Proof. Using the bilinearity of the symmetric product it suffices to prove $\pi^*(a \cdot b) = (\pi^*a)(\pi^*b)$ where $a = a_1 \dotsm a_k$ and $b = b_1 \dotsm b_l$ , for some $a_i, b_j \in C^\infty(M, {\mathrm{Sym}}^1 {\mathrm{T}} M)$ . But this follows from

\begin{align*} (\pi^*a)(\pi^*b) &= \tfrac1{k!}g(a, v^k) \tfrac1{l!} g(b, v^l) = g(a_1, v) \dotsm g(a_k, v) g(b_1, v) \dotsm g(b_l, v)\\ &= \tfrac1{(k + l)!} g(a \cdot b, v^{k + l}) = \pi^*(a \cdot b),\end{align*}

which completes the proof.

The following is standard and is a consequence of the identification of spherical harmonics with harmonic homogeneous polynomials (e.g. see [ Reference Berger, Gauduchon and Mazet3 , chapter C·I]).

Lemma 5·2. The above maps induce pointwise isomorphisms

(5·27) \begin{equation}\pi^*_k\,:\, {\mathrm{Sym}}^k_0 {\mathrm{T}}_x M \xrightarrow{\sim} \Omega_{k}(x),\end{equation}

for every $x\in M$ and for every integer $k\ge 0$ .

If E is any vector bundle over M and $\mathcal{E}$ is its pull-back to SM, the spaces of sections $C^\infty(M,{\mathrm{Sym}} {\mathrm{T}} M\otimes E)$ and $C^\infty(SM,\mathcal{E})$ are modules over the algebras $C^\infty(M,{\mathrm{Sym}} {\mathrm{T}} M)$ and $C^\infty(SM)$ respectively, and we can extend the linear maps above to linear maps

(5·28) \begin{equation}\pi^*_k\,:\, C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M\otimes E) \to C^\infty(SM,\mathcal{E}), \qquad \pi^*_k (K\otimes\xi)(v)\,:\!=\pi_k^*(K)\pi^*\xi\end{equation}

compatible with the module structures in sense that

(5·29) \begin{equation}\pi_k^*(K)\cdot\pi_l^*(K'\otimes \xi)=\pi_{k+l}^*((K\cdot K')\otimes \xi)\end{equation}

for every $K\in C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M)$ , $K'\in C^\infty(M,{\mathrm{Sym}}^l {\mathrm{T}} M)$ and $\xi\in C^\infty(M,E)$ . In particular, since

\begin{align*}\pi_2^*({\mathrm{L}})(v)=\tfrac12g({\mathrm{L}},v\cdot v)=\tfrac12g(v\lrcorner{\mathrm{L}}, v)=\tfrac12g(2v, v)=1,\qquad\forall v\in SM,\end{align*}

we have $\pi_{k+2}^*({\mathrm{L}} K)=\pi_k^*(K)$ for every $K\in C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M)$ .

We now relate the operators $\mathbf{X}$ , $\mathbf{X}_+$ and $\mathbf{X}_-$ with the operators ${\mathrm{d}}$ , ${\mathrm{d}}_0$ and ${\mathrm{d}}_0^*$ defined in Section 2·2.

Lemma 5·3. The following relation holds on sections of ${\mathrm{Sym}}^k{\mathrm{T}} M \otimes E$ :

(5·30) \begin{equation} \mathbf{X} \pi_k^* =\pi_{k+1}^* {\mathrm{d}},\end{equation}

while on sections of ${\mathrm{Sym}}^k_0{\mathrm{T}} M \otimes E$ we have:

(5·31) \begin{align} \mathbf{X}_+ \pi_k^*=&\pi_{k+1}^* {\mathrm{d}}_0,\\[-40pt]\nonumber\end{align}

(5·32) \begin{align}\mathbf{X}_- \pi_k^*=&-\tfrac1{n+2k-2}\pi_{k-1}^* {\mathrm{d}}_0^*.\end{align}

