Classification of Algebraic Subgroups of Lower Order Unipotent Algebraic Groups

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1 Classification of Algebraic Subgroups of Lower Order Unipotent Algebraic Groups AMS Special Session, Boston, MA; Differential Algebraic Geometry and Galois Theory (in memory of J. Kovacic) V. Ravi Srinivasan January 5, 2012

2 Introduction and Preliminaries All fields considered in this talk are of characteristic zero. We consider only ordinary differential fields (one derivation).

3 Introduction and Preliminaries Let F be a differential field with a field of constants C. Let U(n + 1, C) denote the group of upper triangular matrices with 1 s on the diagonal. We ask the following question: Under what conditions on F does there exist a P-V extension (of F) with a Galois group isomorphic to U(n + 1, C)?

4 Introduction and Preliminaries Condition on F (N & S) Let F be a differential field that satisfies the following condition: there are elements f 1, f 2,, f n F such that for any c 1, c 2,, c n C and for any f F if n i=1 c if i = f then c i = 0 for all i. Bialynicki-Birula, J. Kovacic

5 Extensions with unipotent algebraic groups as Galois Groups 1 x 1,1 x 2,1 x n,1 0 1 x 1,2 x n 1,2 Let E = F (g), where g := x 1,n Extend the derivation of F to E by setting g = Ag, where 0 f f 2 0 A := f n Then E is a Picard-Vessiot extension of F with a differential Galois group naturally isomorphic to U(n + 1, C).

6 Extensions with unipotent algebraic groups as Galois Groups We can also compute a linear differential operator over F for E. L(Y ) := w(y, 1, x 1,1, x 2,1,, x n,1 ) w(1, x 1,1, x 2,1,, x n,1 ) L 1 {0} =span C {1, x 1,1, x 2,1,, x n,1 }

7 Computing a differential equation for U(3, C) Let f 1, f 2 F be elements satisfying the condition N & S.

8 Computing a differential equation for U(3, C) Let f 1, f 2 F be elements satisfying the condition N & S. 1 x 1 y Let E = F (g), where g = 0 1 x Extend the derivation on F to E by setting 1 x 1 y 0 f x 1 y 0 1 x 2 = 0 0 f x We have x 1 = f 1, x 2 = f 2 and y = f 1 x 2..

9 Computing a differential equation for U(3, C) The differential field E is a NNC extension of F.

10 Computing a differential equation for U(3, C) The differential field E is a NNC extension of F. Let V :=span C {1, x 1, y} and G be the group of differential automorphisms of E fixing F.

11 Computing a differential equation for U(3, C) The differential field E is a NNC extension of F. Let V :=span C {1, x 1, y} and G be the group of differential automorphisms of E fixing F. Then GV V, F V = E and E G = F.

12 Computing a differential equation for U(3, C) The differential field E is a NNC extension of F. Let V :=span C {1, x 1, y} and G be the group of differential automorphisms of E fixing F. Then GV V, F V = E and E G = F. Thus the differential equation L(y) = w(y, 1, x 1, y)/w(1, x 1, y) has coefficients in F. Clearly L 1 {0} = V.

13 Computing a differential equation for U(3, C) The differential field E is a NNC extension of F. Let V :=span C {1, x 1, y} and G be the group of differential automorphisms of E fixing F. Then GV V, F V = E and E G = F. Thus the differential equation L(y) = w(y, 1, x 1, y)/w(1, x 1, y) has coefficients in F. Clearly L 1 {0} = V. Note that σ(x 1 ) = x 1 + α σ, σ(x 2 ) = x 2 + β σ and σ(y) = y + β σ x 1 + γ σ for some α σ, b σ, γ σ C.

14 Computing a differential equation for U(3, C) A computation shows ( f L(Y ) = Y 2 + 2f 1 f 2 f 1 ) ( Y + f 1 f f 1 f 2 ( f 1 f 1 ) ) 2 f 1 Y. f 1

15 Computing a differential equation for U(3, C) If E is Picard-Vessiot extension over F with a Galois group isomorphic to U(3, C) then there are algebraically independent elements x 1, x 2 and y E such that E = F (x 1, x 2, y) and x i F and y = x 1 x 2.

16 F = C(z), z = 1 Let c 1, c 2 C be distinct complex numbers and let f i = 1 x+c i. Then Y 3z + 2c 2 + c 1 (z + c 1 )(z + c 2 ) Y 1 + (z + c 1 )(z + c 2 ) Y = 0 has differential Galois group U(3, C). } ln(z + c2 ) Solution Space: span C {1, ln(z + c 1 ),. z + c 1

17 F = C(x), x = 1, G = U(4, C) let f i = 1 x+c i, where c i are distinct complex numbers for i = 1, 2, 3, 4. The differential equation d 4 dx 4 + 6x 2 + (3c 1 + 4c 2 + 5c 3 )x + c 2 c 1 + 2c 3 c 1 + 3c 3 c 2 d 3 (x + c 1 )(x + c 2 )(x + c 3 ) dx 3 d 2 + 7x + c 1 + 2c 2 + 4c 3 (x + c 1 )(x + c 2 )(x + c 3 ) dx 2 1 d + (x + c 1 )(x + c 2 )(x + c 3 ) dx. Solution Space: span C {1, ln(x + c 1 ), ln(x + c2 ) ln(x+c 3 } ) x+c, 2. x + c 1 x + c 1

18 Algebraic Subgroups of U(3, C) Let E = F (x 1, x 2, y) with x 1 = f 1, x 2 = f 2 and y = f 1 x 2 and let E K F be an intermediate differential field.

