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Results & Lemmas (18)

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THEOREM 1. THEOREM 1. — Let /,/' be two harmonic mappings M —> Y. If they agree on an open set then they are identical (M being assumed connected,…
THEOREM 1. — Let /,/' be two harmonic mappings M —> Y. If they agree on an open set then they are identical (M being assumed connected, naturally); and indeed the conclusion holds if f and f agree to infinitely high order at some point. In particular, a harmonic mapping which is constant on an open set is a constant mapping.
THEOREM 2. THEOREM 2. — Let S be apiece ofC2 hyper surf ace in X passing through q, at which point we assume that the second fundamental form is…
THEOREM 2. — Let S be apiece ofC2 hyper surf ace in X passing through q, at which point we assume that the second fundamental form is definite. Iffis not a constant mapping, then no neighbourhood of p is mapped entirely to the concave side of S. ANNALES SCIENTIFIQUES DE I/ECOLE NORMALE SUPERIEURE
Theorem 1 Theorem 1 to complete the proof. Q.E.D.
Theorem 1 to complete the proof. Q.E.D.
THEOREM 3. THEOREM 3. — Iff: M —> Y is harmonic and M connected, and ifdfhas rank 1 in an open set, then f maps M into a geodesic arc in Y, and df has…
THEOREM 3. — Iff : M —> Y is harmonic and M connected, and ifdfhas rank 1 in an open set, then f maps M into a geodesic arc in Y, and df has rank 1 in an open, dense set. If M is compact, then the geodesic arc is closed. We note the following immediate corollary: COROLLARY. — IfM has dimension 2 and iff : M —> Y has rank 2 in an open set, then it has rank 2 in an open, dense set.
Theorem 1. Theorem 1. Suppose now that dfhas rank 1 in an open set U. Then every pe has a neighbourhood which is mapped into a regular arc c in Y. As…
Theorem 1. Suppose now that dfhas rank 1 in an open set U. Then every pe\J has a neighbourhood which is mapped into a regular arc c in Y. As usual, let x resp. y denote local coordinates at/? [resp./(/?)]. Since c is regular, we may assume that the ^-system is chosen so that c is the coordinate curve defined by y " = 0 for a > 1. Accordingly we have y9 (x) = 0 for a > 1, x near/?. The harmonic equation (1) then shows that P^ = 0 along c near / (p) for a > 1. Now it is quickly seen that an approp
Theorem 2 Theorem 2 above. Hence we conclude that c is closed and that/(M) = c. Q.E.D. We shall apply the foregoing to prove a sharpened form of a…
Theorem 2 above. Hence we conclude that c is closed and that/(M) = c. Q.E.D. We shall apply the foregoing to prove a sharpened form of a result of Hartmann (cf. [15], Th. H, Cor. 2).
THEOREM 4. THEOREM 4. — Letf: M —> Y be a harmonic mapping, where M is assumed to be compact and Y of non-positive sectional curvature. Iff(M)…
THEOREM 4. — Letf : M —> Y be a harmonic mapping, where M is assumed to be compact and Y of non-positive sectional curvature. Iff(M) contains a point q at which the sectional curvatures of Y are all < 0, and ;//(M) is not contained in a geodesic on Y, then f is the only harmonic mapping in its homotopy class.
THEOREM 5. THEOREM 5. — Let M be compact and of non-positive sectional curvature which is strictly negative at some point. Then the group of…
THEOREM 5. — Let M be compact and of non-positive sectional curvature which is strictly negative at some point. Then the group of isometrics ofM is finite, and no two of its elements are homotopic. For the isometrics are harmonic mappings and are therefore unique in their homotopy classes. The finiteness of the group of isometrics is known from Bochner [5]; but it is also an immediate consequence of the present argument, since obviously the group must be compact and discrete. The conclusion of t
THEOREM 6. THEOREM 6. — Let f: M —> Y be a harmonic mapping, and let V be a complete, totally geodesic submanifold ofY. If an open set ofM is mapped…
THEOREM 6. — Let f : M —> Y be a harmonic mapping, and let V be a complete, totally geodesic submanifold ofY. If an open set ofM is mapped into V, then all ofM is mapped into V.
THEOREM 7. THEOREM 7. - Let dim M = 2 and let f: M —> Y be a harmonic immersion. Denoting the image by V, we assume that the metric carriedbyffrom M…
THEOREM 7. - Let dim M = 2 and let f : M —> Y be a harmonic immersion. Denoting the image by V, we assume that the metric carriedbyffrom M to V is conformally equivalent to the metric induced on Vfrom Y. Then the curvature o/V at any points is ^ the Rieman- nian curvature o/Y at that point on the 1-dimensional section defined by V.