Proof. For the first equation, it is enough to check it on decomposable sections $\Psi=K\otimes \xi$ , with $K\in C^\infty(M,{\mathrm{Sym}}^k {\mathrm{T}} M)$ and $\xi\in C^\infty(M,E)$ . Then

\begin{align*}\mathbf{X}\pi_k^*\Psi=\nabla^{\mathcal{E}}_X(\pi_k^*(K)\pi^*\xi)=X(\pi_k^*(K))\pi^*\xi+\pi_k^*(K)\pi^*(\nabla^{E}_X\xi)\end{align*}

and

\begin{eqnarray*}\pi_{k+1}^* {\mathrm{d}} \Psi&=&\pi_{k+1}^*({\mathrm{d}} K\otimes\xi+\mathbf{e}_i\cdot K\otimes\nabla^E_{\mathbf{e}_i}\xi)=\pi_{k+1}^*({\mathrm{d}} K)\pi^*\xi+g(\mathbf{e}_i,v)\pi_k^*(K)\nabla^{\mathcal{E}}_{\mathbf{e}_i}\pi^*\xi\\&=&\pi_{k+1}^*({\mathrm{d}} K)\pi^*\xi+\pi_k^*(K)\nabla^{\mathcal{E}}_{X}\pi^*\xi=\pi_{k+1}^*({\mathrm{d}} K)\pi^*\xi+\pi_k^*(K)\pi^*(\nabla^E_{v}\xi),\end{eqnarray*}

where we identified $\mathbf{e}_i$ with their horizontal lifts to SM and used that $d\pi(X) = v$ . It remains to prove that $X(\pi_k^*(K))=\pi_{k+1}^*({\mathrm{d}} K)$ . Let $v\in SM$ be any vector and denote by $x\,:\!=\pi(v)$ . The geodesic in M determined by (x,v) will be denoted by $\gamma_t$ . Then the integral curve of X through v is $\dot\gamma_t$ . We can thus compute

\begin{eqnarray*}X(\pi_k^*(K))(v)&=&\frac{d}{dt}\bigg|_{t=0}\pi_k^*(K)(\dot\gamma_t)=\frac1{k!}\frac{d}{dt}\bigg|_{t=0}g(K,\dot\gamma_t^k)\\&=&\frac1{k!}g(\nabla_{\dot\gamma_0}K,\dot\gamma_0^k)\stackrel{2\cdot5}{=}\frac1{(k+1)!}g({\mathrm{d}} K,v^{k+1})=\pi_{k+1}^*({\mathrm{d}} K)(v),\end{eqnarray*}

where in the third equality we used that $\nabla_{\dot\gamma_0} \dot \gamma_0 = 0$ . This proves (5·30). Using this equation applied to some twisted trace-free symmetric tensor $\Psi\in C^\infty(M, {\mathrm{Sym}}^k_0{\mathrm{T}} M \otimes E)$ together with (2·4) we then obtain

\begin{eqnarray*}\mathbf{X}_+ \pi_k^*\Psi+\mathbf{X}_- \pi_k^*\Psi&=&\pi_{k+1}^*{\mathrm{d}} \Psi=\pi_{k+1}^*\left( {\mathrm{d}}_0 (\Psi)-\tfrac{1}{n+2k-2}{\mathrm{L}}{\mathrm{d}}_0^*(\Psi)\right)\\&=&\pi_{k+1}^*( {\mathrm{d}}_0 (\Psi))-\tfrac{1}{n+2k-2}\pi_{k-1}^*( {\mathrm{d}}_0^* (\Psi)).\end{eqnarray*}

Comparing the components in $\Omega_{k+1}\otimes E$ and $\Omega_{k-1}\otimes E$ yields (5·31)–(5·32) at once.

Consider now the operator $\nabla_\mathbb{V}:C^\infty(SM,\mathcal{E})\to C^\infty(SM,\mathcal{N}\otimes\mathcal{E})\subset C^\infty(SM,\pi^*({\mathrm{T}} M)\otimes\mathcal{E})$ and its formal adjoint $\nabla_\mathbb{V}^*:C^\infty(SM,\pi^*({\mathrm{T}} M)\otimes\mathcal{E})\to C^\infty(SM,\mathcal{E})$ . Define the bundle map

