Smoothness of moduli spaces of stable maps

Question

If $X$ is a projective variety the moduli space of stable maps $\overline{M}_{0,0}(X,\beta)$ is a normal variety with finite quotient singularities.

Can the pairs $(X,\beta)$ such that $\overline{M}_{0,0}(X,\beta)$ is smooth as a variety be characterized?

I know that $\overline{M}_{0,0}(\mathbb{P}^2,2)$ is smooth. For instance, are $\overline{M}_{0,0}(G(2,4),2)$ and $\overline{M}_{0,0}(LG(2,4),2)$ smooth? Here $G(2,4)$ is the Grassmannian of planes in $\mathbb{C}^4$ and $LG(2,4)\subset G(2,4)$ the Lagrangian Grassmannian.

Thank you very much.

Answer 1

Here is a partial answer to your question. If $X$ is homogeneous then $\overline{M}_{0,0}(X,\beta)$ is a projective normal variety with at most finite quotient singularities. The singularities arise along the loci parametrizing maps with non trivial automorphisms. However, if such a locus is in codimension one the general point of the divisor is a smooth point of $\overline{M}_{0,0}(X,\beta)$ since it is locally at such a point the quotient of a smooth variety by a group generated by pseudoreflections.

For instance, consider in $\overline{M}_{0,0}(G(2,4),2)$ the locus $\Gamma$ of maps that are $2$ to $1$ onto a line. The Fano variety of lines in $G(2,4)$ is $5$-dimensional and for any line you have to choose two points on it identifying the ramification of the map. So $\Gamma$ has codimension $2$ in $\overline{M}_{0,0}(G(2,4),2)$ and since the general map in $\Gamma$ has a non-trivial involution $\overline{M}_{0,0}(G(2,4),2)$ is singular along $\Gamma$.

A similar argument shows that the locus in $\overline{M}_{0,0}(LG(2,4),2)$ parametrizing $2$ to $1$ maps onto a line is a divisor. So $\overline{M}_{0,0}(LG(2,4),2)$ is smooth.