Cf. [Drazin 1989], p. 121. Let L(t) be a bounded linear operator that acts on a Hilbert space ℋ and depends on a parameter t. Define the adjoint A^{†}of any bounded operator A as usual through the inner product of the Hilbert space, by (u, Av) = (A^{†}u, v) for all u and v in ℋ. If L(t) is unitarily equivalent to L(0), i.e., there exists a U(t) such that U^{†}(t) L(t) U(t) = L(0) U^{†}(t) U(t) = I then there also exists a skewsymmetric M which may depend on t M^{†}(t) = - M(t) such that dU(t)/dt = M(t) U(t) Clearly, in this case, [M(t), U(t)] = 0. Now, if L is selfadjoint, then dL(t)/dt + [L(t), M(t)] = 0. If L(t) and M(t) are nxn complex matrices, and L(t) is Hermitean, and dL(t)/dt + [L(t), M(t)] = 0. then the eigenvalues of L(t) are independent of t. This "derives" a Lax equation; now, in reverse to solve it: If dU(t)/dt = M U(t) and M is NOT dependent on t, this equation integrates immediately to U(t) = exp( M t ) If instead M = M(t) is a continuous function of t, and M(t) is a commutative family of operators, [M(t), M(t')] = 0, then dU(t)/dt = M(t) U(t) integrates to t U(t) = exp( ∫ M(s) ds ) 0 in the sense of a Riemann integral. The sum of skew-symmetric matrices is skew-symmetric, and so U is indeed unitary, and the Riemann integral is easily and well defined. ------------- Checking this solution by computing the first derivative according to definition: dU(t)/dt = t+Δ(t) t lim (1/Δ(t)) [exp( ∫ M(s) ds ) - exp( ∫ M(s) ds )] 0 0 t+Δ(t) t = lim (1/Δ(t)) [exp( ∫ M(s) ds ) - 1 ] exp( ∫ M(s) ds ) t 0 Using the exp power series expansion about zero = lim (1/Δ(t)) X t+Δ(t) t [ 1 + ∫ M(s) ds + O( Δ²(t) ) - 1 ] exp( ∫ M(s) ds ) t 0 = lim (1/Δ(t)) X t+Δ(t) t [ ∫ M(s) ds + O( Δ²(t) ) ] exp( ∫ M(s) ds ) t 0 Then dU(t)/dt = M(t) U(t) When M(t) is skewsymmetric => U(t) is unitary, and also then dU^{†}(t)/dt = + U^{†}(t) M^{†}(t) dU^{†}(t)/dt = - U^{†}(t) M(t) and (d/dt)² U(t) = d( M(t) U(t) )/dt = d( M(t) )/dt U(t) + M(t) d( U(t) )/dt = d( M(t) )/dt U(t) + M²(t) U(t) = ( dM(t)/dt + M²(t) ) U(t) (d/dt)^{3}U(t) = d( dM(t)/dt + M²(t) )/dt U(t) + ( dM(t)/dt + M²(t) ) dU(t)/dt = d²M(t)/dt² U(t) + 2 M(t) dM(t)/dt U(t) + dM(t)/dt dU(t)/dt + M²(t) dU(t)/dt = d²M(t)/dt² U(t) + 2 M(t) dM(t)/dt U(t) + dM(t)/dt M(t) U(t) + M^{3}(t) U(t) = ( d²M(t)/dt² + 2 M(t) dM(t)/dt + dM(t)/dt M(t) + M^{3}(t) ) U(t) Etc., etc., etc. ------------- If U^{†}(t) L(t) U(t) = L(0) L(0) being independent of t, then taking a derivative dU^{†}(t)/dt L(t) U(t) + U^{†}(t) dL(t)/dt U(t) + U^{†}(t) L(t) dU(t)/dt = 0 - U^{†}(t) M(t) L(t) U(t) + U^{†}(t) dL(t)/dt U(t) + U^{†}(t) L(t) M(t) U(t) = 0 multiplying on the left by U(t) - M(t) L(t) U(t) + dL(t)/dt U(t) + L(t) M(t) U(t) = 0 multiplying on the right by U^{†}(t) - M(t) L(t) + dL(t)/dt + L(t) M(t) = 0 L(t) M(t) - M(t) L(t) + dL(t)/dt = 0 [L(t), M(t)] + dL(t)/dt = 0 or dL(t)/dt = [M(t), L(t)] QED PRODUCT INTEGRALS But, M(t) being a commutative family implies that M(t) commutes with with all its derivatives w.r.t. t, if they exist. If M(t) is polynomial, M(t) = Σ m_{k}t^{k}k with matrix coefficients m_{k}, then [m_{k}, m_{j}] = 0, for all k, j. The requirement then of M(t) being a commutative family is a fairly strong requirement. Suppose M(t) is not a commutative family, is there anything that can be done to express a solution to the Lax equation? If one approaches anew the idea of the Riemann integral, there is a method possibly suitable for approximation in the case of continuous t, and also possibly useful as a definition in an analogous case of a discrete parameter t. The method of "Product Integrals" was apparently first introduced in [Volterra 1938]. Cf. [Dollard 1977]. Where the Riemann integral is defined as a limit of successively approximating Riemann sums, the product integral is defined as a limit of approximating products. Start as in the Riemannian case by partitioning the interval [t_{a}, t_{b}] with a sequence of t_{k}∈ [t_{a}, t_{b}], with t_{a}= t_{0}< t_{1}< ... < t_{k-1}< t_{k}< t_{k+1}< ... < t_{n-1}= < t_{b}making n cuts and n-1 subintervals contained in [t_{a}, t_{b}], Δt_{k}:= t_{k}- t_{k-1}If the covering of [t_{a}, t_{b}] can be taken with n large enough, and the Δt_{k}small enough so that M(t_{k}) can reasonably be taken as constant within Δt_{k}, then