Edit: As Dan Petersen pointed out, this question is a duplicate of a  [previous one](https://mathoverflow.net/questions/5211/). 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?



Geometric interpretation of the exact sequence for the cotangent bundle of the projective space
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](http://www.math.lsa.umich.edu/~idolga/631.pdf) , Example 13.2


