ROSE リポジトリいばらき ( 茨城大学学術情報リポジトリ )

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1 ROSE リポジトリいばらき ( 茨城大学学術情報リポジトリ ) Title Liouville type theorem on conformal metrics associate with ultra-hyperb Author(s) SHIMOMURA, Katsunori Citation Mathematical Journal of Ibaraki Uni Issue Date 2016 URL Rights このリポジトリに収録されているコンテンツの著作権は それぞれの著作権者に帰属します 引用 転載 複製等される場合は 著作権法を遵守してください お問合せ先 茨城大学学術企画部学術情報課 ( 図書館 ) 情報支援係

2 Math. J. Ibaraki Univ., 48 (2016), 1-18 Liouville type theorem on conformal mapping for indefinite metrics associate with ultra-hyperbolic equations Katsunori Shimomura Dedicated to Professor Noriaki Suzuki on the occasion of his sixtieth birthday Abstract In [4], we proved that any transformation which preserves solutions of the wave equation is a similarity, an inversion, or a Bateman transformation. This paper gives a generalization of the results to ultra-hyperbolic type equations. 1. Introduction The well-known Liouville s theorem states that every conformal mapping in the n-dimensional Euclidean space (n 3) is a similarity or an inversion with respect to a sphere. In [4], we considered a Liouville type theorem associate with the wave equation, and proved that every conformal mapping for Lorentzian metric is one of the following: a similarity, an inversion composed with similarities, a Bateman transformation composed with similarities. We also determined the transformations which preserve solutions of the wave equation. Let R n be the n-dimensional Euclidean space (n 2), and denote its point by x = (x 1,..., x n ) and the norm of x by x. Let p and q be positive integers such that p + q = n. We consider a second order ultra-hyperbolic type equation p 2 u q 2 u Uu = U p,q u = x 2 j x 2 = 0. (1.1) p+j We introduce an indefinite scalar product (non-degenerate indefinite symmetirc bilinear form) of signature (p, q) on R n by p q x, y = x, y p,q = x j y j x p+j y p+j, (1.2) Received 16 September 2015; revised 23 October Mathematics Subject Classification. 31B99, 35L99, 35A30 Key Words and Phrases. semi-euclidean, conformal mapping, ultra-hyperbolic equation, Bateman transformation Partially supported by Grant-in-aid for Scientific Research (C) No , 15K04934, Japan Society for the Promotion of Science. Ibaraki University, Mito, Ibaraki, , Japan. (katsunori.shimomura.sci@vc.ibaraki.ac.jp)

3 2 Katsunori SHIMOMURA assosiate with the ultra-hyperbolic ( ) type equation (1.1). Ip O Put J = J p,q =, where I O I k denotes the identity matrix of degree k. q Then the scalar product, is written by the Euclidean inner product (, ) as x, y = (Jx, y) = (x, Jy), x, y R n. (1.3) The vector space R n with this scalar product, p,q is denoted by R p,q. Definition 1.1. Let D R n be a domain, f : D R n a C 2 -mapping, and φ a positive C 2 -function on D. A pair (f, φ) is called a transformation which preserves solutions of the ultra-hyperbolic equation, if f and φ satisfy the following conditions: (1) f (x) is non-degenerate for all x D. (2) For every solution u of the ultra-hyperbolic type equation (1.1) on R n, φ (u f) satisfies the ultra-hyperbolic type equation (1.1) on D. In this note, we shall generalize the Liouville type theorem in [4] to R p,q with p + q 3, and determin all the transformations which preserve solutions of the ultrahyperbolic equation. Let D R n and E R k be domains and f = (f 1,..., f k ) : D E a C 1 -mapping. The derivative of f is the linear mapping f (x) : R n R k such that for u R n f f(x + hu) f(x) (x)u = lim. h 0 h Definition 1.2. Let D, E R p,q be domains. A C 1 -mapping f : D E is said to be comformal on D if there exists a function λ(x) = λ f (x) > 0 such that holds for all x D, and u, v R p,q. f (x)u, f (x)v = λ(x) 2 u, v (1.4) Let O(p, q) = {R GL(n, R); Rx, Ry = x, y for all x, y R n }. Our Liouville type theorem is the following: Theorem 1.1. Let n 3 and f be a C 3 -mapping on a domain D R p,q. If f is conformal, then f has one of the following forms. or αr(x a) f(x) = x a, x a + b, λ α f (x) = x a, x a, α f(x) = αrb d (x a) + b, λ f (x) = d, x a, f(x) = αrx + b, λ f (x) = α, (c) where α > 0, R O(p, q), a, b, d R n with d, d = 0, d = 1, and B d (x) = 1 d, x ( x, x 2 ) d Jd + Q d x, Q d x = x x, Jd d x, d Jd. (a) (b)

