$\frac{\partial}{\partial_j}$ is a basis of the tangent space

Question

Let 's say we have a function :

$$f:E \to V$$

where $E$ and $V$ are vector spaces.

Now, what I don't understand is the fact that $\frac{\partial}{\partial x_j}$ is a basis of the tangent space of $f$.

According to me, $\frac{\partial}{\partial x_j}$ is a special operator to whom we give a function $f$, and we get back the derivative of $f(\cdot, \ldots, x_j, \ldots)$.

So what does it mean : $\frac{\partial}{\partial x_j}$ when we don't give this operator a function ?

Answer 1

You can think of the tangent space at a point as the space of linear derivations of smooth functions at that point. In other words, they are objects that map smooth functions to $\mathbb{R}$ while satisfying some type of product rule you’re familiar with from calculus. So a basis for the tangent space consists of linearly dependent derivations. It just so happens that in $\mathbb{R}^n$, the partial derivative operators evaluated at that point give you such derivations. So in a sense, the tangent space is a space of functions of functions.

Answer 2

You can define a vector as a real-valued linear map acting on functions defined (at least) locally on your $E$. More precisely, if $\mathcal{F}(E,p)$ is the space of smooth real-valued functions defined on $E$ around a point $p$, then a vector in $T_pE$ is a map $$X_p\colon \mathcal{F}(E,p) \to \mathbb{R}: f \mapsto X_p(f)$$ where $X_p(f) = \frac{d}{dt}f(x(t))_{|t=0}$ and $(-\varepsilon,\varepsilon) \ni t \mapsto x(t)$ is a smooth curve on $E$ such that $x(0) = p$ and $x'(0) = X_p$. Observe that in the last equation, $X_p$ must be thought of as a vector in the original sense, which you probably already know.

With this definition, you can check that $X_p$ is a derivation, that is $$X_p(fg) = X_p(f)g(p)+f(p)X_p(g)$$ and that the $\frac{\partial}{\partial x^i}_{|p}$ give a basis of $T_pE$. You can extend all of this using $V$-valued functions. You can find this definition and the results I mentioned in any book about Differential geometry. I recommend Kobayashi and Nomizu - Foundations of Differential Geometry.

Answer 3

Just to complete the other answers: Vector spaces over a field $\mathbb K$ can be endowed with a differentiable structure, i.e., can be smooth manifolds. I agree with you when you say we don't give a function to the operator. However, the author of the book, and the other answers also, are considering $E$ and $V$ not as a vector spaces but as smooth manifolds. And it is as follows. Pick basis $\{E_i\}$ in $E$ and $\{v_i\}$ in $V$ and let $\{x^i\}$ and $\{y^i\}$ the dual basis ($x^i(e_j)=\delta_j^i$ and the same for $(y^i,v_j)$). Then, the pairs $(E,\varphi)$ and $(V,\psi)$, where

$$\varphi:E\rightarrow \mathbb K^n\\ a \mapsto \varphi(a) = (x^1(a),\dots,x^n(a))$$

and

$$\psi:V\rightarrow \mathbb K^m\\ c \mapsto \varphi(c) = (y^1(c),\dots,y^m(c))$$

are charts for the manifolds $E$ and $V$, resp. Now you can see the link provided by @counterxample.net, keeping in mind that you haven't only a vector space but also a manifold.

Finally, let me remark that $f:E\rightarrow V$ is a smooth map between manifolds, so the sentence the tangent space of $f$ doesn't make more sense to me. It is better to consider the explanation given by @Gibbs about the derivations of the algebra of functions at a point (vector in our case).