For $U(n)$ the Lie group of $n \times n$ unitary matrices, and $T^n$ its maximal torus subgroup, the homogeneous space $$ U(n)/T^n $$ is called the complete flag variety of order $n$. For the special case of $n=2$ this gives the sphere $S^2$. In this special case the topological $K$-theory group $K^1$ vanishes, i.e. $$ K^1(S^2) = 0. $$ Is this true for higher complete flag varieties? In explicit form, is it true that $$ K^1(U(n)/T^n) = 0, ~~ \forall n? $$
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4$\begingroup$ Yes. For any connected compact group, the quotient by the maximum torus has a CW-complex structure where all of the cells have even dimension, called Schubert cells (see also Bruhat). Compute $K$-theory by induction adding the cells one by one: it is a free abelian group on the cells, all in degree zero. Similarly, the cohomology is a free abelian group in dimensions of the cells. Why are the cells even dimensional? For $U(n)$ do by hand. One reason is that $K/T=G/B$ is an algebraic variety. But you should be able to see it directly from the compact. Maybe use T action eigenspaces are complex? $\endgroup$– Ben WielandJul 24, 2020 at 17:05
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