Prescribed Schouten Tensor in Locally Conformally Flat Manifolds

Abstract

We consider the pseudo-Euclidean space $({\mathbb {R}}^n,g)$, with $n \ge 3$ and $g_{ij} = \delta _{ij} \varepsilon _{i}$, where $\varepsilon _{i} = \pm 1$, with at least one positive $\varepsilon _{i}$ and non-diagonal symmetric tensors $T = \sum \nolimits _{i,j}f_{ij}(x) dx_i \otimes dx_{j}$. Assuming that the solutions are invariant by the action of a translation $(n-1)$- dimensional group, we find the necessary and sufficient conditions for the existence of a metric $\bar{g}$ conformal to g, such that the Schouten tensor $\bar{g}$, is equal to T. From the obtained results, we show that for certain functions h, defined in $\mathbb {R}^{n}$, there exist complete metrics $\bar{g}$, conformal to the Euclidean metric g, whose curvature $\sigma _{2}(\bar{g}) = h$.

1 Introduction

In recent years, problems involving the Ricci curvature have aroused great interest. Among the problems studied, we highlight the Einstein manifolds, Ricci solitons gradient, prescribed Ricci and Schouten tensor, prescribed curvature tensor, and Einstein field equation. For more details see [1, 3, 6, 10, 14, 19]. In [5], Deturck and Yang considered the following problem:

Given a Riemannian manifold $(M^{n}, g_{0})$, with $n\ge 3$, and a symmetric tensor of order 2, T, defined in $M^{n}$, find a Riemannian metric g such that

$$\begin{aligned} Ric_{g} + {\lambda } K_{{g}}g = T \end{aligned}$$

(1.1)

where $\lambda$ $\in$ $\mathbb {R}$ is a constant and $Ric_{g}$ and ${K}_{g}$ are the Ricci tensor and scalar curvature of g, respectively. They have shown that when T is non-singular, the problem (1.1) has always a local solution. In this case, because the problem (1.1) admits local solutions, it makes sense to consider the following: find necessary and sufficient conditions so that problem (1.1) has a global solution; further, once global solutions are found, under what conditions are they complete? When $\lambda = 0$ and $\lambda = -\frac{1}{2}$, the problem (1.1)is known in the literature as the prescribed Ricci and Einstein tensor, respectively. This problem has been studied for a particular family of tensors (see [16,17,18,19,20]). Recently, Pulemotov studied the problem (1.1) for $\lambda = 0$ in homogeneous manifolds (see [21]). When $\lambda = \frac{-1}{2(n-1)}$, the problem (1.1) is equivalent to the prescribed Schouten tensor, because the Schouten tensor of a metric g is defined by

$$\begin{aligned} A_{g} = \dfrac{1}{n-2} \left( Ric_{g} - \dfrac{K}{2(n-1)}g\right) . \end{aligned}$$

Problem (1.1) has also been studied locally by Robert Brayant for any value of $\lambda$, proving that the problem always has local solutions when the components of the tensor are analytic functions. Motivated by the work of Deturck and Yang [5], our goal is to find global solutions to the following problem:

Given a (0, 2)symmetrical tensor T defined in a manifold $(M^{n}, g_{0})$, with $n\ge 3$, is there a metric g such that

$$\begin{aligned} A_{g} = T? \end{aligned}$$

(1.2)

This problem corresponds to studying a system of nonlinear second order partial differential equations. The importance of Schouten tensors in conformal geometry can be seen in the following decomposition of the Riemann curvature tensor

$$\begin{aligned}R_g = W_g + A_g \odot g,\end{aligned}$$

where $R_g$ is the Riemann curvature tensor, $\odot$ is the Kulkarni–Nomizu product, and $W_g$ is the Weyl tensor of g (see [2]). Because the Weyl tensor is conformally invariant, i. e., $g^{-1}W_g$ is invariant in a given conformal class, in a conformal class the Schouten tensor is important, especially when g is locally conformally flat ($W_g=0$). Therefore, if g is locally conformally flat, the Riemann curvature tensor is determined by the Schouten tensor. From the Schouten tensor, curvatures that extend the concept of the scalar curvature can be defined. This study was first conducted by Jeff Viaclovsky in [22]. For an integer $1 \le k \le n$ and $\sigma _k$- or k-scalar curvature, the Schouten curvature is defined by

$$\begin{aligned} \sigma _k(g) := \sigma _k(g^{-1} \cdot A_g), \end{aligned}$$

where $(g^{-1} \cdot A_g)$ is defined locally by $(g^{-1} \cdot A_g)_{ij} = \sum _k g^{ik}(A_g)_{kj}$ and $\sigma _k$ and the k-th symmetrical elementary function. Thus, we define $\sigma _k(g)$ as being the k-th elementary symmetric function of the auto-values of the operator $g^{-1}A_g$, to $1 \le k \le n$, where $\sigma _0(g) = 1$. Considering the eigenvalues of the Schouten tensor $A_{g}$ $(\lambda _{1}, \lambda _{2}, \ldots , \lambda _{n})$ with respect to the metric g, to $1 \le k \le n$, the kth polymorphic elementary symmetric functions $\sigma _{k}$ are given by $\sigma _{k}(A_{g}) = \sigma _{k}(\lambda ) = {\sum _{i_{1}< \cdots < i_{k}}} \lambda _{i_{1}} \cdots \lambda _{i_{k}}$. When $k=1$, a $\sigma _1(g)$-scalar curvature is exactly the scalar curvature (less than one constant). Thus, $\sigma _1(g)$ is constant if and only if $(M^n , g)$ has constant scalar curvature. In [11], the authors considered the problem of classifying compact Riemannian manifolds locally conformally flat with $\sigma _k(g)$ constant for some $k\ge 2$.

