Does a Riemannian metric allow definition of a tangent vector's length?

Question

In Euclidean spaces, we define the Euclidean norm of a vector $\vec{x} = (x_1,x_2,...x_n)$ as $\|\vec{x}\|:=\sqrt{x_1^2+x_2^2+ \cdots +x_n^2 }$

Does the metric tensor field of a Riemannian manifold allow us to establish something similar to a norm in Euclidean space? If so, why is this notion of length preserved from point to point on the manifold.

REMARK: By the definition of a differentiable manifold, I understand that a manifold is locally homeomorphic to $\Bbb{R}^{n}$. But if I take the Euclidean norm of a tangent vector at a point $p$ on the manifold, shouldn't its norm only be valid in the tangent space $T_{p}M$?

Answer 1

If you have a Riemannian manifold then at each point on the manifold you have a matrix $\Sigma$ which can be used to represent the Riemannian metric as a quadratic form. You can define the length of a tangent vector $x$ as $$\sqrt{x^T \Sigma x}$$ In Euclidean space the matrix $\Sigma$ is just the identity matrix. Remember that from point to point the metric must vary smoothly so the definition will carry across the manifold in a nice way. Remember in a smooth manifold (which a Riemannian manifolds is) the overlap maps must be smooth. If the metric does not vary smoothly then it will not behave well with the charts. Essentially we want the lengths of tangent vectors in near by tangent spaces be similar, we don't want a discontinuous jump in length between two near by tangent vectors. The smooth variation of the metric across tangent spaces ensures this.

Answer 2

By definition, a tangent vector at a point $p$ exists only in the tangent space $T_pM$.

What you get from the manifold being Riemannian, however, is exactly a norm on each tangent space (which makes the norms of neighboring tangent spaces vary in a nice and "sufficiently smooth" way compatible with the charts). That's what a Riemannian metric is.

So you can certainly speak about the norm of a tangent vector in a Riemannian manifold -- that's what makes it Riemannian rather than a generic differentiable manifold.