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In complex projective geometry, we have a specified Kähler class $\omega$ and we have a Lefschetz operator $L:H^i(X,\mathbb{C})\to H^{i+2}(X,\mathbb{C})$ given by $L(\eta)=\omega\wedge \eta$. We then define primitive cohomology $P^{n-k}(X,\mathbb{C})=\ker(L^{k+1}:H^{n-k}(X)\to H^{n+k+2}(X))$, and we even have a nice theorem, the Lefschetz decomposition, that says $H^m(X,\mathbb{C})=\oplus_k L^kP^{n-2k}$. Often, in papers, people just prove their result for primitive classes, as they seem to be easier to work with.

So, what exactly ARE primitive classes? Sure, they're things that some power of $\omega$ kills, but what's the intuition? Why are they an interesting distinguished class? Is there a good reason to expect this decomposition?

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Stupid notational confusion: by $L^{k+1}$ do you mean L iterated k+1 times or L applied to $H^{k+1}$? –  Ilya Grigoriev Feb 8 '10 at 18:03
    
Sorry, meant applied $k+1$ times. I'll edit. –  Charles Siegel Feb 8 '10 at 18:04
    
Typo: $L^{k+1}$ has degree $2k+2$, not $2k$. –  VA. Feb 8 '10 at 18:29
    
I believe that fixes it, unless I've butchered the definitions. –  Charles Siegel Feb 8 '10 at 18:31
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up vote 16 down vote accepted

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 $x\in X$, and let $O_x = O(T_x X\otimes \mathbb{C})\cong O(4n,\mathbb{C})$. Then $O_x$ acts projectively on $\Lambda^\bullet (T_x X\otimes \mathbb{C})$ via the spinor representation (which lives inside the Clifford action). The holonomy group $Hol_x\cong U(n)$ 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 $T_x X \otimes\mathbb{C} = T^{1,0} \oplus T^{0,1}$. 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.)

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Sooooo, how do you explain in an invariant way what the group is? :) –  Mariano Suárez-Alvarez Feb 9 '10 at 7:14
    
@Mariano. Since you ask...see added paragraph. –  Tim Perutz Feb 9 '10 at 16:55
    
Tim: Awesome answer as usual. Does the added paragraph have any physics-y explanation? –  Kevin H. Lin Feb 9 '10 at 20:02
    
@Kevin - thanks - maybe, but not one I know... –  Tim Perutz Feb 9 '10 at 20:53
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Another small addition : the kähler differential approach described in the last paragraph is described in the following paper : mathjournals.org/mrl/2008-015-004/2008-015-004-007.pdf , along with extensions to other holonomy groups (Calabi-Yau, Hyperkähler,...) –  Simon Pepin Lehalleur Aug 10 '10 at 20:16
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If you take Poincare duals, then the wedge product becomes intersection, and the Kahler form $\omega$ becomes a hyperplane $H$. Then, the primitive cohomology in dimension $n - k$ consists exactly of those classes whose duals (which are homology classes in dimension $n + k$), when we intersect with $k + 1$ generic hyperplanes gives us zero. In other words, the homology classes which don't intersect with some $(n - k - 1)$-dimensional linear subspace of $\mathbb{CP}^n$. The Lefschetz theorem in this context just says that we can get all of the homology by taking those classes which don't intersect with a linear subspace of complementary dimension and then intersecting with hyperplanes.

In the case $k = 0$, we can choose the hyperplane to be the one at infinity, so the primitive homology consists of the $n$-dimensional homology classes which sit in the "$\mathbb{C}^n$" part of $\mathbb{CP}^n$, which I think is a pretty natural set of classes to look at.

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Is it true that they are dual to homology classes that can't be expressed as $X'\cdot H$, where $H$ stands for the hyperplane in $\mathbb C\mathbb P^n$?

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Wouldn't this be better as a separate question? I feel like using answers to ask new questions is an ill-advised habit. –  Ben Webster Feb 9 '10 at 3:59
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What about "answers given in the form of a question", Jeopardy style, (which is how I interpret Ilya's statement)? –  Charles Rezk Feb 10 '10 at 0:19
    
Actually, I first write this as a statement, but then decided to change to question, since I didn't check it carefully. Will think a bit more and post! –  Ilya Nikokoshev Feb 10 '10 at 18:12
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