In [22], Viaclovsky also noted that $\sigma _2(g)$ has still a variational structure. For $k>2$, $\sigma _k(g)$ has a variational structure, if and only if the manifold considered is locally conformally flat. From this work of Viaclovisky and the work of Chang, Gursky, and Yang in [4],an intensive investigation started for the variational problem related to the Schouten curvature $\sigma _k (g)$, seeking to find a metric g, in the class of $[g_0]$, satisfying

$$\begin{aligned} \sigma _k(g) = c, \end{aligned}$$

(1.3)

where

$$\begin{aligned} g \in [g_0]\bigcap \Gamma _k^+ . \end{aligned}$$

and $\Gamma _k^+$ is a convex open cone (the Garding cone) defined by

$$\begin{aligned} \Gamma _k^+ = \{ \Lambda = (\lambda _1, \lambda _2, \ldots ,\lambda _n) \in R^n \vert \sigma _j(\Lambda )>0, \forall j\le k\}. \end{aligned}$$

Here, $g \in \Gamma _k^+$ represents the Schouten tensor $A_g(x) \in \Gamma _k^+$, for any $x \in M$; an important fact is that $g\in \Gamma _k^+$ guarantees that equation (1.3) is elliptic.

Several authors have recently studied subjects related to the Schouten curvature, see for example [7, 8], and [11]. In [7], the authors consider the problem $\frac{\sigma _2}{\sigma _1}=f$, where f is a given differentiable function.

In [12], Simon et al. showed that, if $(M^{n}; g)$ is a compact locally conformally flat manifold with nonzero curvature $\sigma _{k}(A_{g})$ for some $2 \le k \le n$ and $A_{g}$ defined as semi-positive, then $(M^{n}, g)$ is a space form of positive sectional curvature. In [13], Simon et al. studied the extreme properties of the Schouten function defined in the quotient of the Riemannian metric space by the group of diffeomorphisms.

In [15], the authors considered the pseudo-Euclidean space $(\mathbb {R}^n,g)$, with $n \ge 3$ and $g_{ij} = \delta _{ij} \varepsilon _i, \varepsilon _i=\pm 1$, and tensors of the form $T=\sum _i \varepsilon _i f_i (x) dx_i^2$, and found necessary and sufficient conditions for the existence of a metric $\bar{g}$, conformal to g, such that $A_{\bar{g}}=T$. The solution to this problem was explicitly given for special cases of the tensor T, including a case where the metric $\bar{g}$ is complete in $\mathbb {R}^n$. Similar problems were considered for locally conformally flat manifolds. As an application of these results, the authors considered the problem of finding metrics $\bar{g}$, conformal to g, such that $\sigma _2 ({\bar{g}})$ or $\frac{\sigma _2 ({\bar{g}})}{\sigma _1 ({\bar{g}})}$ are equal to a certain function.

In this work we will consider the pseudo-Euclidean space $({\mathbb {R}}^n,g)$, with $n \ge 3$, coordinates $x=(x_1,..,x_n)$, and metric g, where $g_{ij} = \delta _{ij} \varepsilon _{i}$, with $\varepsilon _{i} = \pm 1$, with at least one positive $\varepsilon _{i}$, and a non-diagonal tensor of order 2 of the form $T = \sum _{i,j}f_{ij}(x) dx_i \otimes dx_{j}$, where $f_{ij}(x)$, $1 \le i,j \le n$, are differentiable functions. We want to find metrics $\bar{g} = \frac{1}{\varphi ^2} g$, such that the Schouten tensor of the metric $\bar{g}$ is T, that is, we want to solve the following problem:

$$\begin{aligned} \left\{ \begin{array}{l} A_{\bar{g}} = T \\ \bar{g} \; = \frac{1}{\varphi ^2} g. \end{array} \right. \end{aligned}$$

(1.4)

To obtain solutions to the problem (1.4), let us assume that the metric $\bar{g}$ is invariant by the action of a $(n-1) -dimensional$ translation group. In this case, we find necessary conditions on the tensor T, so that the problem admits solution (Lemma 2.2). For this special class of metrics, we find necessary and sufficient conditions for the problem to have solutions (Theorem 2.3). As a consequence of the Theorems (2.3) and (2.5) we obtain complete metrics in Euclidean space $\mathbb {R}^n$, with prescribed Schouten tensors. The results obtained were extended to locally conformally flat manifolds (Theorem 2.6).

