Let $C\subset\mathbb{P}^n$ be a smooth curve, and let $Y\subseteq\mathbb{P}^n$ be its tangent developable. 


Given two general points $y\_1,y\_2\in Y$ does there exist a smooth curve $\Gamma\subset Y$ joining $y\_1$ and $y\_2$ ?


Assume that $C$ is rational. In this case does there exist a smooth rational curve in $Y$ connecting $y\_1$ and $y\_2$ ?



Smooth curves in Tangent Developables
**Definition of the Tangent Developable Surface.** Let $k$ be a field. For every $k$-scheme $X$ and for every quasi-coherent $\mathcal{O}\_X$-module $\mathcal{F}$, denote by $\mathcal{P}^1\_X(\mathcal{F})$ the bundle of principal parts $\text{pr}\_{2,\*}(\text{pr}\_1^\*\mathcal{F}\otimes \mathcal{O}/\mathcal{I}^2)$ with respect to the two projections $\text{pr}\_1,\text{pr}\_2:X\times X \to X$ and the ideal sheaf $\mathcal{I}$ of the diagonal, $X\to X\times X$. This fits into a natural short exact sequence, $$0\to \Omega\_{X/k}\otimes \mathcal{F} \xrightarrow{u\_{\mathcal{F}}} \mathcal{P}^1\_X(\mathcal{F}) \xrightarrow{v\_{\mathcal{F}}} \mathcal{F} \to 0.$$ This is functorial in $\mathcal{F}$ and in $X$. The homomorphism of sheaves of rings $\text{pr}\_2^{\#}$ defines a splitting of $\text{pr}\_2^{\#}:\mathcal{O}\_X \to\mathcal{P}^1\_X(\mathcal{O}\_X)$ of $v\_{\mathcal{O}\_X}$. Thus, for every $\mathcal{O}\_X$-module together with a global trivialization, i.e., for $V\otimes\_k \mathcal{O}\_X$, there is an associated splitting $s\_V:V\otimes\_k \mathcal{O}\_X \to \mathcal{P}^1\_X(V\otimes\_k \mathcal{O}\_X)$. In particular, for the projective space $X$ of a finite-dimensional $k$-vector space $V$ together with its canonical invertible quotient $q:V\otimes\_k \mathcal{O}\_X \to \mathcal{O}(1)$, there is a composite morphism, $$V\otimes\_k \mathcal{O}\_X \xrightarrow{s\_V} \mathcal{P}^1\_X(V\otimes\_k \mathcal{O}\_X) \xrightarrow{\mathcal{P}^1(q)} \mathcal{P}^1\_X(\mathcal{O}(1)). $$ This composite is an isomorphism, and the short exact sequence above gives the Euler sequence.


For an unramified morphism $g:Y\to X$, the induced morphism $g^\*\Omega\_X\to \Omega\_Y$ is surjective. Thus the following homomorphism is also surjective, $$\mathcal{P}^1\_g(\mathcal{O}(1)):g^\*\mathcal{P}^1\_X(\mathcal{O}(1)) \to \mathcal{P}^1\_Y(g^\*\mathcal{O}(1)).$$ By the previous paragraph, this is equivalent to a surjection, $$ \mathcal{P}^1\_g(q):V\otimes\_k \mathcal{O}\_Y \to \mathcal{P}^1\_Y(g^\*\mathcal{O}(1)). $$ On the one hand, if $Y$ is smooth of dimension $d$, this gives rise to the first associated map $\widetilde{g}:Y\to \text{Flag}(V;d+1,1)$ to the partial flag variety of flags of quotients of $V$. On the other hand, for the projective bundle $\pi:\mathbb{P}\to Y$ with universal invertible quotient $r:\pi^\*\mathcal{P}^1\_Y(g^\*\mathcal{O}(1)) \to \mathcal{L},$ the associated composite surjection $$V\otimes\_k \mathcal{O}\_{\mathbb{P}} \xrightarrow{\pi^\*\mathcal{P}^1\_g(q)} \pi\* \mathcal{P}^1\_Y(g^\*\mathcal{O}(1)) \xrightarrow{r} \mathcal{L},$$ defines a morphism $g':\mathbb{P} \to X$. There is a tautological cross section $s:Y\to \mathbb{P}$ that pulls back the invertible quotient $r$ to the invertible quotient $v\_{g^\*\mathcal{O}(1)}$. The composition $g'\circ s$ equals $g$. 


