Galois theory of linear ODEs by Berman P.H.

By Berman P.H.

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5. 5 and basic Galois theory yield s V˜i,H (direct sum of GH -modules), where V˜i,H is the solution space of Y˜ = Ai Y˜ . 2 implies that W i=1 V˜i,H . We need to show that πi (W ) Ui . ˜ i /KH be Fix i, 1 ≤ i ≤ s. Consider Y˜ = Ai Y˜ + Bi as an equation over KH and let K ˜ i /KH ) is a quotient of U and K ˜ i /KH is the Picard-Vessiot extension. We have that Gal(K generated by the coordinates of ηi . 5, one ˜ i /KH ) checks that Gal(K πi (W ). Now consider the following diagram: ✚ ✚ ˜i K KH ❩ ❩ Ki,I ◗ ◗ ✑ ✑ KH ∩ Ki,I Ki,H By definition we have Ui = Gal(Ki,I /Ki,H ).

Yn+m : m ym = − a ˜i−1 yi + ym+1 i=1 n yn+m = − ai−1 ym+i i=1 yi = yi+1 for i ∈ / {m, n + m} . Written in matrix form, this system is  Y =  AL2 C0 0 AL1  Y, where C0 ∈ k m×n is the matrix having 1 in the m, 1 position and zero everywhere else. 22) Y = 0 A1 where Y = Ai Y is completely reducible for i = 1, 2 and C is an arbitrary given matrix. 16). Suppose that the associated homogeneous system Y = AY is completely reducible. Let U = Gal(KI /KH ) ⊆ GI . Then the following statements hold: 1.

Given σ ∈ S3 , then we define the permutation matrix Pσ to be the matrix whose (i, j)th entry is 1 if i = σ(j), 0 otherwise. Given an ordered basis E = {e1 , e2 , e3 } and a permutation σ ∈ S3 , then Eσ is the ordered basis given by Eσ = eσ−1 (1) , eσ−1 (2) , eσ−1 (3) . That is, if σ(Ij ) = j for 1 ≤ j ≤ 3, then Eσ = {eI1 , eI2 , eI3 } . Otherwise stated, if Eσ = {˜e1 , ˜e2 , ˜e3 } , then ej = ˜eσ(j) for 1 ≤ j ≤ 3. It is a fact that Pσ = [id]E,Eσ . For example, if σ = (1 2 3), then Eσ = {e3 , e1 , e2 } and   0   Pσ = [id]E,Eσ =  1  0 0 0 1 1   0 .

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