As applications of these results, we show explicit solutions for a second order nonlinear partial differential equation in $\mathbb {R}^n$. The geometric interpretation of this result is equivalent to finding conformal metrics in $\mathbb {R}^n$ with $\sigma _2(\bar{g})$ prescribed. In particular, by considering $f: \mathbb {R}^n \longrightarrow \mathbb {R}$, we find examples of complete metrics $\bar{g}$, conformal to the Euclidean metric, such that $\sigma _2 (\bar{g}) = f$.

2 Main Results

Let $\varphi _{,x_{i}x_{j}}$ and $f_{ij,x_k}$ denote the second order derivatives of $\varphi$ with respect to $x_i x_{j}$ and the derivative of $f_{ij}$ with respect to $x_k$, respectively.

Theorem 2.1

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be the pseudo-Euclidean space, with coordinates $x = (x_{1},\ldots ,x_{n})$ and $g_{ij}=\delta _{ij} \varepsilon _{i}$, and $T = \sum _{i,j=1}^n f_{ij}(x)dx_{i} \otimes dx_{j}$ be a non-diagonal tensor of order 2, where $f_{ij}(x)$ are differentiable functions. Then, there exists a positive function $\varphi$ such that the metric $\bar{g} = \frac{1}{\varphi ^2} g$ satisfies $A_{\bar{g}} = T$ if and only if the functions $f_{ij}$ and $\varphi$ satisfy the following set of equations:

$$\begin{aligned} \left\{ \begin{array}{ll} 2\varphi \varphi _{x_ix_i} - \sum _{k=1}^{n}\epsilon _{k}\left( \varphi _{x_{k}}\right) ^{2}\epsilon _{i} - 2\varphi ^{2}f_{ii} = 0 &{}\quad \forall \ i: 1,\ldots ,n,\\ \varphi _{x_ix_j} - f_{ij}\varphi = 0 &{}\quad \forall \ i \ne j. \end{array} \right. \end{aligned}$$

(2.1)

In an attempt to find solutions to the system (2.1) we will consider the solutions $\overline{g} =\frac{1}{\varphi ^2} g$ to be invariant under the action of a group $(n-1)$-dimensional, where $\xi = \sum _{i=1}^{n}a_{i}x_{i}$, with $a_{i}$ $\in$ $\mathbb {R}$, is a basic invariant of the group.

Initially, we will determine the necessary conditions on the tensor T.

Lemma 2.2

Consider $\varphi = \varphi (\xi )$, where $\xi = \sum _{i}^{n} a_{i}x_{i}$, with $a_{i} \in$ $\mathbb {R}$. If $\varphi = \varphi (\xi )$ is the solution of the system (2.1), then $f_{ij} = a_{i} a_{j} f(\xi )$ $\forall$, with $i \ne j = 1, \ldots , n$, and $f_{ii} = f_{ii}(\xi )$ , $\forall \; \; i = 1,\ldots ,n$, where f is a differentiable function.

From Lemma (2.2), we can state one of the main theorems of this section. We consider $\sum _{k=1}^{n} \epsilon _{k}a_{k}^{2} \ne 0$ and, without loss of generality, we assume that $\sum _{k=1}^{n}\epsilon _{k}a_{k}^{2} = \epsilon = \pm 1$, and the case where $\sum _{k=1}^{n} \epsilon _{k}a_{k}^{2} = 0$ will be dealt with later.

Theorem 2.3

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be the pseudo Euclidean space, with coordinates $x =(x_{1},\ldots ,x_{n})$ and metric $g_{ij}=\delta _{ij} \varepsilon _{i}$. Consider the non-diagonal tensor of order 2 $T = \sum _{i=1}^n f_{ij}(\xi )dx_{i}\otimes dx_{j}$, where $f_{ii} = f_{ii}(\xi )$, $f_{ij} = a_{i}a_{j}f(\xi )$, $\xi = \sum _{i=1}^{n} a_{i}x_{i}$, $a_{i} \ne 0$, $\forall \; i=1, \ldots , n$, and $\sum _{k=1}^n a_{i}^{2}\epsilon _{i} = \epsilon$. Then, there is a metric $\bar{g} = \frac{1}{\varphi ^2} g$ such that $A_{\bar{g}} = T$, with $\varphi = \varphi (\xi )$, if and only if the components of the tensor satisfy the following equations

$$\begin{aligned} \left\{ \begin{array}{ll} \epsilon _{i}(fa_{i}^2 - f_{ii}) = \epsilon _{j}(fa_{j}^2 - f_{jj}) &{}\quad \forall \ i, j = 1, \ldots ,n \\ 2\epsilon \epsilon _{i} (fa_{i}^2 - f_{ii}) > 0 &{}\quad \\ 2 \epsilon \epsilon _{i} (fa_{i}^2 - f_{ii}) + \dfrac{\epsilon _{0} \epsilon _{i} \epsilon (f^{\prime }a_{i}^2 - f_{ii}^{\prime })}{\sqrt{2 \epsilon \epsilon _{i} (fa_{i}^2 - f_{ii})}} = f &{}\quad \forall \ i=1 , \ldots , n, \end{array} \right. \end{aligned}$$

(2.2)

and

$$\begin{aligned} \varphi (\xi ) = Ke^{\epsilon _{0}{\int } \sqrt{2 \epsilon \epsilon _{i} (fa_{i}^2 - f_{ii})}d\xi }, \end{aligned}$$

where $\epsilon _{0} = \pm 1$ and K is a positive constant.

