Exponential map

The exponential map in the context of Lie groups is exp:T_eG→G, or equivalently exp:g→G. We continue with discussion of the Lie group SL_2R.

G = Setup[{(x_1 y_1 + x_2 y_3)/(1 + x_3 y_2), (x_2 + x_1 y_2)/(1 + x_3 y_2), (x_3 y_1 + y_3)/(1 + x_3 y_2)}, {1, 0, 0}] ;

The exponential of a tangent vector {v_1,v_2,v_3}∈T_eG is determined by the integral curve of the left-invariant vector field induced by v through e∈G. Recall the general form of a left-invariant vector field on G.

ΤL[G, Τ★[v, 3], 1]

{v_3 x_2 + x_1 (v_1 - v_2 x_3), v_2 (x_1 - x_2 x_3), v_3 + x_3 (v_1 - v_2 x_3)}

Even for "simple" instances for v∈T_eG, Mathematica has trouble finding exact solutions. We try the three basis vectors e_1,e_2,e_3 of g.

MF[DSolve[E0[ΧIntegral[ΤL[G, #1, 1]], Χ0[0, G[0]]], Χ★[3], ] &/@Id[3]]

DSolve :: bvnul : For some branches of the general solution, the given boundary conditions lead to an empty solution.  More…

( {{{{x_1[] →^, x_2[] →0, x_3[] →0}}}, {{}}, {{{x_1[] →1, x_2[] →0, x_3[] →}}}} )

For e_2 we yield the (unique) solution exp t e_2={1,t,0}, since

sol = Χ★[3] → {1, , 0}

E0[ΧIntegral[ΤL[G, {0, 1, 0}, 1]], Χ0[0, G[0]]]/.Ft[sol, ∂_sol, sol/.→0]

{x_1[] →1, x_2[] →, x_3[] →0}

{True, True, True, True, True, True}


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