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Let $(M, g)$ be a Riemannian manifold. Standard definitions of a Riemannian metric $g$ states that $g$ specifies a symmetric, bilinear, positive definite form on each tangent space $T_{p}M$ that varies smoothly with $p$. Why do we not require $g$ to satisfy the triangle inequality in our definition of a Riemannian metric? Is it simply because all we really need to study geometry is an inner product structure which induces a distance function that satisfies the usual definition of a metric?

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    $\begingroup$ How would it even make sense for "$g$ to satisfy the triangle inequality"? It's a completely different type of object than the kind of metric in metric spaces. $\endgroup$ – user98602 Oct 15 '15 at 0:12
  • $\begingroup$ That $g(v,u) \leq g(v,w) + g(w,u)$ for all $v,u,w \in T_{p}M$. $\endgroup$ – user265669 Oct 15 '15 at 1:57
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    $\begingroup$ That's not true for any bilinear form on a vector space. See if you can prove why. $\endgroup$ – user98602 Oct 15 '15 at 2:16
  • $\begingroup$ Ah, I see why. Thanks. $\endgroup$ – user265669 Oct 15 '15 at 3:22

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