As a direct consequence of the Theorem (2.3), we obtain the following example in the Riemannian case.

Example

Consider the Euclidean space $(\mathbb {R}^n, g)$ $n \ge 3$, with coordinates $x = (x_{1}, \ldots , x_{n})$. Given the tensor $T = \sum _{i=1}^{n}f_{ii}dx_{i}^2 + \sum _{i \ne j=1}^{n}f_{ij} dx_{i}\otimes dx_{j}$, where $f_{ii} = \left( \frac{2-n}{2n}\right) g - \frac{g^{\prime }}{2n\sqrt{g}}$ and $f_{ij} = \frac{1}{n}\left( g - \frac{g^{\prime }}{2\sqrt{g}}\right)$, $a_{i} = a_{j} = a$, $\epsilon = 1$, $\epsilon _{0} = -1$, and g is a positive function. Then, there is a metric $\bar{g} = \frac{1}{\varphi ^{2}}g$ such that $A_{\bar{g}} = T$, where:

$$\begin{aligned} \varphi (\xi ) = Ke^{- {\int } \sqrt{g} d\xi }. \end{aligned}$$

(2.3)

Moreover, if $\mid \int \sqrt{g} d\xi \mid \le C_{1}$, $C_{1}$ is a positive constant, then the metric $\bar{g}$ is complete in $\mathbb {R}^{n}$.

The case where $\xi = a_{k}x_{k}$ for some fixed k, will be treated in the next Theorem. Note that, in this case, in (2.1), $\varphi _{x_{i}} = \varphi ^{\prime }a_{i}$, $\varphi _{x_{i}x_{i}} = \varphi ^{\prime \prime }a_{i}^{2} \; \; \; \forall i = 1, \ldots n$, and $\varphi _{x_{i}x_{j}} = 0, \; \; \forall \; \; i \ne j$. Then, for the second equation, we have $f_{ij} \varphi = 0$, which is equivalent to the given tensor T being diagonal, because, in this case, $f_{ij} = 0$ $\forall$ $i \ne j$. Without loss of generality, let us consider $\xi = a_{1}x_{1}$ and $\epsilon _{1}a_{1}^{2} = \epsilon$.

Theorem 2.4

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be the pseudo Euclidean space, with coordinates $x = (x_1,\ldots ,x_n )$, and metric $g_{ij}=\delta _{ij} \varepsilon _{i}$. Consider $T = \sum _{i=1}^n f_{ii}(\xi )dx_i^{2}$, where $\xi = a_{1}x_{1}$, $\epsilon _{1}a_{1}^{2} = \epsilon = \pm 1$. Then, there is a metric $\bar{g} = \frac{1}{\varphi ^2} g$, with $\varphi = \varphi (\xi )$, such that $A_{\bar{g}} = T$, if and only if the components of the tensor satisfy the equations:

$$\begin{aligned}&\epsilon _{i} f_{ii} = \epsilon _{j} f_{jj} \qquad \forall \; i \ne j= 2, \ldots , n \\&-2\epsilon \epsilon _{i} f_{ii} > 0 \qquad \forall \; i = 2, \ldots , n \\&f_{11} = -2a_{1}^{2} \epsilon \epsilon _{i} f_{ii} - \dfrac{\epsilon \epsilon _{0} \epsilon _{i} f_{ii}^{\prime }a_{i}^2}{\sqrt{-2 \epsilon \epsilon _{i} f_{ii}}} + \epsilon _{i}\epsilon _{1} f_{ii} \end{aligned}$$

and

$$\begin{aligned} \varphi (\xi ) = Ke^{\epsilon _{0}{\int } \sqrt{-2 \epsilon \epsilon _{i} f_{ii}}d\xi } \end{aligned}$$

where K is a positive constant and $\epsilon _{0} = \pm 1$.

As a consequence of Theorem (2.4), we will show an example in the Riemannian case.

Example

Consider the Euclidean space $(\mathbb {R}^n, g)$ $n \ge 3$, with coordinates $x = (x_{1}, \ldots , x_{n})$. Given the tensor $T = f_{11} dx_{1}^{2} + \sum _{i=2}^{n} g(\xi )dx_{i}^2$, where $f_{11} = g(1 + 2a_{1}^{2}) + \frac{g^{\prime }a_{1}^{2}}{\sqrt{-2g}}$, $\epsilon = 1$, $\epsilon _{0} = -1$, and $f_{ii} = f_{jj} = g$ $\forall i \ne j = 2, \ldots , n$, where g is a differentiable function smaller than zero. Then, there is a metric $\bar{g} = \frac{1}{\varphi ^{2}}g$ such that $A_{\bar{g}} = T$, where

$$\begin{aligned} \varphi (\xi ) = Ke^{-{\int } \sqrt{-2g} d\xi }. \end{aligned}$$

(2.4)

Note that in the above theorem, conditions (2.4) are checked. Because the $g < 0$ function is arbitrary, it can be chosen so that $\int \sqrt{-2g}d\xi$ is limited, and, in this case, the metric $\bar{g}$ will be complete in $\mathbb {R}^{n}$.

