Does there exist an irreducible surface, other than Hermitian surface, in $\mathbb{P}^3 (\mathbb{F}_q)$ containing three skew lines?

I know that this is true for Hermitian surface. In fact, at every point of a generator of a Hermitian surface, it admits a tangent plane containing the generator and at every tangent plane, it has $\sqrt{q}+1$ generators. But can this happen for an irreducible surface of degree at least three which is not a Hermitian surface?

Edit I think it may be useful to define the Hermitian surface. It is given by equation $$\sum_{i=0}^3 x_i^{\sqrt{q} + 1} = 0.$$

Thanks in advance!

  • 1
    $\begingroup$ I don't remember what a Hermitian surface is, but since there are surfaces with three skew lines in $\mathbb{P}^3(\mathbb{C})$, by general principles à la Lefschetz, there are such in $\mathbb{P}^3(\overline{\mathbb{F}_p})$ for $p\gg 0$, so there are such in $\mathbb{P}^3(\mathbb{F}_q)$ for $q=p^d$ with $d\gg 0$. $\endgroup$
    – Gro-Tsen
    Oct 31, 2017 at 13:36
  • $\begingroup$ By counting dimensions, there's a unique quadric surface containing three skew lines. (There's a $10$ dimensional space of quadrics, and $9$ conditions to contain $3$ points on each line.) The quadric must be irreducible, because three skew lines aren't contained in a union of two planes. $\endgroup$ Nov 2, 2017 at 18:56

2 Answers 2


Let $K$ be any field of characteristic not 3. (I will leave that case for someone more energetic.)

Let $S$ be the Clebsch diagonal surface defined in $\mathbf P^4_K$ by the equations

$$ \sum_{i=0}^4 x_i=0 \\ \sum_{i=0}^4 x_i^3 =0.$$

(Since the first equation defines a hyperplane in $\mathbf P^4$, this is really a cubic surface in $\mathbf P^3$.)

Then $S$ contains the three skew lines

$$ L_1 = \{ [s,t,-s,-t,0] \mid s, \ t \in K \}\\ L_2 = \{ [s,t,0,-s,-t] \mid s, \ t \in K \} \\ L_3 = \{ [s,-s,t,0,-t] \mid s, \ t \in K \}.$$

  • $\begingroup$ Excellent! I was wondering if we could find something similar like the Hermitian surfaces as I mentioned in the question? In particular, does there exist a surface which has three skew lines contained in three distinct planes that intersect in a common line? What happens if it is given that the common line is contained in the surface? $\endgroup$
    – M. D.
    Oct 31, 2017 at 15:09
  • 1
    $\begingroup$ Dear user116689, unfortunately I don't know either what is a Hermitian surface. If I have an opportunity to think about your further questions, I will. $\endgroup$
    – Pooter
    Oct 31, 2017 at 15:59

I also don't know what a Hermitian surface is, but it seems to me that there are lots of these surfaces for any given set of lines.

Let $L$ be the union of $m$ pairwise disjoint union of copies of $\mathbb P^r$ in $\mathbb P^n$ over a field $k$ and let $\mathscr I\subseteq \mathscr O_{\mathbb P^n}$ denote its ideal sheaf. A degree $d$ hypersurface in $\mathbb P^n$ containing $L$ corresponds to a global section of $\mathscr I(d)\subseteq \mathscr O_{\mathbb P^n}(d)$. So, (it seems) all you need is that $d$ is large enough so this would not be zero. This will follow as soon as $$ \dim_k H^0(\mathbb P^n, \mathscr O_{\mathbb P^n}(d)) > \dim_k \oplus^mH^0(\mathbb P^r, \mathscr O_{\mathbb P^r}(d)), $$ i.e., when $$ {n+d \choose d} > m\cdot{r+d \choose d}. $$ In particular, if $L$ is a disjoint union of $3$ lines in $\mathbb P^3$, then one needs that $$ {d+3\choose d} > 3 (d+1). $$ It seems to me that this happens as soon as $d\geq 2$. If most surfaces are not Hermitian, then this should give you plenty of examples as $d$ grows.

Note that this computation does not use that these are lines, only that they are isomorphic to $\mathbb P^1$ (or $\mathbb P^r$ in the general case), so you could include other rational curves. Furthermore, one can easily develop a formula for arbitrary curves where the actual bound on the degree would depend on the genus of the curves involved, but in any case I would expect that for any fixed genus one would get surfaces containing a fixed set of curves if one allows the degree to be sufficiently large.

EDIT: As M.D. correctly points out this procedure might produce reducible hypersurfaces. However, using the same formula one can count those and conclude that, at least for large $d$, there is still enough dimension left so the general member is indeed irreducible.

For simplicity let's do this for $n=3$, but the method is the same in higher dimensions: A reducible degree $d$ surface is a union of smaller degree surfaces. We can just think of a union of two who themselves may be reducible. So how many of those are there? Let $L_{i,d-i}\subseteq H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(4))$ denote the images of the product maps $$ H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(i))\times H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(6-i)) \longrightarrow H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(6)) $$ for $i=1,\dots,\left[d/2\right]$.

Since (as above) $\dim_k H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(j))={j+3\choose 3}$, we get that
$\dim_kL_{i,d-i}\leq {i+3\choose 3} + {d-i+3\choose 3}$

At the same time, according to the above argument, the subspace of $H^0(\mathbb P^3, \mathscr O_{\mathbb P^3}(d))$ containing three skew lines is at least $$ {d+3\choose 3} - 3 (d+1), $$ so as soon as $$ {d+3\choose 3} - 3 (d+1) - \max_{0<i<d}\left({i+3\choose 3} + {d-i+3\choose 3}\right) >0, \tag{$\star$} $$ the general such surface will be irreducible. It is easy to see that the left hand side of $(\star)$ is a quadratic polynomial with positive leading coefficient, so this will be positive for $d\gg 0$. In fact, I think it is positive for $d\geq 6$.

Note that this was a very rough calculation as we counted all reducible surfaces, not just the ones containing the three lines, so one can probably do better for smaller degrees, but I leave that to the reader.

  • $\begingroup$ Hi, thanks for your answer. I was wondering if your surfaces are irreducible? $\endgroup$
    – M. D.
    Nov 1, 2017 at 12:35
  • $\begingroup$ Right. That's actually a valid point, but by counting the reducible ones you get that, at least for large $d$, most of them are irreducible. Let me add something to the answer along these lines. $\endgroup$ Nov 2, 2017 at 17:13

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