# The notation $\frac{\partial}{\partial x}$

In Jost's Riemannian Geometry and Geometric Analysis (Sect. 1.2, Chap. 1), the tangent space at a point $$x_0$$ in $$\mathbb{R}^d$$ is defined as $$T_{x_0}\mathbb R^d=\{x_0\}\times E$$ where $$E$$ is the vector space spanned by $$\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^d}$$. Then the books says: "Here, $$\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^d}$$ are the partial derivatives at the point $$x_0$$." This is where I get confused. They are the partial derivatives of what? The only partial derivative I know is that of a function, but no function is given here.

Sure, if one wants to argue that $$\frac{\partial}{\partial x^1},\cdots,\frac{\partial}{\partial x^d}$$ are just formal notations here that don't mean anything other than a formal basis of $$E$$, then I can accept that even though I have doubts. But then there comes something that confuses me even more. If $$f:\mathbb R^d\to\mathbb R^c$$ is a differentiable map, then the derivative of $$f$$ at $$x_0$$ is defined to be (Einstein convention is used below) $$df(x_0):T_{x_0}\mathbb R^d\to T_{x_0}\mathbb{R}^c\\ \quad v^i\frac{\partial}{\partial x^i}\mapsto v^i\frac{\partial f^j}{\partial x^i}\frac{\partial}{\partial f^i}$$ So apparently $$\frac{\partial}{\partial f^j}$$ here depend on $$f$$ and are not arbitraily selected, so the notation cannot simply be a formal one, which brings me back to the original question: what does $$\frac{\partial}{\partial x^i}$$ and $$\frac{\partial}{\partial f^j}$$ mean?

The tangent space $$T_pM$$ can be seen as the space of local linear operators acting on the functions $$f: M \rightarrow \mathbb R$$. If you have vector $$v\in T_pM$$ you can define how it acts on a function: $$v(f) = \left.\frac{d f(\gamma_v(t))}{dt}\right|_{t=0}$$ where $$\gamma_v$$ is any curve on $$M$$ such that $$\gamma_v(0) = p$$ and $$\frac{d\gamma_v}{dt}(0) = v$$.

Given a coordinate system $$(x_i)$$ you can find that there exist vectors in $$T_pM$$ that act on functions exactly like the partial derivatives $$\frac{\partial}{\partial x_i}$$, that is $$v_i(f) = \frac{\partial f}{\partial x_i}(p)$$. They are therefore denoted $$v_i = \frac{\partial}{\partial x_i}$$. Such vectors form a basis of $$T_pM$$, so any vector can be written as $$v = v^i \frac{\partial}{\partial x_i}$$

If the point of differentiation is obvious, the vetor can be denoted as $$v_i=\left.\frac{\partial}{\partial x_i}\right|_p$$. Sometimes $$\frac{\partial}{\partial x_i}$$ can also denote the whole vector field, defining a vector at every point of the manifold.

The notation $$\frac{\partial f}{\partial x_j}$$ means: take $$f$$, which is a function on the manifold $$M$$, consider its expression in the coordinate system $$(x_1, x_2, \ldots, x_n)$$, so that $$f$$ is now a function on $$\mathbb R^n$$, take the partial derivative with respect to the $$j$$-th variable.

The part in bold is often not said explicitly and that can become confusing.

• I know what $\frac{\partial f}{\partial x_j}$ means, it is a number. But now I am curious about what $\frac{\partial}{\partial x_j}$ means, and apparently it is not a number nor a operator in my context. Commented Jul 5, 2019 at 8:41
• It is a linear operator, defined on the space of smooth functions on the manifold. Commented Jul 5, 2019 at 8:45
• So $E$ is a vector space formally generated by these linear opeartors? Commented Jul 5, 2019 at 8:47
• Yes, but it is not just formal. E is a vector space spanned by these operators. Commented Jul 5, 2019 at 8:48
• @GiuseppeNegro So if $f$ is a function on the manifold and $X$ is a chart, are you saying that $\partial / \partial x_j$ takes $f$ as an input, or $f \circ X$ as an input? Commented Oct 4, 2023 at 23:17