Mathematics

V. A. Toponogov

The Dependence Between Curvature and the Topological Structure of Riemannian Spaces of Even Dimension

(Presented by Academician P. S. Aleksandrov on 7 IV 1960)

Rauch (¹) proved the following theorem:

Theorem R. Let (M) be a complete (n)-dimensional Riemannian space. If the curvature (K) of the space (M) at every point and in every two-dimensional direction satisfies the inequality (0<hL<K\leq L), where (h\approx 0.74) is the solution of the equation (\sin \pi\sqrt{h}=\sqrt{h}/2), then (M) is homeomorphic either to a sphere or to an elliptic space.

Klingenberg (²) for even (n) strengthened Rauch’s theorem, proving the same assertion for (h\approx 0.54) ((h) is the solution of the equation (\sin \pi\sqrt{h}=\sqrt{h})). In the present paper, for even (n), the value of the constant (h) in Theorem R is lowered to (1/4). In (³) it is shown that this value of (h) cannot be decreased further.

Thus, for even dimensions, the problem posed by Rauch may be regarded as completely solved. Let us also note that the assertion of the theorem of Bochner–Yano ((⁴), p. 67, Theorem 5.1) for even dimension is also covered by our theorem.

We shall prove two lemmas, which are also of independent interest. Consider, in a Riemannian space, triangles composed of two shortest arcs and one geodesic line. If the sum of the lengths of any two sides of a triangle is greater than the third, we shall call such a triangle proper.

Lemma 1. If in a complete Riemannian space (M) the curvature (K) at every point and in every two-dimensional direction is not less than (H), and if the perimeter of a proper triangle (ABC), composed of shortest arcs (AB), (AC) and a geodesic (BC), is equal to (2\pi/\sqrt{H}), then the angles (B) and (C) are equal to (\pi).

Proof. Divide the geodesic (BC) by points (E_p) ((p=0,\ldots,s)) ((E_0=B,\ E_s=C)) in such a way that the segments (E_{i-1}E_i) ((i=1,\ldots,s)) are shortest arcs. In the space (M) construct the triangles (AE_{i-1}E_i), and on the sphere (S_H) of radius (1/\sqrt{H}) construct the triangles ((AE_{i-1}E_i)H) with the same side lengths as the triangles (AEE_i)}E_i). Placing next to one another, along the equal side (AE_i), the triangles ((AE_{i-1H) and ((AE_iE), it is either a great circle or a digon; in both cases the angles (B) and (C) are equal to (\pi). Applying again the theorem on the angles of triangles in Riemannian spaces ((⁵), p. 127), we obtain the proof of Lemma 1.})_H), we obtain on the sphere (S_H) a polygon (A'E'_0\ldots E'_i\ldots E'_s). Relying on the theorem on the angles of triangles composed of shortest arcs ((⁵), p. 127), it is not difficult to show that all the angles of this polygon are not greater than (\pi). But since the perimeter of the polygon (A'E'_0\ldots E'_i\ldots E'_s) is equal to (2\pi/\sqrt{H

Lemma 2. Under the conditions of Lemma 1, the perimeter of a proper triangle does not exceed (2\pi/\sqrt{H}).

Proof. Suppose the contrary. Let the perimeter of the regular triangle (ABC) be equal to (2a), (a>\pi/\sqrt{H}). Take on the side (BC) a point (D) such that the triangle (ACD) is regular and such that its perimeter, equal to (2b), is less than (2a), but not less than (2\pi/\sqrt{H}). Since (\pi^{2}/b^{2}) is not greater than (H), Lemma 1 can be applied to the triangle (ACD). Hence we obtain that the angle (D) is equal to (\pi), but then the shortest path (AD) will go along the line (ABD), and, consequently, the perimeters of the triangles (ACD) and (ACB) coincide, contrary to the choice of the point (D). The contradiction obtained proves Lemma 2.

For what follows we shall need a theorem proved in ((^{2})).

Theorem K. If the curvature (K) of a compact simply connected Riemannian space is, at every point and in every two-dimensional direction, strictly positive and not greater than (L), then every geodesic segment of length (\pi/\sqrt{L}) is a shortest path.

