For a general plane curve of degree $\ge 4$, the number of bitangent lines is known. Also, I found that the number of tritangent planes have been worked out for some space curves given by intersection of two surfaces (e.g., quadric and cubic).

I have the following related questions:

(1) Is it true that every smooth plane curve of degree $\ge 4$ has at least one bitangent line?

(2) Let $C$ be a curve given by a complete intersection in $\mathbb{P}^n$ of large degree, (possibly with mild singularities only), is it true that there is some multi-tangent hyperplane? More precisely, I want to show that there is a hyperplane tangent to the curve at $n \choose 2$ points.

  • $\begingroup$ For a "general" curve in $\mathbb{P}^n$, I would expect that there exist hyperplanes that are tangent at $n-1$ points, but no hyperplanes tangent at $n$ points. Why do you want $\binom{n}{2}$ points of tangency? $\endgroup$ Mar 27 '14 at 9:20
  • $\begingroup$ I want $n \choose 2$ points because I expect `in general'' the tangent vectors span $\wedge^2 V$, where $\mathbb{P}V$ is the hyperplane. $\endgroup$ Mar 27 '14 at 9:52
  • $\begingroup$ Hey, wait a minute! Didn't I already answer another of your questions where I proved that the tangent vectors need not span $\bigwedge^2 V$? $\endgroup$ Mar 27 '14 at 20:51
  • $\begingroup$ Yes, you gave an example. But the curve $C \subset \mathbb{P} W$ itself was tangentially degenerate, in the sense that the span of all the tangent vectors is a proper subspace of $\wedge^2 W$. In my case (complete intersection curve for example) it is tangentially non-degenerate. I think in this case, the question may make sense. $\endgroup$ Mar 27 '14 at 23:53
  • $\begingroup$ In comment above, I was off by $1$: as my answer below explains, there are $n$-tangent hyperplanes, but no $(n+1)$-tangent hyperplanes. $\endgroup$ Mar 29 '14 at 19:22

This is an answer to the second question, for which integers $r$ does a "general" complete intersection curve has an $r$-tangent hyperplane. The computation below show that for every integer $r>n$, for a sufficiently "general" complete intersection curve $C$, there exists no $r$-tangent hyperplane $H$. For $r=n$, the argument below (essentially) proves that a general complete intersection curve has a finite number of $n$-tangent hyperplanes. At any rate, for $n\geq 4$, there is no $\binom{n}{2}$-tangent hyperplane.

Let $V$ be a vector space of dimension $n+1$ so that $\mathbb{P}(V)$ is isomorphic to $\mathbb{P}^n$. Denote by $\mathbb{P}(V^\vee)$ the parameter space for hyperplanes $H\subset \mathbb{P}(V)$. Denote by $\mathcal{H}\subset \mathbb{P}(V^\vee)\times \mathbb{P}(V)$ the universal hyperplane, i.e., the parameter space $\textbf{Flag}(1,n,V)$ of pairs $([H],q)$ of a hyperplane $H\subset \mathbb{P}(V)$ and a point $q\in H$. Finally, denote by $$\rho:\mathbb{P}T \to \mathcal{H},$$ the projective bundle of the relative tangent bundle of $$\text{pr}_1: \mathcal{H} \to \mathbb{P}(V^\vee),$$ i.e., the relative Proj, $$\mathbb{P}T = \textbf{Proj}_{\mathcal{H}}(\text{Sym}^\bullet_{\mathcal{O}_{\mathcal{H}}} \Omega_{\text{pr}_1}).$$ This is the parameter space for triples, $([H],q,L)$ of a hyperplane $H\subset \mathbb{P}(V)$, a point $q\in H$ and a line $L\subset \mathbb{T}_q H$. In fact, $\mathbb{P}T$ is also the partial flag variety $\textbf{Flag}(1,2,n,V)$ together with its obvious projection to $\textbf{Flag}(1,n,V)$.

For every integer $r\geq 0$, denote by $$\mathcal{H}_r=\mathcal{H}\times_{\mathbb{P}(V^\vee)} \dots \times_{\mathbb{P}(V^\vee)} \mathcal{H},$$ the $r$-fold fiber product of $\mathcal{H}$ over $\mathbb{P}(V^\vee)$, i.e., the parameter space for tuples $([H],(q_1,\dots,q_r))$ of a hyperplane $H\subset \mathbb{P}(V^\vee)$ together with $r$ points, $q_1,\dots,q_r\in H$. For $i=1,\dots,r$, denote by $$\pi_i:\mathcal{H}_r \to \mathcal{H},$$ the projection onto the $i^{\text{th}}$ factor, i.e., $\pi_i([H],(q_1,\dots,q_r))$ equals $([H],q_i)$. For every $i=1,\dots,r$, denote by $$ \pi_i^*\rho :\pi_i^*\mathbb{P}T \to \mathcal{H}_r, $$ the fiber product of $\rho$ and $\pi_i$. Denote by $$\rho_r : (\mathbb{P}T)_r \to \mathcal{H}_r,$$ the fiber product from $i=1,\dots,r$ of $\pi_i^*\rho$, i.e., the parameter space for tuples $([H],((q_1,L_1),\dots,(q_r,L_r)))$ of a hyperplane $H\subset \mathbb{P}(V)$, points $q_i\in H$ and tangent line $L_i \subset T_{q_i}H$.

