1. Introduction

Variational problems arise in various areas of mathematics and physics. Suppose $(M, g)$ is a Riemannian manifold with volume form $d v_M$, it is the case that functionals of the form

$I ( \varphi, g) = \int_M \, \sigma ( \varphi, g ) \, d v_M$

(1.1)

very often occur ^[1,2]. Here $\varphi$ could be a mapping between Riemannian manifolds of a vector bundle valued differential form. Given a variational problem starting from (1.1), the stress-energy tensor $S$ can be derived by considering variations of the metric on $M$. If $I$ has a critical point with respect to variations of $\varphi$, then the stress-energy tensor is divergence free, and there are conservation laws.

To provide some motivation, let $(M, g)$ and $(N, h)$ be two smooth Riemannian manifolds which are connected, compact, orientable and without boundary, and $\varphi :(M, g) \to (N, h)$ a smooth map. The differential of $\varphi$ which is $d \varphi$ can be thought of as a section of the bundle $T^* M \otimes \varphi^{-1} TN$, with norm $| d \varphi|$. If $\{ x^i \}$ and $\{u^a \}$ constitute local coordinate systems around $x$ and $\varphi (x)$, respectively, then in terms of coordinates, we can write

$| d \varphi |^2 = g^{ij}\, h_{ab} ( \varphi) \frac{\partial \varphi^a}{\partial x^i} \frac{\partial \varphi^b}{\partial x^j},$

(1.2)

where $(\partial \varphi^a / \partial x^i)$ is the local representation of $d \varphi$. Then the energy density of $\varphi$ can be defined as $e (\varphi) = 1/2 | d \varphi|^2$ and the energy density of the field is given by the positive functional $E (\varphi) = \int_M \, e (\varphi) \, d v_M$.

A large class of maps which come up in physics, especially in gravity, are called harmonic. A mapping $\varphi : M \to N$ is harmonic if and only if it is an extremal of the energy. Consequently, it is the case that a map $\varphi$ is harmonic if and only if it satisifes the Euler-Lagrange equation

$\tau ( \varphi) =- d^{*} d \, \varphi = {\rm{tr}} \nabla \, d \varphi.$

This defines the tension field of $\varphi$, and may be expressed in local coordinates on $M$ and $N$ as follows

$\nabla_{\partial_i} ( d \varphi) = ( \frac{\partial \varphi_j^a}{\partial x^i} ) \, d x^j \frac{\partial}{\partial u^a} + \varphi_j^a ( \nabla_{\partial_i} \, d x^j ) \frac{\partial}{\partial u^a} + \varphi_j^a \, dx^j \, \nabla_{\partial_i} \frac{\partial}{\partial u^a}$

$= \varphi^a_{ij} - \, ^M\Gamma_{ij}^k \varphi^a_k + \, ^N \Gamma_{b \gamma}^b \varphi_j^{\gamma}.$

(1.3)

The tension field is the trace of (1.3),

$\tau (\varphi)^a = g^{ij} ( \nabla \, d \varphi)^a_{ij} =- \Delta \varphi^a + \, ^N \Gamma^a_{b \gamma} \varphi_i^b \varphi_j^{\gamma} g^{ij}.$

(1.4)

Thus (1.4) is a semilinear, elliptic, second-order system. If $N$ is the space $\mathbb R$, a harmonic map is called a harmonic function ^[4,7,8].

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2. Energy Functional and Critical Point

Now let us extend this idea to another object which may be defined on a manifold. Let $M$ be a Riemannian manifold and $E$ a Riemannian vector bundle over $M$, where each fiber carries a positive definite inner product denoted by $\langle \cdot, \cdot \rangle^E$. Let $\Omega^p (E)$ be the space of smooth $p$-forms which have values in $E$, where it is assumed throughout that $p \geq 1$. For $\omega \in \Omega^p (E)$, define the energy functional

$I ( \omega, g) = \int_M \langle \omega( e_{i_1}, \ldots, e_{i_p} ), \omega ( e_{i_1}, \ldots, e_{i_p} ) \rangle ^E \, d v_m.$

(2.1)

where $\{ e_i \}$ is an orthonormal basis on $M$ and repeated indices are summed for $1 \leq i_1, \ldots, i_p \leq m$ and $m =\dim M$. With respect to a local coordinate system $\{ x^i \}$ on $M$ and local frame $\{ s_a \}$ of $E$, the norm of $\omega$, which is the integrand of (2.1) can be written

$| \omega |^2 = \langle \omega ( e_{i_1}, \ldots, e_{i_p}), \omega ( e_{i_1}, \ldots, e_{i_p}) \rangle = g^{i_1 j_1} \cdots g^{i_p j_p} \omega^a_{i_1 \cdots i_p} \omega^b_{j_1 \cdots j_p} h_{ab}.$

(2.2)

Suppose $M$ is compact, then vary the integral (2.1) with respect to metric $g$. If $g (u)$ is a smooth, one-parameter family of metrics such that $g (0)=g$, then the variation $\delta g = \partial g / \partial u |_{u=0}$ is a smooth symmetric tensor on $M$.

