# Understanding a point of the example that the Euclidean metric is a Riemannian metric

A hint that many geometers give for people who start in Riemannian Geometry is associate the definitions of the course of Differential Geometry of curves and surfaces on $$\mathbb{R}^3$$ with the respectively definition in Riemannian Geometry and observe that the definitions in Riemannian Geometry are just generalizations of the definitions of the Differential Geometry of curves and surfaces on $$\mathbb{R}^3$$. Following this hint and based on that I did a course of Differential Geometry of curves and surfaces on $$\mathbb{R}^3$$ by the Do Carmo's book, I'm trying understand how the Riemannian metric generalizes of the First Fundamental Form.

Let be $$M^n$$ a $$n$$-differentiable manifold. In this case is just consider $$g_{ij} = \delta_{ij}$$ for each $$i, j = 1, \cdots, n$$ and $$\frac{\partial}{\partial x_i} = e_i$$ for each $$i = 1, \cdots, n$$, considering $$e_i$$ as the elements of the canonical basis of the $$\mathbb{R}^n$$ and $$\frac{\partial}{\partial x_i}$$ the elements of the coordinate basis of $$T_pM$$ for some $$p \in M^n$$ fixed. My doubt is how can I interpret the equality $$\frac{\partial}{\partial x_i} = e_i$$? Because it doesn't seem that this equality it's true since $$\frac{\partial}{\partial x_i}$$ is a derivation defined at $$p$$, i.e., a linear operator which take a real function defined on $$M$$, which is differentiable on $$p \in M$$, and returns $$\frac{\partial f}{\partial x_i} \in \mathbb{R}$$, which is not an element of $$\mathbb{R}^n$$ for every $$f$$ that the operator can take, while $$e_i \in \mathbb{R}^n$$. Is $$\frac{\partial}{\partial x_i} = e_i$$ considering the trivial isomorphism between the space of derivations in $$p$$ and $$\mathbb{R}^n$$ or is just an abuse of notation for something? If it is an abuse of notation, can you explain what is the abuse of notation? I really didn't understand why this equality it's true.

Yes, the identification $$\left.\frac{d}{dx^i}\right\vert_p \leftrightarrow e_i$$ is made exactly via the canonical isomorphism $$T_p \Bbb R^n \leftrightarrow \Bbb R^n$$ defined for each $$p \in \Bbb R$$ (what you called the "trivial isomorphism"). This isomorphism is recorded, for example, as Proposition 3.2 (and immediately proved) in Lee's Introduction to Smooth Manifolds (at least in the first edition).
In fact, such a canonical isomorphism exists if we replace $$\Bbb R^n$$ with any real or complex finite-dimensional vector space, but NB there is no analogue of this map for general smooth manifolds.
• What you mean when you said "there is no analogue of this map for general smooth manifolds"? Because the isomorphism would be between $T_pM^n$ and $\mathbb{R}^n$ and the canonical isomorphism must exist if we do the identification as you did in the beginning of the answer. – George Apr 11 at 18:35
• I mean exactly that: For a general manifold $M$ and $p \in M$ there is no preferred identification $T_p M \leftrightarrow \Bbb R^n$ (where $n := \dim M$). – Travis Apr 11 at 20:16