Bourbaki

Exercise 6, § 1

General Topology, VII

Bourbaki marks this one as harder than the rest.

Let HH and KK be two closed subgroups of Rn\mathbf{R}^n such that H+KH + K is a closed subgroup.

a) Show that, if VV (resp. WW) is the largest vector subspace contained in HH (resp. KK), then V+WV + W is the largest vector subspace contained in GG, and therefore d(H)+d(K)=d(H∩K)+d(H+K)d(H) + d(K) = d(H \cap K) + d(H + K) [note that the rational rank of G/(V+W)G/(V + W) is finite and consequently that GG cannot contain a line which is not contained in V+WV + W].

b) Let UU be a subspace complementary to V+WV + W, such that GG is the direct sum of V+WV + W and M=G∩UM = G \cap U, and such that H∩KH \cap K is the direct sum of (V+W)∩(H∩K)(V + W) \cap (H \cap K) and L=H∩K∩UL = H \cap K \cap U (Exercise 5). Let H′H' (resp. K′K') be the subgroup of MM consisting of the components of the points of HH (resp. KK) in the decomposition of GG into the direct sum of V+WV + W and MM. Show that M=H′+K′M = H' + K' and that H′∩K′=LH' \cap K' = L.

c) If we put H′′=H∩(V+W)H'' = H \cap (V + W) and K′′=K∩(V+W)K'' = K \cap (V + W), show that r(H′′)+r(K′′)=r(H′′∩K′′)+r(H′′+K′′)r(H'') + r(K'') = r(H'' \cap K'') + r(H'' + K'') (reduce to the case where V∩W={o}V \cap W = \{ o \}, and show that then H′′∩K′′H'' \cap K'' is the direct sum of V∩K′′V \cap K'' and W∩H′′W \cap H''.

d) Show that

r(H)+r(K)=r(H∩K)+r(H+K) r(H) + r(K) = r(H \cap K) + r(H + K)
[use c) and Exercise 4, noting that r(H)=r(H′)+r(H′′)r(H) = r(H') + r(H'') and r(K)=r(K′)+r(K′′)r(K) = r(K') + r(K'')].

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.