Considering now the case where $\xi = \sum a_{i}x_{i}$ and $\sum a_{i}^{2} \epsilon _{i} = 0$, we obtain the following result.

Theorem 2.5

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be the pseudo Euclidean space, with coordinates $x = (x_1,\ldots ,x_n )$ and metric $g_{ij}=\delta _{ij} \varepsilon _{i}$. Consider $T = \sum _{i,j}^n f_{ij}(\xi )dx_i \otimes dx_{j}$, where $f_{ii} = f_{ii}(\xi )$, $f_{ij} = a_{i}a_{j}f(\xi )$, and $\xi = \sum _{i=1}^{n}a_{i}x_{i}$, with $\sum _{i=1}^{n} a_{i}^{2}\epsilon _{i} = 0$ and $a_{i} \ne 0$, for at least one pair of indexes and the differentiable function f. Then, there is a metric $\bar{g} = \frac{1}{\varphi ^2} g$ such that $A_{\bar{g}} = T$ if and only if $f_{ii} = a_{i}^{2}f$, $\forall \; i = 1, \ldots , n$, and $\varphi$ is a solution of the equation

$$\begin{aligned} \varphi ^{\prime \prime } - f\varphi = 0. \end{aligned}$$

Next we present two examples for the Theorem (2.5), considering particular solutions for the equation $\varphi ^{\prime \prime } - f\varphi = 0$.

The study of oscillations is an important part of mechanics because of the frequency with which they occur. The simple swaying of leaves of a tree, radio waves, sound, and light are typical examples where oscillatory motion occurs. Below, we present an application of the physics of the above theorem, with the equation of free oscillations, obtained when $f = k$.

Example

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be a pseudo Euclidean space, with coordinates $x = (x_1,\ldots ,x_n )$ and metric $g_{ij}=\delta _{ij} \varepsilon _{i}$. Consider $T = \sum _{i,j}^n f_{ij} dx_i \otimes dx_{j}$, where $f_{ii} = k a_{i}^2$, $\forall \; i=1, \ldots , n$, and $f_{ij} = ka_{i}a_{j}$, $\forall \; i \ne j=1, \ldots ,n$, and $\sum _{i=1}^{n} a_{i}^{2}\epsilon _{i} = 0$. Then, there exists a $\bar{g} = \frac{1}{\varphi ^{2}}g$ such that $A_{\bar{g}} = T$, if and only if the function $\varphi$ is satisfies:

$$\begin{aligned} \varphi (\xi ) = \left\{ \begin{array}{lll} C_{1}\sinh (\xi \sqrt{|k|}) + C_{2}\cosh (\xi \sqrt{|k|}) &{}\quad if &{}\;\; k < 0 \\ C_{1} + C_{2}\xi &{}\quad if &{}\; \; k =0 \\ C_{1}\sin (\xi \sqrt{k}) + C_{2}\cos (\xi \sqrt{k}) &{}\quad if &{}\;\; k > 0 . \end{array} \right. \end{aligned}$$

Example

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be a pseudo Euclidean space, with coordinates $x = (x_1,\ldots ,x_n )$ and metric $g_{ij}=\delta _{ij} \varepsilon _{i}$. Consider $T = \sum _{i,j}^n f_{ij} dx_i \otimes dx_{j}$, where $f_{ii} = (h^{2}+ h^{\prime }) a_{i}^2$ $\forall \; i=1, \ldots , n$ and $f_{ij} = (h^{2} + h^{\prime })a_{i}a_{j}$ $\forall \; i \ne j=1, \ldots ,n$, and $\sum _{i=1}^{n} a_{i}^{2}\epsilon _{i} = 0$, and a differentiable function h. Then, there exists a $\bar{g} = \frac{1}{\varphi ^{2}}g$ such that $A_{\bar{g}} = T$, if and only if the function $\varphi$ is given by

$$\begin{aligned} \varphi (\xi ) = C_{1}\varphi _{0} + c_{2}\varphi _{0}\int \frac{d\xi }{\varphi _{0}^{2}}, \end{aligned}$$

where $C_{1}$ and $C_{2}$ are arbitrary constants and $\varphi _{0} = e^{\int hd\xi }$, with $h = h(\xi )$.

Next, we present a generalization of the Theorem (2.3), for locally conformally flat manifolds.

Let us now consider a locally conformally flat Riemannian manifold $(M^{n},g)$. We can consider the problem (1.4) for a neighborhood $V \subset M$ with coordinates $x = (x_{1}, \ldots , x_{n})$ such that $g_{ij} = \frac{\delta _{ij}}{F^{2}}$, where F is a non-null, differentiable function in V. Given a tensor $T = \sum _{i,j}f_{ij}(x)dx_{i} \otimes dx_{j}$ defined in V, we want to find a metric $\bar{g} = \frac{1}{\phi ^{2}}g$ such that $A_{\bar{g}} = T$. Considering that g and $\bar{g}$ are translation invariant, where $\xi = \sum a_{i}x_{i}$ is the basic invariant of the action, we have, in a way analogous to Lemma (2.2), that the components of the given tensor T, are necessarily given by $f_{ij}= a_{i}a_{j}f(\xi )$ and $f_{ii} = f_{ii}(\xi )$, where f is a differentiable function.