4 Liouville type theorem for indefinite metrics 3 In [6], we have defined the Bateman mapping B d (x) as a composition of inversions and translations. Thus the following Corollary follows immediately. Corollary 1.1. Let f be a conformal C 3 -mapping on a domain D R p,q. Then f is obtained by composing similarities and inversions. We have also proved that the inversion is a composition of Bateman transformations and similarities in [6]. Thus the following Corollary also follows immediately. Corollary 1.2. Let f be a conformal C 3 -mapping on a domain D R p,q. Then f is obtained by composing similarities and Bateman mappings. We shall also determine the transformations which preserve solutions of the ultrahyperbolic equation. Theorem 1.1 gives the following theorem. Theorem 1.2. Let n 3, f = (f 1,..., f n ) be a C 3 -mapping, and φ a positive function on a domain D R n. Assume (f, φ) be a transformation which preserves solutions of the ultra-hyperbolic equation. If p q, then f is conformal, and f and φ have one of the following forms : f(x) = αr(x a) 2 n + b, φ(x) = C x a, x a 2, (a) x a, x a f(x) = αrb d (x a) + b, φ(x) = C d, x a 2 n 2, (b) f(x) = αrx + b, φ(x) = C, (c) where α > 0, R O(p, q), a, b R n, and d R n with d, d = 0 and (d, d) = 1. If p = q, then either f is conformal and (f, φ) satisfies one of the above (a), (b), and (c), or Sf is conformal and (f, φ) satisfies one of the following: f(x) = αsr(x a) 2 n + b, φ(x) = C x a, x a 2, (a ) x a, x a f(x) = αsrb d (x a) + b, φ(x) = C d, x a 2 n 2, (b ) f(x) = αsrx + b, φ(x) = C, (c ) ( ) O Ip where α, R, a, b, d are the same as above, and S =. O Remark 1. In the case of p = q, (f, φ) satisfies one of (a), (b), and (c), if f j, f j > 0 (1 j p). If f j, f j < 0 (1 j p), then (f, φ) satisfies one of (a ), (b ), and (c ). Corollary 1.3. Let n 3. If p q, every transformation which preserves solutions of the ultra-hyperbolic equation is obtained by composing similarities and inversions. If p = q, every transformation which preserves solutions of the ultra-hyperbolic equation is obtained by composing the linear map S, and similarities and inversions. I p

5 4 Katsunori SHIMOMURA Corollary 1.4. Let n 3. If p q, every transformation which preserves solutions of the ultra-hyperbolic equation is obtained by composing similarities and Bateman transformations. If p = q, every transformation which preserves solutions of the ultra-hyperbolic equation is obtained by composing S and similarities and Bateman transformations. 2. Conformal mapping In the previous section, we introduced the scalar product, =, p,q and the associated space R p,q. A vector u R p,q is called null if u, u = 0 and u 0. For a null vector u, we define a conjugate of u by u = u 2 Ju, so that u, u = 1 and u, u = 0. Clearly, u and u are linearly independent. If a null vector u satisfies u = 1, then u = Ju. Recall that O(p, q) = {R GL(n, R); Rx, Ry = x, y for all x, y R p,q }. O(p, q) is the group of all matrices which preserve the scalar product,. It is clear that R O(p, q) if and only if t RJR = J. Note that R O(p, q) if and only if J = RJ t R. For a subspace X of R p,q, we define a subspace X J of R p,q by X J = {x R p,q ; x, y = 0 for all y X}, while we denote by X the Euclidean orthogonal complement of X in R n. The following lemma will be used in section 3. Lemma 2.1. Let α 0, R O(p, q), and X be a subspace of R p,q. Then X J = (JX) = J(X ), (X J ) J = X, (2.1) αr(x J ) = (αrx) J. (2.2) Proof. Since x, y = (x, Jy), x X J is equivalent to x (JX). Since x, y = 0 is equivalent to (Jx, y) = 0 and J 2 = I, x X J is equivalent to x J(X ). We have the first equality. The second equality follows from (X J ) J = J(((JX) ) ) = J 2 X = X. x αr(x J ) if and only if α 1 R 1 x, y = 0 for all y X. Then if x αr(x J ) αα 1 RR 1 x, αry = x, αry = 0 for all y X, and hence x (αrx) J, and vise versa. We need the assumption n 3 for the following lemma. Lemma 2.2. (1) Let n 3, u R p,q and V = {v; u, v = 0}. Then {v V ; v, v 0} is dense in V. (2) Let n 3. For any u, v R p,q with u, v = 0, there exist sequences {u n }, {v n } in R p,q such that u u n, v v n 0, u n, v n = 0, u n, u n = 0, and v n, v n = 0. Proof. (1) n 3 implies p 2 or q 2. Assume p 2. Then there exists w + R p such that w + u + where u + = (u 1,..., u p ) in R p. Put w = (w +, 0,..., 0) R p,q.