\begin{align*} S_k\,:\,{\mathrm{Sym}}^k{\mathrm{T}} M\otimes E\to {\mathrm{Sym}}^{k-1}{\mathrm{T}} M\otimes(E\otimes {\mathrm{T}} M), \quad S_k(K\otimes \xi)\,:\!= \sum_i(\mathbf{e}_i\lrcorner K)\otimes(\xi\otimes\mathbf{e}_i),\end{align*}

where $(\mathbf{e}_i)$ is some local orthonormal frame of ${\mathrm{T}} M$ . Let $\pi_{\mathcal{N}}:\pi^*{\mathrm{T}} M\to\mathcal{N}$ be the orthogonal projection. By definition, for every section $K\otimes \xi$ of ${\mathrm{Sym}}^k{\mathrm{T}} M\otimes E$ and at any $v\in SM$ we have:

\begin{eqnarray*}\pi_{k-1}^*S_k(K\otimes \xi)&=&\pi_{\mathcal{N}}\pi_{k-1}^*S_k(K\otimes \xi)+\sum_i\tfrac1{(k-1)!}g(\mathbf{e}_i\lrcorner K,v^{k-1})\,(g(\mathbf{e}_i,v)v\otimes\xi)\\&=&\pi_{\mathcal{N}}\pi_{k-1}^*S_k(K\otimes \xi)+k\pi_{k}^*(K\otimes\xi)\otimes v,\end{eqnarray*}

thus showing that for every $\Psi\in C^\infty(M,{\mathrm{Sym}}^k{\mathrm{T}} M\otimes E)$ ,

(5·33) \begin{equation} \pi_{k-1}^*S_k\Psi=\pi_{\mathcal{N}}\pi_{k-1}^*S_k\Psi+k\,\pi_{k}^*\Psi\otimes v.\end{equation}

It is possible to give a formula relating $S_k$ and $\nabla_{\mathbb{V}}$ :

Lemma 5·4. The following relation holds for sections of ${\mathrm{Sym}}^k_0 {\mathrm{T}} M\otimes E$ :

(5·34) \begin{equation} \nabla_\mathbb{V} \pi_k^*=\pi_{\mathcal{N}}\pi_{k-1}^* S_k.\end{equation}

Moreover, for every $K\otimes\xi\in C^\infty(M,{\mathrm{Sym}}^k_0 {\mathrm{T}} M\otimes E)$ , and $w\in C^\infty(M,{\mathrm{T}} M)$ ,

(5·35) \begin{equation} \nabla_\mathbb{V}^* \pi_k^* (K\otimes (w\otimes \xi))=-\pi_{k-1}^* ((w\lrcorner K)\otimes \xi)+k\pi_{k+1}^*((w\cdot K)\otimes \xi).\end{equation}

Proof. Let $v, w\in S_xM$ with $w \perp v$ . We denote by $v_t\,:\!=\cos t\,v+\sin t\, w$ the curve in $S_xM$ which satisfies $v_0=v$ and $\dot v_0=w$ . We then compute

(5·36) \begin{equation}\begin{split} w(\pi_k^*(K))&=\tfrac{d}{dt}\big|_{t=0} \pi_k^*(K)(v_t) = \tfrac1{k!}\tfrac{d}{dt}\big|_{t=0}g(K,v_t^k) = \tfrac1{(k-1)!}g(K,w\cdot v^{k-1})\\ &=\tfrac1{(k-1)!}g(w\lrcorner K,v^{k-1}),\end{split}\end{equation}

whence for $\Psi\,:\!= K \otimes \xi$ we have

\begin{align*} \nabla_\mathbb{V} \pi_k^*(\Psi)(w)=\nabla^{\mathcal{E}}_w(\pi_k^*(K)\pi^*\xi)=w(\pi_k^*(K))\pi^*\xi=\tfrac1{(k-1)!}g(w\lrcorner K,v^{k-1})\pi^*\xi,\end{align*}

where we identified w with its vertical lift. Then, computing the right-hand side at the point v yields

\begin{eqnarray*}\pi_{\mathcal{N}}\pi_{k-1}^* S_k (\Psi)(w)&=&\pi_{k-1}^* \left(\sum_i(\mathbf{e}_i\lrcorner K)\otimes(\mathbf{e}_i\otimes\xi)\right)(w) = \sum_i \pi_{k-1}^*(\mathbf{e}_i\lrcorner K) g(\mathbf{e}_i,w)\pi^*\xi\\&=&\tfrac1{(k-1)!}\sum_i g(\mathbf{e}_i\lrcorner K, v^{k-1}) g(\mathbf{e}_i,w)\pi^*\xi=\tfrac1{(k-1)!}g(w\lrcorner K, v^{k-1})\pi^*\xi,\end{eqnarray*}

thus proving (5·34).

