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Let $V$ be a four-dimensional complex vector space and $\mathbb{P}^3=\mathbb{P}(V)$. There are two interesting bundles $N$ and $T$ on $\mathbb{P}^3$, both of rank 2, called respectively a null-correlation bundle and a Tango bundle. Let me briefly remind the construction.

Taking symplectic form $\omega$ on $V$, the null-correlation bundle $N$ is defined by the exact sequence

$$0\to\mathcal{O}_{\mathbb{P}(V)}(-1)\stackrel{\varphi}\to\Omega_{\mathbb{P}(V)}(1)\to N\to0,$$ where $\varphi$ is constructed in the following way: there is a complex $$0\to\mathcal{O}_{\mathbb{P}(V)}(-1)\to V\otimes\mathcal{O}_{\mathbb{P}(V)}\stackrel{\pi}\to\mathcal{O}_{\mathbb{P}(V)}(1)\to0$$ in which the map $\pi$ is given by the composition $V\otimes\mathcal{O}_{\mathbb{P}(V)}\stackrel{\omega}\to V^*\otimes\mathcal{O}_{\mathbb{P}(V)}\to\mathcal{O}_{\mathbb{P}(V)}(1)$ and thus we obtain the map $\varphi:\mathcal{O}_{\mathbb{P}(V)}(-1)\to\text{ker}(V^*\otimes\mathcal{O}_{\mathbb{P}(V)}\to\mathcal{O}_{\mathbb{P}(V)}(1))=\Omega_{\mathbb{P}(V)}(1)$.

To define the Tango bundle $F$ we start from the Euler exact sequence

$$0\to\mathcal{O}_{\mathbb{P}(V)}(-1)\to V\otimes\mathcal{O}_{\mathbb{P}(V)}\to T_{\mathbb{P}(V)}(-1)\to0$$ and the second exterior power of it, which is $$0\to T_{\mathbb{P}(V)}(-2)\to\bigwedge^2V\otimes\mathcal{O}_{\mathbb{P}(V)}\to\bigwedge^2T_{\mathbb{P}(V)}(-2)\to0.$$

We have that $E:=\bigwedge^2T_{\mathbb{P}(V)}(-2)$ is rank-3 vector bundle generated by global sections, $K:=H^0(\mathbb{P}^3, E)\cong\bigwedge^2V$ and $c_3(E)=0$. Consider $Y=\{(s,x)\in\mathbb{P}(K)\times\mathbb{P}(V)| s(x)=0\}$, which is a 5-dimensional subvariety in $\mathbb{P}(K)\times\mathbb{P}(V)$. Thus by the proof of Lemma4.3.2 from the book "Vector Bundles on Complex Projective Spaces" of Okonek, Schneider and Spindler, the projection map $p:Y\to\mathbb{P}(K)$ is not a surfection, thus any $s\in\mathbb{P}(K)\setminus p(Y)$ gives an injective map of bundles $\mathcal{O}_{\mathbb{P}(V)}\to E$, thus we obtain the exact sequence

$$0\to\mathcal{O}_{\mathbb{P}(V)}\to\bigwedge^2T_{\mathbb{P}(V)}(-2)\to F\to0.$$

Now, since $\bigwedge^2T_{\mathbb{P}^3}(-2)\cong\Omega_{\mathbb{P}^3}(2)$ it seems that $F\cong N(1)$.

How to prove this?

  • What is the map $\mathcal{O}{\mathbb{P}(V)}\to\bigwedge^2T{\mathbb{P}(V)}(-2)$ and why does it define the Tango bundle, which I have never seen described this way ? – Georges Elencwajg Jan 09 '16 at 23:20
  • Dear Georges Elencwajg, I made an edit. Basically this is the construction from the book of Okonek, Schneider and Spindler, but taking the second exterior power of the Euler sequence, instead of $(n-1)$-th. However for $n=3$ these constructions are absolutely the same. But if you provide the proof by using some different definition your answer will be welcomed also. – mahavishnu Jan 10 '16 at 01:06
  • Ah, thank you very much for your edit, mahavishnu. Unfortunately, I cannot answer your interesting question. Why not ask on MathOverflow (and explaining that you asked the question here with no result) ? By the way, you write mathematics very clearly, in the great Indian tradition: bravo! – Georges Elencwajg Jan 10 '16 at 08:52
  • Why is $H^0(\mathbb{P}^3,E) \cong \Lambda^2V$? – u184 Apr 20 '19 at 00:11

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