ag.algebraic geometry - Intuition for Primitive Cohomology

The primitive classes are the highest weight vectors.

Hard Lefschetz says that the operator $L$ (which algebraic geometers know as intersecting with a hyperplane) is the "lowering operator" $rho(F)$ in a representation $rho
colon mathfrak{sl}_2(mathbb{C})to End (H^ast(X;mathbb{C}))$. The raising operator $rho(E)$ is $Lambda$, the restriction to the harmonic forms of the the formal adjoint of $omega wedge cdot$ acting on forms. The weight operator $rho(H)$ has $H^{n-k}(X;mathbb{C})$ as an eigenspace (= weight space), with eigenvalue (=weight) $k$.

The usual picture of an irreducible representation of $mathfrak{sl}_2(mathbb{C})$ is of a string of beads (weight spaces) with $rho(F)$ moving you down the string and decreasing the weight by 2, and $rho(E)$ going in the opposite direction. The highest weight is an integer $k$, the lowest weight $-k$.

From this picture, it's clear that the space of highest weight vectors in a (reducible) representation is $ker rho(E)$. It's also clear that, of the vectors of weight $k$, those which are highest weights are the ones in $ker rho(F)^{k+1}$. So the highest weight vectors in $H^{n-k}(X; mathbb{C})$ are those in $ker L^{k+1}$.

Of course, all this ignores the rather subtle question of how to explain in an invariant way what this $mathfrak{sl}_2(mathbb{C})$, or its corresponding Lie group, really is.

Added, slipping Mariano an envelope. But here's what that group is. Algebraic geometers, brace yourselves. Fix $xin X$, and let $O_x = O(T_x Xotimes mathbb{C})cong O(4n,mathbb{C})$. Then $O_x$ acts projectively on $Lambda^bullet (T_x Xotimes mathbb{C})$ via the spinor representation (which lives inside the Clifford action). The holonomy group $Hol_xcong U(n)$ also acts on complex forms at $x$, and the "Lefschetz group" $mathcal{L}$ is the centralizer of $Hol_x$ in $O_x$. One proves that $mathcal{L}cong GL(mathbb{C}oplus mathbb{C})$. Not only is this the right group, but its Lie algebra comes with a standard basis, coming from the splitting $T_x X otimesmathbb{C} = T^{1,0} oplus T^{0,1}$. Now, $mathcal{L}$ acts on complex forms on $X$, by parallel transporting them from $y$ to $x$, acting, and transporting back to $y$. Check next that the action commutes with $d$ and $*$, hence with the Laplacian, and so descends to harmonic forms = cohomology. Finally, check that the action of $mathcal{L}$ exponentiates the standard action of $mathfrak{gl}_2$ where the centre acts by scaling. (This explanation is Graeme Segal's, via Ivan Smith.)