Theorem 2.6

Let $(M^n, g)$, with $n\ge 3$, be a locally conformally flat Riemannian manifold. Let V be an open subset of M with coordinates $x=(x_1,x_2, \ldots , x_n)$ with $g_{ij} = \dfrac{1}{F^{2}(\xi )}\delta _{ij}$. Consider a non-diagonal tensor $T = \sum _{i=1}^nf_{ij}(\xi )dx_{i}\otimes dx_{j}$. Then, there is a metric $\bar{g} = \frac{1}{\phi ^2} g$, with $\phi = \phi (\xi )$, such that $A_{\bar{g}} = T$ if and only if the functions $f_{ij}$ and $\varphi$ are given in Theorem (2.3) and $\phi = \frac{\varphi }{F}$.

Analogously, the Theorems (2.4) and (2.5) can be extended to locally conformally flat manifolds.

In [22], Vioclovisk extended the concept of scalar curvature using the Schouten tensor, by introducing the $\sigma _{k}$ curvatures, which are obtained from the eigenvalues of the Schouten tensor. Recall that $\sigma _{0}(g)$ has been defined as 1 and $\sigma _{1}(g)$ is the scalar curvature less than constant. Recently, many works have considered the $\sigma _{2}(g)$ prescribed problem. For more details see [7, 11], and [22].

We know that, given a function $h: M^{n} \longrightarrow \mathbb {R}$, finding a metric $\bar{g} = \frac{1}{\varphi ^{2}}g$ such that $\sigma _{2}(\bar{g}) = h$ is equivalent to studying the following partial differential equation (ver [9]).

$$\begin{aligned} \left[ \left( \Delta \varphi \right) ^2 -\vert Hess_g \varphi \vert ^2 \right] \varphi ^2 - (n-1)\Delta \varphi \vert \nabla \varphi \vert ^2 \varphi + \frac{n(n-1)}{4}\vert \nabla \varphi \vert ^4 = 2h. \end{aligned}$$

(2.5)

Let us show that, for certain functions h, the above equation admits infinite solutions.

Corollary 2.7

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be the Euclidean space, with coordinates $x = (x_1,\ldots ,x_n )$ and metric $g_{ij}=\delta _{ij}$. Given the function g, which can be any positive function, K is a positive constant and a $\in$ $\mathbb {R}$

$$\begin{aligned} h(\xi ) = \frac{K(n-1)}{2}n^{2}a^{4}ge^{-4 {\int } \sqrt{g}d\xi }\left[ \frac{ng}{4} - \left( g - \frac{g^{\prime }}{2\sqrt{g}}\right) \right] , \end{aligned}$$

(2.6)

Then, the partial differential equation,

$$\begin{aligned} \left[ \left( \Delta \varphi \right) ^2 -\vert Hess_g \varphi \vert ^2 \right] \varphi ^2 - (n-1)\Delta \varphi \vert \nabla \varphi \vert ^2 \varphi + \frac{n(n-1)}{4}\vert \nabla \varphi \vert ^4 = 2h, \end{aligned}$$

has infinite solutions, globally defined in $\mathbb {R}^{n}$, given by

$$\begin{aligned} \varphi (\xi ) = Ke^{ - \int \sqrt{g} d\xi }. \end{aligned}$$

The geometric interpretation of this result is presented below. We show an example of a metric $\bar{g}$, conformal to the Euclidean metric, complete in $\mathbb {R}^{n}$, with curvature $\sigma _{2}(\bar{g})$ prescribed for h.

Corollary 2.8

Let $(\mathbb {R}^n, g)$, with $n\ge 3$, be the Euclidean space, with coordinates $x = (x_1,\ldots ,x_n )$ and metric $g_{ij}=\delta _{ij} \varepsilon _{i}$. Given a function h, in (2.6) there exists a metric $\bar{g} = \frac{1}{\varphi ^{2}}g$ with $\varphi$ given by (2.3) such that $\sigma _{2}(\bar{g}) = h$.

By choosing the function g so that $\varphi$ is limited, the metrics obtained in the corollary (2.8) will be complete in $R^ {n}$.

3 Proof of the Main Results

Before proving our results, it follows from [15], that if $(\mathbb {R}^n,g)$ is a pseudo-Euclidean space and $\bar{g}=g/\varphi ^2$ Is a conformal metric, then the Ricci tensor $\bar{g}$ is given by

$$\begin{aligned} \text{ Ric } \;\bar{g} = \frac{1}{\varphi ^2}\left\{ (n - 2)\varphi Hess_g \varphi + \left( \varphi \Delta _g \varphi - (n-1)|\nabla _g\varphi |^2\right) g \right\} \end{aligned}$$

(3.1)

and the scalar curvature of $\bar{g}$ is given by

$$\begin{aligned} \bar{K} =(n - 1)\left( 2\varphi \Delta _g \varphi - n|\nabla _g\varphi |^2 \right) . \end{aligned}$$

(3.2)

In the rest of this section, we demonstrate the main results of this article.

