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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?

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2 Answers 2

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – trisct
    Commented Jul 5, 2019 at 8:41
  • $\begingroup$ It is a linear operator, defined on the space of smooth functions on the manifold. $\endgroup$ Commented Jul 5, 2019 at 8:45
  • $\begingroup$ So $E$ is a vector space formally generated by these linear opeartors? $\endgroup$
    – trisct
    Commented Jul 5, 2019 at 8:47
  • $\begingroup$ Yes, but it is not just formal. E is a vector space spanned by these operators. $\endgroup$ Commented Jul 5, 2019 at 8:48
  • $\begingroup$ @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? $\endgroup$
    – user56202
    Commented Oct 4, 2023 at 23:17

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