We now remark that since $SM\to M$ is a Riemannian submersion, the formal adjoint of the operator $\nabla_\mathbb{V}$ can be written as $ \nabla_\mathbb{V}^*(\sigma\otimes\psi)=-\sum_i \mathbf{f}_i\lrcorner \nabla^{\mathcal{E}}_{\mathbf{f}_i}(\sigma\otimes\psi)$ for all sections $\sigma\in C^\infty(SM,\pi^*{\mathrm{T}} M)$ , and $\psi\in C^\infty(SM,\mathcal{E})$ , where $(\mathbf{f}_i)$ denotes a local orthonormal frame of $\mathbb{V}\subset {\mathrm{T}}(SM)$ and the interior product is taken with respect to the bilinear form $\mathbb{V}\otimes \pi^*{\mathrm{T}} M\to \mathbb{R}$ determined by the metric g, after identification of $\mathbb{V}_v$ with the orthogonal complement of v in $\pi^*({\mathrm{T}} M)_v$ for every $v\in SM$ . We then denote by $w^\bot\,:\!= w-g(w,v)v\in\mathbb{V}_v$ at some $v\in SM$ and compute:

\begin{align*}\nabla_{\mathbb{V}}^* \pi_k^* (K\otimes (w\otimes \xi))&=-\sum_i \mathbf{f}_i\lrcorner \nabla^{\mathcal{E}}_{\mathbf{f}_i}\big(\pi_k^* (K\otimes (w\otimes \xi))\big)=-\sum_i \mathbf{f}_i\lrcorner\big( \mathbf{f}_i(\pi_k^* (K))\pi^*(w\otimes \xi)\big)\\&=-w^\bot(\pi_k^* (K))\pi^*(\xi)\stackrel{\eqref{deriv}}{=}-\tfrac1{(k-1)!}g(w^\bot\lrcorner K,v^{k-1})\pi^*(\xi)\\&=-\pi_{k-1}^*(w\lrcorner K)\pi^*(\xi)+\tfrac1{(k-1)!}g(K,v^{k})\pi_1^*(w)\pi^*(\xi)\\&=-\pi_{k-1}^* ((w\lrcorner K)\otimes \xi)+k\pi_{k+1}^*((w\cdot K)\otimes \xi).\end{align*}

Finally, we compute the action of the operator $P_3^{E}$ pulled back to the unit sphere bundle.

Lemma 5·5. For every $\Psi\in C^\infty(M,{\mathrm{Sym}}^k_0 {\mathrm{T}} M\otimes E)$ , and $w\in C^\infty(M,{\mathrm{T}} M)$ ,

(5·37) \begin{equation}Z_k \pi_k^* \Psi=\pi_k^*P^E_3 \Psi,\qquad Z_k^* \pi_k^*(w\otimes\Psi)=\pi_k^*((P^E_3)^*(w\otimes\Psi))\end{equation}

where $Z_k:C^\infty(M,\Omega_k \otimes E)\to C^\infty(SM,\mathcal{N}\otimes \mathcal{E})\subset C^\infty(SM,\pi^* {\mathrm{T}} M \otimes\mathcal{E})$ is the operator defined by

(5·38) \begin{equation} Z_k f\,:\!=\nabla_\mathbb{H} f-\tfrac1{k+1}\nabla_\mathbb{V}\mathbf{X}_+f+\tfrac{1}{n+k-3}\nabla_\mathbb{V}\mathbf{X}_-f\end{equation}

and

\begin{align*}P_3^E\,:\,C^\infty(M,{\mathrm{Sym}}^k_0{\mathrm{T}} M\otimes E)\to C^\infty(M,{\mathrm{Sym}}^k_0{\mathrm{T}} M\otimes({\mathrm{T}} M\otimes E))\end{align*}

is the first order differential operator appearing in (3·22).