Theorem 1. For $\omega \in \Omega^p (E)$ and $p \geq 1$,

$\frac{d I}{d u} |_{u=0} = \int_M \, \langle S ( \omega), \frac{\partial g}{\partial u} |_{u=0} \rangle \, d v_M,$

(2.3)

where $S (\omega)$ is the symmetric two-tensor defined by

$S ( \omega) = \frac{1}{2} | \omega |^2 g - p \sum\limits_{i_2, \cdots, i_p} \langle \omega ( \cdot, e_{i_2}, \ldots, e_{i_p} ), \omega ( \cdot, e_{i_2}, \ldots, e_{i_p}) \rangle^E.$

(2.4)

where $\{ e_i \}$ is an orthonormal basis on $M$.

Proof: Let $\{ x^i \}$ be a local coordinate system on $M$, and $\{ s_a \}$ a local frame for $E$,

$\frac{ d I (\omega)}{ d u} |_{u=0} = \int_M \, \frac{\partial | \omega|^2} {\partial g_{ij}} \delta g_{ij} \, d v_M + \int_M \, |\omega|^2 \frac{\partial ( d v_M)}{\partial g_{ij}} \, \delta g_{ij}.$

(2.5)

The volume form on $M$ is given by

$d v_M = (\det g)^{1/2} \, d x^1 \wedge \ldots \wedge d x^m,$

where $\det g$ is the determinant of the metric tensor $g_{ij}$ and therefore,

$\frac{\partial}{\partial g_{ij}} d v_M = \frac{1}{2} ( \det g)^{-1/2} \frac{\partial}{\partial g_{ij}} ( \det g) \, d x^1 \wedge \ldots \wedge d x^m = \frac{1}{2} g^{ij} \, d v_M.$

Differentiating the expression for the metric tensor $g^{ij} g_{jk} = \delta^i_k$, the following relation holds

$\frac{\partial g^{i_s j_s}}{\partial g_{ij}} =- g^{i_s i} g^{j_s j}.$

The first term in (2.5) is

$\frac{\partial |\omega|^2}{\partial g_{ij}} = \frac{\partial}{\partial g_{ij}} \big( g^{i_1 j_1} \cdots g^{i_p j_p} \omega^a_{i_1 \cdots i_p} \omega^b_{j_1 \cdots j_p} h_{ab} \big)$

$= - \sum\limits_{s=1}^p \, g^{i_1 j_1} \cdots g^{i_s i} g^{j_s j} \cdots g^{i_p j_p} \omega^a_{i_1 \cdots i_s \cdots i_p} \omega^b_{j_1 \cdots j_s \cdots j_p} \cdot h_{ab}.$

Therefore,

$\frac{\partial | \omega|^2}{\partial g_{ij}} g_{ik} g_{jl} =- \sum\limits_{s=1}^p \, g^{i_1 j_1} \cdots \delta_k^{i_s} \delta^{j_s}_l \cdots g^{i_p j_p} \omega_{i_1 \cdots i_s \cdots i_p}^a \omega_{j_1 \cdots j_s \cdots j_p}^b h_{ab}$

$=-p g^{i_2 j_2} \cdots g^{i_p j_p} \omega^a_{k i_2 \cdots i_p} \omega^b_{l j_2 \cdots j_p} \cdot h_{ab}.$

Definition 1. Let $M$ be an arbitrary not necessarily compact Riemannian manifold, and let $E$ be a Riemannian vector bundle over $M$. Let $\omega \in \Omega^p (E)$ and define the stress-energy tensor of the form $\omega$ to be the following symmetric $2$-tensor

$S ( \omega) = \frac{1}{2} | \omega|^2 g - p \sum\limits_{j_1, \cdots, j_p} \, \langle \omega ( \cdot, e_{j_2}, \cdots, e_{j_p}), \omega ( \cdot, e_{j_2}, \cdots, e_{j_p} ) \rangle^E$

(2.6)

at each point $x$ where there is an orthonormal basis $\{ e_i \}$.

Let the vector bundle $E$ be endowed with a Riemannian connection denoted by $\nabla^E$ so

$X \langle s, t \rangle^E = \langle \nabla^E_X s, t \rangle^E + \langle s, \nabla^E_X t \rangle^E,$

(2.7)

Theorem 2. Let $\omega \in \Omega^p (E)$ and $S (\omega)$ be the stress-energy tensor associated with $\omega$, then for all $x \in M$ and each $X \in T_x M$,

$\div \, S ( \omega ) (X) = \langle \omega, d \omega \rfloor X \rangle + p \langle d^* \omega, \omega \rfloor X \rangle$

(2.8)

where the contraction of a $p$-form with a vector field $X$ is given by

$( \omega \rfloor X) ( X_1, \ldots, X_{p-1}) = \omega ( X, X_1, \ldots, X_{p-1}).$

Proof: Let $\{ e_i \}$ be an orthonormal basis at $x$ and extend objects to a neighborhood of $x$. Suppose that $\nabla_{e_i} \, e_j =0$ at $x$ for all $i, j$. The tensorial property allows one to evaluate at $X = e_k$ without loss of generality. Consequently,

$\div S (X) = \sum\limits_j \, (\nabla_{e_j} S) ( e_j, e_k ) = \sum\limits_j \, e_j S ( e_j, e_k) = \sum\limits_j e_j \{ \frac{1}{2} | \omega| g ( e_j, e_k) - p \langle \omega \rfloor e_j, \omega \rfloor e_k \rangle \}$

$= \langle \nabla_{e_k} \omega, \omega \rangle - p \sum\limits_j \, \langle \nabla_{e_j} ( \omega \rfloor e_j), \omega \rfloor e_k \rangle - p \sum\limits_j \langle \omega \rfloor e_j, \nabla_{e_j} ( \omega \rfloor e_k ) \rangle.$

