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Let $X=G/P$ be a homogeneous space where $G$ is a connected, simply connected, simple (in the sense that the root system $R$ is irreducible), complex, linear algebraic group and $P$ is a parabolic subgroup of $G$. Let $\overline{M}_{0,2}(X,d)$ be the (coarse) moduli space of stable maps to $X$ (of genus zero and degree $d$ with $2$ marked points). Let $\overline{M}(1)=\mathrm{ev}_1^{-1}(1.P)$ be the space of all stable maps which pass trough $1.P$ where $\mathrm{ev}_1$ is the first evaluation map on $\overline{M}_{0,2}(X,d)$. Let $$(\mathbb{P}^1,p_1,p_2,f\colon\mathbb{P}^1\to X)$$ be a point in $\overline{M}(1)$, i.e. we have $f_*[\mathbb{P}^1]=d$ and $f(p_1)=1.P$. We can consider the orbit $Pf$ of $f$ in $\overline{M}(1)$ under the action of $P$. Intuitively, we have $$ \mathrm{dim}(Pf)=\#\{\text{tangent directions in }\mathfrak{g}/\mathfrak{p}\text{ of }f\text{ at the second marked point }p_2\} $$ where $\mathfrak{g}$ and $\mathfrak{p}$ are the Lie algebras of $G$ and $P$ respectively.

Q1: How can I compute $\mathrm{dim}(Pf)$, i.e. how can I count the number of tangent directions of $f$ at $p_2$? If you like, you can assume that $f$ is of the form $$ t\mapsto\exp(tx_\theta)1.P $$ for some $x_\theta$ which is the sum of distinct root vectors.

Q2: From the above formula for $\mathrm{dim}(Pf)$, it is immeadiate that $\mathrm{dim}(Pf)\leq\mathrm{dim}(X)$. On the other hand, it is well-known that $$ \mathrm{dim}\left(\overline{M}(1)\right)=\int_d c_1(T_X)-1\,. $$ If we have $$ \dim(X)<\int_d c_1(T_X)-1 $$ (which can well happen, for example if $R$ is of type $\mathsf{A}_4$ and $P$ is the maximal parabolic associated to the third simple root, in which case this inequality becomes $6<2\cdot 5-1=9$), how can it still happen that $f$ has a dense orbit in $\overline{M}(1)$? (This is a "paradox" which I cannot explain to myself.)

I would appreciate it very much if someone can help me with these issues. Thank you in advance.

Remark: Concerning Q1, I have some clues how to count the tangent directions, but Q2 proves them to be false - hence the question.

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  • $\begingroup$ For some choices of curve class $d$ (when $d$ is sufficiently "positive"), the morphism of evaluation at both marked points, $ev_{1,2}:\overline{M}_{0,2}(X,d)\to X \times X$ can be surjective with positive dimensional fibers. Thus, it can certainly happen that $\text{dim}(\overline{M}(1))$ has strictly larger dimension that $X$. Consider, for instance, the case that $X$ is $\mathbb{P}^n$ and $d\geq 2$. $\endgroup$ – Jason Starr Dec 2 '16 at 17:41
  • $\begingroup$ @JasonStarr: I totally agree with your comment, thanks. The "paradox" I was speaking about rather occurs for $\mathbb{G}(3,5)$ and $d=2$, in which case I highly expect a dense orbit, although I have a strict inequality $6<9$. The question was not very precise about this. $\endgroup$ – user66288 Dec 3 '16 at 10:23
  • $\begingroup$ To be more concrete, in this case ($\mathbb{G}(3,6)$, $d=2$), I expect a dense orbit in $\overline{M}_{0,3}(X,2)$, via fixing two points and counting tangent directions at the third, which I thought somehow to be equivalent to the case of two marked points and one fixed, by dropping one point and still producing a dense orbit. $\endgroup$ – user66288 Dec 3 '16 at 10:40
  • $\begingroup$ I am a little bit confused about the case you are considering, but there is a theorem of Buch, Kresch, Tamvakis that $\overline{M}_{0,3}(\text{Grass}(\mathbb{C}^k,\mathbb{C}^{2k}),k)\to \text{Grass}(\mathbb{C}^k,\mathbb{C}^{2k})^3$ is birational. $\endgroup$ – Jason Starr Dec 3 '16 at 10:47
  • $\begingroup$ This is true, I am more or less aware of this theorem. As a consequence of this birational map, or in the course of its proof, you see that $\overline{M}_{0,3}(\mathbb{G}(k,2k),k)$ has a dense orbit under the action of $G$. Now, I wanted to generalize similar statements to other spaces and predicted the "right" degrees for each space (i.e. those who give a dense orbit). By fixing points under $\mathrm{ev}$ you translate, according to my understanding, the problem to another moduli space and a different group acting (like $P$ for one fixed point). How does this match the counting argument? $\endgroup$ – user66288 Dec 3 '16 at 11:06

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