Proof. It is enough to check the first relation, the second following by taking the metric adjoints. By definition we have $P^E_3=\nabla^E-P_1^E-P_2^E$ . Let us first explicit the last two operators. Using (3·14)–(3·16) we compute

\begin{align*}P_1^E\Psi=\frac1{k+1}({\mathrm{q}}_1^E)^*{\mathrm{q}}_1^E(\nabla^E\Psi)=\frac1{k+1}\sum_i(\mathbf{e}_i\lrcorner {\mathrm{d}}_0\Psi)\otimes\mathbf{e}_i=\frac1{k+1}S_{k+1}{\mathrm{d}}_0\Psi.\end{align*}

From (5·31), (5·33) and (5·34) we thus get at any $v\in SM$ :

(5·39) \begin{equation} \pi_k^*P_1^E\Psi=\frac1{k+1}\pi_{\mathcal{N}}\pi_k^*S_{k+1}{\mathrm{d}}_0\Psi+\pi_{k+1}^*{\mathrm{d}}_0\Psi\otimes v=\frac1{k+1}\nabla_\mathbb{V}\mathbf{X}_+\pi_k^*\Psi+\mathbf{X}_+\pi_k^*\Psi\otimes v. \end{equation}

Similarly, from (3·17)–(3·19) we obtain

\begin{eqnarray*} P_2^E\Psi&=&\tfrac{n+2k-4}{(n+2k-2)(n+k-3)} ({\mathrm{q}}_2^E)^* {\mathrm{q}}_2^E(\nabla^E\Psi)=-\tfrac{n+2k-4}{(n+2k-2)(n+k-3)} ({\mathrm{q}}_2^E)^* {\mathrm{d}}_0^*\Psi\\ &=&-\tfrac{n+2k-4}{(n+2k-2)(n+k-3)} \sum_i\left((\mathbf{e}_i\cdot {\mathrm{d}}_0^*\Psi)\otimes\mathbf{e}_i-\tfrac{1}{n+2k-4} {\mathrm{L}}(\mathbf{e}_i\lrcorner {\mathrm{d}}_0^*\Psi)\otimes\mathbf{e}_i\right). \end{eqnarray*}

Applying this equation at some $v\in SM$ and using (5·32), (5·33) and (5·34) we get:

(5·40) \begin{align} \pi_k^*P_2^E\Psi&=-\tfrac{n+2k-4}{(n+2k-2)(n+k-3)} \sum_i\left((\pi_1^*\mathbf{e}_i\cdot \pi_{k-1}^*{\mathrm{d}}_0^*\Psi)\otimes\mathbf{e}_i-\tfrac{1}{n+2k-4}\pi_{k-2}^*(\mathbf{e}_i\lrcorner {\mathrm{d}}_0^*\Psi\otimes\mathbf{e}_i)\right) \nonumber \\ &= \tfrac{n+2k-4}{n+k-3} \sum_i(g(\mathbf{e}_i,v)\cdot \mathbf{X}_-\pi_{k}^*\Psi)\otimes\mathbf{e}_i + \tfrac{1}{(n+2k-2)(n+k-3)}(\pi_{k-2}^*S_{k-1}{\mathrm{d}}_0^*\Psi) \nonumber \\ &=\tfrac{n+2k-4}{n+k-3}\mathbf{X}_-\pi_{k}^*\Psi\otimes v +\tfrac{1}{(n+2k-2)(n+k-3)}(\nabla_\mathbb{V}\pi_{k-1}^*{\mathrm{d}}_0^*\Psi+(k-1)\pi_{k-1}^*{\mathrm{d}}_0^*\Psi\otimes v) \nonumber\\ &=\mathbf{X}_-\pi_{k}^*\Psi\otimes v-\tfrac{1}{n+k-3}\nabla_\mathbb{V}\mathbf{X}_-\pi_{k}^*\Psi. \end{align}

Finally, using the fact that $\pi\,:\,SM\to M$ is a Riemannian submersion, we readily obtain at any $v\in SM$ :

(5·41) \begin{equation} \pi_k^*(\nabla^E\Psi)=\nabla_\mathbb{H}\pi_k^*\Psi+\mathbf{X}\pi_k^*\Psi\otimes v. \end{equation}

