## When is a variety not rational?

I also have an excuse for my absence… I’ve also been working on grant applications, job applications, my thesis, etc!  I’ll take a little time now to write more about rational varieties.  In general, it can be quite difficult to decide if a given variety is rational or not.  Charles mentioned in a comment that he would like to see the proof that cubic threefolds are not rational.  This is perhaps a good topic for the future, but maybe it requires more work to explain than I have time for right now.  In any case, in this post I’ll talk about one way to verify that a variety is not rational.  We won’t be distinguishing any unirational varieties from rational ones today though.

Suppose $X$ is a smooth projective variety defined over a field $k$.  Let $\Omega_X$ denote the sheaf of differential forms.  The main theorem uses the existence of a generically finite map from projective space to the given variety to force numerical constraints.

Theorem: If $X$ (as above) is rational, then it has no non-trivial global one forms.  In general, the group $\Gamma(X, \Omega_X^m) = 0$ for all positive integers m.  In case the ground field has characteristic zero, the same is true for unirational varieties.

Proof: Suppose that there is a generically finite, dominant map $f: \mathbb{P}^n \dashrightarrow X$.  Let $U \subset \mathbb{P}^n$ be an open set where this map is defined.  Because $X$ is projective, the complement of this open set may be assumed to have codimension two.  The claim is that non-zero differential forms on $X$ pull back to be non-zero on $U$, or in other words, that there is an inclusion $f^*\Omega_X^m \rightarrow \Omega_U^m$.  When the map is birational, this is obvious.  When the map is only finite though, this is only true in characteristic zero!  Indeed, the pull back map on differential forms can be zero, even for finite maps (consider Frobenius).  Now, because the complement has codimension two, differential forms extend from $U$ to all of $\mathbb{P}^n$.  The theorem will be complete if we can prove that $\Gamma(\mathbb{P}^n, \Omega_{\mathbb{P}^n}^m) = 0$.  To see this, note that $\Omega_{\mathbb{P}^n}$ injects into $\mathcal{O}(-1)^{n+1}$ and this sheaf has no global sections so that there can be no global one forms on projective space.  By taking m-th powers, we get the same result for all differential forms.

One sees that in positive characteristic the notion of being unirational is not well behaved.  So we make the definition that a variety is separably unirational if there is a dominant, generically etale map from projective space to our variety.  This means that the induced inclusion of function fields is separable and exactly implies that the pullback on differentials is injective.  Using the same argument, there are no global differential forms on a smooth projective separably unirational variety in any characteristic.

As a corollary, we have that all the plurigenera for a separably unirational variety (which includes rational varieties) must be zero.  Sometimes these are easier to calculate because they just involve working with invertible sheaves.  For a smooth variety $X$, recall that the m-th plurigenus of $X$ is the dimension of the space of global sections of the invertible sheaf $\mathcal{O}(mK_X)$

Proof: There is an injection $\mathcal{O}(K) \rightarrow \Omega_X^{dim X}$.  Since these latter have no global setions, neither will $\mathcal{O}(K)$.  Taking m-th powers again gives the result that all the plurigenera vanish.

In fact, we don’t even need the strength of our assumptions to prove that these numbers vanish.  All we need is that our variety is separably uniruled.  A variety is (separably uni)ruled, if there exists a variety $Y$ and a birational (resp. generically finite seperable map) $Y \times \mathbb{P}^1 \rightarrow X$.  Loosely, this means that the variety has many rational curves on it.  Ruled varieties are separably uniruled, and unirational varieties are also uniruled.  These implications are strict.  In any case, we can prove that

Theorem 2:  The plurigenera are zero for any smooth projective separably uniruled variety $X$.

Proof:  Let $F: Y \times \mathbb{P}^1 \rightarrow X$ be the separable uniruling.  As in the first theorem, we will prove that $\Gamma(X, \mathcal{O}(mK_X)) = 0$.  To do this, we again get an injection from the pullback of the differential forms into those of $Y \times \mathbb{P}^1$.  So it will be enough to prove that $\Gamma(Y \times \mathbb{P}^1, \mathcal{O}(mK)) = 0$.  However, $K_{Y \times \mathbb{P}^1} = p_1^*K_Y + p_2^*K_{\mathbb{P}^1}$ where $p_i$ is the projection map.  So finally, $\Gamma(Y \times \mathbb{P}^1, \mathcal{O}(mK)) = \Gamma(Y, \mathcal{O}(mK_Y)) \otimes \Gamma(\mathbb{P}^1, \mathcal{O}(mK_{\mathbb{P}^1}))$.  The last term though is zero (when m is positive) because $\mathcal{O}(mK) = \mathcal{O}(-2m)$ which has global sections.

Finally, we can give an example of smooth varieties which are rational.  Suppose $X$ is a smooth hypersurface in $\mathbb{P}^n$ of degree $d > n$.  By adjunction, the canonical class of $X$ is $K_X = (K_{\mathbb{P}^n} + X)|_X$.  The first term is $-(n+1)H$ where $H$ is the hyperplane class on projetive space restricted to $X$.  The second term is the normal bundle of $X$ in projective space;  in this case it is just a line bundle and it is $dH$ where d is the degree.  Putting this all together then, we see that $K_X = \mathcal{O}_X((-n - 1 + d)H)$.  Since $d > n$ we get that all powers of $K_X$ have global sections, and so cannot be separably uniruled, and in particular, not rational!  The same argument works for smooth complete intersections in projetive space where the degrees of the defining equations add up to more than the dimension of the projective space.

This is really the only numerical (that is, easy!) conclusion we can make in the case of smooth hypersurfaces.  Low degree hypersurfaces (the Fano ones) are harder to analyze, but not impossible!  Perhaps I can talk about it more in the future.

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### 2 Responses to When is a variety not rational?

1. anon says:

In the Corollary: given a d-dimensional separably unirational variety X, why does the canonical bundle inject into d-th tensor power of the cotangent bundle in positive characteristic? The usual argument requires semisimplicity of Rep(S_d) (one has to divide by m! in the base field), and that’s true only in char > d.

2. Matt DeLand says:

Ah. Yes. It’s still true in all characteristics though: because the map is separable, there’s an injection $f^*\mathcal{O}_X(mK_X) \rightarrow \mathcal{O}_U(mK_U)$ as in the theorem. Then the rest of the proof of the theorem applies.

It’s also true when the variety is separably rationally connected (which wasn’t defined above). This means there is a map from $\mathbb{P}^1$ to X (for simplicity assume X is smooth) such that the pull back of the tangent bundle is ample on $\mathbb{P}^1$. This property guarantees the existence of many rational curves on the variety, in fact covering all of X. Any differential form on X will restrict to be zero on each of these rational curves, forcing the form to be zero.