(2.9)

At $x$, it is the case that

$d^{*} \omega ( e_{j_1}, \ldots, e_{j_{p-1}} ) =- \nabla_{e_j}^E \, ( \omega ( e_j, e_{j_1}, \ldots, e_{j_{p-1}} )),$

and

$d \omega ( e_k, e_{i_1}, \ldots, e_{i_p}) = \nabla_{e_k}^E ( \omega ( e_{i_1}, \ldots, e_{i_p})) + \sum\limits_{k=1}^p \, (-1)^k \nabla_{e_{i_k}}^E \, ( \omega (e_k, e_{i_1}, \ldots, \hat{e}_{i_k}, \ldots, e_{i_p} )).$

Solving for $\nabla_{e_k}^E$ and substituting it into the divergence, we get

$\div S (X) = \langle d \omega \rfloor e_k, \omega \rangle$

$- \sum\limits_{i_1, \cdots, i_p} \sum\limits_{k=1}^p \, (-1)^k \langle \nabla_{e_{i_k}}^E ( \omega ( e_k, e_{i_1}, \ldots, \hat{e}_{i_k}, \ldots, e_{i_p})), \omega ( e_{i_1}, \ldots, e_{i_p} ) \rangle^E$

$+ p \langle d^* \omega, \omega \rfloor X \rangle -p \sum\limits_j \langle \omega \rfloor e_j, \nabla_{e_j} ( \omega \rfloor e_k ) \rangle.$

The double sum in this can be simplified to the form

$\sum\limits_{i_1, \cdots, i_p} \, \sum\limits_{k=1}^p \, (-1)^{k+1} \langle \nabla_{e_{i_k}}^E ( \omega ( e_k, e_{i_1}, \ldots, \hat{e}_{i_k}, \ldots, e_{i_p} )), \omega ( e_1, \ldots, e_{i_p}) \rangle$

$= p \sum\limits_{j=1}^n \, \langle \nabla_{e_j} ( \omega \rfloor e_k), \omega \rfloor e_j \rangle.$

The claim follows as a result of these

$\div \, S( \omega ) = \langle \omega, d \omega \rfloor e_k \rangle + p \langle d^* \omega, \omega \rfloor e_k \rangle + p \sum\limits_j \langle \nabla_{e_j} ( \omega \rfloor e_k ), \omega \rfloor e_j \rangle_j \langle \omega \rfloor e_j, \nabla_{e_j} ( \omega \rfloor e_k) \rangle$

$= \langle \omega, d \omega \rfloor e_k \rangle + p \langle d^* \omega, , \omega \rfloor e_k \rangle.$

The form $\omega \in \Omega^p (E)$ is called harmonic if and only if $d \omega = d^* \omega =0$. Substituting these derivatives into the right-hand side of (2.8), we prove the following result.

Corollary 1. If $\omega \in \Omega^p (E)$ is a harmonic form, then $\div \, S(\omega) =0$.

Clearly, $S$ is symmetric and so given a symmetric, divergence-free $2$-tensor $S$, conservation laws can be formulated by contracting with a Killing vector field. If $X$ is a Killing vector field, then $S \rfloor X$ is also divergence free. However, the converse of Corollary 1 is not true. If $\div \, S (\omega)=0$, it may not be concluded that $\omega$ is harmonic. However, when $\omega$ is a differential of a submersive mapping, there is equivalence.

Corollary 2. Let $\varphi : M \to N$ be a smooth mapping between Riemannian manifolds and let $\omega = d \varphi$ be the corresponding $\varphi^{-1} TN$-valued one-form on $M$. Then for each vector field $X$,

$\div S ( d \varphi) (X) = \langle \tau ( \varphi), \, d \varphi (X) \rangle.$

(2.10)

In (2.10), $\tau (\varphi)$ is called the tension field of $\varphi$ and is defined as

$\tau ( \varphi ) =- d^* d \varphi.$

(2.11)

If $\varphi$ is a submersive almost everywhere, then $\tau (\varphi) =0$ if and only if $\div S (d \varphi)=0$.

Proof: Substitute $\omega = d \varphi$ into (2.8) and use the identity $d (d \varphi) =0$ to obtain,

$\div \, S ( d \varphi) (X) = \langle d \varphi, d d \varphi \rfloor X \rangle + \langle d^* d \varphi, d \varphi \rfloor X \rangle =- \langle \tau ( \varphi ), d \varphi \rfloor X \rangle.$

(2.12)

Clearly, if $\tau (\varphi) =0$, then the right side of (2.12) vanishes, $\div S (d \varphi) (X) =0$, and conversely.

The form $\omega$ is a critical form with respect to variations of the metric, so the conditions under which the stress-energy tensor vanishes should be studied.

Definition 2. The form $\omega \in \Omega^p (E)$ is conformal if the map $X \to \omega \rfloor X$ is conformal for each $x \in X$, or equivalently, $\omega$ is conformal if and only if there is a real $\lambda$ such that

$\langle \omega \rfloor X, \omega \rfloor Y \rangle = \lambda^2 \, \langle X, Y \rangle, \qquad X, Y \in T_x M.$

(2.13)

To obtain an expression for $\lambda^2$, evaluate $S (\omega)$ in (2.13) on the vectors $X=Y= e_i \in T_x M$ and carry out the trace on both sides

$m \lambda^2 = | \omega|^2.$

(2.14)

If $\varphi : M \to N$ is a smooth map between Riemannian manifolds, then $d \varphi \in \Omega^1 (\varphi^{-1} TN)$ is a conformal form if and only if the map is conformal as well.