6 Liouville type theorem for indefinite metrics 5 For v V with v, v = 0, put v t = v + tw (t R). Then v t V for all t R because v t, u = v, u + t w, u = 0 + t(w +, u + ) = 0. On the other hand, v t, v t = 2t v, w + t 2 w, w = 0 except for at most 2 values of t because w, w = w Since lim v t = v, the assersion holds in the case p 2. In the case q 2, the assertion t 0 follows similarily. (2) If u, u 0, put u n = u for all n. Then by (1), there exists a sequence {v n } such that u n, v n = u, v n = 0, v n, v n = 0, and v v n 0. Let u, u = 0. Then there exists a sequence {u n } such that u u n 0 and u n, u n = 0. Consider the hyperplanes V n = {v : u n, v = 0} and let P n be the Euclidean orthogonal projection to V n. For each n, (1) implies that there exists a vector v n such that v n V n, v n P n v < 1/n, v n, v n = 0. Since u n u, P n v v and hence v v n 0. Lemma 2.3. (1) Let max(p, q) 3. Then for any u, v R p,q with u, v = 0, there exists w R p,q with w, w = 0 such that u, w = v, w = 0. (2) Let n = 3 or p = q = 2. Then for any u, v R p,q with u, u = 0, v, v = 0, and u, v = 0, there exists w R p,q with w, w = 0 such that u, w = v, w = 0. Proof. (1) Let p 3, and u, v R p,q with u, v = 0. There exists (w 1,..., w p ) 0 such that (w 1,..., w p ) (u 1,..., u p ) and (w 1,..., w p ) (v 1,..., v p ). Then the vector w = (w 1,..., w p, 0,..., 0) R p,q satisfies the conditions. The case q 3 follows similarly. (2) Let n = 3, u, v R p,q, and assume u, u 0, v, v = 0, and u, v = 0. Then u and v are linearly independent. Since dim(ru + Rv) J = 1, there exists a non-zero vector w (Ru + Rv) J. The vectors u, v, w are linearly independent. In fact, if au + bv + cw = 0, a = 0 and b = 0 follow by taking the scalar product with u and v, u, x respectively. Hence c = 0 because w 0. Then the linear mapping Cx = v, x is w, x u, w injective. Therefore Cw = v, w = and hence w, w = 0. w, w w, w Finally, let p = q = 2, u, v R p,q, u, u 0, v, v 0, and u, v = 0. Then u and v are linearly independent. Since dim(ru + Rv) J = n 2 = 2, there exists a pair of linearly independent vectors x, y (Ru + Rv) J. The vectors u, v, x, y are linearly independent. In fact, if au + bv + cx + dy = 0, a = 0 and b = 0 follow by taking the scalar product with u and v, respectively. Then c = d = 0 follows from the linear independence of x and y. If x, x = 0 or y, y = 0 holds, we can choose w = x or w = y, respectively. Assume that x, x = y, y = 0 hold. Since the linear u, z u, y 0 mapping Cz = v, z x, z is injective and y 0, Cy = v, y x, y = 0 x, y 0. y, z y, y 0 Then w = x + y satisfies w, w = 2 x, y = 0, u, w = 0, and v, w = 0. The following proposition follows easily from the chain rule.

7 6 Katsunori SHIMOMURA Proposition 2.1. (1) If f : D E and g : E F are conformal mappings, then the composition mapping g f : D F is also a conformal mapping, and λ g f (x) = λ g (f(x))λ f (x), x D. (2.3) (2) If f : D E is a conformal mapping, then f has a local inverse f 1 at each point of f(d), f 1 is also conformal, and λ f 1(f(x)) = 1 λ f (x). (2.4) In the following, we list fundamental conformal mappings. Example 2.1 (Similarity). The mapping f(x) = αrx + a (α R, α > 0, R O(p, q), a R n ) is a conformal mapping defined on R p,q satisfying λ f (x) = α. We call such mapping similarity (mapping). We introduce the notation T a (x) = x a for later use. λ Ta (x) = 1. Example 2.2 (Inversion). The mapping K(x) = x x, x is a conformal mapping defined on each connected component of {x; x, x = 0}. We call K the inversion (mapping). By simple calculation, we have K 1 = K and λ K (x) = 1 x, x. (2.5) Example 2.3 (Bateman mapping [6]). For each null vector d, the mapping defined on each connected component of {x; d, x = 0} B d (x) = 1 d, x ( x, x 2 ) d d + Q d x, where Q d x = x x, d d x, d d, is a conformal mapping because B d = T d K T 1 2 d K T d. We call B d the Bateman mapping. As we have shown in [6], B 1 d = B d. (2.6) By the chain rule and (2.5), λ Bd (x) = 1 d, x. (2.7)

