3
$\begingroup$

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 $y_2$ ?

$\endgroup$
  • $\begingroup$ For the curve $C$, form the extension $E_C$ of the structure sheaf $\mathcal{O}_C$ by the dualizing sheaf $\omega_C$ coming from $1\in \text{Ext}_{\mathcal{O}_C}^1(\mathcal{O}_C,\omega_C)$. For every morphism of curves $f:C'\to C$, the pullback map $f^*:f^*\omega_C \to \omega_{C'}$ induces a pullback map of extensions $f^*E_C\to E_{C'}$. Thus, a splitting of $E_C\to \mathcal{O}_C$ after pullback by $f$ would induce a splitting of $E_{C'}\to \mathcal{O}_{C'}$. Therefore the tautological cross section of $\mathbb{P}(E_C)$ is ample. This maps to the singular locus of the tangent developable. $\endgroup$ – Jason Starr Jun 18 '16 at 18:56
  • 1
    $\begingroup$ I re-read the question, and now I see that you are not asking whether there is a curve $\Gamma$ that is disjoint from the singular locus of $Y$, you are asking whether $\Gamma$ itself is smooth. This should "usually" be true. For a smooth curve $\Gamma$ in $\mathbb{P}(E_C)$, so long as the tangent direction of $\Gamma$ is transverse to the antidiagonal at each of the finitely many points of intersection with the tautological cross section, then the morphism from $\Gamma$ to $\mathbb{P}^n$ is unramified. $\endgroup$ – Jason Starr Jun 18 '16 at 19:05
  • $\begingroup$ I want to correct my first comment: the argument with extensions classes does prove that every curve $\Gamma$ in $\mathbb{P}(E_C)$ intersects the tautological cross section. However, that does not mean that the tautological cross section is an ample divisor on $\mathbb{P}(E_C)$, and usually it is not. The normal bundle of that divisor is $T_C$. This is ample if and only if $C$ is a rational curve. $\endgroup$ – Jason Starr Jun 18 '16 at 19:19
  • $\begingroup$ There are some further corrections in positive characteristic; please see the answer below. $\endgroup$ – Jason Starr Jun 19 '16 at 16:47
1
$\begingroup$

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 $\Gamma$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.