For the projective line $CP^1$, its cohomology ring has a single generator. Moreover, this generator is given by the cohomology class of the fundamental form associated associated to the Fubini--Study Kahler metric on $CP^1$. If I have understood correctly this also holds for all the higher projective spaces.

Now the well-known Borel theorem for the more flag varieties says that in general the cohomology ring of a flag variety has multiple generators. Do these generators come from Kahler metrics too? Or is there any sense in which the $CP^n$ case generalises to a more general result?



The answer to first question is no. Let $F$ be the variety of full flags in $\mathbb{C}^3$. This can be viewed as variety of pairs $F=\lbrace (p,\ell)\in \mathbb{C}\mathbb{P}^2\times \mathbb{C}\check{\mathbb{P}}^2\mid p\in \ell\rbrace$. There are two generators for cohomology given by pulling back the Kaehler class from $\mathbb{C}\mathbb{P}^2$, or its dual, under both projections. Neither of these are Kaehler forms on $F$, because they aren't positive along the fibres of the projection. What is special about projective space is that the second Betti number is $1$, so $H^2$ has to be generated by a Kaehler class.

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  • $\begingroup$ Great, thanks for your answer. Just one question though, does this mean that the (complex) Grassmannian cohomology rings are generated by a Kahler class? $\endgroup$ – Antonio Nogueria Feb 23 '13 at 18:44
  • $\begingroup$ Yes, Grassmanians should work. $\endgroup$ – Donu Arapura Feb 23 '13 at 19:55
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    $\begingroup$ @Donu: How come it should work? For the Grassmannian, Picard group is $\mathbb Z$, so there is only one Kahler class (up to proportionality), which does not generate the cohomology ring. $\endgroup$ – Serge Lvovski Feb 24 '13 at 9:58
  • $\begingroup$ Serge, you're right! The higher Chern classes of the universal bundle won't be in the ring generated by $H^2$ of the Grassmanian in general. I wasn't really thinking when I made the comment. $\endgroup$ – Donu Arapura Feb 24 '13 at 13:51
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    $\begingroup$ @Antonio: Hardly ever. If cohomology ring is generated by one element, then all Betti numbers are 1 (or zero:) Needless to say, this situation is pretty rare for flag varieties. $\endgroup$ – Serge Lvovski Feb 25 '13 at 15:50

If you look at the second question, the OP is asking when the cohomology ring can be generated by some set of Kähler forms, not by a single Kähler form. Thus, while the variety $F$ of full flags in $\mathbb{C}^3$ has two generators, say $\omega_1$ and $\omega_2$ that are not Kähler forms, the forms $\eta_1 = 2\ \omega_1+\omega_2$ and $\eta_2=\omega_1+2\ \omega_2$ are both positive and hence Kähler forms. They generate the cohomology ring over the rationals.

In general, if the space is of the form $X = G/P$ where $G$ is complex semi-simple and $P$ is a minimal parabolic, then $X = U/T$ where $U$ is a compact form of $G$ and $T = U\cap P$ is a maximal torus in $U$. The cohomology ring of $X$ will be generated in degree $2$ by a vector space $V$ of $U$-invariant $(1,1)$-forms, and one can find, in the positive cone in this vector space, a basis of generators of the cohomology ring of $X$ (over the rationals), and these generators will all be Kähler forms.

Thus, for example, when $F$ is the full flag variety of $\mathbb{C}^n$, $G=\mathrm{SL}(n,\mathbb{C})$, $P$ is the upper triangular elements in $G$, $U=\mathrm{SU}(n)$, and $K=\mathbb{T}^{n-1}$ is the maximal torus of diagonal elements in $G$. The vector space $V$ has dimension $n{-}1$ and there is a positive 'orthant' in $V$ that consists of Kähler forms that are $\mathrm{SU}(n)$-invariant. These suffice to generate (rationally) the cohomology of $F$ (with, of course, some relations).

Obviously, the case of minimal parabolics is not the only case in which this happens, but I'd have to think about this more to 'remember' the criterion, which I think I did know, at one time.

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