8 In fact, since T 1 2 d (K(y)), T 1 2 d (K(y)) = Liouville type theorem for indefinite metrics d, y, we have y, y 1 + d, x 1 (T 1 2 d K T d )(x), (T 1 2 d K T d )(x) = T d (x), T d (x) = d, x T d (x), T d (x). By (2.3), λ Bd (x) = 1 λ K (T 1 2 d K T d (x)) 1 λ K (T d (x)) 1 and hence λ Bd (x) = 1 (T 1 2 d K T d )(x), (T 1 2 d K T d )(x) 3. Liouville type theorem for conformal mapping 1 T d (x), T d (x) = 1 d, x. In this section, we shall prove Theorem 1.1. The first derivative of a C 1 -mapping f : R n R k is the linear mapping f (x) : R n R k f (x)u = lim h 0 f(x + hu) f(x) h (u R n ). The second derivative of a C 2 -mapping f : R n R k is the symmetric bilinear mapping f (x) : R n R n R k such that f 1 ( ) (x)[u, v] = lim f (x + hu)v f (x)v h 0 h (u, v R n ). The third derivative of a C 3 -mapping f : R n R k is the symmetric trilinear mapping f (x) : R n R n R n R k such that f 1 ( ) (x)[u, v, w] = lim f (x + hu)[v, w] f (x)[v, w] h 0 h (u, v, w R n ). Lemma 3.1. u, v, w R p,q, If f is a conformal C 2 -mapping on a domain D R p,q, then for each f (x)[u, v], f (x)w + f (x)[u, w], f (x)v = 2λ(x) v, w λ (x)u. (3.1) Note that λ (x) is a linear map from R n to R. Proof. Taking the derivative in u direction of the conformality condition (1.4) f (x)v, f (x)w = λ(x) 2 v, w, we have (3.1). Lemma 3.2. Let f be a conformal C 2 -mapping on a domain D R p,q. If u, v, w R p,q and u, v = v, w = w, u = 0, then f (x)[u, v], f (x)w = 0.

9 8 Katsunori SHIMOMURA Proof. We put A[u, v, w] = f (x)[u, v], f (x)w for u, v, w R p,q. Then A satisfies A[v, u, w] = A[u, v, w], (3.2) A[u, v, w] = A[u, w, v] if v, w = 0. (3.3) In fact (3.2) is clear from the symmetry of the bilinear map f (x), and (3.3) immediately follows from Lemma 3.1. If u, v, w R n and u, v = v, w = w, u = 0, then we have A[u, v, w] = A[v, u, w] by (3.2), A[v, u, w] = A[v, w, u] by (3.3) and u, w = 0, A[v, w, u] = A[w, v, u] by (3.2), A[w, v, u] = A[w, u, v] by (3.3) and u, v = 0, A[w, u, v] = A[u, w, v] by (3.2), A[u, w, v] = A[u, v, w] by (3.3) and v, w = 0. Therefore A[u, v, w] = 0. This proves the lemma. Lemma 3.3. Let f be a conformal C 2 -mapping on a domain D R p,q. If λ is equal to a constant α on D, then f is a similarity f(x) = αrx + b, with R O(p, q) and b R n. Proof. Since λ is constant, (3.1) implies f (x)[u, v], f (x)w = 0 for all u, v, w R p,q. Then A[u, v, w] = f (x)[u, v], f (x)w satisfies (3.2) and A[u, v, w] = A[u, w, v] (u, v, w R p,q ). (3.4) From the same argument as in the proof of Lemma 3.2, A[u, v, w] = 0 follows for all u, v, w R n. Since f (x) is surjective, f (x)[u, v] = 0 for all u, v R n, and hence f (x) = 0. Then α 1 f (x) is equal to a constant matrix R satistying Ru, Rv = u, v. Therefore R O(p, q) and f(x) = αrx + b, where b R n. Lemma 3.4. Let f be a conformal C 2 -mapping on a domain D R p,q. If λ f (x) = α x, x with some constant α > 0 on D, then there exist R O(p, q) and b R n such that Proof. By the assumption and (2.5), λ f K (x) = λ f (K(x)) λ K (x) = f(x) = αrk(x) + b. α K(x), K(x) 1 α x, x 2 = x, x x, x 1 x, x = α. Then by Lemma 3.3, (f K)(x) = αrx + b, with R O(p, q) and b R n. Hence the lemma follows from K 1 (x) = K(x). Lemma 3.5. Let f be a conformal C 2 -mapping on a domain D R p,q. If λ f (x) = α d, x with some null vector d and a constant α > 0 on D, then with some R O(p, q) and b R n. f(x) = αrb d (x) + b,

10 Liouville type theorem for indefinite metrics 9 Proof. By (2.6), (2.7), and the assumption, λ f B 1(x) = λ f (B d (x)) λ B d (x) = d Since ( d) = d and Q d x, d = 0, and hence d, B d (x) = 1 d, x α d, B d (x) 1 d, x. x d, x, d + d + Q d x = d, d 2 d, x = λ f B 1(x) = α. d 1 d, x, Therefore (f B 1 d )(x) = αrx+b with R O(p, q) and b Rn. The lemma follows. By Lemmas 3.3, 3.4 and 3.5, to prove Theorem 1.1, it suffices to prove the following: Theorem 3.1. Let n 3 and f be a conformal C 3 -mapping on a domain D R p,q. Then λ f has one of the following forms: α λ f (x) = x a, x a, (3.5) α λ f (x) = d, x a, (3.6) λ f (x) = α, (3.7) where α > 0 is a constant, a R n, and d R n is a null vector. In the following, we prepare several lemmas to prove Theorem 3.1. The following lemma is well-known in linear algebra. Lemma 3.6. Let l(u, v) is a symmetric bilinear form on R n. If l(u, v) = 0 for all u, v R n with u, v = 0, then there exists a constant c R such that l(u, v) = c u, v (u, v R p,q ). Lemma 3.7. x D, Let f be a conformal C 2 -mapping on D. If u, v = 0, then for each f (x)[u, v] Rf (x)u + Rf (x)v. Proof. By Lemma 3.2, if u, v = 0 and w (Ru + Rv) J, Then f (x)[u, v] f (x)[u, v], f (x)w = 0. ( ) f (x)(ru + Rv) J ( J holds, and (2.2) implies f (x)(ru + Rv) J ) J = f (x) ( (Ru + Rv) J ) J = f (x)(ru + Rv) = Rf (x)u + Rf (x)v.

