Harmonic Maps, Loop Groups, and Integrable Systems by Martin A. Guest

By Martin A. Guest

This can be an obtainable advent to a couple of the elemental connections between differential geometry, Lie teams, and integrable Hamiltonian structures. The textual content demonstrates how the idea of loop teams can be utilized to check harmonic maps. via targeting the most principles and examples, the writer leads as much as issues of present learn. The e-book is acceptable for college students who're starting to research manifolds and Lie teams, and will be of curiosity either to mathematicians and to theoretical physicists besides.

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Example text

Im are independent integrals, and if they are in involution, then (under suitable conditions) we obtain a new Hamiltonian system on a phase space M , with dim M = dim M − 2m. (We say that I1 , . . , Im are independent if dI1 , . . , dIm are linearly independent at each point, and that I1 , . . ) In general, integrals (apart from H) do not exist (and even if they exist, they are not easy to find). If there exist n independent integrals in involution, where dim M = 2n, then we say that the Hamiltonian system (M, ω, H) is completely integrable.

Here, SLn R = N n Exercise: ˆ N2 . 1) Verify that SL2 R = N 2 ˆ Nn ˆ n ⊕ nn and the corresponding “partial decomposition” SLn R ⊇ N The decomposition sln R = n n are called Gauss decompositions. ) Modification of the phase space 8 Concluding remarks on one-dimensional Lax equations 37 As a second generalization, let us consider the Lax equation X˙ = [X, π1 X] for X : R → g (or more generally X˙ = [X, π1 (∇f )X ]). Here we assume that g = g1 ⊕ g2 , but we do not choose a co-adjoint orbit. The argument of Chapter 5 (and the previous paragraph) shows that we have a local solution X(t) = Ad(exp tV )−1 > 0.

We use the following version of the harmonic map equation (see Chapter 9): (∗) (Aλ )z¯ − (Bλ )z = [Aλ , Bλ ] for all λ ∈ C with |λ| = 1.  0 0   0 .  −1 0 50 Part II Two-dimensional integrable systems This is equivalent to (∗∗) Fλ−1 (Fλ )z = 12 (1 − λ1 )A, Fλ−1 (Fλ )z¯ = 12 (1 − λ)B. The original harmonic map equation (φ−1 φz¯)z + (φ−1 φz )z¯ = 0 obviously admits any constant function φ(z) = g as a solution (where g ∈ G). A corresponding solution of (∗) is simply the zero solution: Aλ = Bλ = 0.

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