19 Algebraic Subgroups of U(3, C) Let E = F (x 1, x 2, y) with x 1 = f 1, x 2 = f 2 and y = f 1 x 2 and let E K F be an intermediate differential field. Assume that C is algebraically closed. We will first classify the differential subfields of E and then use the Galois correspondence to classify the algebraic subgroups of U(3, C).

20 Algebraic Subgroups of U(3, C) Let E = F (x 1, x 2, y) with x 1 = f 1, x 2 = f 2 and y = f 1 x 2 and let E K F be an intermediate differential field. Assume that C is algebraically closed. We will first classify the differential subfields of E and then use the Galois correspondence to classify the algebraic subgroups of U(3, C). If tr.d (K F ) = 3 then K = E and G(E K) is the trivial group. If tr.d (K F ) = 2 then we have three cases to discuss.

21 Algebraic Subgroups of U(3, C) E = F (x 1, x 2, y) with x 1 = f 1, x 2 = f 2 and y = f 1 x 2 and let E K F. First note that there is an element s K such that 0 s span C {x 1, x 2 }.

22 Algebraic Subgroups of U(3, C) E = F (x 1, x 2, y) with x 1 = f 1, x 2 = f 2 and y = f 1 x 2 and let E K F. First note that there is an element s K such that 0 s span C {x 1, x 2 }. Case 1. K F (x 1, x 2 ). Then K = F (x 1, x 2 ). 1 0 c G(E K) = c C 0 0 1

23 Algebraic Subgroups of U(3, C) E = F (x 1, x 2, y) with x 1 = f 1, x 2 = f 2 and y = f 1 x 2 and let E K F. First note that there is an element s K such that 0 s span C {x 1, x 2 }. Case 1. K F (x 1, x 2 ). Then K = F (x 1, x 2 ). 1 0 c G(E K) = c C Case 2. Suppose that x 2 / K. Then s = x 1 + cx 2 K for some c 2 K. Then one can deduce that K(x 2 ) = E. Thus y K(x 2 ).

24 Algebraic Subgroups of U(3, C) A computation will show that y = αx βx 2 + γ, for some α C, β, γ K. Furthermore, one can conclude, α = c/2, β = x 1 + cx 2 + d K for some d C, γ = y c 2 x2 2 x 1x 2 + cx 2 K.

25 Algebraic Subgroups of U(3, C) A computation will show that y = αx βx 2 + γ, for some α C, β, γ K. Furthermore, one can conclude, α = c/2, β = x 1 + cx 2 + d K for some d C, γ = y c 2 x2 2 x 1x 2 + cx 2 K. Since tr.d (K F ) = 2, we obtain K = F (β, γ).

26 Algebraic Subgroups of U(3, C) K = F (x 1 + cx 2, y c 2 x2 2 x 1x 2 + cx 2 ) G(E K) = 1 cδ c 2 δ2 + dδ 0 1 δ δ C

27 Algebraic Subgroups of U(3, C) Case 3. Suppose that x 2 K. Then K(x 1 ) = E and thus y K(x 1 ). Now one can conclude that y + dx 1 K for some d C.

28 Algebraic Subgroups of U(3, C) Case 3. Suppose that x 2 K. Then K(x 1 ) = E and thus y K(x 1 ). Now one can conclude that y + dx 1 K for some d C. Thus K = F (x 2, y + dx 1 ).

29 Algebraic Subgroups of U(3, C) Case 3. Suppose that x 2 K. Then K(x 1 ) = E and thus y K(x 1 ). Now one can conclude that y + dx 1 K for some d C. Thus K = F (x 2, y + dx 1 ). G(E K) = 1 δ dδ δ C

30 Algebraic Subgroups of U(3, C) If tr.d (K F ) = 1 then K = F (c 1 x 1 + c 2 x 2 ), where c 1, c 2 C and 1 a c G(E K) = 0 1 b c 1 a + c 2 b = 0, a, b, c C If tr.d (K F ) = 0 then K = F and G(E K) = U(3, C).

31 Algebraic Subgroups of U(3, C) Three dimensional subgroups: U(3, C). Two dimensional subgroups: Let c 1, c 2 C. 1 α γ 0 1 β c1 α + c 2 β = 0, α, β, γ C 0 0 1

32 Algebraic Subgroups of U(3, C) One dimensional subgroups: Let c, d C. 1 0 γ cδ c 2 δ2 + dδ 0 1 δ γ C δ C Let d C. 1 δ dδ δ C

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