11 10 Katsunori SHIMOMURA Lemma 3.8. Let n 3. Let f be a conformal C 2 -mapping on D R p,q. If u, v = 0, then for each x D, f (x)[u, v] = λ(x) 1 λ (x)vf (x)u + λ(x) 1 λ (x)uf (x)v. (3.8) Proof. Let u, v R p,q. Applying Lemma 3.1 with w = v, we have f (x)[u, v], f (x)v = λ(x) v, v λ (x)u. (3.9) Fix arbitrary u R p,q. If v R p,q satisfies u, v = 0, then f (x)[u, v] = af (x)u + bf (x)v (3.10) with some a = a(x, u, v) and b = b(x, u, v) by Lemma 3.7. From (3.9), we have a f (x)u, f (x)v + b f (x)v, f (x)v = λ(x) v, v λ (x)u. Since u, v = 0 and f is conformal, Similarily we have Thus we have bλ(x) 2 v, v = λ(x) v, v λ (x)u. aλ(x) 2 u, u = λ(x) u, u λ (x)v. a = λ(x) 1 λ (x)v, b = λ(x) 1 λ (x)u (3.11) for all u, v with u, v = 0, u, u = 0, and v, v = 0, because λ(x) > 0. By Lemma 2.2 and continuity, we can drop u, u = 0 and v, v = 0. Lemma 3.9. Let n 3. If f is a conformal C 3 -mapping, then ρ(x) = 1 λ(x) and where c is a constant. is C ρ (x)[u, v] = c u, v (3.12) Proof. Since f is C 3, ρ = 1/λ is C 2 and ρ = λ /λ 2. Then by (3.8), if u, v = 0, ρ(x)f (x)[u, v] + ρ (x)vf (x)u + ρ (x)uf (x)v = 0. Taking the derivative in direction w, we have ρ (x)wf (x)[u, v] + ρ(x)f (x)[u, v, w] + ρ (x)[v, w]f (x)u + ρ (x)vf (x)[u, w] + ρ (x)[u, w]f (x)v + ρ (x)uf (x)[v, w] = 0. (3.13) Assume w, v = 0 and interchange u and w in the equation. Then ρ (x)uf (x)[w, v] + ρ(x)f (x)[w, v, u] + ρ (x)[v, u]f (x)w + ρ (x)vf (x)[w, u] + ρ (x)[w, u]f (x)v + ρ (x)wf (x)[v, u] = 0. (3.14)

12 Liouville type theorem for indefinite metrics 11 Subtract (3.14) from (3.13). Then ρ (x)[v, w]f (x)u ρ (x)[v, u]f (x)w = 0 holds for all u, v, w R p,q satisfying u, v = w, v = 0. Taking a scalar product with f (x)w, we have and hence if u, v = w, v = 0, ρ (x)[v, w]f (x)u ρ (x)[v, u]f (x)w, f (x)w = 0, ρ (x)[v, w]λ(x) 2 u, w ρ (x)[v, u]λ(x) 2 w, w = 0. By Lemma 2.3, we can choose w so that u, w = w, v = 0 and w, w = 0, if u, u 0 and v, v = 0. Then we obtain ρ (x)[u, v] = 0. By continuity and Lemma 2.2, we can drop u, u = 0 and v, v = 0. By Lemma 3.6, ρ (x)[u, v] = c(x) u, v (3.15) holds for all u, v R p,q with some c(x) R. Then for j = 1,..., n, 2 ρ x 2 (x) = ρ (x)[e j, e j ] = c(x)ε j, j where ε j = 1 (1 j p), ε j = 1 (p + 1 j n). Hence ρ is C because ρ satisfies an elliptic equation p q 2 ρ q x 2 (x) + p 2 ρ j x 2 (x) = 0. p+j and then To show c(x) is constant, take the derivative of (3.15) in direction w. We have ρ (x)[u, v, w] = c (x)w u, v, (c (x)w)u (c (x)u)w, v = c (x)w u, v c (x)u w, v for all u, v, w R p,q. That implies and hence = ρ (x)[u, v, w] ρ (x)[w, v, u] = 0 (c (x)w)u (c (x)u)w = 0, c (x)u = c (x)w = 0 for each linearly independent pair of u and w. Therefore c (x) = 0.

