Why are the partial derivatives a basis of the tangent space?

Question

Why are $\frac{\partial }{\partial x_i}$ , $i=1,...,n$, a basis for $T_x\mathbb{R}^n$?

My understanding is that the tangent space at $x$ is the set of all vectors beginning at $x$. I would be led to believe that all vectors beginning at $x$ would have the form $\vec{v}-\vec{x}$.

My question is three fold:
1. Why are the basis vectors partial derivatives and not the vector above?
2. How do they span all vectors originating at $x$?
3. Are elements of the tangents space thus linear functions?

Answer 1

The tangent space is not the set of all vectors beginning at $x$. It is a different vector space that is "attached" to the original vector space at $x$, and the vectors in this space are not elements of the original vector space. You have one of these tangent spaces attached at each point, and the tangent bundle for a vector space is the space together with all of the attached tangent spaces.

The tangent space is viewed as the space of infinitesimal displacements starting at $x$, since it does not bleed into other points but in a sense does contain $x$. That is why they are identified as directional derivative operators.

Answer 2

You're right that on a plane, the tangent vectors at $x$ have the form $v-x$, but note that forming vectors by subtracting points only makes sense in Euclidean space. The points of an abstract manifold are not necessarily elements of a vector space, and if you embed the manifold and do subtraction in the ambient space, the vector between two points is not necessarily tangent to the surface (think of the sphere, say).

You could try to generalize the idea as follows: pick a point $v\neq x$ and draw the shortest curve possible that travels along the manifold from $x$ to $v$, and replace $v-x$ with the derivative of the curve at the starting point. This idea will lead to manifold exponential and logarithm maps. The issue is that defining "shortest possible" requires additional structure (e.g. a metric) beyond differential structure.

For this reason the tangent space is usually defined in terms of linear operators (as in Giuseppe's linked answer) or equivalence class of tangent vectors of curves through $x$.

Answer 3

To distinguish between points and tangent vectors, let $p=(p_1,...,p_n)\in \mathbb{R}^n$ a point of $\mathbb{R}^n$ and $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ a point of the tangent space $\mathbb{R}^n$.

The line through $p=(p_1,...,p_n)\in \mathbb{R}^n$ with direction $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ has parametrization $a(t)=(p_1+tv_1,...,p_n+tv_n)$.

If f is $C^\infty$ in a neighborhood of $p\in \mathbb{R}^n$ and $v$ is a tangent vector at $p$, define the directional derivative of $f$ in the direction of $v$ at $p$ as $$D_vf=\lim\limits_{t \to 0} \frac{f(a(t))-f(p)}{t}.$$

By the multi-variable chain rule, we have $$D_vf=\sum_{i=1}^{n} \frac{da^i}{dt}(0)\frac{\partial f}{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial f}{\partial x^i}(p).$$

Of course in the above notation $D_vf$, the partial derivatives are evaluated at $p$, since v is a vector at $p$. Now, we can define a map $D_v$ (which assigns to every f which is $C^\infty$ the real number $D_v(f)$ ) with the natural way $$D_v=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}\Bigr\rvert_{p}.$$

This map $D_v\in \mathcal{D}_p(\mathbb{R}^n)$ is in fact a derivation at $p$. Finally, you can show that the map \begin{align} \phi :T_p(\mathbb{R}^n) &\to \mathcal{D}_p(\mathbb{R}^n) \\ v &\to D_v \end{align}

is a linear isomorphism of vector spaces (for surjectivity, you can use Taylor's theorem).

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So the answers to your questions are:

3) Υes, you can see every tangent vector $v \in T_p(\mathbb{R}^n)$ as a derivation $D_v\in \mathcal{D}_p(\mathbb{R}^n)$ using the isomorphism $\phi$.

2) Since $e_1,...,e_n$ is the canonical basis of $T_p(\mathbb{R}^n)$ and $\phi$ is an isomorphism, then $\phi(e_1),...,\phi(e_n)$ is a basis of $D_v\in \mathcal{D}_p(\mathbb{R}^n)$. But $\phi(e_i)=\frac{\partial }{\partial x^i}\Bigr\rvert_{p}$, hence $\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$ is a basis of the tangent space $\mathcal{D}_p(\mathbb{R}^n)\simeq T_p(\mathbb{R}^n)$.

1) As a result, you can say that the basis of $T_p(\mathbb{R}^n)$ is $\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$. You can say all that because of $\phi$.

Answer 4

About the basis of $T_p M$;

Let $α:(-ε,ε)\to M$ a curve in differential manifold $M$ such that $α(0)=p\in M$. If $\mathscr D_p:=\{f:M\to \Bbb R |f$ differentiable at $p\}$, then we define the tangent space to $M$ at $p$ to be the set

$$T_p M:=\{α'(0):\mathscr D_p\to \Bbb R|f\mapsto \frac {d}{dt}(f\circ α)(0)\},$$

which is a linear space (the proof is easy).

We get

$$α'(0)f=\frac {d}{dt}(f\circ α)(0)=\nabla f(α(0))\cdot α'(0)=(\sum_i x_i'(0)\frac {\partial}{\partial x_i}|_0)f,$$ where $α(t)=(x_1(t), ...,x_n(t))$ and $x:U\to M, p\in U,$ a parametrization of $M$.

We just proved that function $α'(0)$ is a linear combination of $\{\frac {\partial}{\partial x_i}|_0\}_{i=1,...n}$, which means that $span\{\frac {\partial}{\partial x_i}|_0\}_{i=1,...n}=T_p M.$

We need to show that $\{\frac {\partial}{\partial x_i}|_0\}_{i=1,...n}$ is linear independent; thus for $λ_i\in \Bbb R$ and for $π_j:M\to \Bbb R:x\mapsto x_j$ we have :

$$\sum_iλ_i \frac {\partial}{\partial x_i}|_0=0\implies (\sum_iλ_i \frac {\partial}{\partial x_i}|_0)π_j=0\implies \sum_iλ_i δ_{ji}=0\implies λ_j=0.$$