Geometric interpretation of the exact sequence for the cotangent bundle of the projective space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:53:38Z http://mathoverflow.net/feeds/question/98634 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98634/geometric-interpretation-of-the-exact-sequence-for-the-cotangent-bundle-of-the-pr Geometric interpretation of the exact sequence for the cotangent bundle of the projective space auniket 2012-06-02T04:44:47Z 2012-06-02T19:47:39Z <p>Edit: As Dan Petersen pointed out, this question is a duplicate of a <a href="http://mathoverflow.net/questions/5211/" rel="nofollow"> previous one</a>. 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!).</p> <p>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 $\mathcal{O}_X(1)$. </p> <p>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?</p> http://mathoverflow.net/questions/98634/geometric-interpretation-of-the-exact-sequence-for-the-cotangent-bundle-of-the-pr/98642#98642 Answer by Martin Brandenburg for Geometric interpretation of the exact sequence for the cotangent bundle of the projective space Martin Brandenburg 2012-06-02T08:58:35Z 2012-06-02T08:58:35Z <p>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 <code>$\bigoplus_{u&lt;v} \mathcal{O}(-2) \twoheadrightarrow \Omega^1$</code>, which is given on $D_+(x_i)$ by mapping</p> <p>$$x_i^{-2} e_{uv} \mapsto \dfrac{x_u}{x_i} \cdot d\left(\dfrac{x_v}{x_i}\right)- \dfrac{x_v}{x_i} \cdot d\left(\dfrac{x_u}{x_i}\right).$$</p> <p>You can also compute the relations between these elements and arrive at the exact sequence</p> <p><code>$$\bigoplus_{u&lt;v&lt;w} \mathcal{O}(-3) \to \bigoplus_{u&lt;v} \mathcal{O}(-2) \to \Omega^1 \to 0.$$</code></p> <p>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</p> <p><code>$$\dotsc \to \bigoplus_{u&lt;v&lt;w} \mathcal{O}(-3) \to \bigoplus_{u&lt;v} \mathcal{O}(-2) \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0.$$</code></p> <p>These combine to the Euler sequence $0 \to \Omega^1 \to \bigoplus_{u} \mathcal{O}(-1) \to \mathcal{O} \to 0$.</p> http://mathoverflow.net/questions/98634/geometric-interpretation-of-the-exact-sequence-for-the-cotangent-bundle-of-the-pr/98646#98646 Answer by Georges Elencwajg for Geometric interpretation of the exact sequence for the cotangent bundle of the projective space Georges Elencwajg 2012-06-02T10:00:59Z 2012-06-02T10:06:35Z <p>By dualizing and twisting we obtain the <em>equivalent</em> exact sequence of vector bundles<br> $$0\to \tau\to \mathbb P^n_k\times k^{n+1} \to T_{\mathbb P^n}(-1)\to 0 \quad (*)$$ The first morphism is just the inclusion of the tautological vector bundle $\tau$ into the trivial bundle and is geometrically transparent.<br> 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) .<br> At $p$ the exact sequence $(*)$ becomes the exact sequence of vector spaces$$0\to l\to k^{n+1} \to T_{\mathbb P^n}[p]\otimes l\to 0$$<br> Exactness then translates into the canonical isomorphism $$T_{\mathbb P^n}[p] = \mathcal L(l,k^{n+1}/l) \quad (**)$$ </p> <p>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 $p=(a_0....:a_n)\in \mathbb P^n$.<br> Here is the idea inspired by differential geometry. </p> <p>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 $t\in T_{\mathbb P^n}[p]$.<br> The canonically associated linear map $\lambda _t:l\to k^{n+1}/l$ is then characterized by the condition $$\lambda _t(a_0,...,a_n)=\overline {(t_0,...,t_n)}$$<br> [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 $(t_0',...,t_n')$]<br> The details are in <a href="http://www.math.lsa.umich.edu/~idolga/631.pdf" rel="nofollow">Dolgachev's online notes</a> , Example 13.2</p>