From (5·39)–(5·41) we thus get:

\begin{equation*} \pi_k^*P^E_3 \Psi=\pi_k^*(\nabla^E\Psi-P_1^E\Psi-P_2^E\Psi)=\nabla_\mathbb{H}\pi_k^*\Psi-\frac1{k+1}\nabla_\mathbb{V}\mathbf{X}_+\pi_k^*\Psi+\tfrac{1}{n+k-3}\nabla_\mathbb{V}\mathbf{X}_-\pi_{k}^*\Psi, \end{equation*}

which proves the lemma.

We note that as a consequence of the preceding lemma, the operator $Z_k$ defined in (5·38) does not change the degree of the section it acts on (since $P_3^E$ does not change the degree).

6. Twisted Pestov identity

The Pestov identity is a classical identity in Riemannian geometry, see [ Reference Croke and Sharafutdinov14, Reference Guillemin and Kazhdan16, Reference Guillarmou, Paternain, Salo and Uhlmann19, Reference Paternain, Salo and Uhlmann32 ] for the twisted version. Our aim is to obtain a pointwise version of this identity from the twisted Weitzenböck formula. Let us start with introducing the relevant curvature operators in our setting.

If $(E,\nabla^{E})$ is a vector bundle with metric connection, we denote by

\begin{align*}R^{E} \in C^\infty(M, \Lambda^2{\mathrm{T}}^*M \otimes {\mathrm{End}}(E)),\end{align*}

its curvature. Let $\mathcal{E}\,:\!=\pi^*E$ denote as before the pull-back of E to SM endowed with the pull-back connection $\nabla^{\mathcal{E}}\,:\!= \pi^*\nabla^{E}$ and curvature

\begin{align*}R^{\mathcal{E}}\in C^\infty(SM, \Lambda^2{\mathrm{T}}^*M \otimes {\mathrm{End}}(\mathcal{E})),\end{align*}

satisfying $R^{\mathcal{E}}_{X,Y}(\pi^*\xi)=\pi^*(R^{E}_{X,Y}\xi)$ for all $X,Y\in{\mathrm{T}} M$ (identified with their horizontal lifts) and $\forall\xi\in C^\infty(M,E)$ . Consider the vector bundle morphism $\mathcal{F}^{\mathcal{E}}:\mathcal{E}\to \mathcal{N} \otimes \mathcal{E}$ defined by:

(6·42) \begin{equation}\langle \mathcal{F}^{\mathcal{E}}(\psi),w\otimes\psi' \rangle\,:\!= \langle R^{\mathcal{E}}_{v,w}\psi,\psi'\rangle,\end{equation}

for every $v \in SM, \ w \in \mathcal{N}_v $ and $\psi,\psi' \in \mathcal{E}_v$ . The value of $\mathcal{F}^{\mathcal{E}}$ on pull-backs of sections of E can be explicitly computed as

(6·43) \begin{equation} \mathcal{F}^{\mathcal{E}}(\pi^*\xi)=\sum_i \mathbf{e}_i^\bot\otimes\pi^*( R_{v,\mathbf{e}_i}^E\xi),\end{equation}

where $(\mathbf{e}_i)$ is a local orthonormal frame. We also define a vector bundle morphism $\mathcal{R}:\mathcal{N} \otimes\mathcal{E}\to \mathcal{N} \otimes \mathcal{E}$ by:

(6·44) \begin{equation} \mathcal{R}(w\otimes\psi)\,:\!=( R_{w,v}v)\otimes\psi,\end{equation}

for every $v \in SM, \ w \in \mathcal{N}_v $ and $\psi\in \mathcal{E}_v$ , where R is the Riemann curvature tensor of (M,g).

We will now give the relations between the operators $\mathcal{R}$ and $\mathcal{F}^{\mathcal{E}}$ on one side, and q(R) and $R^{E}$ on the other side.

