I'm trying to understand the use of coordinates in the calculation of the differential of a function between manifolds. There was a certain problem in a past differential geometry exam, and while I understand it geometrically, I'm having trouble understanding the general principles involved in the calculation. My confusion arises from the fact that we get two different expressions that could be called the "differential" of f in local coordinates. One in (i) and another one in (ii)-(iii). However, the expression in (i) doesn't really allow us to get the result that we expect geometrically and so I mostly guessed the right coordinate expression through trial and error. (meaning $\phi \;\circ f$).
The problem:
Let $f:P\mathbb{R}^{n} \to P\mathbb{R}^{n}$ be a function defined as $f([x_1,...,x_n]) = [x_1^2,...,x_n^2]$.
(i) Prove it is differentiable,
(ii) Calculate $d_pf$ where $p=[1,...,1]$,
(iii) For another $p$ in projective space, find the points at which $d_p f$ is not injective.
My attempt:
(i) Standard, by the book calculation: Assume the first coordinate doesn't vanish. Since $x_1\neq0$ implies $x_1^2\neq 0$, we can take the same chart $(U, \phi)$ at both $p$, and $f(p)$. Then, we can give local coordinates as:
$$\phi([x_1,...,x_n]) = \left(\frac{x_2}{x_1},...,\frac{x_n}{x_1} \right), $$ $$\phi^{-1}(y_1...,y_{n-1}) ) = [1,y_1,...,y_{n-1}]. $$
And so, $$ \phi \;\circ\; f \;\circ\; \phi^{-1}(y_1,....y_{n-1}) = \phi \;\circ\; f([1,y_2,...,y_{n-1}]) $$ $$ = \phi ([1,y_1^2,...,y_{n-1}^2]) = (y_1^2,...,y_{n-1}^2).$$
Which is clearly differentiable.
(ii) So here's the source of the confusion. I'm assuming that, since the differential must be evaluated at a point in projective space, the coordinate expression should actually be
$$(\phi \;\circ \; f) ([x_1,...,x_n]) = \phi ([x_1^2,...,x_n^2]) =\left(\left(\frac{x_2}{x_1}\right)^2,...,\left( \frac{x_n}{x_1} \right)^2 \right),$$
and so, the differential I calculated was
$$d_p f([h_1, ..., h_n]) = 2\left(\frac{x_2}{x_1}\left( \frac{x_1 h_2 - x_2 h_1}{x_1^2}\right),...,\frac{x_n}{x_1}\left( \frac{x_1 h_n - x_n h_1}{x_1^2}\right) \right)$$ $$ = \frac{2}{x_1^3}(x_2 ( x_1 h_2 - x_2 h_1),...,x_n ( x_1 h_n - x_n h_1)).$$
At $p = [1,...,1]$, we have:
$$ d_p f = 2(h_2 - h_1, ..., h_n - h_1).$$
Moreover, $d_p f$ is zero when $h_1 = h_k$ for $1\leq k \leq n$, this implies that
$$ [h] = [h_1, ..., h_1] = [1,...,1],$$
and so, $d_p f$ is not injective at this point.
(iii) The expression we previously calculated for a generic point $p$ is zero only if
$$x_1 h_k = x_k h_1 \quad\text{or}\quad x_k=0 \quad\text{for all $1<k\leq n$},$$
but notice that $x_k=0$ implies $x_1 h_k = 0$, which only happens if $h_k = 0$. We now look for other non-trivial solutions to $d_p f = 0$.
$$ \frac{x_k}{x_1} = \frac{h_k}{h_1}, $$
which means that $p$ and $h$ are proportional and $[p]=[h]$. Explicitly, because of how we picked the coordinate neighborhood, we can assume $x_1 = h_1 = 1$ and
$$ x_k = h_k,$$
and so the differential is injective whenever $[h] \neq [p]$.
Is this allright? Is there some definition, principle or theorem that distinguishes between the two different coordinate expressions used in (i) and (ii)-(iii) ? Also, any feedback on style and writing would be greatly appreciated.
EDIT: Corrected some indexes.
Coordinates can often make things obscure when dealing with projective spaces. For example, how are the tangent spaces to ${\rm P}(\Bbb R^n)$ described? You should be computing the differential of $\phi\circ f\circ \phi^{-1}$, not of $\phi\circ f$. The latter is a function ${\rm P}(\Bbb R^n) \to \Bbb R^{n-1}$, and you don't know how to directly compute the derivative of that since, at this stage, you don't know what is the velocity vector of a curve of lines in $\Bbb R^n$. The derivative ${\rm d}f_p$ takes as inputs vectors tangent to ${\rm P}(\Bbb R^n)$, not points of ${\rm P}(\Bbb R^n)$! This means that your expression $d_pf[h_1,\ldots, h_n]$ doesn't even compile... tangent spaces to ${\rm P}(\Bbb R^n)$ are not naturally identified with ${\rm P}(\Bbb R^n)$ itself like what happens in $\Bbb R^n$ (hell, ${\rm P}(\Bbb R^n)$ is not even a vector space). You are also sometimes writing $(y_1,\ldots, y_n)$ instead of $(y_1,\ldots, y_{n-1})$ (the former is wrong since $\dim {\rm P}(\Bbb R^n) = n-1$), etc. etc. etc...