Proof

[Theorem 2.1]. Using the expressions (3.1) and (3.2), we can write the Schouten tensor of the metric $\bar{g}$ as:

$$\begin{aligned}A_{\bar{g}} = \frac{Hess_g \varphi }{\varphi } - \frac{ \Vert \nabla _g\varphi \Vert ^2}{2\varphi ^2} g,\end{aligned}$$

where $\nabla _g$ denotes the gradient of the pseudo-Euclidean metric g.

In this case, studying the problem (1.4) with $T = \sum _{ij}^n f_{ij}({x})dx_i \otimes dx_j$, is equivalent to studying the following system of equations:

$$\begin{aligned} \left\{ \begin{array}{ll} 2\varphi \varphi _{,x_ix_i} - \sum _{k=1}^{n}\epsilon _{k}\left( \varphi _{,x_{k}}\right) ^{2}\epsilon _{i} - 2\varphi ^{2}f_{ii} = 0 , &{}\quad \forall \ i: 1,\ldots ,n,\\ \varphi _{,x_ix_j} - f_{ij}\varphi = 0, &{}\quad \forall \ i \ne j, \end{array} \right. \end{aligned}$$

(3.3)

$\square$

Proof

[Lemma 2.2]. Because $\varphi = \varphi (\xi )$, we have that:

$\varphi _{x_{i}} = \varphi ^{\prime }a_{i}$, $\varphi _{x_{i}x_{i}} = \varphi ^{\prime \prime }a_{i}^{2} \; \; \forall \ i = 1, \ldots n$, $\varphi _{x_{i}x_{j}} = \varphi ^{\prime \prime }a_{i}a_{j} \; \; \; \forall \ i \ne j$, and $||\nabla _{g}\varphi ||^{2} = \sum _{i=k}^{n}\epsilon _{k}(\varphi _{x_{k}})^{2} = (\varphi ^{\prime })^2 \epsilon$.

Substituting these in the system (3.3), we obtain:

$$\begin{aligned} \left\{ \begin{array}{ll} 2\varphi \varphi ^{\prime \prime }a_{i}^{2} - (\varphi ^{\prime })^{2}\epsilon \epsilon _{i} = 2\varphi ^{2}f_{ii}, &{}\quad \forall \ i: 1,\ldots ,n,\\ \varphi ^{\prime \prime }a_{i}a_{j} - f_{ij}\varphi = 0, &{}\quad \forall \ i \ne j. \end{array} \right. \end{aligned}$$

(3.4)

In this case, it follows directly from the second equation of (3.4) that $f_{ij} = f(\xi )a_{i}a_{j}$, for $i \ne j = 1, \ldots , n$, where $\frac{\varphi ^{\prime \prime }}{\varphi } = f(\xi )$. Similarly, it follows from the first equation of the system (3.4) that $f_{ii} = f_{ii}(\xi )$, because

$$\begin{aligned} f_{ii} = \frac{\varphi ^{\prime \prime }}{\varphi }a_{i}^2 - \left( \frac{\varphi ^{\prime }}{\sqrt{2} \varphi }\right) ^2 \epsilon \epsilon _{i} \qquad \forall \ i=1, \ldots , n. \end{aligned}$$

$\square$

Proof

[Theorem 2.3]. It follows by Lemma (2.2) that finding $\bar{g} = \frac{1}{\varphi ^2} g$ such that $A_{\bar{g}} = T$, with $\varphi = \varphi (\xi )$, is equivalent to studying the following system of equations:

$$\begin{aligned} \left\{ \begin{array}{ll} 2\varphi \varphi ^{\prime \prime }a_{i}^{2} - (\varphi ^{\prime })^{2}\epsilon \epsilon _{i} = 2\varphi ^{2}f_{ii}, &{}\quad \forall \ i: 1,\ldots ,n,\\ \varphi ^{\prime \prime } - f \left( \xi \right) \varphi = 0, &{}\quad \forall \ i \ne j. \end{array} \right. \end{aligned}$$

From the second equation we have $\varphi ^{\prime \prime } = f \varphi$; substituting this in first equation above, we obtain

$$\begin{aligned} 2 a_{i}^{2}f - \left( \frac{\varphi ^{\prime }}{\varphi } \right) ^{2} \epsilon = 2 f_{ii}, \quad \forall \ i: 1,\ldots ,n. \end{aligned}$$

Because $f_{ii}$ depends only on $\xi$, we also have:

$$\begin{aligned} \left( \frac{\varphi ^{\prime }}{\varphi } \right) ^{2} = 2(fa_{i}^2 - f_{ii}) \forall \ i: 1,\ldots ,n \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{\varphi ^{\prime }}{\varphi } = \pm \sqrt{2\epsilon (fa_{i}^2 - f_{ii})} \end{aligned}$$

(3.5)

And it follows directly from Eq. (3.5) that

$$\begin{aligned} \left\{ \begin{array}{ll} 2\epsilon (fa_{i}^2 - f_{ii}) > 0 , &{}\quad \forall \ i: 1,\ldots ,n,\\ fa_{i}^2 - f_{ii} = fa_{j}^2 - f_{jj}, &{}\quad \forall \ i \ne j. \end{array} \right. \end{aligned}$$

