Convex varieties that are not homogeneous

Question

A projective variety $X$ is convex, if for any $f:\mathbb{P}^1 \to X$, the group $H^1(\mathbb{P}^1, f^*(T_X))$ vanishes. A big group of examples of convex varieties is made of homogeneous varieties. An homogeneous variety is the quotient variety $G/P$ of a Lie group $G$ by a parabolic subgroup $P$ of it.

My question is: is there an easy example of a projective variety, possibly smooth, that is convex but that does not admit such a description?

In the lectures: http://www.math.uic.edu/~coskun/utah-notes.pdf, the author asks whether every rationally connected smooth convex projective variety must also be homogeneous, thus suggesting that it should be rather easy to find such an example in the realm of non-rationally connected varieties.

Answer 1

Of course every variety containing no rational curves is convex, by default. For a less trivial example, take the product of one such variety (for example, an abelian variety) with a homogeneous variety.

Answer 2

This is a variation of Angelo's first example. Consider a surface without rational curves and blow-up a finite number of distinct points. If $E$ is one of the exceptional divisors and $f \colon E \to X$ is the inclusion in the blow-up, since

$H^0(E, N_{E|X})=H^1(E, N_{E|X})=0$,

we obtain $H^1(E, f^*T_X)=H^1(E, T_E)=0$, so $X$ is convex. On the other hand, $X$ is not homogeneous, since its automorphism group does not act transitively.

ADDED. As remarked by mdeland, this example does not really work since we can take finite covers of $E$ in order to obtain curves that violate the convexity condition. In order to avoid this problem, we must require that the splitting type of the tangent bundle of $X$ over any rational curve does not contain summands of negative degree. Let us give an example where this condition is satisfied.

Let $C$ be any curve of genus $g(C) \geq 1$ and let $\mathcal{E} = \mathcal{O}_C \oplus \mathcal{L}$, where $\mathcal{L}$ is a line bundle of negative degree $-e$. Then $X = \mathbb{P}(\mathcal{E}) $ is a ruled surface over $C$ which contains a unique section $C_0$ such that $C_0^2 = -e$, in particular $X$ is not homogeneous.

Now let us show that $X$ is convex. Let $F \cong \mathbb{P}^1$ be any fibre of $p \colon X \to C$. We have a short exact sequence

$0 \to T_F \to (T_X)|_F \to N_F \to 0$.

Since $T_F=\mathcal{O}_{\mathbb{P^1}}(2)$ and $N_F=\mathcal{O}_{\mathbb{P}^1}$, it follows $\textrm{Ext}^1(N_F, T_F)=0$. Therefore the sequence above actually splits and we obtain

$(T_X)|_F = \mathcal{O}(2) \oplus \mathcal{O}$.

On the other hand, since $g(C) \geq 1$ the unique rational curves on $X$ are the fibres of $p$, so every non-constant holomorphic map $f \colon \mathbb{P}^1 \to X$ is given by the inclusion of a fibre composed with a finite cover. If $d$ is the degree of such a cover, we obtain

$f^*T_X = \mathcal{O}(2d) \oplus \mathcal{O}$.

It follows $H^1(\mathbb{P}^1, f^*T_X)=0$, so $X$ is convex. Notice that $X$ is uniruled, but not rationally connected.