Lemma 1. For $\omega \in \Omega^p (E)$ which is not identically zero, the stress-energy tensor (2.6) vanishes identically if and only if $m= 2p$ and $\omega$ is conformal.

Proof: Evaluating $S (\omega)$ on the pair $(e_i, e_i)$ we obtain,

$S ( \omega) ( e_i, e_i) = \frac{1}{2} | \omega|^2 -p \langle \omega \rfloor e_i, \omega \rfloor e_i \rangle.$

Tracing on both sides gives

${\rm{tr}} \, S ( \omega ) = (\frac{m}{2} - p) | \omega |^2.$

This vanishes for $\omega \neq 0$ when $m = 2p$. The definition of conformal form (2.13) gives equivalence.

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3. Weizenböck Formulas and Applications to the Stress-Energy Tensor

Now consider the Laplacian of the tensor $S (\omega)$. The Laplacian on $p$-forms was extended to arbitrary tensors on a Riemannian manifold by Lichnerowicz ^[5,6]. For a $2$-tensor $Q$, the Laplacian on $Q$ is given by

$\Delta \, Q (X, Y ) =- {\rm{tr}} \nabla^2 Q (X, Y) + S ( {\rm{Ricci}}^M (X), Y) + S ( X, {\rm{Ricci}}^M (Y) )$

$-2 \sum\limits_{i, j} \langle R^M (X, e_i) Y, e_j \rangle S (e_i, e_j ).$

(3.1)

where $\{ e_i \}$ is an orthonormal basis. The curvature tensor on $M$ is represented by $R^M$ with sign convention

$R^M (X, Y) Z = - \nabla_X \nabla_Y Z + \nabla_Y \nabla_X Z + \nabla_{[X, Y]} Z.$

(3.2)

In (3.2), ${\rm{Ricci}}^M$ is the Ricci tensor defined by ${\rm{Ricci}}^M (X, Y) = {\rm{tr}} (Z \to R^M (X, Z) Y)$.

The following theorem given first by Lichnorowicz ^[5,6] is very useful and is presented without proof.

Proposition 1. (Lichnerowicz) Let $M$ be a Riemannian manifold with $\nabla {\rm{Ricci}}^M =0$. Let $Q$ be a $2$-tensor on $M$, then the divergence commutes with the Laplacian

$\div ( \Delta Q) = \Delta ( \div Q),$

(3.3)

where the right-hand side is now the standard Laplacian on one-forms.

Define the symmetric $2$-tensor $L (Q)$ to be

$L (Q) = Q ( {\rm{Ricci}}^M (X), Y) + Q(X, {\rm{Ricci}}^M (Y)) -2 \sum\limits_{i, j} \, \langle R^M (X, e_i) Y, e_j \rangle \, Q ( e_i, e_j ).$

(3.4)

Suppose $E \to M$ is a vector bundle endowed with a metric and Riemannian connection and associated curvature $R^E$. The Ricci operator on a $p$-form $\omega$ written as ${\rm{Ricci}}^p = {\rm{Ricci}}$ is defined to be

$( {\rm{Ricci}} (\omega) ) ( X_1, \ldots, X_p) = \sum\limits_{i, k} ( R^p ( e_i, X_k ) \omega ) ( e_i, X_1, \ldots, \hat{X}_k, \ldots, X_p).$

(3.5)

In (3.5), $R^p$ is the canonical curvature and is defined to be ^[4]

$(R^p (X, Y) \omega ) ( X_1, \ldots, X_p ) = R^E (X, Y) ( \omega (X_1, \ldots, X_p))$

$- \sum\limits_k \, \omega (X_1, \ldots, R^M (X, Y) X_k, \ldots, X_p).$

(3.6)

The following theorem is due to Weizenböck ^[3].

Proposition 2. (Weitzenböck) Let $E \to M$ be a Riemannian vector bundle over a Riemannian manifold $M$. Then for any $\omega \in \Omega^p (E)$,

$\begin{array}{lc} (i) & \quad \Delta \omega = - {\rm{tr}} \nabla^2 \omega + {\rm{Ricci}} (\omega). \\ & \\ (ii) & \qquad \frac{1}{2} \Delta | \omega |^2 = \langle \Delta \omega, \omega \rangle - | \nabla \omega |^2 - \langle {\rm{Ricci}} (\omega), \omega \rangle . \\ \end{array}$

(3.7)

Proof: Let us prove $(ii)$ by using $(i)$ for $-{\rm{tr}} \nabla^2 \omega$,

$\frac{1}{2} \Delta | \omega |^2 =- \frac{1}{2} {\rm{tr}} \nabla \, d | \omega |^2 =- {\rm{tr}} \nabla \langle \, \nabla \omega, \omega \rangle =- {\rm{tr}} \langle \nabla \nabla \omega, \omega \rangle - \langle \nabla \omega, \nabla \omega \rangle$

$= \langle \Delta \omega, \omega \rangle - \langle {\rm{Ricci}} (\omega ), \omega \rangle - | \nabla \omega |^2.$

It is worth noting a few important applications of this proposition. If $\omega \in \Omega^0 (E)$, then (3.7) reduces to

