5
$\begingroup$

The SageMath documentation of toric varieties reads:

The integral cohomology of smooth toric varieties is torsion-free, so in this case there is no loss of information when going to rational coefficients.

The above sentence may be interpreted as the following implication: if

  • $X$ and $Y$ are manifolds
  • $H^*(X,\mathbb Z)$ and $H^*(Y,\mathbb Z)$ have no torsion
  • $H^*(X,\mathbb Q)$ is isomorphic to $H^*(Y,\mathbb Q)$ as graded rings

then

  • $H^*(X,\mathbb Z)$ is isomorphic to $H^*(Y,\mathbb Z)$ as graded rings?

Question: Is this implication true?

$\endgroup$
3
  • $\begingroup$ I'm not sure that's quite true if you want to state it for graded rings. What will probably be true however is if you start from a map $X\to Y$, then this map induces an isomorphism rationally if and only if it does so integrally (given that $X,Y$ have no torsion in their cohomology) $\endgroup$ Aug 25, 2020 at 14:55
  • $\begingroup$ @MaximeRamzi I've edited the question: what matters is the discrimination power of cohomology as a topological invariant. $\endgroup$ Aug 25, 2020 at 15:04
  • 1
    $\begingroup$ As stated, I don't think it's true. I don't have an example, but you could imagine an example where the multiplication of two generators $x_0,x_1$ yields $2y$ for some generator $y$, as opposed to one where the multiplication of two generators is just a third generator. Then they would be isomorphic over $\mathbb Q$, but not over $\mathbb Z$. But I don't have examples in mind $\endgroup$ Aug 25, 2020 at 18:30

1 Answer 1

2
$\begingroup$

The answer is no.

Let X be a Hirzebruch surface $H_n$. It can be viewed as a (smooth) toric variety generated by the fans

z0=(0,1)
z1=(0,-1)
z2=(1,0)
z3=(-1,-n)
f0=[z2,z0]  
f1=[z2,z1]
f2=[z3,z0]
f3=[z3,z1]

, shown in this picture for $n=2$.

Fans that generate the Hirzebruch surface as a toric variety

According to p.106 of Fulton's book "Introduction to toric varieties", the $\mathbb Z$-cohomology ring of $X$ is

$\mathbb Z(z_0,z_1,z_2,z_3)/(z_0z_1,z_2z_3,z_0-z_1-nz_3,z_2-z_3)$, and the $\mathbb Q$-cohomology

$\mathbb Q(z_0,z_1,z_2,z_3)/(z_0z_1,z_2z_3,z_0-z_1-nz_3,z_2-z_3)$. $^\text{[remark]}$

Therefore all the Hirzebruch surfaces $H_n$ have the same $\mathbb Q$-cohomology when $n\geq 1$, but their $\mathbb Z$-cohomology are different for even and odd $n$.

Remark: In fact, the Fulton book only provides $\mathbb Z$-cohomology directly, but the proof also works for $\mathbb Q$-cohomology. A more direct method of obtaining $\mathbb Q$-cohomology is by a SageMath computation:

n = 4  #try different values of n
cone1 = Cone([(1,0), (0,1)])
cone2 = Cone([(1,0), (0,-1)])
cone3 = Cone([(-1,-n), (0,-1)])
cone4 = Cone([(-1,-n), (0,1)])
H = Fan([cone1, cone2, cone3, cone4])
T = ToricVariety(H)
print(T.cohomology_ring().defining_ideal())
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .