I recently had a chance to study a little bit about this subject and presented in a ‘graduate seminar’ course. I am attaching the note to prepare for this presentation. There are many ways to prove this interesting fact, but I took the one of the easiest way to prove it, namely, using Complex manifolds theory. Here is the note.

**1. What is a Smooth Cubic Surface? **

We first define what a smooth cubic surface is. For simplicity, we will work on the complex field .

**Definition 1** *An -dimensional projective space is a space consists of lines in -dimensional complex space which pass through the origin. In other words,*

*where the equivalence is given by*

For a projective space , we use the homogeneous coordinate. That is, the point corresponding to is where the equivalence induces the relation for .

**Remark 1** *For , there is that has nonzero . Then, there is a homeomorphism between and given by . Thus, is covered by copies of . Thus, is a (smooth) manifold. *

From now on, we will mostly consider the case when . That is, . Also, we will denote the coordinate by instead of .

Observe that a polynomial function is not well-defined on . However, if the polynomial is homogeneous, i.e. degree of each monomial is all identical, the inverse image of zero(=zero locus) is well-defined.

**Definition 2** *A smooth cubic surface is a degree 3 hypersurface in that is smooth. More precisely, a smooth cubic surface is the zero locus of a degree 3 homogeneous polynomial with the non-zero jacobian*

**Remark 2** *In fact, the more precise definition would be the zero locus of a degree 3 homogeneous polynomial where for each point such that , the jacobian at is of rank 1. However, the previous definition suffices in this case because*

*and thus if any point in has rank 0 jacobian matrix, then it must be in the zero locus.*

Additionally, we remark that irreducibility of is guaranteed by the Jacobian criterion. If for some non-constant homogeneous polynomials and , then by BΓ¨zout’s theorem, there is such that . This guarantees that the Jacobian of at is zero.

The simplest example of a smooth cubic surface is when . We call it the Fermat cubic.

**2. Cayley-Salmon theorem **

**Definition 3** *By a line in a smooth cubic surface , we mean a line in the projective space that is contained entirely in . *

We are now ready to state the main theorem.

**Theorem 4 (Cayley-Salmon)** *Any smooth cubic surface contains exactly 27 lines. *

In case of the Fermat cubic, we can actually formulate all those 27 lines:

Note that a line in the projective space is given by two hyperplanes(=hypersurface of degree 1). Choose a line in . Up to the coordinate exchange(permutation of ), if necessary, those two hyperplanes can be given as

Therefore the line is contained in the Fermat cubic if and only if

Observe that, if , then from relation 3 and 4, , thus it contradicts to the first relation. Therefore, at least one of those coefficients is zero. Again, by the coordinate exchange, we may assume that .

If , then we get and . Therefore and . The candidate for where is the primitive 3rd root of unity and . That is, in this case, the lines are of the form

and by the permutation, we get also get

This shows that there are exactly 27 lines on the Fermat cubic.

**3. Proof **

The tactic for the proof is the following: We will show that the number of lines contained in a smooth cubic surface is invariant as we vary the cubic polynomial that has non-zero jacobian in . From the fact that the Fermat cubic has 27 lines in it, the theorem is proved.

**3.1. Preparation for the proof **

We introduce spaces that take important roles to prove this theorem.

First, the space that parametrise cubic surfaces in is , because there are different degree 3 monomials with 4 variables, and the constant multiplication does not affect the zero locus, which, in fact, is the cubic surface. From now on, we will identify an element of with a cubic surface in as long as the polynomial is irreducible.

**Lemma 5** *The subset of that consists of elements corresponding to SMOOTH cubic surfaces forms a Zariski open subset in . That is, is an open dense subset and the complement consists of finite union of closed sets of positive codimension. *

Second, the space that parametrise lines in is the Grassmannian , because a line in corresponds to a 2-dimensional subspace in . An element in consists of two linearly independent vectors in , and an equivalence relation given by the subspace they span. Therefore, elements of the Grassmannian are RREF matrices of rank 2. Another characterisation of the Grassmannian space gives the following property:

**Lemma 6** *The (complex) Grassmannian space is homeomorphic to , with the obvious quotient action. In particular, is compact. *

Finally, we are ready to describe the space we want. Consider the subspace of the product space defined by

where is the open subset consisting of smooth cubic surfaces.