13 12 Katsunori SHIMOMURA Lemma Let n 3. If f is conformal and C 3 on D, then λ = λ f satisfies one of the following on D with c, b, l R, c 0, l > 0, a, d R n, d 0: 1/λ f (x) = c x a, x a + b, (3.16) 1/λ f (x) = d, x a, (3.17) 1/λ f (x) = l. (3.18) Proof. By Lemma 3.9, ρ(x) = λ(x) 1 is equal to a polynomial of degree at most 2. Integrating the equation (3.12), we have the lemma. Now we are ready to prove Theorem 3.1. Proof of Theorem 3.1. By Lemma 3.10, λ f satisfies (3.16), (3.17), or (3.18). Fix arbitrary x D. Since f is conformal, f has a conformal local inverse f 1 defined in a neighborhood W of x. Then Lemma 3.10 implies that f 1 satisfies one of the following in W with c, b, l R, c 0, l > 0, a, d R n, d 0: λ f 1(x) 1 = c x a, x a + b, λ f 1(x) 1 = d, x a, λ f 1(x) 1 = l. By (2.4), we have the following possibilities: (c x a, x a + b)(c f(x) a, f(x) a + b ) = 1, (3.19) (c x a, x a + b) d, f(x) a = 1, (3.20) d, x a (c f(x) a, f(x) a + b ) = 1, (3.21) d, x a d, f(x) a = 1, (3.22) λ f (x) = l, (3.23) because λ f is constant if and only if λ f 1 is constant. We begin with the case (3.19). Assume b 0. We can choose a point z D so that v = z a satisfies v, v = 0. The line l : v(t) = a + tv (t R) (3.24) is orthogonal to every hypersurface S η = {x; c x a, x a + b = η} for η R and η b with respect to the indefinite metric. Since f is conformal, the image curve f(l) is orthogonal to every image hypersurface f(s η ). By (3.19), f(s η ) is contained in S 1/η = {y; c y a, y a + b = 1/η}. Then f(l) is orthogonal to every S η, and hence f(l) = {f(v(t)); t R} must be in a line passing through a. Therefore, f(v(t)) = a + β(t)v (3.25) where v R n and β(t) is a real valued C 3 -function. Differentiating (3.25), we have f (v(t))v = β (t)v.

14 Liouville type theorem for indefinite metrics 13 The conformality of f implies that Since b 0, β (t) 2 v, v = f (v(t))v, f (v(t))v = λ f (v(t)) 2 v, v = β(t) = v, v (c v, v t 2 + b) 2. { b p1 arctan p 2 t if c v,v > p 1 log p 2t 1 b p 2 t+1 if c v,v < 0 (3.26) with some p 1, p 2 R. Especially, β(t) is a transcendental function. On the other hand, substituting (3.24) and (3.25) into (3.19), we have c β(t) 2 v, v + b = 1 t 2 + b, which means β(t) 2 is a rational function. This is a contradiction. Therefore b = 0 and thus we have (3.5) by putting α = 1/ c. Next we show that the case of (3.20) does not occur. Let v, l, and S η be as above. Equation (3.20) shows that f(s η ) (η R, η 0) is contained in the hyperplane P 1/η = {y; d, y a = 1/η} with d 0. Since f is conformal, f(l) is orthogonal to every f(s η ) and hence to every hyperplane P 1/η. Therefore, f(l) is a line of direction d, and there exist w R n and a real valued function β(t) such that Differentiating the both sides, we have The conformality of f implies that f(v(t)) = w + β(t)d. (3.27) f (v(t))v = β (t)d. β (t) 2 d, d = f (v(t))v, f (v(t))v = λ f (v(t)) 2 v, v = v, v (c v, v t 2 + b) 2. Then d, d = 0 because c 1 0. Assume b 0. Then { b p1 arctan p 2 t if c v,v β(t) = > p 1 log p 2t 1 b p 2 t+1 if c v,v < 0 (3.28) with some p 1, p 2 R. Especially, β(t) is a transcendental function. On the other hand, substituting (3.27) into (3.20), we have (c v, v t 2 + b)( d, d β(t) + d, w a ) = 1, and hence β(t) is a rational function. This is a contradiction. Therefore b = 0. Then β (t) 2 = d, d 1 c 2 v, v t 4 and β(t) + b = d, d 1 c v, v t 2, where b = d, w a d, d, which is also a contradiction. Thus the case (3.20) does not occur.