$\Delta \omega =- {\rm{tr}} \nabla d \omega, \qquad \frac{1}{2} \Delta | \omega |^2 = \langle \Delta \omega, \omega \rangle - | \nabla \omega |^2.$

(3.8)

If $\omega \in \Omega^1 (E)$ and $X \in \Gamma (TM)$, then

$\Delta \omega (X) =- {\rm{tr}} \nabla^2 \omega (X) - \sum\limits_s \, R^E ( e_s, X) \omega ( e_s) + \sum\limits_s \, \omega (R^M ( e_s, X) e_s),$

(3.9)

where $\{ e_s \}$ is an orthonormal basis for $TM$ and

$\frac{1}{2} \Delta | \omega |^2 = \langle \Delta \omega, \omega \rangle - | \nabla \omega |^2 + \sum\limits_{i, j} \langle R^E ( e_i, e_j ) \omega ( e_i ), \omega ( e_j ) \rangle - \sum\limits_k \, \langle \omega ( {\rm{Ricci}} (e_k )), \omega ( e_k) \rangle.$

(3.10)

Theorem 3. Let $X, Y \in \Gamma (TM)$, it holds that

$( {\rm{tr}} \nabla^2 \langle \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle ) (X, Y) = \langle ({\rm{tr}} \nabla^2 \omega ) \lrcorner X, \omega \lrcorner Y \rangle + \langle \omega \lrcorner X, ( {\rm{tr}} \nabla^2 \omega ) \lrcorner Y \rangle$

$+ 2 \sum\limits_i \langle ( \nabla_{e_i} \omega ) \lrcorner X, ( \nabla_{e_i} \omega ) \lrcorner Y \rangle.$

(3.11)

Proof: Differentiating once gives,

$\nabla_{e_j} \langle \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle (X, Y) = \langle \nabla_{e_j} \omega \lrcorner X, \omega \lrcorner Y \rangle + \langle \omega \lrcorner X, \nabla_{e_j} \omega \lrcorner Y \rangle.$

Differentiating a second time gives

$\nabla_{e_i} \nabla_{e_j} \langle \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle (X, Y) = \langle \nabla_{e_i} \nabla_{e_j} \omega \lrcorner X, \omega \lrcorner Y \rangle + \langle \nabla_{e_j} \omega \lrcorner X, \nabla_{e_j} \omega \lrcorner Y \rangle + \langle \nabla_{e_i} \omega \lrcorner X, \nabla_{e_j} \omega \lrcorner Y \rangle$

$+ \langle \omega \lrcorner X, \nabla_{e_i} \nabla_{e_j} \omega \lrcorner Y \rangle.$

Tracing on both sides of this equation yields (3.11).

Theorem 4. Let $\omega \in \Omega^p (E)$ be a vector bundle valued $p$-form with stress-energy tensor $S (\omega)$. The Laplacian of $S (\omega)$ can be written in the following way,

$\Delta S ( \omega ) = 2 S ( \langle \Delta \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle ) - 2 S ( \sum\limits_j \langle \nabla_{e_j} \omega \lrcorner \cdot, \nabla_{e_j} \omega \lrcorner \cdot \rangle) - 2 S ( \langle {\rm{Ricci}} ( \omega) \lrcorner \cdot, \omega \lrcorner \cdot \rangle ) + L (S (\omega)).$

(3.12)

where ${\rm{Ricci}}^M$ is the Ricci tensor and the symmetric $2$-tensor $L (S)$ was introduced in (3.4). Substituting Weitzenböck formula (3.7) $(i)$ into (3.11), it follows that

${\rm{tr}} \nabla^2 \langle \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle (X, Y) = \langle ( \Delta \omega - {\rm{Ricci}} ( \omega)) \lrcorner X, \omega \lrcorner Y \rangle + \langle \omega \lrcorner X, ( \Delta \omega - {\rm{Ricci}} ( \omega )) \lrcorner Y \rangle$

$- 2 \sum\limits_i \, \langle \nabla_{e_i} \omega \lrcorner X, \nabla_{e_i} \omega \lrcorner Y \rangle$

$= \langle \Delta \omega \lrcorner X, \omega \lrcorner Y \rangle - \langle {\rm{Ricci}} ( \omega) \lrcorner X, \omega \lrcorner Y \rangle + \langle \omega \lrcorner X, \Delta \omega \lrcorner Y \rangle - \langle \omega \lrcorner X, {\rm{Ricci}} ( \omega ) \lrcorner Y \rangle$

$- 2 \sum\limits_i \, \langle \nabla_{e_i} \omega \lrcorner X, \nabla_{e_i} \omega \lrcorner Y \rangle.$

(3.13)

Substituting (3) into (3.12), it follows by symmetry and linearity that,

$\Delta S ( \omega) (X, Y) = 2 S ( \langle \Delta \omega \lrcorner X, \omega \lrcorner Y \rangle ) - 2 S ( \sum\limits_i \, \langle \nabla_{e_i} \omega \lrcorner X, \nabla_{e_i} \omega \lrcorner Y \rangle )$

$-2 S ( \langle {\rm{Ricci}} ( \omega) \lrcorner X, Y \rangle ) + L (S ( \omega )) (X, Y).$

This is the required result.

