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Let $X$ be a smooth projective variety. By Definition 24.4.2 in the 2003 book Mirror Symmetry, $X$ is called convex if $h^1\left( \Sigma, f^*T_X \right) = 0$ for every genus zero stable map $f:\Sigma \to X$.

Question: Why is a convex variety called convex?

On page 505 of the book, it was said that $H^1\left( \Sigma, f^*T_X \right) $ measures obstructions to the deformations of the map $f$ (when the structure of the source curve $\Sigma$ is fixed). But I don't know what the obstructions mean, or what zero obstruction means. So I'm also asking the following

Related question: What do zero and nonzero obstructions to the deformations of the map $f$ mean?

My background and what answer I'm looking for: I have close-to-zero background in deformation theory. So I don't expect to fully understand the precise mathematical meaning over one night. I would appreciate it very much if someone could explain what zero and nonzero obstructions mean in an intuitive way. If this can be further supported by concrete examples for zero and nonzero obstructions, then it will be perfect.

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If there are no obstructions to deformation, i.e. $h^1=0$, then Kodaira's deformation theorem says that the map admits a family of deformations, forming a manifold, near that given map, with tangent space given by $H^0$. You think of genus zero curves as like lines, or line segments, and deforming a line while staying in the variety should be like trying to move around on a set of points in Euclidean space while still being able to move your line segment with you, something you can do when points of the set are connected by line segments, i.e. in a convex set.

For example, if $f$ is the identity map of the projective line, then the space of deformations of $f$ is the space of maps of degree 1, i.e. the space of projective linear transformations of the projective line. The tangent space at the identity element of the group of projective linear transformations is the Lie algebra of vector fields on the projective line, i.e. $H^0$ is the space of linear vector fields, i.e. the space of $2 \times 2$ traceless matrices, so 3 dimensional.

Another example: if $f$ is constant, then the deformations are constant maps, and if the target $X$ is smooth at the image point $x=f(\mathbb{P}^1)$, then $H^0=T_x X$.

Kodaira's book on deformation theory is where I learned about this stuff. Maybe there are better references, especially in the algebraic category.

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  • $\begingroup$ Thanks for the explanation. It is very intuitive and helpful. Also thanks for the reference. Can you say something about $H^1(\Sigma, f^*T_X)$ in the following two cases? Case 1: The map $f$ is a constant, i.e., it maps the source curve to a point. Case 2: The map $f$ is an identity from the line $\mathbb{CP}^1$ to itself. $\endgroup$
    – Yuhang Chen
    Commented Jan 7, 2020 at 15:24
  • $\begingroup$ So both cases have zero obstructions, right? $\endgroup$
    – Yuhang Chen
    Commented Jan 7, 2020 at 16:40
  • $\begingroup$ In case one, $f^*T_X$ is trivial, so $H^1 = g(\Sigma) = 0$. In case 2, $f^* T_X = \mathcal O_{\mathbb P^1}(2)$ so $H^1 = 0$ by standard calculations. So yes, both of these are unobstructed. $\endgroup$
    – Tabes Bridges
    Commented Jan 7, 2020 at 20:37
  • $\begingroup$ @TabesBridges In case one, did you get the result from the Riemann-Roch theorem, i.e., $h^0(\Sigma,L) -h^1(\Sigma,L) = \deg L -g-1$ where the line bundle $L = f^*T_X \cong \Sigma \times \mathbb{C}$? In case 2, excuse my ignorance but what did you mean by standard calculations? $\endgroup$
    – Yuhang Chen
    Commented Jan 7, 2020 at 23:49
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    $\begingroup$ @YuhangChen this is a trick you should get familiar with. The only way I can remember the full calculation of cohomology for line bundles on $\mathbb P^n$ is via the slogan "line bundles of non-negative degree have global sections. the only other cohomology is $H^n$ forced by those global sections and Serre duality." $\endgroup$
    – Tabes Bridges
    Commented Jan 8, 2020 at 4:11

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