15 14 Katsunori SHIMOMURA Interchanging the role of f and f 1, the same argument shows the case (3.21) does not occur. Finally, we treat the case of (3.22). For each hyperplane P η = {x; d, x a = η}, f(p η ) is contained in the hyperplane P 1/η = {y; d, y a = 1/η} by (3.22). For each w R n, the line l : v(t) = w + td (t R) (3.29) is orthogonal to all P η. Since f is conformal, the image curve f(l) is orthogonal to every P 1/η. Hence f(l) has direction d. Therefore, there exist w R n and a C 3 -function β(t) such that Differentiating by t, we have f(v(t)) = w + β(t)d. (3.30) f (v(t))d = β (t)d. Substituting (3.29) and (3.30) into (3.22), we have ( d, d t + d, w a )( d, d β(t) + d, w a ) = 1, and hence β(t) is a rational function. On the other hand, the conformality of f implies that β (t) 2 d, d = λ f (v(t)) 2 d, d = d, d ( d, d t + b) 2, where b = d, w a. If d, d = 0, then d, d = 0 and β(t) is a logarithm function, This is a contradiction. Therefore d, d = 0 and thus we have (3.6) by putting α = 1/ d and normalizing d. 4. Transformation which preserves solutions of the ultra-hyperbolic equation In this section, we shall give the proof of Theorem 1.2. By definition, it is easy to see that if (f, φ f ) and (g, φ g ) are transformations which preserves solutions of the ultra-hyperbolic equation such that the image of f is contained in the domain of g, then (g f, φ f (φ g f)) is also a transformation which preserves solutions of the ultra-hyperbolic equation. We call this transformation (g f, φ f (φ g f)) the composition of (f, φ f ) and (g, φ g ). First, we remark that all the transformations in Theorem 1.2 are the composition of the following fundamental transformations Examples Example 4.1 (similarity). Let α, C R, α, C > 0, R O(p, q), a R n. The pair of a similarity f(x) = αrx + a and a positive constant function φ(x) = C is a transformation which preserves solutions of the ultra-hyperbolic equation. Example 4.2 (inversion [6]). The pair of the inversion mapping K and the function φ K (x) = x, x 2 n 2 is a transformation which preserves solutions of the ultrahyperbolic equation on each connected component of {x R p,q ; x, x = 0}.

16 Liouville type theorem for indefinite metrics 15 Example 4.3 (Bateman transformation [6]). Let d be a null vector. Then the pair of the Bateman mapping B d and the function φ Bd (x) = d, x 2 n 2 is a transformation which preserves solutions of the ultra-hyperbolic equation on each connected component half space of {x R p,q ; d, x = 0}. We also remark that all the mappings of the above three transformations are conformal. In the case p = q, we have a transformation of which the mapping is not conformal. ( ) O Ip Example 4.4. Let p = q and S =. The pair of the mapping f(x) = Sx I p O and the constant function φ(x) = 1 is a transformation which preserves solutions of the ultra-hyperbolic equation on R p,q. The following theorem gives a characterization of the transformations which preserves solutions of the ultra-hyperbolic equation. Theorem 4.1. The pair (f, φ) is a transformation which preserves solutions of the ultra-hyperbolic equation, if and only if f = (f 1, f 2,..., f n ) and φ satisfy the following equations on D: Uφ = 0, (4.1) φ Uf j + 2 φ, f j = 0 (j = 1, 2,..., n) (4.2) f i, f j = 0 (i, j = 1, 2,..., n, i j) (4.3) f i, f i = f j, f j = 0 (1 i p, p + 1 j n), (4.4) where f j = ( f j x 1, f j x 2,..., f j x n ), j = 1, 2,..., n. Proof. By the chain rule, the following holds for any C 2 -function u: U(φ (u f)) = k=1 n 2 ( ) ε j φ (u f) x 2 j = (Uφ) (u f) (4.5) n ( u ) n ( 2 u ) + (φ Uf k + 2 φ, f k ) f + φ f k, f l f. y k y k y l k,l=1 Assume that (f, φ) is a transformation which preserves solutions of the ultra-hyperbolic equation. First we take u = 1. Since u = 2 u = 0 (1 k, l n), we have (4.1): y k y k y l Uφ = 0. Take u(y) = y j (1 j n). Since u = δ jk, y k (4.2): φuf j + 2 φ, f j = 0. 2 u y k y l = 0 (1 k, l n), we have