Lemma 6·1. For every $K\in C^\infty(M,{\mathrm{Sym}}^k_0 {\mathrm{T}} M)$ and $\xi \in C^\infty(M,E)$ , the following relations hold:

(6·45) \begin{align} \nabla_\mathbb{V}^* \mathcal{R} \nabla_\mathbb{V} \pi_k^* (K\otimes\xi)&=\pi_{k}^*((q(R)K)\otimes\xi),\end{align}

(6·46) \begin{align}\nabla_\mathbb{V}^* \mathcal{F}^{\mathcal{E}} \pi_k^* (K\otimes\xi)&=\frac12\pi_{k}^*\left(\sum_{i, j}(\mathbf{e}_i\wedge \mathbf{e}_j)_*K\otimes R^E_{\mathbf{e}_i,\mathbf{e}_j}\xi\right).\end{align}

Proof. Using (5·34) we compute at some $v\in SM$ the left-hand side of (6·45) as:

\begin{align*}\nabla_{\mathbb{V}}^* \mathcal{R} \nabla_{\mathbb{V}} \pi_k^* (K\otimes\xi) &=\sum_i \nabla_{\mathbb{V}}^* \mathcal{R}\left(\pi_{\mathcal{N}}\left(\pi_{k-1}^*( \mathbf{e}_i\lrcorner K)\otimes(\mathbf{e}_i\otimes \xi)\right)\right) \\& =\sum_{i} \nabla_{\mathbb{V}}^* \mathcal{R}\left(\pi_{k-1}^*( \mathbf{e}_i\lrcorner K)\pi_{\mathcal{N}}(\mathbf{e}_i)\otimes \pi^*\xi\right) \\& =\sum_{i} \nabla_{\mathbb{V}}^*\left(\pi_{k-1}^*( \mathbf{e}_i\lrcorner K)R_{\mathbf{e}_i,v}v\otimes \pi^*\xi\right) \\&=\sum_{i, j, l} \nabla_{\mathbb{V}}^*\left((\pi_1^*\mathbf{e}_j)(\pi_1^*\mathbf{e}_l)\pi_{k-1}^*( \mathbf{e}_i\lrcorner K)R_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_l\otimes \pi^*\xi\right)\\&=\sum_{i, j, l}\nabla_{\mathbb{V}}^*\left(\pi_{k+1}^*(\mathbf{e}_j\cdot \mathbf{e}_l\cdot( \mathbf{e}_i\lrcorner K))R_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_l\otimes \pi^*\xi\right).\end{align*}

Using (5·35) we can rewrite this last sum as

\begin{align*}-\sum_{i, j, l}\pi_k^*\left(R_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_l\lrcorner(\mathbf{e}_j\cdot \mathbf{e}_l\cdot( \mathbf{e}_i\lrcorner K))\otimes \xi\right) + (k+1)\sum_{i, j, l}\pi^*_{k+2}\left(R_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_l\cdot \mathbf{e}_j\cdot \mathbf{e}_l\cdot( \mathbf{e}_i\lrcorner K)\otimes \xi\right).\end{align*}

By Lemma 3·2 the first summand is equal to $\pi_{k}^*((q(R)K)\otimes\xi)$ . The second summand vanishes since $\sum_lR_{\mathbf{e}_i,\mathbf{e}_j}\mathbf{e}_l\cdot \mathbf{e}_l = 0$ . This proves (6·45). Similarly, using (6·43) we compute at $v\in SM$ :

\begin{eqnarray*} \nabla_\mathbb{V}^* \mathcal{F}^{\mathcal{E}} \pi_k^* (K\otimes\xi) & =&\sum_i \nabla_\mathbb{V}^*\left(\pi_k^* (K) (\mathbf{e}_i^\bot\otimes \pi^*(R_{v,\mathbf{e}_i}^E\xi))\right) \\ &=&\sum_i \nabla_\mathbb{V}^*\left(\pi_k^* (K) (\mathbf{e}_i\otimes \pi^*(R_{v,\mathbf{e}_i}^E\xi))\right)\\&=&\sum_{i, j} \nabla_\mathbb{V}^*\left(\pi_k^* (K)\pi_1^*(\mathbf{e}_j)(\mathbf{e}_i\otimes \pi^*(R_{\mathbf{e}_j,\mathbf{e}_i}^E\xi))\right) \\&=&\sum_{i, j} \nabla_\mathbb{V}^*\left(\pi_{k+1}^* (\mathbf{e}_j\cdot K)(\mathbf{e}_i\otimes \pi^*(R_{\mathbf{e}_j,\mathbf{e}_i}^E\xi))\right)\\&\stackrel{\eqref{31}}{=}&-\sum_{i, j}\pi_{k}^*(\mathbf{e}_i\lrcorner(\mathbf{e}_j\cdot K))\pi^*(R_{\mathbf{e}_j,\mathbf{e}_i}^E\xi)\\&& +(k+1)\sum_{i, j}\pi_{k+2}^* (\mathbf{e}_i\cdot \mathbf{e}_j \cdot K)\pi^*(R_{\mathbf{e}_j,\mathbf{e}_i}^E\xi).\end{eqnarray*}

