# An example of smooth but not Riemannian [duplicate]

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

• Every smooth manifold admits a Riemannian metric. The proof uses partitions of unity. – 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 $$M$$