Every non-singular complex projective cubic surface has $27$ lines. Many such surfaces contain points where three lines intersect (called Eckardt points). There are even surfaces with many Eckardt points, like the Fermat cubic, which has $18$. Is there any such non-singular complex projective cubic surface where four, five, or six lines intersect at a point?
Algebraic Geometry – Non-Singular Cubic Surface with Four Intersecting Lines
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Related Solutions
It turns out that condition (T) is, indeed, sufficient for the $27$ lines (distinct and intersecting as expected) to lie on a cubic surface.
To see this, consider the lines $a_1,a_2,a_3,a_4,a_5$ and $b_6$, where the labeling is as in note (2) of the question: $a_1$ through $a_5$ are pairwise skew, and $b_6$ intersects all of them. Choose $4$ distinct points on $b_6$, and $3$ distinct points on each of $a_1$ through $a_5$ also distinct from the intersection point with $b_6$: this makes $4+5\times3=19$ points in total; since there are $20$ coefficients in a cubic form on $4$ variables, there exists a cubic surface $S$ passing through these $19$ points. Since $S$ contains four distinct points on $b_6$, it contains $b_6$ entirely, and since it contains four points (viz., the intersection with $b_6$ and the three additional chosen points) on each of $a_1$ through $a_5$, it contains these also.
Now $b_1$ intersects the four lines $a_2$ through $a_5$ in four distinct points since $a_2$ through $a_5$ are pairwise skew; and since these four points lie on $S$, it follows that $b_1$ lies entirely on $S$; similarly, $b_2$ through $b_5$, and finally $a_6$ (which intersects $b_1$ through $b_5$ in distinct points), all lie on $S$. So $S$ contains the "double-six" $\{a_1,\ldots,a_6,b_1,\ldots,b_6\}$.
So far, property (T) has not been used, only the incidence relation. Now it remains to see that the $c_{ij}$ lie on $S$. Consider the intersection line $\ell$ of the planes $a_i\vee b_j$ (spanned by $a_i$ and $b_j$) and $a_j\vee b_i$ (spanned by $a_j$ and $b_i$; these planes are well-defined since $a_i$ meets $b_j$ and $a_j$ meets $b_i$, and they are distinct since $a_i$ and $a_j$ are skew): this line $\ell$ must be equal to the $c_{ij}$ of the given configuration, because $c_{ij}$ intersects both $a_i$ and $b_j$ so property (T) implies that it lies in the plane $a_i\vee b_j$, and similarly it lies in the plane $a_j\vee b_i$. But $\ell$ also lies on $S$ for similar reasons¹, in other words, $c_{ij}$ lies on $S$, and all the given lines lie on $S$.
Finally, $S$ must be smooth because it contains the configuration of $27$ lines expected of a smooth cubic surface (it is easy to rule out the case where $S$ is a cubic cone, a reducible surface or a scroll by considering the intersecting and skew lines in the double-six; and every configuration where $S$ has double point singularities has fewer than $27$ lines).
- Let me be very precise here, because at this stage we don't know whether $S$ is smooth (so we can't invoke (T) directly on $S$). We have four distinct lines $a_i,a_j,b_i,b_j$ on $S$ such that $a_i$ meets $b_j$ and $a_j$ meets $b_i$ and all other pairs are skew. Call $\pi := a_i\vee b_j$ and $\pi' := a_j \vee b_i$ the planes generated by the two pairs of concurrent lines, and $\ell := \pi\wedge\pi'$ their intersection. We want to show that $\ell$ lies on the surface $S$. If $\pi$ or $\pi'$ is contained in $S$ (reducible) then the conclusion is trivial, so we can assume this is not the case. So the (schematic) intersection of $S$ with the plane $\pi$ is a cubic curve containing two distinct lines ($a_i$ and $b_j$), so it is the union of three lines: $a_i$, $b_j$ and a third line $m$ (a priori possibly equal to one of the former). Consider the intersection point of $a_j$ and $\pi$ (which is well-defined since $a_j$ is skew with $a_i$ so does not lie in $\pi$): it lies on $\ell$ because it is on both $\pi$ and $\pi'$; and it must also lie on $m$ since it is on $\pi$ but neither on $a_i$ nor on $b_j$ (as the two are skew with $a_j$); similarly, the intersection point $b_i\wedge\pi$ is well-defined and lies on both $\ell$ and $m$; so $\ell=m$ lies on $S$ (and we are finished) unless perhaps the two intersections considered are equal, i.e., $a_j,b_i,\pi$ concur at a point $P$, necessarily on $m$. Assume the latter case: $S$ must be singular at $P$ because the line $m$ through $P$ does not lie on the plane $\pi'$ generated by two lines ($a_j,b_i$) through $P$ (i.e., we have three non-coplanar tangent directions at $P$). Now symmetrically, if we call $m'$ the third line of the intersection of $S$ with $\pi'$ (besides $a_j$ and $b_i$), we are done unless $a_i,b_j,\pi'$ concur at a point $P'$, necessarily on $m'$ and necessarily singular on $S$. The points $P$ and $P'$ are distinct because $a_i$ and $a_j$ are skew; and they are on $\ell$ because they are on $\pi$ and $\pi'$; and the line joining two singular points on a cubic surface lies on the surface, so $\ell = P\vee P'$ lies on $S$ in any case. (Phew!)
What I still don't know is whether there are configurations of $27$ distinct lines with the expected incidence relations and which satisfy neither condition (T) nor its dual (viz., whenever three lines pairwise meet, all three meet at a common point), and in particular, what are the irreducible components of the space of configurations. (I also don't know if there is a way to substantially simplify the tedious argument given in note (1) above.)
By the classification theorem of cubic surfaces (p.6 in this paper), a cubic surface belongs to the following classes
Has at worst ADE singularities.
Has an elliptic singularity, i.e., the surface is cone over a smooth cubic curve.
Non-normal or non-integral, and singular along a curve.
So if $X$ has two singularities, $X$ belongs to case 1 and all singularities are rational. We can compute the cohomology by the minimal resolution $\tilde{X}\to X$. The surface $\tilde{X}$ is called a weak del Pezzo surface, which is still blowup of 6 points on $\mathbb P^2$, but in less general positions, so $H^{2}(\tilde{X})=\mathbb Z^7$.
Now let's do some topology: Let $E$ be the exceptional divisor. Then the long exact sequence of the pair $(\tilde{X},E)$ reads $$H^1(E)\to H^2(X)\to H^2(\tilde{X})\xrightarrow{r} H^2(E),$$
where we used $H^*(\tilde{X},E)\cong H^*(X)$ because $\tilde{X}/E\cong X$ as CW complex.
$E$ is the disjoint union of two bunches of rational curves over the two singularities, so $H^2(E)$ has rank $\mu_1+\mu_2$, where $\mu_i$ is the Milnor number of the singularity. Also, $H^1(E)=0$ and $r$ is surjective, so
$$H^2(X)=\mathbb Z^{7-\mu_1-\mu_2}.$$
Cohomologies at other degrees are easy to compute.
Best Answer
No, this is not possible. If p is a smooth point on any surface S, and is contained in a line l on S, then l is contained in the tangent plane at p, call it T_p. Now if S is a cubic then it intersects T_p in a cubic curve (with some singularity at p, even though S is smooth at p); and a cubic curve can contain at most three lines.