Now we can prove our theorem:

Theorem. Let (M) be a complete Riemannian space of even dimension. If the curvature (K) of the space (M) at every point and in every two-dimensional direction satisfies the inequality (0<{}^{1}/_{4}\,L<K\leq L), then (M) is homeomorphic either to a sphere or to an elliptic space.

Proof. By the conditions of the theorem, the space (M) is compact and either simply connected itself, or its two-sheeted covering is simply connected ((^{6})). In the first case we shall consider (M) itself, and in the second, its two-sheeted covering.

Thus, we shall always consider a compact simply connected Riemannian space.

Denote by ({}^{1}/_{4}H) the minimum of the curvature of the space (M); (H>L), (H<4L). Choose the number (\rho) so that (\rho<\pi/\sqrt{L}) and (\rho>\pi/\sqrt{H}). On some geodesic lay off a segment (AB) of length (2\rho). From Theorem K and from the choice of the number (\rho) it follows that the points (A) and (B) are distinct.

(\alpha)) Consider a segment (AE) of a geodesic (q) with origin at the point (A) and of length (\rho). By Lemma 2, the distance from the point (E) to (B) does not exceed (\rho=AE); therefore on the segment (AE) there exists a point (D(q)) equidistant from (A) and (B). It is not difficult to show that the point (D(q)) is unique. An analogous assertion holds for geodesics passing through the point (B).

(\beta)) Denote by (C(A)), respectively (C(B)), respectively (C(A,B)), the set of points (P) of the space (M) whose distances to the points (A) and (B) satisfy the relations (PA<PB), respectively (PB<PA), respectively (PA=PB). From the assertion of item (\alpha)) and from Theorem K it follows that any point (P\in C(A)) lies inside the segment (AD(q)) of a geodesic (q) with origin at the point (A) and length not exceeding (\rho); consequently, every point of the set (C(A)) is joined with (A) by a unique shortest path.

An analogous assertion also holds for points of the set (C(B)).

(\gamma)) We now define a homeomorphic mapping (\varphi(P)) of the space (M) onto the sphere (S_R) of radius (R). Let (A') be an arbitrary point of (S_R); by (B') denote the point diametrically opposite to (A'). At the points (A) and (A') construct orthonormal frames (T) and (T'). Put the vectors of the frames (T) and (T') in one-to-one correspondence. To each geodesic (q) of the space (M) passing through the point (A) put in correspondence the geodesic (q') of the sphere (S_2), passing through the point (A') and making the same angles with the vectors of the frame (T') as the geodesic (q) makes with the vectors of the frame (T). Denote by (D'(q')) the point of the geodesic (q') lying in the equatorial plane. The segment (AD(q)) of the geodesic (q) we map onto the segment (A'D'(q')) of the geodesic (q'), assigning to the point (P\in AD(q)) the point (P'\in A'D'(q')), at a distance from the point (A') equal to (\pi R\cdot AP/2AD(q)). Further, to the shortest path (D(q)B) we put in correspondence the shortest path (D'(q')B'), and to the point (P\in D(q)B)—the point (P'\in D'(q')B'), at a distance

from (B') at a distance (\pi R\cdot PB/2D(q)B). From item (\beta)), the choice of the number (\rho), and Theorem K it follows that the mapping (\varphi(P)) thus obtained is a homeomorphism of the space (M) onto the sphere (S_R). The theorem is proved.

Institute of Radiophysics and Electronics
Siberian Branch
of the Academy of Sciences of the USSR

Received
1 IV 1960

References

1. H. E. Rauch, Ann. of Math., 54, 38 (1951).
2. V. Klingenberg, Ann. of Math., 69, 654 (1959).
3. S. Bochner, Bull. Am. Math. Soc., 53, 778 (1947).
4. K. Yano, S. Bochner, Curvature and Betti Numbers, 1957.
5. V. A. Toponogov, UMN, 14, 1 (85), 87 (1959).
6. J. L. Synge, Quart. J. Math., ser. 7, 316 (1936).