17 16 Katsunori SHIMOMURA Take u(y) = y i y j (1 i, j n, i j). Since 2 u y k y l = δ ik δ jl + δ il δ jk (1 k, l n), we have (4.3): f i, f j = 0 (1 i, j n, i j). Finally, take u(y) = y 2 i + y2 j (1 i p, p + 1 j n). Since 2 u y 2 i = 2 u y 2 j 2 u = 0 (k i, j), we have f i, f i + f j, f j = 0 (1 i p, p + 1 j n). y 2 k Together with (4.3), we have f J t f = f 1, f 1 J. Since f is non-degenerate, the left hand side is a regular matrix. Therefore, f 1, f 1 = 0. Conversely, assume that f and φ satisfy (4.1), (4.2), (4.3), and (4.4). Substituting (4.1), (4.2) and (4.3) into (4.5), we have U(φ (u f)) = φ Putting λ 1 = f 1, f 1, (4.4) gives n ( 2 u f j, f j y 2 j ) f. U(φ (u f))(x) = φ(x)λ 1 (x)((uu) f)(x). This shows that Uu = 0 implies U(φ (u f)) = 0. Thus we have the theorem. Theorem 4.2. Let (f, φ) be a transformation which preserves solutions of the ultrahyperbolic equation defined on D R p,q. If p q, then f is conformal on D such that λ f (x) 2 = f 1, f 1. If p = q, then either f is conformal on D such that λ 2 f = f 1, f 1, or Sf(x) is conformal on D such that λ 2 Sf = f 1, f 1. Proof. Put λ 1 = f 1, f 1 and let x D be fixed. (4.3) and (4.4) is equivalent to f (x)j t f (x) = λ 1 (x)j. Since f is non-degenerate, λ 1 (x) 0 and hence t f (x) 1 Jf (x) 1 = λ 1 (x) 1 J. Therefore t f (x)jf (x) = λ 1 (x)j and hence f (x)u, f (x)v = λ 1 (x) u, v for all u, v R p,q. If λ 1 (x) > 0 for all x D, then f is conformal and λ f (x) 2 = λ 1 (x). Next, assume λ 1 (x) < 0. Put e 1 = (1, 0,..., 0), e 2 = (0, 1,..., 0),..., e n = (0,..., 0, 1). We shall show that the set of 2p vectors {e 1,..., e p, f 1 (x),..., f p (x)} and the set of 2q vectors {e p+1,..., e p+q, f p+1 (x),..., f p+q (x)} are both linearly independent. Assume α 1 e α p e p + α p+1 f 1 (x) + + α 2p f p (x) = 0 and β 1 e p β q e p+q + β q+1 f p+1 (x) + + β 2q f p+q (x) = 0. Put u = α 1 e α p e p = α p+1 f 1 (x) α 2p f p (x) and v = β p+1 e p β q e p+q = β q+1 f p+1 (x) β 2q f p+q (x). Then by (4.3), (4.4) and f 1 (x), f 1 (x) = λ 1 (x) < 0, p p 0 αj 2 = u, u = λ 1 (x) αp+j 2 0, 0 q βj 2 = v, v = λ 1 (x) q βq+j 2 0, = 2,

18 Liouville type theorem for indefinite metrics 17 that imply α 1 = = α 2p = 0 and β 1 = = β 2q = 0. Therefore 2p n and 2q n, which imply p = q because p + q = n. Consequently, p q implies λ 1 > 0. Moreover, (Sf) (x)j t (Sf) (x) = Sf (x)j t f (x) t S = λ 1 (x)sjs = λ 1 (x)j. Thus Sf is conformal on D and λ 2 Sf = f 1, f 1. Next lemma shows that φ is uniquely determined by f except a constant multiple. Lemma 4.1. If (f, φ 1 ) and (f, φ 2 ) are two transformations which preserve solutions of the ultra-hyperbolic equation with same mapping f on a domain. Then φ 2 is a constant multiple of φ 1. Proof. By (4.2), we have and then Uf j = 2 log φ 1, f j, Uf j = 2 log φ 2, f j, (j = 0, 1,..., n). f j, log φ 2 φ 1 = f j, log φ 2 log φ 1 = 0, (j = 0, 1,..., n). Since the matrix f (x) is non-degenerate, log φ 2(x) φ 1 (x) φ 2 is a constant multiple of φ 1. = 0 for all x D. Therefore, Now we are ready to prove Theorem 1.2. Proof of Theorem 1.2. Let p q, or p = q and f 1, f 1 > 0. Then by Corollary 4.2, f is conformal on D. Then by Theorem 1.1, f has the form (a), (b), or (c). αr(x a) In the case (a) : f(x) = + b, put h(x) = αrx + b and g(x) = x a. x a, x a Then f(x) = (h K g)(x). Since (h, 1) and (g, 1) are transformations which preserve solutions of the ultra-hyperbolic equation, Lemma 4.1 implies φ is a constant multiple of 1 φ K (g(x)) 1 = x a, x a 2 n 2. Therefore φ(x) = C x a, x a 2 n 2 with C > 0. In the case (b) : f(x) = αrb d (x a) + b, put h(x) = αrx + b and g(x) = x a. Then f(x) = (h B d g)(x). Since (h, 1) and (g, 1) are transformations which preserve solutions of the ultra-hyperbolic equation, φ is a constant multiple of 1 φ Bd (g(x)) 1 = d, x a 2 n 2 by Lemma 4.1. Therefore φ(x) = C d, x a 2 n 2 with C > 0. In the case (c) : f(x) = αrx + b is the similarity. So φ is a positive constant. If p = q and f 1, f 1 < 0, then Sf is conformal by Corollary 4.2. Since S(Sf) = S 2 f = f, we have (a ), (b ), and (c ).

19 18 Katsunori SHIMOMURA References [1] H. Bateman, The conformal transformations of a space of four dimensions and their applications to geometrical optics, Proc. London Math. Soc., 7 (1909) [2] M. Berger, Geometry, Springer-Verlag, [3] B. O Neill, Semi-Riemannian geometry, Academic Press, [4] K. Shimomura, Liouville type theorem associate with the wave equation, Math. J. Ibaraki Univ., 43 (2011), [5] K. Shimomura, A relation between the Lorentzian inversion and the Bateman transformation, Math. J. Ibaraki Univ., 44 (2012), 1 5. [6] K. Shimomura, Generalizations of Bateman s transformation for general indefinite metrics, Math. J. Ibaraki Univ., 45 (2013), 7 13.