THEOREM 8. THEOREM 8. — Letf: M —> Y be a harmonic Riemannian embedding (or immersion), and again call the image V. IfY has Riemannian curvature ^ 0…
THEOREM 8. — Letf : M —> Y be a harmonic Riemannian embedding (or immersion), and again call the image V. IfY has Riemannian curvature ^ 0 in all directions at a point o/V, then V has Ricci curvature ^ 0 at that point. ANNALES SCIENTIFIQUES DE L'ECOLE NORMALE SUPERIEURE
THEOREM 9. THEOREM 9. — The space derivatives o// of all orders are of class C1 in s. Here we understand that we take / = ^ in the case M = circle. If…
THEOREM 9. — The space derivatives o// of all orders are of class C1 in s. Here we understand that we take / = ^ in the case M = circle. If we differentiate (4) with respect to s, we obtain a new equation for the derivative i;' = D^; and the first two terms are the same except that!; is replaced by i;'. It follows that y is C°° in the space variables. The foregoing analysis can be repeated, with only minor changes. In this way it is not difficult to show that/ is in fact of class C°° in all argu
THEOREM 10. THEOREM 10. — Let p dz dz and p' dz dz be two conformal metrics on the compact Riemann surface X, and suppose that they have the same…
THEOREM 10. — Let p dz dz and p' dz dz be two conformal metrics on the compact Riemann surface X, and suppose that they have the same curvature x ^ 0 (x ^ 0). Then they are equal.
PROPOSITION 1. PROPOSITION 1. — IfX and Y have the same genus, and if deg/= 1, then ^f > 0 everywhere. Our original proof was restricted to x < 0. The…
PROPOSITION 1. — IfX and Y have the same genus, and if deg/= 1, then ^f > 0 everywhere. Our original proof was restricted to x < 0. The fundamental step here is the index theorem of Eells-Wood [10]: Let Jf have zeroes at p^, ..., /?,.. Then there are integers n, > 0 such that zf^Jf + 0 at p^ (i = 1, ..., r), z, being a local uniformizing parameter at p^ According to [10], Proposition 1, n^ + ... +n, = 0. Hence r = 0. Q.E.D. We now look at the Jacobian / . If f ^ 0 in a region, i. e. 50 ^ Jf, the
PROPOSITION 2. PROPOSITION 2. - Under the hypotheses of Proposition 1, ifK ^ 0, ^A^ ^ ^ 0 wz X. Proo/ - By (20), A log (H/L) ^ 0 wherever / ^ 0, and so…
PROPOSITION 2. - Under the hypotheses of Proposition 1, ifK ^ 0, ^A^ ^ ^ 0 wz X. Proo/ - By (20), A log (H/L) ^ 0 wherever / ^ 0, and so log (^f/J^f) is superharmonic in any region where / ^ 0. Therefore it is not possible to have / < 0 at a point of X. Q.E.D. Combining our results, we shall obtain:
THEOREM 11. THEOREM 11. — Iff: X —» Y is a harmonic mapping of degree 1 of compact surfaces of the same genus, and if the curvature ofY is ^ 0, then…
THEOREM 11. — Iff : X —» Y is a harmonic mapping of degree 1 of compact surfaces of the same genus, and if the curvature ofY is ^ 0, then fis a diffeomorphism. This follows directly from Proposition 2 and from fundamental local results of Wood [29]. Let us turn briefly to the situation when X has genus 1. Then on X, up to a constant multiple, there is but one holomorphic quadratic differential, and it vanishes nowhere. Thus, if/: X -^ Y is harmonic (Y of any genus), then either/is ± holomorphic,
THEOREM 12. THEOREM 12. - Let g andg' be two metrics on X of curvature -1. Iff: X —> X and "0 " ^X^y -» X^, are harmonic mappings of degree 1 which…
THEOREM 12. - Let g andg' be two metrics on X of curvature -1. Iff: X —> X and "0 " ^X^y -» X^, are harmonic mappings of degree 1 which induce the same quadratic differen- tial on Xg , then Xg and Xg, are conformally equivalent. ANNALES SCDBNTIFIQUES DE L'ECOLE NORMALE SUPERIEURE
THEOREM 12. THEOREM 12. — Let Y be a Kdhler manifold of sectional curvature < 0 everywhere; and let V, V be compact complex submanifolds ofY of…
THEOREM 12. — Let Y be a Kdhler manifold of sectional curvature < 0 everywhere; and let V, V be compact complex submanifolds ofY of dimension > 0. Suppose that there is a holomorphic mapping f : V —> V which is homotopic in Y to the identity mapping V —> Y. Then V = V andf = identity.
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