I have been reading a book about Riemannian geometry and at some point the author says that since the metric of the manifold defines an inner product on $T_pM$ we can think of $T_pM$ as the Euclidean n-space $\mathbb{R}^n$. When he says what does he really mean, is there some kinda of isomorphism or something?.Thanks in advance.
Identifying the tangent space with $R^n$
differential-geometryriemannian-geometry
Best Answer
Yes, there is nothing deep here; the vector space $T_p M$ has dimension $n$, thus you can identify it with $\mathbb R^n$. Concretely, if you have a coordinate system $x_1, x_2, \ldots, x_n$, then you have a basis of $T_p M$ given by $$\tag{1} \frac{\partial}{\partial x_1}, \ldots, \frac{\partial }{\partial x_n}; $$ so you can identify the vector $$ v_1\frac{\partial}{\partial x_1}+\ldots + v_n\frac{\partial}{\partial x_n} $$ with the $n$-uple $$ (v_1, v_2, \ldots, v_n)\in \mathbb R^n.$$ Of course, if you change coordinates, the identification will change accordingly.
Steffen remarks in comments that this is only a algebraic isomorphism and that it needs not preserve the scalar product. This means that it needs NOT be the case that $$ \langle v, w\rangle_{T_p M}=\sum_{j=1}^n v_jw_j.$$ For this to be true, the basis of $T_p M$ should be orthonormal, and there is no guarantee that (1) is. However, as in any scalar product space, given an arbitrary basis we can always construct an orthonormal one via the Gram-Schmidt algorithm. (This orthonormal basis might not come from a coordinate system, though).