Let $X\subset \mathbb P^n$ be a smooth projective variety which is Fano and $M_{0,0}(X,e)$ the (projective) Kontsevich moduli space of rational curves of degree $e>1$. Is it possible (or are there examples) for an irreducible component whose general member is a generically injective morphism $\mathbb P^1\rightarrow X$ to contain no reducible rational curve?
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1$\begingroup$ There are silly examples where you embed $X$ by a positive integer multiple $e>1$ of an ample invertible sheaf. Then the degree of every curve is divisible by $e$. So even the "minimal" curves have degree $\geq e$. $\endgroup$– Jason StarrCommented Mar 19, 2016 at 18:46
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$\begingroup$ On the other hand, it seems almost certain (based on parameter counts) that for every Fano complete intersection in $\mathbb{P}^n$, for every $e>1$, every irreducible component of $\overline{M}_{0,0}(X,e)$ intersects the locus parameterizing reducible rational curves. Maybe the question has a positive answer if $\mathcal{O}_{\mathbb{P}^n}(1)$ generates the Picard group of $X$. $\endgroup$– Jason StarrCommented Mar 20, 2016 at 21:04
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$\begingroup$ Is it sufficient to say that $\bar{M}_{0,0}(X,e)\subset \bar{M}_{0,0}(\mathbb P^n,e)$ and that $\bar{M}_{0,0}(\mathbb P^n,e)$ is irreducible with divisors (hence with nonempty intersection with a complete curve) corresponding to maps with reducible domains? $\endgroup$– user3001Commented Mar 23, 2016 at 9:59
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$\begingroup$ "Is it sufficient to say ..." To what are you referring? As my example shows, it is not true that every complete curve intersects the boundary of $\overline{M}_{0,0}(\mathbb{P}^n,e)$. In fact, there exists a (projective) contraction of the boundary of $\overline{M}_{0,0}(\mathbb{P}^n,e)$ discovered independenty by Anca and Andrei Mustata, by Adam Parker, and by Izzet Coskun, Joe Harris and myself. So, even for Fano complete intersections, a general point of $\overline{M}_{0,0}(X,e)$ is contained in a curve that does not intersect the boundary. $\endgroup$– Jason StarrCommented Mar 23, 2016 at 10:16
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