Distance difference functions on nonconvex boundaries of Riemannian manifolds. (English. Russian original) Zbl 1491.53044
St. Petersbg. Math. J. 33, No. 1, 57-64 (2022); translation from Algebra Anal. 33, No. 1, 81-92 (2021).
Suppose \(M\) is a complete Riemannian manifold with non-empty boundary \(F=\partial M\) and with arc-length distance \(d_M:M\times M\rightarrow \mathbb{R}\). For each \(x\in M\), the author considers the distance difference function \(D_x:F\times F\rightarrow \mathbb{R}\) defined by \(D_x(y,z) = d_M(x,y)- d_m(x,z)\), which is then collected into a subset \(\mathcal{D}(M) = \{D_x: x\in M\}\subseteq C(F\times F)\) of the continuous functions on \(F\times F\)
If \((M,g)\) and \((M',g')\) are Riemannian manifolds both with boundary homeomorphic to \(F\), a choice of homeomorphism allows one to view both \(\mathcal{D}(M)\) and \(\mathcal{D}'(M')\) as subsets of \(C(F\times F)\). The main result of the article is that if \(\mathcal{D}(M) = \mathcal{D}'(M')\), then \(M\) and \(M'\) are isometric by an isometry which fixes \(F\) pointwise.
Prior results include establishing an analogous result in the boundaryless case where \(F\) is replaced by an arbitrary open subset of \(M\) [M. Lassas and T. Saksala, Asian J. Math. 23, No. 2, 173–200 (2019; Zbl 1419.53038); S. Ivanov, Geom. Dedicata 207, 167–192 (2020; Zbl 1478.53068)]. In the case of non-empty boundary, an analogous result was established [M. V. de Hoop and T. Saksala, J. Geom. Anal. 29, No. 4, 3308–3327 (2019; Zbl 1428.53047)] under certain geometric assumptions on \(F\). In the article under review, no assumptions on the boundary are needed.
The idea of the proof is as follows. In [S. Ivanov, Geom. Dedicata 207, 167–192 (2020; Zbl 1478.53068)], the author had already established that the map \(\mathcal{D}\) is a locally bi-Lipschitz homeomorphism onto its image. Then, since \(\mathcal{D}(M) = \mathcal{D}'(M')\) one obtains a map \(\phi = (\mathcal{D}')^{-1}\circ \mathcal{D}:M\rightarrow M'\) which is a locally bi-Lipschitz homeomorphism.
The goal then becomes to show that \(\phi\) is a Riemannian isometry. At any point \(p\) where \(\phi\) is differentiable the author establishes this by exploiting the geometry of shortest paths from \(p\) to \(F\). As the set of differentiable, points is dense in \(M\), this is sufficient to prove that \(\phi\) is a metric isometry. Finally, one applies the classical Myers-Steenrod result that a metric isometry is a Riemannian isometry.
If \((M,g)\) and \((M',g')\) are Riemannian manifolds both with boundary homeomorphic to \(F\), a choice of homeomorphism allows one to view both \(\mathcal{D}(M)\) and \(\mathcal{D}'(M')\) as subsets of \(C(F\times F)\). The main result of the article is that if \(\mathcal{D}(M) = \mathcal{D}'(M')\), then \(M\) and \(M'\) are isometric by an isometry which fixes \(F\) pointwise.
Prior results include establishing an analogous result in the boundaryless case where \(F\) is replaced by an arbitrary open subset of \(M\) [M. Lassas and T. Saksala, Asian J. Math. 23, No. 2, 173–200 (2019; Zbl 1419.53038); S. Ivanov, Geom. Dedicata 207, 167–192 (2020; Zbl 1478.53068)]. In the case of non-empty boundary, an analogous result was established [M. V. de Hoop and T. Saksala, J. Geom. Anal. 29, No. 4, 3308–3327 (2019; Zbl 1428.53047)] under certain geometric assumptions on \(F\). In the article under review, no assumptions on the boundary are needed.
The idea of the proof is as follows. In [S. Ivanov, Geom. Dedicata 207, 167–192 (2020; Zbl 1478.53068)], the author had already established that the map \(\mathcal{D}\) is a locally bi-Lipschitz homeomorphism onto its image. Then, since \(\mathcal{D}(M) = \mathcal{D}'(M')\) one obtains a map \(\phi = (\mathcal{D}')^{-1}\circ \mathcal{D}:M\rightarrow M'\) which is a locally bi-Lipschitz homeomorphism.
The goal then becomes to show that \(\phi\) is a Riemannian isometry. At any point \(p\) where \(\phi\) is differentiable the author establishes this by exploiting the geometry of shortest paths from \(p\) to \(F\). As the set of differentiable, points is dense in \(M\), this is sufficient to prove that \(\phi\) is a metric isometry. Finally, one applies the classical Myers-Steenrod result that a metric isometry is a Riemannian isometry.
Reviewer: Jason DeVito (Martin)
MSC:
| 53C20 | Global Riemannian geometry, including pinching |
References:
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