Geometric meaning of the Euler sequence on $\mathbb{P}^n$ (Example 8.20.1 in Ch II of Hartshorne)

Question

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:

$0\to O_{\mathbb{P}^n}\to O_{\mathbb{P}^n}(1)^{n+1}\to T_{\mathbb{P}^n}\to 0$

Answer 1

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".

Answer 2

To me, at least, this is a manifestation of the fact that for homogeneous polynomials, we have $\frac{1}{d}\sum_{i=0}^n x_i\partial_i f=f$. The map $O(1)^{n+1}\to T$ tells you that every vector field is a linear combination of the $\partial_i$ with linear coefficients. The map $O\to O(1)^{n+1}$ sends the section 1 to the vector $(x_0,\ldots,x_n)$, which says that the vector field $\frac{1}{d}\sum x_i\partial_i$ acts trivially on functions. Thus, the quotient must be the actual tangent vector fields, giving us the tangent bundle.

EDIT: Apparently this package doesn't know mathscr. Have made it readable.

Answer 3

Let us consider a vector space $E$ of dimension $n+1$ and a line $L\subset E$, which corresponds to the point $x\in \mathbb {P(E)}$. The tangent space $T_x\mathbb P (E)$ is canonically isomorphic to the space of linear maps $\mathcal{L}(L,E/L)$ [Harris,Algebraic Geometry, page 200, where it is even done for Grassmannians].

Hence we get a canonical isomorphism $T_x\mathbb P (E)=L^\ast \otimes E/L$ , transformed into $T_x \mathbb P(E)\otimes L=E/L$

which we write as an exact sequence $0 \to L \to E \to T_x \mathbb P E \otimes L \to 0$

This was just over the point $x\in\mathbb P(E)$. If we globalize this over the whole of $\mathbb P(E)$ we get the exact sequence of locally free sheaves [Recall that the fibre at x of the tautological line bundle $\mathcal O (-1)$ is precisely $L$]

$0\to \mathcal O(-1) \to \mathcal O ^{n+1} \to T \mathbb P (E)\otimes \mathcal O (-1) \to 0$

By tensoring this exact sequence by the invertible sheaf $\mathcal O (1)$ we obtain the euler sequence.

Answer 4

Since ${\bf P}^n$ is the quotient of ${\bf A}^{n+1} - 0$ by the action of ${\bf G}_m$, the tangent bundle of ${\bf P}^n$ is the quotient of the tangent bundle of ${\bf A}^{n+1} - 0$ by the action of the tangent bundle of ${\bf G}_m$:

$T {\bf P}^n = T ({\bf A}^{n+1} - 0) / T {\bf G}_m$ .

As a group, the tangent bundle of ${\bf G}_m$ is the product of ${\bf G}_m$ and a 1-dimensional vector space V. Therefore we can take the quotient of everything on the right side above by ${\bf G}_m$. Note that $T({\bf A}^{n+1}-0)$ is the product of ${\bf A}^{n+1} - 0$ with the direct sum of (n+1) copies of the weight one representation of ${\bf G}_m$. Therefore its quotient by ${\bf G}_m$ is $\mathcal{O}_{{\bf P}^n}(1)^{n+1}$. We get

$T {\bf P}^n = \mathcal{O}_{{\bf P}^n}(1)^{n+1} / V$ .

A (linear) action of a one-dimensional vector space on a vector bundle is the same thing as an exact sequence, so this gives the Euler sequence.

Answer 5

After three excellent answers, here's a bad one. This isn't a geometric way to understand the sequence, just another way to look at it.

Let $R = k[x_0,...,x_n]$, and consider the graded module $M = R^{n+1}(1)$ : the $x_i$ are homogeneous of degree in 0 in $M$. So the associated Koszul complex looks like

$0 \rightarrow R \stackrel{f}{\rightarrow} R^{n+1}(1) \rightarrow \wedge^2 R^{n+1}(2) \rightarrow ...$ .

Now take the cokernel of $f$; you get $0 \rightarrow R \rightarrow R^{n+1}(1) \rightarrow \mathrm{coker }f \rightarrow 0$ .

The Euler sequence is the sequence of associated sheaves (so in particular $T_{\mathbb{P}^n} \simeq \widetilde{\mathrm{coker }f}$).

Eisenbud talks about this connection in `Commutative Algebra with a view...'.

Answer 6

Not the answer you were looking for, but fun nonetheless: if you take global sections of the Euler exact sequence, you get a short exact sequence since $H^1(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(1)) = 0$. The exact sequence that you get is $$0 \to H^0(\mathbb{P}^n, \mathcal{O}) \to H^0(\mathbb{P}^n, \mathcal{O}(1))^{\oplus (n+1)} \to H^0(\mathbb{P}^n, T_{\mathbb{P}^n}) \to 0.$$ The cohomology of line bundles on projective space tells us that $h^0(\mathbb{P}^n, T_{\mathbb{P}^n}) = (n+1)^2 - 1$. However $H^0(\mathbb{P}^n, T_{\mathbb{P}^n})$ has a natural Lie algebra structure (since sections of the tangent bundle correspond to vector fields) and in fact is $\mathfrak{sl}_{n+1}$. We can interpret this as follows: $$H^0(\mathbb{P}^n, \mathcal{O}(1))^{\oplus (n+1)} \cong ((k^{n+1})^{\vee})^{\oplus(n+1)} \cong M_{(n+1) \times (n+1)}(k).$$ The quotient should be viewed as the $((n+1)^2 - 1)$-dimensional vector space of traceless matrices while the kernel is generated by any matrix $A$ having nonzero trace (let's choose the identity). The surjection is then given by $\displaystyle M \mapsto M - \frac{tr(M)}{tr(A)}A.$ This is clearly a map of $k$-vector spaces.

I think the ambiguity in the choice of $A$ corresponds to a choice of coordinates, but I'm not 100% certain on this point.

Answer 7

Very concretely, now a vector field on $\mathbb{A}^{n+1}$ wants to act on $\mathcal{O}_{\mathbb{P}^{n}}$. But if you have a degree 0 section, namely secion of $\mathcal{O}$ and take any derivative, you get a degree -1 section, which says vector field on $\mathbb{A}^{n+1}$ cannot act directly on $\mathbb{P}^{n}$. Then how to compensate? You put some degree 1 coefficients, namely twist $\mathcal{O}^{n+1}$ by 1. Then sections in $\mathcal{O}(1)^{n+1}$ can act, and you figure out the kernel is exactly this $\sum x_{i}\partial x_{i}$, which is identified as basis of $\mathcal{O}$.