For any $2$-tensor $Q \in \Gamma (\otimes^2 T^* M)$, define the $p$-th stress-energy tensor associated to $Q$ to be the $2$-tensor

${\cal S}_p (Q) = \frac{1}{2} ( {\rm{tr}}_g \, Q ) g - p \, {\rm{sym}} \, Q,$

(3.14)

where ${\rm{tr}} Q = \sum_i \, Q (e_i, e_i)$, and $\{ e_i \}$ is orthonormal with respect to the metric $g$. Moreover, define

$( {\rm{sym}} \, Q ) (X, Y) = \frac{1}{2} ( Q (X, Y) + Q ( Y, X)).$

(3.15)

Then ${\rm{sym}} \, Q$ in (3.15) is called the symmetrization of $Q$. For a $p$-form $\omega \in \Omega^p (E)$, this is simply

$S ( \omega ) = {\cal S}_p ( \langle \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle ).$

(3.16)

The previous result can be written in completely symmetric form in terms of ${\cal S}_p$.

Corollary 3. Let $\omega \in \Omega^p (E)$ be a vector-bundle valued $p$-form with associated stress-energy tensor $S (\omega)$. The Laplacian of $S (\omega)$ is given by

$\Delta S ( \omega ) = 2 {\cal S}_p ( \langle \Delta \omega \lrcorner \cdot, \omega \lrcorner \cdot \rangle ) - 2 {\cal S}_p ( \sum\limits_i \, \langle \nabla_{e_i} \omega \lrcorner \cdot \nabla_{e_i} \omega \lrcorner \cdot \rangle ) - 2 {\cal S}_p ( \langle {\rm{Ricci}} ( \omega ) \lrcorner \cdot, \omega \lrcorner \cdot \rangle ) + L ( S ( \omega) ).$

(3.17)

It is also worth writing this out for the special case of a $1$-form.

Corollary 4. Let $\omega \in \Omega^1 (E)$ be a vector bundle valued $1$-form, then (3.17) takes the form

$\Delta S ( \omega) = 2 {\cal S} ( \langle \Delta \omega, \omega \rangle ) - 2 {\cal S} ( \sum\limits_i \langle \nabla_{e_i} \omega, \nabla_{e_i} \omega \rangle ) + 2 {\cal S} ( \sum\limits_i \langle R^E ( \omega ) \lrcorner \cdot, \omega \lrcorner \cdot \rangle )$

$- 2 {\cal S} \Big( \langle \omega \big( {\rm{Ricci}}^M ( \cdot) \big), \omega \rangle \Big) + L (S ( \omega )).$

(3.18)

Moreover, for $\omega \in \Omega^1 (E)$, a closed vector bundle valued $1$-form,

$\Delta S ( \omega ) = 2 {\cal S} ( \langle \Delta \omega, \omega \rangle ) - 2 S ( \nabla \omega) + 2 {\cal S} ( \sum\limits_i \, \langle R^E ( e_i, \cdot) \omega ( e_i), \omega (\cdot ) \rangle )$

$- 2 {\cal S} ( \langle \omega ( {\rm{Ricci}}^M ( \cdot), \omega \rangle ) + L (S ( \omega)).$

(3.19)

where $S (\nabla \omega)$ denotes the stress-energy tensor of the $T^* M \otimes E$ valued one-form $\nabla_X \omega$.

Proof: First (3.18) is a consequence of (3.17). For (3.19), since $\omega$ is closed, $d \omega =0$, and it follows that $(\nabla_{e_i} \omega)(X) = (\nabla_X \omega) (e_i)$, consequently $\sum_i \, \langle \nabla_{e_i} \omega, \nabla_{e_i} \omega \rangle = \langle \nabla \omega, \nabla \omega \rangle$.

Theorem 5. Let $\varphi : (M, g) \to (N, h)$ be a smooth map between Riemannian manifolds and let $d \varphi \in \Omega^1 (\varphi^{-1} TN)$ be its exterior derivative.

$\Delta S ( d \varphi) =- 2 {\cal S} ( \langle \nabla \tau ( \varphi), d \varphi \rangle ) - 2 S ( \nabla d \varphi) - 2 {\cal S} ( \langle d \varphi ( {\rm{Ricci}}^M ( \cdot)), d \varphi ( \cdot) \rangle )$

$- 2 {\cal S} ( \sum\limits_i \langle R^N ( d \varphi ( e_i), d \varphi( \cdot))\, d \varphi ( e_i), d \varphi ( \cdot) \rangle ) + L ( S ( d \varphi)),$

(3.20)

where $\tau (\varphi) =-d^* d \varphi$ denotes the tension field of the map $\varphi$.

Proof: For the one-form $\omega$, substitute $\omega = d \varphi$ into (3.19). Since it follows that $\Delta d \varphi = (d^* d + d d^*) d \varphi = -d \tau (\varphi) =-\nabla \tau (\varphi)$, and $R^N \circ d \varphi$ is the induced curvature in the bundle $E= \varphi^{-1} TN$.

Theorem 6. Let $\varphi : (M^2, g) \to (N^2, h)$ be a harmonic map between surfaces where $M^2$ has Gaussian curvature $c^M$, then the folowing holds,

$\Delta S ( d \varphi) =- 2 S ( \nabla d \varphi ) + 2 c^M S ( d \varphi).$

(3.21)

In particular, if $M$ has constant curvature, then it follows that $\div S (\nabla d \varphi)=0$.

