Smooth cubic surfaces are known to have many wonderful properties in common. However, this means I cannot easily show any two to be non isomorphic. Presumably there are non isomorphic (over C) pairs of smooth cubic surfaces... Can someone exhibit a pair with proof?

  • 1
    $\begingroup$ Pick $6$ points in general position on $\mathbb P^2$. The blow-up in these six points is a smooth cubic surface. Now, let $P_1,\ldots,P_6$ be six points in general position and let $Q_1,\ldots, Q_6$ be six points in general position. Under which condition on the points are the associated cubic surfaces isomorphic? $\endgroup$ – Ariyan Javanpeykar Feb 16 '17 at 16:59
  • $\begingroup$ The fact that there exist non-isomorphic smooth cubic surfaces can also be seen via a moduli count. Let $H$ be the space of homogeneous polynomials in four variables of degree 3. (What is its dimension?) This comes equipped with an action of $GL_4$ (which is of dimension $16$). The dimension of the moduli space of smooth cubic surfaces is therefore $\dim H - 16$. $\endgroup$ – Ariyan Javanpeykar Feb 16 '17 at 17:01
  • $\begingroup$ @AriyanJavanpeykar I was asking myself the first question, but couldn't see anything. Do you have a hint? Also, for the moduli count, is it clear that isomorphic smooth cubics must be projectively equivalent? $\endgroup$ – Lorenzo Najt Feb 16 '17 at 18:43
  • $\begingroup$ @AriyanJavanpeykar I mean, what I can say is that there is an automorphism carrying the first four of one to the first four of the other. So we in general have a space that is 4 dimensional. It's not clear to me that an isomorphism between the surfaces will descend to an isomorphism between the arrangements of points on the plane after blowing... though perhaps that follows? Will think... if so then that proves it. $\endgroup$ – Lorenzo Najt Feb 16 '17 at 20:23

This is just an expansion on one of Ariyan's suggestions above.

We can indeed show that cubic surfaces up to isomorphism form a 4-dimensional family just by counting moduli. The tricky point is precisely what the OP asked in comments: why must an isomorphism of cubic surface come from a projective equivalence? Let me answer that.

Every smooth cubic surface $S$ is anticanonically embeddeded in $\mathbf P^3$, meaning that $O_{\mathbf P^3}(1)_{\vert S} = -K_S$.

Moroever, since the codimension is 1, this embedding is linearly normal, meaning that that the restriction map $H^0(\mathbf P^3, O(1)) \rightarrow H^0(S,-K_S)$ is an isomorphism.

An isomorphism $S \cong S'$ of two cubic surfaces will induce an isomorphism $$H^0(S,-K_S) \cong H^0(S',-K_{S'}),$$ hence an automorphism of $H^0 (\mathbf P^3,O(1))$, and hence a projective equivalence of

$$ \mathbf P^3 = \mathbf P \left( H^0 (\mathbf P^3,O(1))^\ast \right).$$

Keeping tabs on the various maps, we see that this projective equivalence induces exactly the original isomorphism between $S$ and $S'$.

Remark: notice that this proof doesn't use the description of cubic surfaces as blowups of $\mathbf P^2$.

| cite | improve this answer | |
  • $\begingroup$ It's clear to me why $H^0(P^3, O(1)) \to H^0(S, -K_S)$ is injective ($S$ is contained in no hyperplane), but its not clear why it is surjective - I don't know how to compute its dimension. Can you suggest an approach? I think the rest is fairly clear -- I think is the same reasoning that shows isomorphic elliptic curves are projectively equivalent. $\endgroup$ – Lorenzo Najt Feb 17 '17 at 3:13
  • $\begingroup$ Surjectivity comes from the long exact sequence associated to the short exact sequence $O(-S) \otimes O(1) \rightarrow O(1) \rightarrow O(1)_{\vert S}$. This is the ideal sheaf sequence associated to $S$, twisted by $O(1)$. $\endgroup$ – bertram Feb 17 '17 at 8:52
  • $\begingroup$ @AreaMan I think you might find it to be an interesting exercise to show that the group of projective automorphisms of a smooth hypersurface $X$ of degree $d\geq 3$ in $\mathbb P^n$ ($n>2$) is in fact equal to the group of automorphisms of $X$, if we assume $(d,n)\neq (4,2)$. $\endgroup$ – Ariyan Javanpeykar Feb 18 '17 at 21:56

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.