Page 49 - Textos de Matemática Vol. 36

2.4. Flows on Graßmann manifolds 39
(ii) Let C ∈ Gζ(ak) and for μ ̸= k let D(aμ,C) ̸= 0. Then ( λ ν 1k , . . . , λ ν nk ) = ( λ ν 1μ , . . . , λ ν nμ ) .
Proof. ThematrixD0 =(vj,vν2k,...,vνnk)hasranknandrankD1 =rank(vν2k, ..., vνnk) = n − 1. Let the first n − 1 rows of D1 be linearly independent – otherwise we proceed analogously – then we can transform D0 by elementary column operations into the form
0 F m+1 b D , whereb∈C
\{0} andF∈C
(n−1)×(n−1)
isregular.
This proves (i), since rank D0 = n.
To prove (ii) we observe from the definition of Gζ(ak) that ζ(ak) = ζ(aμ).
Let μ ̸= k and let us assume that the assertion in (ii) does not hold. Then there exists a q with λνiμ = λνik for q + 1 ≤ i ≤ n and (without loss of generality)
R(λνqμ eiφ) < R(λνqk eiφ)
for 0 < φ < ε0. Since D(ak,C) ̸= 0 the rows cν1k,...,cνnk
dent. Therefore we can expand cνqμ as
n
c ν qμ = γ i c ν ik . i=1
are linearly indepen-
Since, together with cνik = cνiμ for q + 1 ≤ i ≤ n, by our assumption c ν μ c ν μ
.1 .1
. .
cμ cμ cνμ νq−1 νq−1
1n q
.
μ
D(a ,C)= . = γicνik = γicνik ̸=0,
c i=1 νnμ c ν k
i=1 c ν k
q+1
. .
q+1 .. cνnk cνnk
it follows that (γ1,...,γq) ̸= 0. Let γp ̸= 0 for some p ∈ {1,...,q}. Then
ckcνk ν.1 1 . . .
ck cνpk−1
νp−1 n .
cνqμ = γicνik =γp . ̸=0
cνk n cνnk
i=1
cνk ck
p+1 cνk
. p+1 . .
νn
cνk
1