The second summand vanishes because of the skew-symmetry of $R_{\mathbf{e}_j,\mathbf{e}_i}^E$ in i and j, whereas the first summand is equal to

\begin{align*} -\sum_{i, j} \pi_{k}^*\left(\mathbf{e}_i\lrcorner(\mathbf{e}_j\cdot K)\otimes R_{\mathbf{e}_j,\mathbf{e}_i}^E\xi\right) &= \sum_{i, j}\pi_{k}^*\left(\mathbf{e}_j\cdot(\mathbf{e}_i\lrcorner K)\otimes R_{\mathbf{e}_i,\mathbf{e}_j}^E\xi\right)\\ &= \frac12\sum_{i, j} \pi_{k}^*\left((\mathbf{e}_i\wedge \mathbf{e}_j)_* K\otimes R_{\mathbf{e}_i,\mathbf{e}_j}^E\xi\right).\end{align*}

Combining (3·11) with Lemma 6·1, we obtain for every section of ${\mathrm{Sym}}^k_0 {\mathrm{T}} M\otimes E$ :

(6·47) \begin{equation}\pi_k^*q(R)^E=(\nabla_\mathbb{V}^* \mathcal{R} \nabla_\mathbb{V}+ \nabla_\mathbb{V}^* \mathcal{F}^{\mathcal{E}}) \pi_k^*.\end{equation}

Then, using Lemma 5·3 we compute for every section of ${\mathrm{Sym}}^k_0 {\mathrm{T}} M\otimes E$ :

(6·48) \begin{equation}\pi_k^*{\mathrm{d}}_0^*{\mathrm{d}}_0=-(n+2k)\mathbf{X}_-\pi_{k+1}^*{\mathrm{d}}_0=-(n+2k)\mathbf{X}_-\mathbf{X}_+\pi_{k}^*,\end{equation}

and similarly

(6·49) \begin{equation}\pi_k^*{\mathrm{d}}_0{\mathrm{d}}_0^*=\mathbf{X}_+\pi_{k-1}^*{\mathrm{d}}_0^*=-(n+2k-2)\mathbf{X}_+\mathbf{X}_-\pi_{k}^*.\end{equation}

Finally, by (5·37) we obtain

(6·50) \begin{equation}\pi_k^*(P^E_3)^*P^E_3=Z_k^*\pi_{k}^*P^E_3=Z_k^*Z_k\pi_{k}^*.\end{equation}

Altogether, we obtain the following:

Proposition 6·2 (Pointwise localised Pestov identity). On $C^\infty(M, \Omega^k\otimes E) \subset C^\infty(SM, \mathcal{E})$ , the following relation holds:

(6·51) \begin{equation} \nabla_\mathbb{V}^* \mathcal{R} \nabla_\mathbb{V} + \nabla_\mathbb{V}^* \mathcal{F}^{\mathcal{E}}= \tfrac{k(n+2k)}{k+1} \mathbf{X}_-\mathbf{X}_+ \,- \, \tfrac{(n+k-2)(n+2k-4)}{(n+k-3)}\,\mathbf{X}_+\mathbf{X}_-\,+ \,Z_k^*Z_k.\end{equation}

Applying (6·51) to $\Psi \in C^\infty(M,\Omega^k \otimes E)$ , pairing with $\Psi$ and then integrating over SM with respect to the Liouville measure, we retrieve the localised Pestov identity in its integrated version [Reference Cekić, Lefeuvre, Moroianu and Semmelmann11, lemma 2·3].