As we have (for trivial reasons!) that $${\rm d}(\phi\circ f\circ \phi^{-1})_{(y_1,\ldots, y_{n-1})}(h_1,\ldots, h_{n-1}) = (2y_1h_1,\ldots, 2y_{n-1}h_{n-1}),$$ we immediately see that setting $y_1=\cdots =y_{n-1}=1$ (which corresponds to $\phi^{-1}(1,\ldots ,1) = [1:\cdots : 1]$) leads to ${\rm d}f_{[1:\cdots : 1]}$ being twice the identity mapping of $T_{[1:\cdots : 1]} {\rm P}(\Bbb R^n)$.
Now, even though it is not what OP is looking for, I think it is worth it to register a coordinate-free analysis of the situation.
(1) If $\widetilde{f}\colon \Bbb R^n\smallsetminus \{0\} \to \Bbb R^n \smallsetminus \{0\}$ is given by $\widetilde{f}(x_1,\ldots, x_n) = (x_1^2,\ldots, x_n^2)$ and $\pi\colon \Bbb R^n\smallsetminus \{0\} \to {\rm P}(\Bbb R^n)$ is the quotient (span) projection, then $\pi\circ \widetilde{f} = f\circ \pi$ says that $\widetilde{f}$ is smooth since $f$ is smooth and $\pi$ is a surjective submersion.
(2) Then, using the chain rule, we have that $${\rm d}\pi_{(x_1^2,\ldots, x_n^2)}(2x_1v_1,\ldots, 2x_nv_n) = {\rm d}f_{[x_1:\cdots : x_n]}({\rm d}\pi_{(x_1,\ldots, x_n)}(v_1,\ldots, v_n))$$for every $[x_1:\cdots : x_n]\in {\rm P}(\Bbb R^n)$ and $(v_1,\ldots, v_n)\in \Bbb R^n$. Since every element in $T_{[x_1:\cdots : x_n]} {\rm P}(\Bbb R^n)$ has the form ${\rm d}\pi_{(x_1,\ldots, x_n)}(v_1,\ldots, v_n)$, the above formula completely describes the linear map ${\rm d}f_{[x_1:\cdots: x_n]}\colon T_{[x_1:\cdots:x_n]} {\rm P}(\Bbb R^n) \to T_{[x_1^2:\cdots : x_n^2]} {\rm P}(\Bbb R^n)$.
When $[x_1:\cdots : x_n] = [1:\cdots : 1]$, it follows that $$2 {\rm d}\pi_{(1,\ldots, 1)}(v_1,\ldots, v_n) = {\rm d}f_{[1:\cdots : 1]}({\rm d}\pi_{(1,\ldots, 1)}(v_1,\ldots, v_n)),$$which is to say that ${\rm d}f_{[1:\cdots : 1]}$ equals twice the identity mapping of $T_{[1:\cdots : 1]}{\rm P}(\Bbb R^n)$.
(3) How to find the points $[x_1:\cdots : x_n]$ for which ${\rm d}f_{[x_1:\cdots : x_n]}$ is not injective? Here, we need to know that $\ker {\rm d}\pi_{(a_1,\ldots, a_n)}$ is the line spanned by $(a_1,\ldots, a_n)$.
So, assuming the existence of $(v_1,\ldots, v_n) \in \Bbb R^n$ such that
(i) $(v_1,\ldots, v_n)$ and $(x_1,\ldots, x_n)$ are linearly independent, and
(ii) $(x_1v_1,\ldots, x_nv_n)$ and $(x_1^2,\ldots, x_n^2)$ are linearly dependent,
what can we say about $(x_1,\ldots, x_n)$? Condition (i) gives us $i \neq j$ such that $x_iv_j - x_jv_i \neq 0$, and for the same indices we have $x_iv_ix_j^2 - x_jv_jx_i^2 = 0$, which reads as $x_ix_j(x_iv_j-x_jv_i) = 0$, and hence $x_ix_j = 0$. Therefore, if ${\rm d}f_{[x_1:\cdots : x_n]}$ is singular, then $x_i = 0$ for at least one $i$.
It now becomes natural to try and prove the converse: if there is at least one $i$ such that $x_i = 0$, then ${\rm d}f_{[x_1:\cdots : x_n]}$ is singular. Permuting coordinates and normalizing, we may simply consider $[1:0: x_3:\cdots : x_n]$, and note that $${\rm d}f_{[1:0:x_3:\cdots : x_n]}({\rm d}\pi_{(1,0,x_3,\ldots, x_n)}(1,1,x_3,\ldots, x_n)) = 2 {\rm d}\pi_{(1,0,x_3^2,\ldots, x_n^2)}(1,0,x_3^2,\ldots, x_n^2) = 0,$$despite ${\rm d}\pi_{(1,0,x_3,\ldots, x_n)}(1,1,x_3,\ldots, x_n) \neq 0$.
Conclusion: the set of points in $p \in {\rm P}(\Bbb R^n)$ for which ${\rm d}f_p$ is singular is precisely $$\{ [x_1:\cdots : x_n] : x_i=0\mbox{ for some i}\}.$$You can probably go further and try to relate $\dim \ker {\rm d}f_{[x_1:\ldots, x_n]}$ with $|\{ i =1,\ldots, n : x_i=0\}|$.
(My knee-jerk conjecture is that they're equal.)