1
$\begingroup$

I am using primitivie cohomology more or less as a black box right now. The very little intuition I have comes from this answer on MathOverflow and appendix C in 3264 and all that, by Eisenbud and Harris.

If $X$ is an $n$-dimensional (complex dimension) smooth projective variety and $h\in H^{2}(X)$ is the [complex de Rahm] cohomology [= singular cohomology tensor $\mathbb{C}$, I hope] class corresponding to a hyperplane section, then the cup product induces isomorphisms $$ (-) \cup h^{k} \colon H^{n-k}(X) \to H^{n+k}(X) $$

We can take Poincaré duals and visualize this as intersecting with an hyperplane. It is (slightly) intuitive to have an isomorphism because the dimensions are complementary and we are in projective space.

For $k\leqslant n$ we define the $n-k$ primitive cohomology group $H^{n-k}_{pr}(X)\subseteq H^{n-k}(X)$ to be the kernel of the map induced by cupping one more time with $h$: $$ (-) \cup h^{k+1} \colon H^{n-k}(X) \to H^{n+k+2}(X) $$

Again, for dimensional reasons, it is intuitive to expect the possiblity of a non-trivial kernel now.

My question: it is often used in the notes I am following that the middle primitive Betti and Hodge numbers are the usual Betti and Hodge numbers minus 1. I think this should also be slightly intuitive by dimensional arguments in projective space as above, but I couldn't make this precise and I think part of the problem is that complex and real dimensions are getting mixed.

Can anyone make these arguments more precise and explain intuitively why we expect the middle primitive Betti numbers to be one less that the usual middle Betti numbers?

I suspect it is also related to the Lefschetz decomposition, but I also don't have an intuitive understanding of this fact. Any hints are appreciated, thanks.

$\endgroup$
  • $\begingroup$ I am a bit late, but if you know the theory of Lie algebras representation, then you can see that $H^*(X)$ is a $\mathfrak{sl}_2$-module, and that the primitve cohomology corresponds to the lowest-weight modules. This is mentionned in Serre's book on complex semisimple Lie algebras. $\endgroup$ – Nicolas Hemelsoet Apr 1 '18 at 22:02
2
$\begingroup$

Here is how I though about primitive cohomology :

We have the operation $\cup h:H^*(X)\rightarrow H^{*+2}(X)$. Hence, every class $x\in H^*(X)$ defines classes in higher degree. So first define the non-primitive cohomology : a class $\alpha$ is non primitive if it can be written $\alpha=x\cup h$. In other words, the non-primitive are those who are not new, but come from a lower degree cohomology group. They form a subspace $NonPrim^{n-k}=Im(\cup h:H^{n-k-2}\rightarrow H^{n-k})$

Now the primitive cohomology is exactly the opposite : a class is primitive if it is brand new. The definition is $Prim^{n-k}=\ker(\cup h^{k+1}:H^{n-k}\to H^{n+k+2})$

Claim : $H^{n-k}=Prim^{n-k}\oplus NonPrim^{n-k}$.

Indeed if $\alpha$ belong to the intersection, then $\alpha=x\cup h$ and $\alpha\cup h^{k+1}=0$. But this means that $x\cup h^{k+2}=0$. This implies that $x=0$ since $h^{k+2}:H^{n-k-2}\to H^{n+k+2}$ is an isomorphism. Moreover, if $\alpha\in H^{n-k}$ is any class, then $\alpha\cup h^{n+1}\in H^{n+k+2}$, so it is of the form $x\cup h^{n+k+2}$. Hence $\alpha=(\alpha-x\cup h)+x\cup h$ is a decomposition into a primitive class and a non primitive one.


Thus, the primitive classes span the cohomology together with the operation $\cup h$. In other words, every class can be written as $\sum \alpha_i\cup h^i$ where $\alpha_i$ are primitive. You can also prove that this decomposition is unique (as long as none of the $\alpha_i\cup h^i$ are zero...). This is the Lefschetz decomposition.


Using the non-primitive cohomology, your claim is now that the middle non-primitive cohomology $NonPrim^n$ is of dimension 1.

Now as stated, this is clearly false : for example, $H^1$ is obviously primitive (nothing can be non primitive). For curves it is the middle cohomology, so for curves, the middle cohomology is purely primitive. More generally, for odd dimensional varieties of dimension $n$, you can have the equality $Prim^n=H^n$. (For even dimensional varieties, this cannot happen since $h^{n/2}$ is always non-primitive). So these are example where the primitive part is big : it is all of $H^n$.

In even and odd dimension, you can also have a very small primitive part $Prim^n$ inside $H^n$ ! In fact, as long as you have a primitive class $\alpha$ in even degree $2k<n$, then $\alpha\cup h^{m-k}$ will be a non-zero non-primitive class in $H^n$.


I suspect you meant that $X$ is an hypersurface of even dimension $n=2m$ in some projective space. Then this is indeed true that the non-primitive cohomology is of dimension 1. Indeed, from Lefschetz theorems, every $H^{2m-2k}(X)=H^{2m-2k}(\mathbb{P}^{n+1})=\mathbb{C}.h^{m-k}$. So there is no primitive part in even degree less than $n$. It follows that the only non-primitive class in $H^n$ is simply $h^n$. In other words $NonPrim^n(X)=\mathbb{C}.h^m$.

$\endgroup$
  • $\begingroup$ Thanks a lot for your answer and your clarification! Yes, indeed, I forgot the condition of $n$ being even. Thanks for pointing it out and for the counterexamples in the odd case. $\endgroup$ – Pedro Mar 19 '18 at 17:56

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.