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Let $X = G/K$ be a Riemannian symmetric space of compact type and consider the "Weil homomorphism" $$w^\bullet: H^\bullet(BK; \mathbb R) \to H^\bullet(X; \mathbb R),$$ i.e. the map in cohomology induced by the classifying map of the $K$-principal bundle $G \to G/K$.

Then $H^\bullet(BK;\mathbb R) = \mathbb R[\alpha_1, \dots, \alpha_r]$ is a polynomial algebra generated by even degree classes $\alpha_1, \dots, \alpha_r$ (where $r = {\rm rk}(K)$) which transgress through the universal bundle to the primitive elements of $H^\bullet(K; \mathbb R)$.

I am interested in the kernel of $w^\bullet$, i.e. the relations that the classes $w^\bullet(\alpha_1), \dots, w^\bullet(\alpha_r)$ satisfy in $H^\bullet(X; \mathbb R)$. There are some obvious relations in high degrees, since $w^n$ vanishes for $n>\dim X$. Also, all classes in degree $\dim X$ are proportional. But what about low degrees? The map $w^2$ is always injective. Is it true that $w^4$ is injective provided $\dim X > 4$? (For some reason I am particularly interested in whether $w^4 $ is injective if $\dim X \geq 8$, but never mind.) What are the lowest degree relations between the $w^\bullet(\alpha_j)$ in general?

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  • $\begingroup$ Apparently I was mislead by some unfortunately chosen examples. Relations can appear in very small degrees already. In the equal rank case one can easily compute the kernel explicitly, and there are relations already in degree 4, no matter how large the dimension o $X$. Thus, the question does not make sense. $\endgroup$ Mar 12, 2015 at 12:32
  • $\begingroup$ I must say, I am not happy that my answer was converted into a comment. My question makes sense a priori - just the answer is, that one cannot say anything general. That is a perfectly valid answer. Deleting the question would make this information unavailable. $\endgroup$ Mar 12, 2015 at 19:05

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