# Understanding the Euler sequence on $\mathbb{P}^n$

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.

• 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 $\Bbb P^n$. Apr 9, 2021 at 4:15
• @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 $\mathbb{C}^{n+1} \setminus {0}$? Apr 9, 2021 at 10:09
• 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 Jan 11, 2022 at 12:38
• @nonreligious simple question: What's the defintion for $\mathcal{O}_{\mathbb{P}^n}(d)$? Feb 2, 2023 at 21:37
• @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"
– lou
Jun 27, 2023 at 19:02