**3.2. Proof **

**Proposition 7** *The space is locally a zero locus of polynomials in . *

*Proof:* Observe first that with change of coordinate on , we can represent any point with . Therefore, the open set around the point is of the form

with the case corresponds to .

Write to denote a cubic polynomial corresponding to a cubic surface . Then

This shows that is locally zero locus of ‘s.

**Proposition 8** *Choose a point . Then for the projection , there is a neighbourhood of in such that for each , there is a unique (continuously differentiable) section on . That is, each connected component of is homeomorphic to by the map restricted to the connected component. *

*Proof:* To prove this proposition, we will use the implicit function theorem:

**Lemma 9 (Implicit Function theorem)** *Suppose is a continuously differentiable function. Write and , so that . If the Jacobian matrix is invertible at , then there is a neighbourhood of and the unique continuously differentiable function s.t and . *

We apply this theorem to where , which is defined on the open neighbourhood of . That is, if we find out that is invertible, then there is an open neighbourhood of and the unique continuously differentiable function such that and .

Consider

Similarly, \begin{align*} \frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial b}(0,0,0,0)&=s\cdot \frac{\partial f_X}{\partial w}(s,t,0,0),

\frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial c}(0,0,0,0)&=t\cdot \frac{\partial f_X}{\partial z}(s,t,0,0),

\frac{\partial (\sum s^i t^{3-i} F_{X,i})}{\partial d}(0,0,0,0)&=t\cdot \frac{\partial f_X}{\partial w}(s,t,0,0). \end{align*}

Since formally distinguishes the row of the Jacobian matrix, if this Jacobian matrix is not invertible, we must have

for some .

Since is a homogeneous polynomial, and are both homogeneous polynomial with 2 variables over the algebraically closed field , thus they are factored into linear terms. Notice that there are at least two zeros for and therefore to get the above equation for both zeros, we must have a common zero so that .

Since for all , we get . Therefore, is a singular point in a smooth cubic surface , which is the contradiction. Therefore the Jacobian matrix is invertible and we are done.

*Proof:* For the projection , fix a point . From the propositions above, if the fibre is of cardinality , then there is a neighbourhood of in such that any fibre in the neighbourhood is of cardinality larger of equal to . Also, since is locally closed, if , then there is a neighbourhood of in such that the neighbourhood is disjoint from . Since is compact, we can find a neighbourhood of in so that . That is, the cardinality is locally constant.

Since is connected, and the Fermat’s cubic contains 27 lines, we conclude that all smooth cubic surfaces contains 27 lines.

**4. Furthermore, and References **

There are many other ways to prove this theorem. One another way is purely algebro-geometric. You can find out the result in many reasonably thick algebraic geometry books. There is also a proof using Chern class of the 3-symmetric line bundle of the tautological line bundle over the Grassmannian , and a proof using representation theory – which I do not know well, but it is related to the Weyl group .

Moreover, there is a theorem saying that all smooth cubic surfaces are 2-dimensional projective spaces blown up at 6 points.

References for this mini note is the following:

A lecture note of Andreas Gathmann

http://www.mathematik.uni-kl.de/gathmann/class/alggeom-2014/chapter-11.pdf

A lecture note of Ravi Vakil

http://math.stanford.edu/vakil/216blog/FOAGfeb0717public.pdf

Jack Huizenga’s Quora Answer

https://www.quora.com/Algebraic-Geometry/Algebraic-Geometry-Why-are-there-exactly-27-straight-lines-on-a-smooth-cubic-surface/answer/Jack-Huizenga

AMS Blog on this topic

http://blogs.ams.org/visualinsight/2016/02/15/27-lines-on-a-cubic-surface/