RICCI CURVATURE OF CONFORMAL DEFORMATION ON COMPACT 2-MANIFOLDS

. In this paper, we consider Ricci curvature of conformal deformation on compact 2-manifolds. And we prove that, by the conformal deforma- tion, the resulting manifold is an Einstein manifold.


Introduction.
A Riemannian manifold (M, g) with a metric g is an Einstein manifold provided Ricci curvature Ric = cg for some constant c. It is well known that if M is connected, n = dim M ≥ 3, and Ric = k(p)g, then M is Einstein( [1, p.7], [3], [6]).
In general, if M is connected and 2-dimensional, then Ric = k(p)g for some function k(p). In this case, we have the following question: Question A. Is it possible that (M,g) is an Einstein manifold (Ricg = Cg, where C is a constant ) by conformal deformation wheng = e 2f g for some function f ?
Now if a given metric g on M , where dimM = 2, has Ricci curvature such that Ric g = k(p)g for some function k(p), and we seek K(p) as the Ricci curvature of the metricg = e 2f g pointwise conformal to g such that Ricg = K(p)g = K(p)e 2f g, then f must satisfy f − k(p) + K(p)e 2f = 0, (1.1) where is the Laplacian in the metric g. Several authors have studied the solutions of equation (1.1) (cf. [2], [4], [5], etc.).
In this paper, to solve Question A, when K(p) ≡ C for some constant, instead of equation (1.1), we consider the solvability of the following equation In particular, using the change of variables, instead of equation (1.1), Kazdan and Warner consider the following form, which has a similar form with (1.2), where c is a constant, and h is some prescribed function, with neither c nor h depending on the geometry of (M, g) ( [5]). Kazdan and Warner discussed the ii) Case c = 0. Then, excluding the trivial case h ≡ 0, a solution of equation ( iii) Case c > 0. Then there is a constant 0 < c + (h) ≤ ∞, possibly depending on M, such that a solution exists if h is positive somewhere and if 0 < c < c + (h). They had shown that, in the case c > 0, there exists some obstructions of the solvability of (1.3). They proved that, in the case c = 2 on the sphere S 2 , equation (1.3) has no solutions for any function h such that ∇h · ∇F 0 has a fixed sign for some spherical harmonic F 0 of degree 1, in particular, for all functions h of the form F 0 + constant. And they proved also that, in the case c > 2 on S 2 , if h = F 0 is a spherical harmonic of degree 1, then (1.3) has no solutions.
In this paper, for the given Ricci curvature k(p), we prove the solvability of equation (1.2), using the variational method, for some constant C. The aspect of the solvability of eqation (1.2) is different from that of equation (1.3). In equation(1.3), they consider the solvability of (1.3) for h according to c, but we consider the solvability of (1.2) for some constant C according to k.
Let M be a compact connected 2-dimensional manifold, which is not necessarily orientable and possesses a given Riemannian structure g. We denote the volume element of this metric by dV , the gradient by ∇, and the associated Laplacian by (we use the sign convention which gives u = u xx + u yy for the standard metric on R 2 ). The mean value of a function f on M is written f , that is, We let H s,p (M ) denote the Sobolev space of functions on M whose derivatives through order s are in L p (M ). The norm on H s,p (M ) will be denoted by || || s,p . In the special case s = 0, H s,p (M ) is just L p (M ), and we denote the norm by || || p . If dim M = 2 and u ∈ C ∞ (M ) with u = 0 , then for any p ≥ 1 there is a constant c 1 independent of p and u such that The point here is the sharp control of the dependence of the right side on p ( [5, p.21-22] ). Another immediate consequence of the Poincaré inequality (1.4) is that there is a constant c 2 such that for any u ∈ C ∞ (M ) with u = 0, one has (1.5) , then e uj → e u strongly in L 2 (M ).

Preminaries.
In this section, we consider the Ricci curvature of conformal deformation on a 2-dimensional manifold.
Theorem 2.1. If (M, g) is connected and 2-dimensional, then there exists k(p) such that R ij = k(p)g ij , i, j = 1, 2, where R ij is a Ricci tensor and g ij is a metric tensor.
Proof. Using the normal coordinate system, let for all i, j, k = 1, 2, then by boring computation, And since for all i, j, k, l = 1, 2, the boring computation leads that Hence If we let so R ij = k(u, v)g ij for all i, j = 1, 2, which is our desired one.
In fact, the above k(u, v) is a Gaussian curvature on (M, g).
, Γ k ij = 0, otherwise, and Since the Ricci tensors are R ij = k R k ikj for i, j = 1, 2, then Hence Let us consider the conformal metricg = e 2f g with f ∈ C ∞ (M ). By Definition 1.19 in [1], ifΓ l ij and Γ l ij denote the Christoffel symbols relating tog and g, respectively, for i, j, l = 1, 2, theñ And the Ricci tensors arẽ for i, j = 1, 2,. IfR ij = K(p)g ij = K(p)e 2f g ij and R ij = k(p)g ij for i, j = 1, 2 on a 2-dimensional manifold M , then In other words, f − k(p) + K(p)e 2f = 0.  Proof. If f is a solution of equation (2.2) for C and C C = e 2b for some constant b, (2.2) for C . Proof. If v ≡ 0 and v = 0, we put v + (x) = max{v(x), 0} and v − (x) = max{−v(x), 0}.  Proof. Put f (c) = M e cv dM for some v ∈ H 1,2 (M )(v ≡ 0 and v = 0). Then f (c) = M ve cv dM . Theorem 2.6 implies that if c > 0, then f (c) > 0 and if c < 0, then f (c) < 0. Therefore f (c) has the minimum at c = 0, which means that