MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Edit: As Dan Petersen pointed out, this question is a duplicate of a previous one. I would leave it for the moderators to decide if this should be closed. On the other hand, may be this should be left open on the merit of the excellent answers and comments (@Emerton: Thanks!).

I was trying to understand the following exact sequence (for $X := \mathbb{P}^n_k$, where $k$ is an algebraically closed field): $$0 \to \Omega_X \to \mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X \to 0$$ The explanation (as in the proof of Theorem II.8.13 of Hartshorne) is given by some algebraic formulae, which I am having trouble to digest. I was trying to see in more geometric terms what is going on, and was somewhat successful in the case of the surjection $\mathcal{O}_X(-1)^{n+1} \to \mathcal{O}_X$, namely: we can regard $\mathcal{O}_X(1)$ (respectively $\mathcal{O}_X(-1)$) as the normal bundle $\mathcal{N}$ of (respectively conormal bundle) of $X$ in $Z := \mathbb{P}^{n+1}_k$. Any global section of $\mathcal{O}_X(1)$ therefore induces a map (via evaluation) from $\mathcal{O}_X(-1)$ to $\mathcal{O}_X$. The above surjection comes from taking $n+1$-linearly independent global sections of $\mathcal{O}_X(1)$.

But I do not understand how to interpret the injection $\Omega_X \to \mathcal{O}_X(-1)^{n+1}$. How would someone 'naturally' come up with the algebraic formula?

share|cite|improve this question
I am definitely not an expert, but what happens if you realize the conormal bundle as a sub bundle of the cotangent bundle? Don't you find that the cotangent bundle of the embedded variety gets realized as a piece of the conormal? – Filippo Alberto Edoardo Jun 2 '12 at 4:53
There is a quotient map $\pi:\mathbb A^{n+1} \setminus \{0\} \to \mathbb P^n$, and so we can pull back a differential form $\omega$ on $U \subset \mathbb P^n$ to a differential form $\pi^* \omega$ on $\pi^{-1}(U)$. The sheaf of differential forms on $\mathbb A^{n+1} \setminus \{0\}$ is globally free, with a basis $dx_0, \ldots, dx_n$. If you think about it, you can then interpret the pull-back map as an embedding $\Omega_{\mathbb P^n} \hookrightarrow \mathcal O(-1)^{n+1}$. Not every section of $\mathcal O(-1)^{n+1}$ will give rise to a differential form, though --- when you interpret such ... – Emerton Jun 2 '12 at 4:58
... a section as a differential form on $\mathbb A^{n+1}\setminus \{0\}$, to come from $\Omega_{\mathbb P^n}$ is has to be invariant under the scaling action of $\mathbb G_m$. This can be tested by pairing with the Euler class $x_0\partial_{x_0} + \cdots + x_n\partial_{x_n}$, and this gives the map $\mathcal O(-1)^{n+1} \to \mathcal O$ which appears in the exact sequence. – Emerton Jun 2 '12 at 5:00
You might be interested to know that Ravi Vakil has a short but sweet exposition of this exact sequence and its various generalizations in section 23.3 of his notes ( – Sam Lichtenstein Jun 2 '12 at 5:58
This is an exact duplicate of – Dan Petersen Jun 2 '12 at 12:17
up vote 5 down vote accepted

By dualizing and twisting we obtain the equivalent exact sequence of vector bundles
$$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0 \quad (*) $$ The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.
To understand the second morphism geometrically, fix a point $p\in \mathbb P^n_k$ and the corresponding line $l\subset \mathbb P^n_k$ (I forgot to say I'm using the pre-Grothendieck definition of projective space as a set of lines) .
At $p$ the exact sequence $(*)$ becomes the exact sequence of vector spaces$$0\to l\to k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$
Exactness then translates into the canonical isomorphism $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$

So the whole problem boils down to understanding $(**)$, i.e.understanding in a canonical way the fiber of the tangent bundle to $\mathbb P^n$ at a point $p=(a_0....:a_n)\in \mathbb P^n$.
Here is the idea inspired by differential geometry.

The "curve" $\epsilon \mapsto (a_0+\epsilon t_0,....,a_n+\epsilon t_n)\; (\epsilon^2=0)$ [algebraic geometers consider very short curves!] gives rise to a tangent vector $t\in T_{\mathbb P^n}[p]$.
The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is then characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)} $$
[Be careful that if you change the vector $(a_0,...,a_n)$ representing $p$ to a colinear vector $(a_0',...,a_n')$, you also have to change $(t_0,...,t_n)$ to another $(t_0',...,t_n')$]
The details are in Dolgachev's online notes , Example 13.2

share|cite|improve this answer
Thanks! This is pretty close to what I wanted. – auniket Jun 2 '12 at 19:38

Here is another (unknown?) way of optaining the Euler sequence (though not really geometric): Since $\Omega^1_{\mathbb{P}}$ is a coherent sheaf; by Serre it has a "twisted presentation". For that one has to find some $k > 0$ such that $\Omega^1_{\mathbb{P}}(k)$ is generated by global sections. You will find that $k=2$ suffices, namely there is an epimorphism $\bigoplus_{u<v} \mathcal{O}(-2) \twoheadrightarrow \Omega^1$, which is given on $D_+(x_i)$ by mapping

$$x_i^{-2} e_{uv} \mapsto \dfrac{x_u}{x_i} \cdot d\left(\dfrac{x_v}{x_i}\right)- \dfrac{x_v}{x_i} \cdot d\left(\dfrac{x_u}{x_i}\right).$$

You can also compute the relations between these elements and arrive at the exact sequence

$$\bigoplus_{u<v<w} \mathcal{O}(-3) \to \bigoplus_{u<v} \mathcal{O}(-2) \to \Omega^1 \to 0.$$

But now the (graded) Koszul resolution of $R[x_0,\dotsc,x_n]/(x_0,\dotsc,x_n)$ (here $R$ is an arbitrary base ring; it doesn't have to be an algebraically closed field) yields the long exact sequence

$$\dotsc \to \bigoplus_{u<v<w} \mathcal{O}(-3) \to \bigoplus_{u<v} \mathcal{O}(-2) \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0.$$

These combine to the Euler sequence $0 \to \Omega^1 \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.