Denote by $$\mathcal{H}_r^o\subset \mathcal{H}_r$$ the open set that is the complement of all of the diagonals, i.e., the parameter space for $([H],(q_1,\dots,q_r))$ such that $(q_1,\dots,q_r)$ are pairwise distinct points. Denote by $(\mathbb{P}T)_r^o$ the inverse image $\rho_r^{-1}(\mathcal{H}_r^o)$. Of course $\text{dim}\mathbb{P}(V^\vee)$ equals $n$, the fiber dimension of the smooth, Zariski locally trivial fiber bundle $\mathcal{H} \to \mathbb{P}(V^\vee)$ equals $n-1$, the fiber dimension of $\mathbb{P}T$ over $\mathcal{H}$ equals $n-2$, and thus the dimension of $(\mathbb{P}T)_r^o$ equals $$n + r((n-1) + (n-2)) = (2n-3)r + n.$$

Denote by $Z \subset (\mathbb{P}T)_r^o \times \mathbb{P}(V)$ the unique closed subscheme such that (i) the projection $\phi$ of $Z$ to $(\mathbb{P}T)_r^o$ is finite and flat of length $2r$, and (ii) for every (geometric) point $([H],((q_1,L_1),\dots,(q_r,L_r)))$ in $(\mathbb{P}T)_r^o$, the fiber of $\phi$ is the disjoint union from $i=1,\dots,r$ of the length $2$ closed subscheme supported at $q_i$ with tangent direction $L_i$. Denote by $\psi:Z\to \mathbb{P}(V)$ the second projection. For every integer $d\geq 0$, denote by $\alpha_{r,d}$ the natural "evaluation" homomorphism of $\mathcal{O}_{(\mathbb{P}T)_r^o}$-modules, $$ \alpha_{r,d}:\text{Sym}^d(V^\vee)\otimes \mathcal{O}_{(\mathbb{P}T)_r^o} \to \phi_*\psi^*\mathcal{O}_{\mathbb{P}(V)}(d).$$ Since the relative Hilbert polynomial of $Z$ over $(\mathbb{P}T)_r^o$ is the constant polynomial $2r$, there exists an integer $d_o(r)$ such that for every $d\geq d_0(r)$, $\alpha_{r,d}$ is surjective with locally free target of rank $2r$. Concretely, every fiber of $\phi$ imposes $2r$ independent linear conditions on polynomials of degree $d\geq d_0(r)$.

Let $\mathbf{d}$ be an $(n-1)$-tuple of integers, $(d_1,\dots,d_{n-1})$ such that every $d_i\geq d_0(r)$. Denote by $\mathbb{P}_{\mathbf{d}}$ the product of projective spaces, $$\mathbb{P}_{\mathbf{d}} = \mathbb{P}(\text{Sym}^{d_1}(V^\vee))\times \dots \times\mathbb{P}(\text{Sym}^{d_{n-1}}(V^\vee),$$ parameterizing $(n-1)$-tuples, $([Y_1],\dots,[Y_{n-1}])$ of hypersurfaces $Y_i \subset \mathbb{P}(V)$ of degree $d_i$. Denote by $\mathbb{P}_{\mathbf{d}}^o\subset \mathbb{P}_{\mathbf{d}}$ the open subset such that the intersection $C=Y_1\cap \dots \cap Y_{n-1}$ has codimension $(n-1)$ in $\mathbb{P}^n$, i.e., such that $C$ is a complete intersection curve. Denote by $I_{\mathbf{d},r}\subset \mathbb{P}_{\underline{d}}^o\times (\mathbb{P}T)_r^o$ the closed subset of pairs, $$I_{\mathbf{d},r} =\{([Y_1],\dots,[Y_{n-1}]),([H],((q_1,L_1),\dots,(q_r,L_r)))) : Z\subset C\}.$$ Of course the condition is equivalent to the condition that every $Y_i$ contains $Z$. Since every $d_i\geq d_0(r)$, for every $i=1,\dots,n-1$, it is $2r$ independent linear conditions for $Y_i$ to contain $Z$. Thus the total codimension of $I$ is $(n-1)2r = (2n-2)r$. On the other hand, the relative dimension of the projection, $$ \text{pr}_1:\mathbb{P}_{\mathbf{d}}^o \times (\mathbb{P}T)_r^o \to \mathbb{P}_{\mathbf{d}}^o,$$ equals $(2n-3)r+n$.

Thus, whenever $r>n$ so that $(2n-2)r>(2n-3)r+n$, then the codimension of $I_{\mathbf{d},r}$ is greater than the relative dimension of $\text{pr}_1$ so that the morphism $$\text{pr}_1|_I:I_{\mathbf{d},r}\to \mathbb{P}_{\mathbf{d}}^o,$$ cannot be dominant. Therefore, for $r>n$, for a "general" $([Y_1],\dots,[Y_{n-1}])$ in $\mathbb{P}_{\mathbf{d}}^o$, there is no $([H],((q_1,L_1),\dots,(q_r,L_r)))$ such that $C$ contains $Z$, i.e., there is no hyperplane $H$ that is tangent to $C$ at each of $r$ distinct points $q_1,\dots,q_r$. Similarly, when $r=n$, the argument proves that the relative dimension of $\text{pr}_1|_I$ is at most $0$, i.e., a general complete intersection has at most finitely many $n$-tangent hyperplanes.

Edit. I just want to "head off" one possible objection, as well as confirming that $I_{\mathbf{d},n}$ is indeed dominant over the parameter space $\mathbb{P}_{\underline{d}}^o$. The possible objection is that, perhaps, there is a hyperplane $H$ that is $(n+1)$-tangent to $C$ in some "degenerate" sense, even though there is no hyperplane that is tangent to $C$ at $n+1$ distinct points. Of course that is possible for "special" $C$. However, if $C$ is smooth and linearly nondegenerate, then the intersection of $H$ with $C$ is a zero-dimensional curvilinear scheme contained in $H$. Denote by $$\widetilde{\rho}_r: \text{Hilb}_{\mathcal{H}/\mathbb{P}(V^\vee)}^{2r,\text{curv}} \to \mathbb{P}(V^\vee),$$ the open subset of the relative Hilbert scheme whose fiber over $[H]$ parameterizes curvilinear closed subschemes $Z\subset H$ of length $2r$. Then there is a natural morphism of $\mathbb{P}(V^\vee)$-schemes, $$ i : (\mathbb{P}T)_r^o \to \text{Hilb}_{\mathcal{H}/\mathbb{P}(V^\vee)}^{2r,\text{curv}}, $$ whose image is just the quotient of $(\mathbb{P}T)_r^o$ by the natural action of the symmetric group $\mathfrak{S}_r$. Thus, the closure of the image is still flat (in fact a Zariski locally trivial fiber bundle) over $\mathbb{P}(V^\vee)$ of the same relative dimension $(2n-3)r$. Up to replacing $(\mathbb{P}T)_r^o$ by the closure of this image, the rest of the argument goes through precisely as bfore. Thus, even for this "degenerate" notion of $(n+1)$-tangent, for $C$ a "general" complete intersection curve, there is no $(n+1)$-tangent hyperplane.

The second comment concerns the case that $r=n$. The argument above proves that the dimension of $I_{\textbf{d},n}$ equals the dimension of $\mathbb{P}_{\mathbf{d}}^o$, so that the morphism $\text{pr}_1|_I$ has fiber dimension $0$ if it is dominant. However, the argument does not quite prove that the morphism is dominant. Notice though, that the argument does prove that $I_{\textbf{d},n}$ is smooth of the given dimension: it is cut out in the smooth product space $\mathbb{P}_{\mathbf{d}}^o\times (\mathbb{P}T)_r^o$ by a system of independent linear equations. Thus, the morphism $\textbf{pr}_1|_I$ is flat of relative dimension $0$ at every point where the fiber has dimension $0$.

Now let each $Y_i$ be a union of $d_i$ hyperplanes, such that all the $d_1+\dots + d_{n-1}$ hyperplanes are "general". The corresponding curve $C$ is a union of $d_1\cdot \dots \cdot d_{n-1}$ lines with only double point singularities. Certainly there will be hyperplanes $H$ where the fiber has positive dimension, for instance the hyperplanes that contain one irreducible component of $C$. However, if you choose $n$ general double points, no two of which are contained in a common irreducible component of $C$, then the hyperplane they span will be an isolated point of the fiber. Thus the fiber is locally zero dimensional. Thus the flat locus of $\text{pr}_1|_I$ is a nonempty open subset of $I$. Since $I$ is irreducible, it is a dense open subset of $I$. Therefore $\text{pr}_1|_I$ is dominant.

  • $\begingroup$ Thanks a lot! I'm still reading the details, but the argument seems to provide enough tools to work out for the question. I'll come back again when I read through the details. $\endgroup$ Mar 31 '14 at 8:36

1) Yes. The number of bitangents is given by the Plücker formula, see http://en.wikipedia.org/wiki/Plucker_formula.

2) That would imply that the degree of your curve is $\geq n(n-1)$, which is obviously not always the case.

Edit: My previous answer was wrong, sorry!

  • $\begingroup$ Thanks. For 2), I assume that the degree of $C$ is large. For instance, I assume $C$ is a non-degenerate complete intersection, so $\deg C \ge 2^{n-1}$. $\endgroup$ Mar 27 '14 at 9:55
  • $\begingroup$ Thank you for the answer. According to my computation, the expression $(n-1)(n-2)$ seems to be $(n-1)(2n-1)$, so there are a little more exceptions than the case $(2,2)$. $\endgroup$ Mar 27 '14 at 11:05
  • $\begingroup$ Yes, you are right. $\endgroup$
    – abx
    Mar 27 '14 at 11:15
  • $\begingroup$ Oops! Stupid computational mistake. My previous answer was definitely wrong, sorry! $\endgroup$
    – abx
    Mar 27 '14 at 12:37

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