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Michael Hardy
Michael Hardy
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We consider a Riemannian homogeneous space $(R\times S^n, g)$. Suppose that the Lie algebra of the Killing fields has a natural splitting (compatible with the product) as $isom(R)\oplus isom(S^n)$$\operatorname{isom}(R)\oplus \operatorname{isom}(S^n)$. Can we claim that the homogeneous Riemannian metric has a natural splitting $g = dt^2 \oplus h$, where $h$ is a homogeneous metric on $S^n$?

We consider a Riemannian homogeneous space $(R\times S^n, g)$. Suppose that the Lie algebra of the Killing fields has a natural splitting (compatible with the product) as $isom(R)\oplus isom(S^n)$. Can we claim that the homogeneous Riemannian metric has a natural splitting $g = dt^2 \oplus h$, where $h$ is a homogeneous metric on $S^n$?

We consider a Riemannian homogeneous space $(R\times S^n, g)$. Suppose that the Lie algebra of the Killing fields has a natural splitting (compatible with the product) as $\operatorname{isom}(R)\oplus \operatorname{isom}(S^n)$. Can we claim that the homogeneous Riemannian metric has a natural splitting $g = dt^2 \oplus h$, where $h$ is a homogeneous metric on $S^n$?

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ZZZ
ZZZ
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Homogeneous Riemannian metrics

We consider a Riemannian homogeneous space $(R\times S^n, g)$. Suppose that the Lie algebra of the Killing fields has a natural splitting (compatible with the product) as $isom(R)\oplus isom(S^n)$. Can we claim that the homogeneous Riemannian metric has a natural splitting $g = dt^2 \oplus h$, where $h$ is a homogeneous metric on $S^n$?