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Is there a smooth variety $X$ which is a one point compactification of the tangent bundle of $\mathbb P^1$?

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    $\begingroup$ No. Consider $SO(3) \cong T^1(S^2) \subset X$. This splits $X$ into two pieces. The piece containing infinity is homeomorphic to the one-point compactification of $SO(3) \times [0,\infty)$, also known as the cone on $SO(3)$. This space is not even a manifold, which it would be if $X$ were smooth. $\endgroup$
    – mme
    Jan 14 '20 at 18:22
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Did you mean $P^1({\mathbb C})$?

Then, no, because it would a smooth rational projective surface, and we know all of them: blow-ups of $P^2$ or $n$-th Hirzebruch surface.

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    $\begingroup$ This is a bit nonaccurate --- there are other smooth rational surfaces, obtained by blowups. But a smooth one-point compactification would also have Picard group of rank 1, and this property leaves only one possibility --- $\mathbb{P}^2$, $\endgroup$
    – Sasha
    Jan 14 '20 at 18:46
  • $\begingroup$ Right! I forgot blow-ups... $\endgroup$
    – Bugs Bunny
    Jan 15 '20 at 11:40
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No, there is not; one way to see this is from an alternative description of the 1 point compactification $T \mathbb{P}^1 \cup \{\infty\}$.

Notation/convention: For a locally free sheaf $\mathcal{F}$ on a variety $X$, the associated geometric vector bundle is $V(\mathcal{F}) := \mathrm{Spec}_X \mathrm{Sym}^* \mathcal{F}^\vee$.

Claim: $T \mathbb{P}^1 \cup \{\infty\}$ is isomorphic to the cone over a conic $C \subset \mathbb{P}^2$. Explicitly, $T \mathbb{P}^1 \cup \{\infty\} \simeq V(xy - z^2) \subset \mathbb{P}_{xyzw}^3$ (which is singular at the vertex $[0, 0, 0, 1]$).

Step 1: $T \mathbb{P}^1 \simeq V(\mathcal{O}(2))$. Since $T \mathbb{P}^1$ is a line bundle on $\mathbb{P}^1$ it must be $\mathcal{O}(d)$ for some $d$. To compute $d =2$ one can use the Euler sequence $$ 0 \to \mathcal{O} \to \mathcal{O}(1)^{\oplus 2} \to \mathcal{T} \mathbb{P}^1 \to 0 $$

Step 2: In general, if $X \subset \mathbb{P}^n$ is a projective variety, and $C(X) \subset \mathbb{P}^{n+1}$ is the projective cone over $X$ with vertex $v = [0, \dots, 0, 1]$, then $C(X) \setminus \{ v \} \simeq V(\mathcal{O}_X (1))$.

When $X = \mathbb{P}^n$ itself the isomorphism can be described geometrically as follows: over a point $L \in \mathbb{P}^n$ corresponding to a line $L \subset \mathbb{C}^{n+1}$, the fiber of $V(\mathcal{O}(1))$ is the space of linear functionals $\lambda: L \to \mathbb{C}$. For each such $\lambda$ we get an embedding $\iota_L \times \lambda: L \to \mathbb{C}^{n+2}$ (here $\iota_L : L \to \mathbb{C}^{n+1}$ is the inclusion). So, we obtain a morphism $\varphi: V(\mathcal{O}(1)) \to \mathbb{P}^{n+1}$ sending $(L, \lambda) \mapsto \mathrm{im}(\iota_L \times \lambda)$. One can check that $\mathrm{im} \varphi = \mathbb{P}^{n+1} \setminus \{ [0, \dots, 0, 1] \} $.

Step 3: Now apply step 2 to a conic $C \subset \mathbb{P}^2$. Since $C$ is the image of the 2nd Veronese (a.k.a. 2-uple) $\mathbb{P}^1 \to \mathbb{P}^2$ coming from the complete linear system of $\mathcal{O}_{\mathbb{P}^1}(2)$, under the isomorphism $\mathbb{P}^1 \simeq C$ we have $\mathcal{O}_C (1) \simeq \mathcal{O}_{\mathbb{P}^1}(2)$.

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