Is there any geometric way to understand the exact sequence in Example 8.20.1 in Ch II of Hartshorne (p. 182), or its dual from theorem 8.13?


Here is the sequence:


$0\to O\_{\mathbb{P}^n}\to O\_{\mathbb{P}^n}(1)^{n+1}\to T\_{\mathbb{P}^n}\to 0$



Geometric meaning of the Euler sequence on $\mathbb{P}^n$ (Example 8.20.1 in Ch II of Hartshorne)
Yes! The geometric picture is very nice and very easy. It is explained on pages 408-409 of Griffiths-Harris.


Here is roughly how it works:


Let's work over $\mathbb{C}$ for simplicity. Think of $\mathbb{P}^n$ as being the quotient of $X := \mathbb{C}^{n+1} - 0$ by the action of $\mathbb{C}^\ast$. On $X$ we have the vector fields $d/dx\_i$, where the $x\_i$ are the standard coordinates on $\mathbb{C}^{n+1}$. Check that if $v\_i$ are linear functionals on $\mathbb{C}^{n+1}$, then the vector field $\sum v\_i d/dx\_i$ on $X$ descends to a vector field on $\mathbb{P}^n$. The surjection $\mathcal{O}(1)^{n+1} \to \mathcal{T}$ corresponds to taking $n+1$ linear functionals $v\_i$ and projecting the vector field $\sum v\_i d/dx\_i$ down to $\mathbb{P}^n$.
The kernel $\mathcal{O}$ corresponds to the vector field $E = \sum\_i x\_i d/dx\_i$. Intuitively, $E$ is a "radial" vector field on $X$, and if you pretend that $\mathbb{P}^{n}$ is a "sphere" in $X$, then $E$ is "normal" to this "sphere", so it vanishes when we project it down.


Jonathan Wise gives a nice (and rigorous) explanation of this below.


Aside: I think the reason why this is called the Euler sequence is because the vector field $E$ is known as the Euler vector field. And perhaps the reason why $E$ is called the Euler vector field is because its flow is exponential, and $e = 2.718\dots$ is also known as Euler's number. But I'm not sure, and someone should correct me if I'm wrong about this. Edit: Today somebody told me that the relation $E f = d f$ for $f$ a homogeneous degree $d$ polynomial (as in Charles' answer) was discovered by Euler and is known as "Euler's relation".


