Let $X$ be a compact Kähler manifold with $L$ denoting the Lefschetz operator $L(\bullet) = \bullet \wedge \omega$. The primitive cohomology groups are defined, for $k \in \mathbb{N}$, by $$P^k(X, \mathbb{C}) = \ker(L^{n-k+1}: H^{k-1}(X, \mathbb{C}) \longrightarrow H^{2n-k+1}).$$


How do I remember that the $k$th primitive cohomology is the kernel of the $(n-k+1)$th power of the Lefschetz operator?



How do I remember which power of the Lefschetz operator $L$ corresponds to the $k$th Primitive cohomology group?
The formula you give for the primitive cohomology group is not correct. I think you meant to write


$$P^k(X, \mathbb{C}) = \ker(L^{n-k+1}: H^{k}(X, \mathbb{C}) \longrightarrow H^{2n-k+2}(X, \mathbb C)).$$


What's going on here is that the hard Lefschetz theorem says that the map $L^{n-k}: H^k \to H^{2n-k}$ is an isomorphism (as is the map you wrote down, which is the same map shifted in $k$ by $1$). To get an interesting kernel, you want to take one more power of $L$ than that, or $L^{n-k+1}$.


How do we remember that exactly $n-k$ powers gives an isomorphism? Well Poincare duality says $H^k $ and $H^{2n-k}$ have the same dimension, so it's very plausible that they are isomorphic. It takes $n-k$ powers of $L$ to get from $H^k$ to $H^{n-k}$.


So: Think of Poincare duality, then remember the hard Lefschetz isomorphism, then shift by 1.


(This is not the fastest route, but remembering how hard Lefschetz works is good for a number of other things.)


Why shift only by $1$? We want to take only the most interesting cohomology classes, so we want our condition to be as strict as possible while still containing nonzero vectors. Since we're taking a kernel, we want to take the kernel of a map that just barely fails to be injective, so we shift by $1$ instead of a larger number.


Why increase by $1$ and not decrease? The cohomology groups increase towards the center and decrease towards the end, so applying one fewer Lefschetz operator would give a map that is injective but usually not surjective, and applying one more Lefschetz operator would give a map that is surjective but usually not injective. We're looking at the kernel, so we want the surjective but not injective one.




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The other approach is to look via the representation theory of $SL\_2$. The operator $L$ is one part of an action of the Lie algebra $\mathfrak{sl}\_2$ on $\bigoplus\_{i=0}^{2n} H^i ( X, \mathbb C)$. Specifically, it's a nilpotent element.


We can break $\bigoplus\_{i=0}^{2n} H^i ( X, \mathbb C)$ into a sum of irreducible representations, and think about how $L$ acts on each one.


The $r+1$-dimensional representation of $SL\_2$ is the sum of $r+1$ copies of $\mathbb C$, where $L$ sends each copy to the next one in the sequence. Viewed as cohomology classes, the degree of each copy is shifted from the last by $2$. Since representations of $SL\_2$ are self-dual, they're symmetric around the central degree $n$. So the degrees are $$n-r, n+2-r, \dots, n+r-2, n+r.$$


The goal of the primitive cohomology is to detect a single representation. Specifically, in degree $k$, we want to detect the representation that just barely intersects degree $k$, which is why we take $r= n-k$ so that $n-r = k$. The other representations that contribute to cohomology in degree $k$ have larger values $r>n-k$. We want to choose a map that zeroes out or chosen representation and then take the kernel. $L^{r}$ is an isomorphism on lowest-degree piece of the $r$-dimensional representation, so we take $L^{r+1} = L^{n-k+1}$, which vanishes.


