I've started reading Bott and Tu's text Differential Forms in Algebraic Topology and I think my understanding is quite poor when it comes to the standard coordinates and covectors. Let me explain my understanding and if you could confirm/clarify it that would be great.
Suppose we are in $\mathbb{R}^{n}$ for the moment. Then Bott and Tu defines $y_{1},y_{2},\ldots,y_{n}$ as the standard coordinates on $\mathbb{R}^{n}$. I take this to mean that the $y_{i}$ are the standard basis vectors for $\mathbb{R}^{n}$. However on page 19, Bott and Tu define a pullback map $f^{*}:\Omega^{0}(\mathbb{R}^{n}) \to \Omega^{0}(\mathbb{R}^{m})$ (where $f:\mathbb{R}^{m} \to \mathbb{R}^{n}$ is smooth, $\Omega^{0}(\mathbb{R}^{n})$ is the space of $0$-forms on $\mathbb{R}^{n}$, and similarly for $\mathbb{R}^{m}$) defined by
$$f^{*}(g) = g \circ f.$$
This definition is then extended to a map on all forms such that it commutes with the exterior derivative $d$ and the extended map $f^{*}$ is given by
$$f^{*}(\sum g_{I}dy_{{i_{1}}} \cdots dy_{i_{q}}) = \sum(g_{I} \circ f)df_{{i_{1}}} \cdots df_{i_{q}}.$$
where $f_{i} = y_{i} \circ f$ and the $y_{i}$ are the standard coordinates for $\mathbb{R}^{n}$. But if the standard coordinates are basis vectors then how is $y_{i}$ even defined as a function? Moreover, we already have standard coordinates $x_{1},x_{2},\ldots,x_{m}$ for $\mathbb{R}^{m}$ (this is implicit in the defintion above) so how are the $f_{i}$ standard coordinates for $\mathbb{R}^{m}$? Also, aren't the covectors $dx_{i}$ and $df_{i}$ different?
Now for covectors. From sources other than Bott and Tu, I understand that the covectors $dy_{i}$ in $\mathbb{R}^{n}$ are linear functionals from $\mathbb{R}^{n}$ to $\mathbb{R}$ such that $dy_{i}$ sends a vector to its $i$-th coordinate. Is this the correct understanding or am I missing something?
Best Answer
I am comparing the notation used in Bott and Tu to that used in John M. Lee's "Introduction to Smooth Manifolds", since they are similar (if you have access to a copy of Lee's book, check out page 4-5).
For now consider a general $n$-dimensional smooth manifold $M$ with a chart $(U,\varphi)$, where $\varphi:U \rightarrow \mathbb{R}^{n}$. The smooth map $\varphi$ is the (local) coordinate map, and the component functions $(x_{1},\ldots,x_{n})$ of $\varphi$ are defined by $\varphi(p) = (x_{1}(p),\ldots, x_{n}(p))$. Despite looking straight-forward so far I want to apply this to an example (stereographic projection). Let $U = S^{2} - \{(0,0,1)\}$ be the sphere without the North pole, then $\varphi:U \rightarrow \mathbb{R}^{2}$ is given by $$ \varphi(x,y,z) = \bigg(\frac{x}{1-z}, \frac{y}{1-z}\bigg). $$ So now here, $p = (x,y,z)$ and $x_{1}(x,y,z) = x/(1-z)$, and similarly $x_{2}(x,y,z) = y/(1-z)$.
The reason why I gave an example first is because when the manifold is as simple as $\mathbb{R}^{n}$, what is actually going on is much more subtle since unlike with a more general manifold $M$, there is actually just one chart, $\mathbb{R}^{n}$ itself. And the coordinate map $\varphi:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is in fact just the identity map.
As for $f_{j} = y_{j} \circ f$, here $y_{j}$ just represents the $j$-th element of the identity map on $\mathbb{R}^{n}$, i.e. it ends up being the projection onto the $j$-th factor: $$ y_{j} \circ f(p) = y_{j}( f_{1}(p),\ldots, f_{n}(p)) = f_{j}(p), $$ so $f_{j} = y_{j} \circ f$. Please let me know if anything is unclear or if I missed anything.