I just considered reading this paper https://arxiv.org/pdf/1505.01994v1.pdf about spherical metrics. One of the first conditions to the metric is that it has "constant curvature 1".
Now I am asking whether someone could please explain to me how a metric can have a curvature since it is just an inner product on the manifold. Does it somehow induce a curvature in that context?
What does it mean for a metric to have a curvature?
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riemannian-geometry
curvature
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6https://en.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds. Sorry, but a full answer would take a full course about differentiable manifolds, so it's simply not feasible as a Math.SE answer. – 2017-01-02
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1The comment of @Alex is the best answer you're going to get here, but here's just one further comment on your question. A metric is not "just an inner product on a the manifold". Instead a (Riemannian) metric $g$ on a manifold $M$ is a separate inner product $g_p$ given on the tangent space $T_p M$ to each point $p$ of the manifold $M$; furthermore that metric $g_p$ varies continuously as a function of $p$, when expressed in any coordinate chart of $M$. That's a lot of structure. The study of that structure is what Riemannian geometry is all about, with curvature as a central concept. – 2017-01-02
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0@AlexM. The article doesn't answer my question at all, but thanks for your try. – 2017-01-02
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0@LeeMosher That are all things I kinda knew, but in every sense, how can a metric **itself** be curved? I never heard about that. – 2017-01-02
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0The metric determines a geometry, if you say the metric is curved you "really" mean the geometry determined by it is curved. So curvature $1$ means the Ricci scalar is constant $1$. – 2017-01-02
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0@s.harp **That** is the klind of information I wanted to gain. Thanks for it. – 2017-01-02