I want to get an intuition for the Euler sequence, by understanding the explicit construction of maps between terms. I prefer to use this version:$$ 0 \longrightarrow \mathcal{O}_{\mathbb{P}^n} \longrightarrow \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)} \longrightarrow T\mathbb{P}^n \longrightarrow 0 $$rather than its dual or any twisted version
Conventions:$[x_0 : x_1 : \dots :x_n]$ are projective coordinates on $\mathbb{P}^n$. I will slightly abuse notation and describe sections of degree $d$ line bundles with the same notation, e.g. $x_0^d + 2x_1^{d-1}x_2$ is a section of $\mathcal{O}_{\mathbb{P}^n}(d)$.
First map
If $c$ is a locally constant function on $\mathbb{P}^n$, then
$$
f: \mathcal{O}_{\mathbb{P}^n} \longrightarrow \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)},
\quad
c \mapsto (c\cdot x_0, c \cdot x_1, \dots, x_n)
$$i.e. multiplication the linear monomials by $c$ (which is usually taken to be 1).
Second map
For a set of linear (homogeneous degree 1) functions $l_i(x)$, $i=0,\dots,n$ on $\mathbb{P}^n$, we have$$ g: \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)} \longrightarrow T\mathbb{P}^n , \\ (l_0(x), l_1(x), \dots, l_n(x)) \longmapsto l_0(x) \frac{\partial\,}{\partial x_0} + l_1(x) \frac{\partial\,}{\partial x_1} + \dots + l_n(x) \frac{\partial\,}{\partial x_n} $$
Showing $\mathrm{im}(f) = \mathrm{ker}(g)$
If we apply $(g\circ f)$ onto our locally constant $c$, we arrive at the vector$$ c \cdot x_0 \frac{\partial\,}{\partial x_0} + c \cdot x_1 \frac{\partial\,}{\partial x_1} + \dots + c \cdot x_n \frac{\partial\,}{\partial x_n}, $$which is known as the "Euler vector field" or EVF (or at least, it is $c$ times the usual definition of the EVF). It should be straightforward to see that the EVF acting on a homogeneous polynomial $q(x)$ of degree $d$ will return $d \cdot q(x)$. In particular, if $q$ is homogeneous of degree $0$, i.e. constant, then the EVF($q$) returns 0.
This is where my understanding gets a little shaky:
The above makes sense, but the statement I see in the literature jumps from saying "the EVF annihlates degree 0 functions" to saying that "the EVF is the kernel of $g$" and concluding the description. This is a little hard for me to parse, because I feel $\mathrm{ker}(g)$ lies in $\mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)}$, but the EVF seems to lie in $T\mathbb{P}^n$.
Another point -- which may be central to the whole issue -- is if $\frac{\partial}{\partial x_i}$ are a proper set of basis vectors for $\mathbb{P}^n$. The $x_i$ are homogeneous coordinates after all, so is the resolution to my question that:$$ x_j \frac{\partial}{\partial x_i} $$(note the $j$ index, $j=1,\dots,n$are a basis of $T\mathbb{P}^n$, but the EVF$$ x_i \frac{\partial}{\partial x_i} $$is actually the $\vec{\mathbf{0}}$ vector? Should we be working with affine coordinates on a patch, e.g. $y_i = x_i/x_0$ on the patch $x_0 \neq 0$?
I hope to have picture that intuitively maps $c$ to the zero vector field in $T\mathbb{P}^n$, and the above steps don't quite get me there.
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