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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:

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

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I think you'll get a lot more people to think about that question if you mention what the exact sequences actually are. –  Alon Amit Nov 12 '09 at 17:51
The sequence in question is that on $\mathbb{P}^n$, we have $0\to \mathscr{O}\to \mathscr{O}(1)^{n+1}\to T_{\mathbb{P}^n}\to 0$. –  Charles Siegel Nov 12 '09 at 18:07
Ok, as I mentioned below, no mathscr, so the sequence is $0\to O\to O(1)^{n+1}\to T_{\mathbb{P}^n}\to 0$. –  Charles Siegel Nov 12 '09 at 18:09
\mathcal works. –  Ben Webster Nov 12 '09 at 18:18
Would it be useful to change the title of this question to 'Geometric meaning of the Euler sequence on $\mathbb{P}^n$?' This might be more meaningful to those browsing through the question titles, but without a copy of Hartshorne on hand.... –  1-- Nov 12 '09 at 19:05

6 Answers 6

up vote 24 down vote accepted

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".

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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.

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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.

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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.

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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...'.

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This is not a bad answer at all! It's very nice. –  Kevin H. Lin Nov 16 '09 at 17:47
Is it straightforward to see the isomorphism $T_{P^n}=$coker $f$? –  J.C. Ottem Apr 24 '10 at 9:22
@J. C.: Not globally. Essentially one has to repeat the local proof in Hartshorne. –  Martin Brandenburg Jun 3 '12 at 9:14

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.

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Notice that the identity matrix sometimes has zero trace —you need to be careful with the characteristic. –  Mariano Suárez-Alvarez Apr 13 '14 at 5:07
This is a good point! I guess you can just work with the elementary matrix $E_{11}$ to avoid characteristic issues. –  Ashwath Rabindranath Apr 13 '14 at 14:56

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