Proof: If $\varphi$ is harmonic, then $\nabla \tau (\varphi)=0$, so the first term in (3.20) vanishes. Suppose $\sigma (Q) = \frac{1}{2} ({\rm{tr}} Q) g -{\rm{sym}} \, Q$ is a stress-energy tensor associated to the $2$-tensor $Q$ on $M$. Suppose $z$ is an isothermal coordinate, then $\sigma (Q)$ can be expressed in diagonal form $\sigma (Q) = a \, dz \otimes dz + \bar{a} \, d \bar{z} \otimes d \bar{z}$ for some function $a= a (z, \bar{z})$. If $\sigma (Q)$ is divergence free, then $a_{\bar{z}}$ vanishes, so $\sigma (Q)=a \, dz \otimes dz$ defines a quadratic differential form on $M^2$. When $\varphi$ is a smooth map, then locally $\varphi^* h$ can be diagonalized to the form $\varphi^* h = \lambda_1 \, \omega_1 \otimes \omega_1 + \lambda_2 \omega_2 \otimes \omega_2$, where $\{ \omega_1, \omega_2 \}$ is an orthonormal basis of $1$-forms. If $M$ and $N$ have Gaussian curvatures $c^M$ and $c^N$, then letting $\{ e_1, e_2 \}$ be the basis dual to $\{ \omega_1, \omega_2 \}$, summing on $i=1, 2$ we have $\langle R^N (d \varphi (e_i), d \varphi (\cdot)) d \varphi (e_i), d \varphi (\cdot) \rangle = c^N \lambda_1 \lambda_2 g$, and consequently $\sigma (\langle R^N \big(d \varphi (e_i), d \varphi (\cdot) \big) d \varphi (e_i), d \varphi (\cdot) \rangle)=0$. Moreover, $L (S (\varphi)) = 4 c^M S (d \varphi)$ and $\sigma (\langle d \varphi ({\rm{Ricci}}^M (\cdot)), d \varphi (\cdot) \rangle) = c^M S (d \varphi)$, so substituting these calculations into (3.20), equation (3.21) is the result.

Recall that if $\omega \in \Omega^p (E)$ is harmonic, then $\div \, S (\omega) =0$. If we use $\omega = d \varphi \in \Omega^1 (E)$, then $\div S (d \varphi)=0$. Take the divergence of both sides of (3.21) and note that $c^M$ is constant, then the result follows since divergence and Laplacian commute be Proposition 1.

Now $\tau (\varphi)$ defines an energy functional which is called the biharmonic energy functional

$E_2 ( \varphi) = \frac{1}{2} \int_M \, | \tau ( \varphi) |^2 \, d v_M,$

(3.22)

and has the associated stress-energy tensor $S_2 (d \varphi)$,

$S_2 ( d \varphi) = \frac{1}{2} | \tau ( \varphi) |^2 g + 2 {\cal S} ( \langle \nabla \tau ( \varphi), d \varphi \rangle ).$

(3.23)

Theorem 7. Let $\varphi : (M, g) \to (N, h)$ be a smooth map between Riemannian manifolds and let $d \varphi \in \Omega^1 (\varphi^{-1} TN)$ be the derivative.

$(i)$ The Laplacian of $S (d \varphi)$ is given as

$\Delta S ( d \varphi) = \frac{1}{2} | \tau ( \varphi) |^2 g - S_2 ( d \varphi) - 2 S ( \nabla d \varphi) -2 {\cal S} ( \langle d \varphi ( {\rm{Ricci}}^M ( \cdot), d \varphi ( \cdot) \rangle )$

$+ 2 {\cal S} ( \sum\limits_i \, \langle R^N ( d \varphi ( e_i), d \varphi ( \cdot)) \, d \varphi ( e_i), d \varphi ( \cdot) \rangle^{TN} ) + L ( S ( d \varphi)).$

(3.24)

$(ii)$ If $\varphi : M \to N$ is an isometric immersion,

$\frac{1}{2} | \tau ( \varphi) |^2 g - S_2 ( d \varphi) - 2 S ( \nabla d \varphi) -2 {\cal S} ( {\rm{Ricci}}^M)$

$+ 2 {\cal S} ( \sum\limits_i \, \langle R^N ( d \varphi (e_i), d \varphi ( \cdot)) \, d \varphi (e_i), d \varphi (\cdot) \rangle ) =0.$

(3.25)

$(iii)$ If $N$ is a space form of constant curvature $K^N =-1, 0, +1$,

$\frac{1}{2} | \tau ( \varphi) |^2 g - S_2 ( d \varphi) -2 {\cal S} (d \gamma) - 2 {\cal S} ( {\rm{Ricci}}^M) + 2 (m-1) K^N S ( d \varphi) =0.$

(3.26)

Proof: $(i)$ Let $\omega= d \varphi$ then solve for the second term in (3.23) and substitute it into the previous result (3.21) which gives (3.24). $(ii)$ If $\varphi : (M, g) \to (N, h)$ is an isometric immersion, then

$S ( d \varphi) = \frac{1}{2} (m-2) \, g,$

(3.27)

hence the left-hand side of (3.24) as well as $L (S d \varphi))$ both vanish. $(iii)$ Finally, if we map into a space form, the second fundamental form $\nabla d \varphi$ can be identified with the derivative of the associated Gauss map $\Gamma$ and (3.26) follows.

As an application of this theorem, suppose $\varphi$ is a minimal immersion of a surface into Euclidean space, then $\tau (\varphi) =0$, $S_2 (d \varphi) =0$ and $\sigma ({\rm{Ricci}}^M) =0$. Then equation (3.26) implies that $S (d \gamma)=0$.

