Hard Lefschetz Thereom and the Cohomology of Flag Varieties Let $G$ be a compact connected connected Lie group $G$, and $B$ a Borel subgroup containing a maximal torus. Moreover, let $F = G/B$ be the associated flag manifold. Now $F$ is a compact Kahler manifold, and from the Borel--Bott--Weil theorem, we know that dim($H^{(0,k)}(F))$ is non-zero only if $k = 1$, whereupon it is equal to $1$. Since $F$ is Kahler, we can also apply the hard Lefschetz theorem to show that $H^{(a,b)} \simeq H^{(a+2,b+2)}$, for all $a,b$ less than the dimension of $F$. Moreover, complex conjugation and Hodge decomposition show that $H^{(a,b)} \simeq H^{(b,a)}$. Taking these three results together tells us that $H^{(a,b)} = 0$, if $a \neq b$, and $H^{(a,a)} \simeq {\mathbb C}$. THis is all very good until it contradicts Borel's calculation of the cohomology of the flag manifolds! Can someone tell me where the mistake in my reasoning is please? Danke!

There is something wrong already with your statement of Borel--Weil--Bott.

E.g. the simplest example of a flag variety is $\mathbb P^1$ (which is the flag variety for $SL_2$), and its Hodge numbers are $h^{0,0} = h^{1,1} = 1$, and all other $h^{a,b} = 0.$

More generally, as you write, it is true, for any flag variety, that $h^{a,b} \
eq 0$ only if $a = b$.

So maybe when you assert that $h^{0,1} = 1$, you actually mean to assert that $h^{0,0} = 0$. This is certainly true, since flag varieties are connected (and it _is_ a special case of Borel--Weil; it is the special case corresponding to the trivial rep'n of $G$).

Finally, hard Lefschetz doesn't say what you claim; in general $h^{a,b} \
eq h^{a+2,b+2}.$ (Note also that the shift $(a,b) \mapsto (a+2,b+2)$ is a little odd, since cupping with the Kahler class sends $(a,b)$ to $(a+1,b+1)$. But in any case, it's not normally the case that $h^{a,b} = h^{a+1,b+1}$ either.)