Let $\{e^1,...,e^m\}$ standard basis of $(\Bbb R^m)^*$ , $(U,\varphi)$ a chart. Why is $d(x^i\circ \varphi^{-1})_{\varphi(p)}=d(e^i)_{\phi(p)}=e^i$?

Question

Let $\{e^1,...,e^m\}$ be the standard basis of $(\Bbb R^m)^*$ . Let $(U,\varphi)$ be a chart of a smooth manifold $M$ of dimM=m. Let $\varphi(p)=(x^1(p),...,x^m(p))$.

In the middle of some proof I found the following equalities that I am having trouble checking:

$d(x^i\circ \varphi^{-1})_{\varphi(p)}=d(e^i)_{\varphi(p)}=e^i$

So my question is why are these two last equalities true?

1. why is $x^i\circ \varphi^{-1}=e^i$ ?

2. why is $d(e^i)_{\varphi(p)}=e^i$ ? In this case shouldn't the differential of a constant function be 0?

Answer 1

1. $x^i$ is by definition the $i^{th}$ component function of the map $\phi$, i.e $x^i=\text{pr}^i\circ\phi$, where $\text{pr}^i:\Bbb{R}^n\to\Bbb{R}$ is the projection map. So, $x^i\circ\phi^{-1}$ is obviously just $\text{pr}^i$, (well, ok technically, it is the restriction $\text{pr}^i|_{\phi[U]}:\phi[U]\subset\Bbb{R}^n\to\Bbb{R}$). The dual basis of the standard basis serves exactly this purpose. The map $e^i:\Bbb{R}^n\to\Bbb{R}$ is by definition the same linear map as $\text{pr}^i:\Bbb{R}^n\to\Bbb{R}$: they both take as input the tuple $(a^1,\dots, a^n)$ and spit out the number $a^i$.
2. Note that $e^i$ is a linear map, so its derivative at every point is that same linear map. Perhaps this will ease your mind: the thing that is constant is that $d(e^i)_{a}=e^i$ for all points $a\in \phi[U]$. This should sound reasonable because the function $f:\Bbb{R}\to\Bbb{R}$, $f(x)=3x$ for example is linear and has constant derivative $f’(x)=3$ for all $x$.

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Regarding (2), one other thing I guess you might be confusing things with is that while what I wrote is true, sometimes people are not considering the map $e^i$ itself, but rather the constant mapping $a\mapsto e^i$ from an open subset of $\Bbb{R}^n$ into $(\Bbb{R}^n)^*$ (well really a more common scenario is to consider the constant vector-valued map $a\mapsto e_i$, where $e_i\in\Bbb{R}^n$ is the $i^{th}$ standard basis vector). The derivative of this map does vanish. By abuse of notation, one might refer to the map $a\mapsto e^i$ simply as $e^i$ and consequently write $de^i=0$, but you should be mindful of the overload of notation, and not confuse a given map with its constant value (especially when that constant value is itself a linear function). But, let me be clear, the answer to your question is what I wrote above in the second paragraph.

Answer 2

First, the left side of the equation is a scalar function and the right side is a $1$-form, so the equation is obviously incorrect.

Here $x^i$ has two separate but closely related definitions. One is simply the $i$-th coordinate of $\newcommand\R{\mathbb{R}}\R^n$. The other is the function $x^i: U \rightarrow \R$. To keep track of things more easily, let's rename one of them.

Let's use $x^i$ as the $i$-th coordinate of a point in $\R^n$. In other words, denote a point in $\R^n$ by $$x = (x^1, \dots, x^n).$$ On the other hand, let's denote the coordinate map by $$\phi = (\phi^1, \dots, \phi^n),$$ where each $\phi^i: U \rightarrow \R$.

The equation can now be written as $$ d(\phi^i\circ\phi^{-1}) = e^i. $$ This can be proved trivially as follows: For any $x \in \phi(U)$, $$ \phi\circ\phi^{-1}(x) = x, $$ and therefore $$ \phi^i(\phi^{-1}(x)) = x^i. $$ Therefore, $$ d(\phi^i\circ\phi^{-1}(x)) = dx^i = e^i. $$ I agree that $d(e^i) = 0$.

ADDED COMMENT:

Why I consider $e^i$ to be an abuse of notation:

There is a distinction worth making between the manifold $\mathbb{R}^n$ and the vector space of tangent vectors $\mathbb{R}^n$. On manifold, the differential of a function is a basic concept. On the other hand, the tangent space is a vector space, which belongs to the world (i.e., categories) of linear algebra, where differentiation is never useful (for obvious reasons). Even though technically you can differentiate an element of $T_p^*M$, it's pointless because all you get is the element itself.

There is a tendency (especially by physicists) to get sloppy about the distinction between the manifold $\mathbb{R}^n$ and the tangent space $\mathbb{R}^n$. When the authors write $(\mathbb{R}^n)^*$, it indicates that they are aware of and want to account for this distinction. But then they violate it by conflating the two different meanings of $e^i$.

Too many people treat the notation of modern differential geometry as just that, convenient index-free notation. However, it's much more than that. It elucidates the functorial aspects of differential geometry, especially the parallels with the functorial aspects of linear algebra. For that reason, I like to call differential geometry "parameterized linear algebra".