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I've been trying to understand the difference between the notion of smooth (which I understand well) and Riemannian (which I am newly acquainted with).

The definition in the tag for 'riemannian-geometry' is as follows: Riemannian manifolds are smooth manifolds with an inner product smoothly attached to the tangent space of each point.

If $X$ is a a smooth $n$ manifold, then we always have $T_p(X)\cong\mathbb{R}^n$ (as vector spaces) which therefore, by pulling back pushing forward and borrowing the dot product on $\mathbb{R}^n$, it follows that $T_p(X)$ has perfectly good inner product. But, I am supposed to believe that this does not always vary smoothly with $p$. I'm having trouble believing this. Can someone show me a relatively simple manifold which is smooth but not Riemannian?


marked as duplicate by Community Feb 4 at 19:47

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    $\begingroup$ Every smooth manifold admits a Riemannian metric. The proof uses partitions of unity. $\endgroup$ – Max Feb 4 at 19:08

Every smooth manifold can be a Riemannian manifold because there exists always a metric tensor for the manifold.

A Riemannian Manifold M is a smooth manifold with an adjoint structure, a (0,2)-tensor

$g: M\to T^*M\otimes T^*(M)$

Such that for every $p\in M$

$g(p): T_p(M)\times T_p(M)\to \mathbb{R}$

is a metric on $T_p(M)$

with the theory of partition of unity it is possible to prove that there exists always a metric tensor on $M$


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