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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
$$ f: \mathcal{O}_{\mathbb{P}^n} \longrightarrow \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus(n+1)}, \quad c \mapsto (c\cdot x_0, c \cdot x_1, \dots,c\cdot x_n) $$ i.e. multiplication of 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 annihilates 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 a 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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    $\begingroup$ This is explicitly an approach in which you said you were not interested, but it might give you some insight, anyhow. Yes, the Euler vector field on $\Bbb C^{n+1}$ descends to the $0$ vector tangent to $\Bbb P^n$. $\endgroup$ Apr 9, 2021 at 4:15
  • $\begingroup$ @TedShifrin Thanks - I did come across this earlier and think I follow but it's still a tad too slick for me. I guess you are also pointing out that $\mathcal{O}_{\mathbb{P}^n}(1)^{\oplus (n+1)}$ is isomorphic to $\mathbb{C}^{n+1} \setminus {0}$? $\endgroup$ Apr 9, 2021 at 10:09
  • $\begingroup$ You are right that a lot of small details are usually missing. The EVF is a vector field on $\mathbb{C}^{n+1}$. The sections of $\mathcal{O}(1)$ are seen as functionals on lines in $\mathbb{C}^{n+1}$ and your map $g$ actually gives a vector field on $\mathbb{C}^{n+1}$. The map you are looking for is actually $d\pi\circ g$ which gives a vector field on $\mathbb{P}^n$ (because it is invariant on the lines). Everything is more or less explained in Griffiths and Harris page 409 $\endgroup$
    – BinAcker
    Jan 11, 2022 at 12:38
  • $\begingroup$ @nonreligious simple question: What's the defintion for $\mathcal{O}_{\mathbb{P}^n}(d)$? $\endgroup$ Feb 2, 2023 at 21:37
  • $\begingroup$ @BVquantization $\mathcal{O}(d)=\mathcal{O}(1)^{\otimes d}$ and $\mathcal{O}(1)=\mathcal{O}(-1)^{*}$, where the last bundle ${\mathcal{O}(-1)}$ is known as "the tautological bundle" $\endgroup$
    – lou
    Jun 27, 2023 at 19:02

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