If the divergence of both sides of (3.26) is worked out, since $\div {\cal S} ({\rm{Ricci}}^M) =0$ as a consequence of the Bianchi identity, (3.26) implies that if $\varphi : M \to N$ is an isometric immersion into a space form, it follows that

$\frac{1}{2} d | \tau (\varphi) |^2 - \div S_2 ( d \varphi) -2 \div \, S ( d \gamma )=0.$

(3.28)

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4. Further Results Including an Integral Theorem

Let us examine the influence of conformal vector fields to finish ^[3]. A monotonicity formula describes the growth properties of extremals of the functional (2.1) and may be used to establish their regularity in appropriate situations.

Let $(M, g)$ be a Riemannian manifold of dimension $m$. A vector field $X$ on $M$ is called conformal if it satisfies

${\cal L}_X \, g = \xi g$

(4.1)

for some function $\xi : M \to \mathbb R$ and ${\cal L}$ is the Lie derivative in the direction of $X$. Equivalently, $X$ is conformal if and only if

${\rm{sym}} \: g (\nabla_i) = \frac{1}{m} ( \div X) g,$

(4.2)

in which case, we have $\xi = (2/m) \div X$.

Theorem 8. Let $T$ be a symmetric $2$-tensor, and let $X$ be a conformal vector field, then

$\div ( T \lrcorner X) = ( \div \, T) (X) + \frac{1}{m} ( \div X) \cdot {\rm{tr}} T.$

(4.3)

Proof: Let $\{ e_i \}_1^m$ be a locally defined frame field. If $X$ is a conformal vector field, the divergence can be calculated as follows,

$\div ( T \lrcorner X) = \nabla_{e_i} ( T \lrcorner X) ( e_i) = ( \nabla_{e_i} T)( X, e_i) + T ( \nabla_{e_i} X, e_i)$

$= ( \div T)(X) + T ( \langle \nabla_{e_i} X, e_j \rangle e_j, e_i ).$

Since $T$ is a symmetric $2$-tensor, the definition of conformal implies that

$\div ( T \lrcorner X) = ( \div T) (X) + \frac{1}{2} ( \langle \nabla_{e_i} X, e_j \rangle + \langle \nabla_{e_j} X, e_i \rangle ) T ( e_i, e_j)$

$= ( \div T) (X) + \frac{1}{m} \div X \, g ( e_i, e_j) \, T( e_i, e_j) = ( \div T) + \frac{1}{m} \div X \, {\rm{tr}} \, T,$

since we have ${\rm{tr}} T = g(e_i, e_j) \, T(e_i, e_j)$.

Integrate (4.3) over a compact region, $U$, to obtain the monotonicity formula. The divergence theorem permits us to write

$\int_{U} \, \div ( T \lrcorner X) \, d v_M = \int_{\partial U} \, T (X, {\bf n}) \, d a_M.$

(4.4)

where $d a_M$ is the volume form on $\partial U$.

Theorem 9. Let $(M, g)$ be an oriented Riemannian manifold and let $X$ be a conformal vector field on $M$. Suppose that $U \subset M$ is a compact region with a smooth boundary $\partial U$. Then for any symmetric $2$-tensor $T$ on $M$,

$\int_{\partial U} \, T (X, {\bf n}) \, d a_M = \int_{U} ( \div T) (X) \, d v_M + \frac{1}{m} \int_M \div X \, {\rm{tr}} T \, d v_M,$

(4.5)

where $d v_M$ is the volume element on $M$ and $d a_M$ is the volume element on $\partial U$, ${\bf n}$ the outward pointing normal on $\partial U$.

This is a remarkable theorem as there are numerous applications of it, but only one will be presented here. Let $\varphi : B^m \to (N, h)$ be a harmonic map from the Euclidean $m$-ball of radius $R$ and $S = e (\varphi) g -\varphi^* h$ the corresponding stress-energy tensor. Clearly, $\div S=0$ and ${\rm{tr}} S = e \, {\rm{tr}} g -{\rm{tr}} \varphi^* h = e \, m -2 e = (m-2) \, e (\varphi)$. Take $X$ to be the conformal vector field $X =r \partial_r$ where $r = |{\bf x}|$, ${\bf x} \in \mathbb R^m$. By direct calculation, $\div X = m$, hence substituting these facts into (4.5), the following result holds for $r < R$,

$\int_{\partial B^m} S ( r \partial_r, \partial_r) \, d a = \int_{B^m} (m-2) \, e(\varphi) \, d v_M.$

(4.6)

Substituting for $S$ yields

$(m-2) \int_{B^m} \, e ( \varphi) \, d v_M = \int_{\partial B^m} \, r \big\{ e (\varphi) - h ( \frac{\partial \varphi}{\partial r}, \frac{\partial \varphi}{\partial r}) \big\} \, d a_M.$

(4.7)

The following conclusion can be drawn from (4.7). If $m>2$ and $\varphi |_{\partial B^m_r} =c$ for some $r < R$ and constant $c$, then $\varphi$ must be constant, that is, for $e (\varphi)|_{\partial B^m} = \frac{1}{2} h (\partial_r \varphi, \partial_r \varphi)$ gives the upper bound

$\int_{B^m} \, e ( \varphi ) \, d v_M \leq 0,$

which in turn implies that $e (\varphi) \equiv 0$.

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Conflict of Interest

The author declares no conflicts of interest in this paper.

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