**The Ramification Locus of the Tangent Developable.** Now assume that $Y$ is a smooth projective curve over an algebraically closed field $k$ (much of what follows is fine in the quasi-projective case, but some care is required in the analysis below of "contracted curves"). Since $g$ is unramified, for every $k$-point $y$ of $Y$, there exist two elements of $V$ whose associated global sections of $g^\*\mathcal{O}(1)$ vanish to order $0$ and to order $1$ at $y$. Assume that the image of $g$ is not a line, and assume that for a general $y$, there also exists an element that vanishes to order $2$ at $y$. This is automatic in characteristic $0$, but it can fail in characteristic $p$ cf. Exercise IV.2.4, p. 305 of Hartshorne's "Algebraic Geometry". Call a $k$-point $y$ of $Y$ a "flex point" if every element that vanishes to order $2$ vanishes to order $3$ (this is probably the wrong name, but I do not remember the correct name).


The ramification locus of $g'$ is the support of the cokernel of $(g')^\*\Omega\_X \to \Omega\_{\mathbb{P}}$. This is the set of points at which $g'$ is ramified. The ramification locus of $g'$ equals the union of $s(Y)$ and every fiber $F=\pi^{-1}(\{y\})$ over a flex point $y$. The restriction $s^\*\Omega\_{\mathbb{P}}$ is canonical isomorphic to $\Omega\_Y\oplus \Omega\_Y$, and the cokernel of $\Omega\_X$ is the antidiagonal quotient $\Omega\_Y$. Similarly, the image of $\Omega\_X$ in $\Omega\_{\mathbb{P}}|\_F$ does surject onto $\Omega\_F$. 


Thus, for a smooth curve $\Gamma\subset \mathbb{P}$, so long as for each of the finitely many intersection points of $\Gamma$ with $s(Y)$, the tangent direction of $\Gamma$ is not an antidiagonal tangent direction, and so long as for each of the finitely many intersection points of $\Gamma$ with "flex fibers" $F$, the tangent direction of $\Gamma$ is not in the $1$-dimensional kernel of the derivative map, then the restriction of $g'$ to $\Gamma$ is unramified.


**The Non-Injective Locus of the Tangent Developable.** There is still the issue of whether $g'$ is injective on $\Gamma$. In positive characteristic this is a serious issue. For instance, for the standard plane conic $g:\mathbb{P}^1 \to X=\mathbb{P}^2$ by $[x,y]\mapsto [x^2,xy,y^2]$, in characteristic $2$ there is a cross section $t$ of $\pi$ that gets contracted under $g'$ to the point in $\mathbb{P}^2$ with coordinates $[0,1,0]$. Thus, if $\Gamma$ intersects this cross section in $>1$ point, then $g'$ is not injective on $\Gamma$. This is essentially the only example: every curve of $\mathbb{P}$ that is contracted by $g'$ is a cross section $t$ of $\pi$ (since $g'$ cannot contract any closed subscheme of a fiber of $\pi$ of length $2$ or more). There exists a contracted cross section $t$ if and only if $Y$ is a rational curve, the characteristic equals $2$, and the degree of $g^\*\mathcal{O}(1)$ equals $2$. Thus, assume either that $Y$ is not a conic in degree $2$, so that no curve in $\mathbb{P}$ is contracted by $g'$.


If the image of $g$ is contained in a $2$-plane, then the image of $g'$ is a $2$-plane Thus, assume that the image of $g$ is contained in no $2$-plane. In that case, to general fibers of $\pi$ map to disjoint lines in $X$. Thus the non-injectivity locus in $\mathbb{P}\times \mathbb{P}$ is the union of the diagonal and finitely many curves. Each curve component $B$ maps finitely to its images $B\_1$ and $B\_2$ under both projections to $\mathbb{P}$. Thus, so long as for each of the finitely many curves $B$ in the non-injectivity locus, for each of the finitely many intersection points $p\_1$ of $\Gamma$ with $B\_1$, for each corresponding point $p\_2$ in $B\_2$, $p\_2$ is not one of the finitely many intersection points of $\Gamma$ with $B\_2$, then $g'$ is injective on $\Gamma$. Thus, for a sufficiently general, sufficiently ample curve $\Gamma$ on $\mathbb{P}$, $g'$ restricts to an embedding on $\Gamma$.


Finally, in the case when $Y$ is a conic in degree $2$, every line in the plane gives a smooth curve $\Gamma$.