Considering $\epsilon _{0} = \pm 1$ and integrating Eq. (3.5), we obtain

$$\begin{aligned}\varphi (\xi ) = Ke^{\epsilon _{0}\int \sqrt{2\epsilon (fa_{i}^2 - f_{ii})}d\xi },\end{aligned}$$

where K is a positive constant. Because $\varphi ^{\prime \prime } = f \varphi$, we get:

$$\begin{aligned} 2\epsilon _{0}^2 \epsilon \epsilon _{i} (fa_{i}^2 - f_{ii}) + \dfrac{\epsilon \epsilon _{i}(f^{\prime }a_{i}^2 - f_{ii}^{\prime })}{\sqrt{2 \epsilon \epsilon _{i} (fa_{i}^2 - f_{ii})}} = f \; \; \; \; \forall \ i=1 , \ldots , n \end{aligned}$$

which concludes the proof. $\square$

Proof

[Theorem 2.4]. If $\varphi = \varphi (\xi )$ and $\xi = a_{1}x_{1}$, then $\varphi _{,x_{i}} = \varphi a_{i}$, $\varphi _{,x_{i}x_{i}} = \varphi ^{\prime \prime } a_{i}^{2}$, $\forall \ i=1,\ldots ,n$, and $\varphi _{,x_{i}x_{j}} = 0, \; \; \forall \ i \ne j$. It follows from (3.3) that $f_{ij}\varphi = 0$, that is, $f_{ij} = 0 \; \; \; \forall \ i \ne j$, this is the tensor diagonal. In this case, the system (3.3) is equivalent to

$$\begin{aligned} \left\{ \begin{array}{ll} 2\varphi \varphi ^{\prime \prime }a_{1}^2 - \left( \varphi ^{\prime }\right) ^{2}\epsilon _{i}\epsilon - 2\varphi ^{2}f_{11} = 0 , &{}\quad \forall \ i: 1,\ldots ,n\\ \left( \varphi ^{\prime }\right) ^{2} \epsilon _{i}\epsilon + 2\varphi ^{2}f_{ii} = 0, &{}\quad \forall \ i \ne 1, i = 1, \ldots , n. \end{array} \right. \end{aligned}$$

(3.6)

From the second equation, we have

$$\begin{aligned} \left( \frac{\varphi ^{\prime }}{\varphi }\right) ^2 = -2\epsilon \epsilon _{i}f_{ii}. \end{aligned}$$

(3.7)

and follows from (3.7) that:

$$\begin{aligned} \left\{ \begin{array}{ll} -2 \epsilon \epsilon _{i}f_{ii} > 0 , &{}\quad \forall \ i: 2,\ldots ,n,\\ \epsilon _{i}f_{ii} = \epsilon _{j}f_{jj}, &{}\quad \forall \ i \ne j, \end{array} \right. \end{aligned}$$

Considering $\epsilon _{0} = \pm 1$ and integrating Eq. (3.7), we obtain

$$\begin{aligned} \varphi ( \xi ) = Ke^{\epsilon _{0} \int \sqrt{-2 \epsilon \epsilon _{i} f_{ii}}d\xi }. \end{aligned}$$

where K is a positive constant. Substituting $\varphi (\xi )$ in the first equation in (3.6), we obtain

$$\begin{aligned} f_{11} = -2a_{1}^{2}\epsilon _{0}^2 \epsilon \epsilon _{i} f_{11} - \dfrac{\epsilon \epsilon _{0} \epsilon _{i}f^{\prime }a_{i}^2}{\sqrt{-2 \epsilon \epsilon _{i}f_{ii}}} + \epsilon _{0}^{2}\epsilon ^{2}\epsilon _{i}f_{ii} \; \; \; \; \; \; \forall \ i=1 , \ldots , n, \end{aligned}$$

which concludes the proof. $\square$

Proof

[Theorem 2.5]. Because $\sum _{k=1}^{n}\epsilon _{k}a_{k}^{2} = \epsilon =0$, it follows that the system (3.4) can be reduced to $\varphi ^{\prime \prime } - f\varphi = 0$ and $f_{ii} = a_{i}^{2}f$, $\forall \ i=1, \ldots ,n$. In this case, the tensor T is given by:

$$\begin{aligned} T = f \begin{pmatrix} a_{1}^{2} &{}\quad a_{1}a_{2} &{}\quad \cdots &{}\quad a_{1}a_{n} \\ \vdots &{}\quad a_{2}^{2} &{}\quad \ddots &{}\quad \vdots \\ -a_{n}a_{1} &{}\quad \cdots &{}\quad &{}\quad a_{n}^{2} \end{pmatrix}\end{aligned}$$

$\square$

Proof

[Theorem 2.6]. Consider $\phi = \varphi F$ and apply Theorem (2.3). $\square$

Proof

[Corollary 2.7]. Note that the functions $\varphi (\xi )$ and $h(\xi )$ clearly satisfy the given partial differential equation. $\square$

Proof

[Corollary 2.8]. The proof is an immediate consequence of equation (2.5) and Corollary (2.7). $\square$