we can write an approximating expression U(t_{k+1}) = exp( M(t_{k+1}) Δt_{k+1}) U(t_{k}) Collecting this recursiveness of expression, n U(t) = Π exp( M(t_{k}) Δt_{k}) U(0) k=0 where, of course, the ordering of the exp factors is important, because it is generally true that [M(t_{k}), M(t_{j})] ≠ 0. This is the product analog of the Riemannian sum. Convergence of the Riemann Product Let m{t} := sup Δt_{k}With M(t) a continuous function of t, the limit will actually exist in the limit as the implied mesh goes to zero. [Volterra 1938] Since the limit is actually independent of the initial U(0), we can write then, for an arbitrary such interval [y, x] of the domain of M(t), x Π exp( M(t) dt ) := y n lim Π exp( M(t_{k}) Δt_{k}) m{t}→0 k=0 Properties of the Product Integral In [Dollard 1977], eleven properties of the product intergal are listed and proved. Restating these properties: 1. For M(t) continuous and and [y, x] in the domain of M(t) x x d/dx Π exp( M(t) dt ) = M(x) Π exp( M(t) dt ) y y 2. x z x Π exp( M(t) dt ) = Π exp( M(t) dt ) Π exp( M(t) dt ) y y z for z ∈ [y, x] 3. x Π exp( M(t) dt ) is non singular. y Use det exp(A) = exp( Tr A ) to prove. 4. Relying on 3 and defining, x Π exp( M(t) dt ) := I, we have x x y Π exp( M(t) dt ) = [ Π exp( M(t) dt ) ]^{-1}y x 5. If the operator family {M(t)} is commutative, then x x Π exp( M(t) dt ) = exp( ∫ M(t) dt ) y y 6. x x ‖ Π exp( M(t) dt ) ‖ ≤ ‖ exp( ∫ M(t) dt ) ‖ y y for norms on the Hilbert space. 7. x x ‖ Π exp( M(t) dt ) - I ‖ = exp( ∫ ‖ M(t) ‖ dt ) - 1 y y 8. A "sum rule" for A(t) and B(t) both continuous on their common domain, and [y, x] as before within the domain, define x P(x) := Π exp( A(t) dt ) y Then, x x Π exp( (A(t) + B(t)) dt ) = P(x) Π exp( P^{-1}(t) B(t) P(t) dt ) y y 9a. For a constant (independent of t) operator T with inverse, x x T^{-1}Π exp( A(t) dt ) T = Π exp( T^{-1}A(t) T ) dt y y 9b. The "similarity rule". For M as usual, and T with continuous first derivative, T'(t) x T^{-1}(x) Π exp( A(t) dt ) T(y) y x = Π exp( (T^{-1}(t) A(t) T(t) - T^{-1}(t) T'(t)) dt ) y 10, 11 deal with the definitions of improper product integrals when x → ∞; all is well there as well. Define, ∞ y Π exp( A(t) dt ) := lim Π exp( A(t) dt ) y y→∞ x provided the limit exists. The integral can be singular. E.g., take A = -I for all t, then from 5., the limit of the product integral vanishes. Interest is then in the conditions for which the limit exists and is nonsingular. 10. For nxn A(t) defined on [y, ∞), continuous A(t) ∈ L^{1}(a, ∞), i.e., ∫ ‖ A(t) ‖ dt < ∞, The limit of the product integral exists and is nonsingular. Also, if the A and B are both continuous and in L^{1}(a, ∞), then the product integral exists for A + B exists and is nonsingular. 11. If the improper integral ∞ x h(y) := ∫ A(t) = lim ∫ A(t) dt y x→∞ y exists (without A having to be in L^{1}), then if (h A) is in L^{1}), the product integral of A exists and is nonsigular. MODELS FOR (NON)COMMUTATIVE FAMILIES OF M(t_{k}): Let T be a normal operator acting on a finite dimensional complex Hilbert space. T is then diagonalizable and possesses an orthonormal eigenbasis |t_{μ}> that spans the Hilbert space, and there exists a cyclic operator C_{T}, easily constructible: C_{T}|t_{μ}> = |t_{μ-1}> with μ understood as discrete and (mod n), n being the dimension of the Hilbert space. Both Hermitean and unitary operators are normal, and the spectrum of T is in general complex. T can always be diagonalized by a unitary operator, and it also has a generator that is not necessarily Hermitean, but is necessarily normal, i.e., exp maps normal operators to normal operators. Also, then, C_{T}^{k}|t_{μ}> = |t_{μ-k}> C_{T}^{†k}|t_{μ}> = |t_{μ+k}> Now define as an analytic continuation from integral k, M(s) := C_{T}^{s}M C_{T}^{†s}for arbitrary complex s. C_{T}has a generator K so that we can write, C_{T}= exp( +K ) C_{T}^{†}= exp( -K ) so K is skewhermitean; then also express M(s) by M(s) := exp( +K s ) M exp( -K s ) For such M(s), [M(s), M(s')] = 0. If s is real, M(s) is skewhermitean if M is; for general s, it is not skewhermitean even if M is. Taking a derivative w.r.t. s, dM(s)/ds = exp( +K s ) [K, M] exp( -K s ) dM(s)/ds = [K, M(s)] Suggesting that such a construction is repeatable if K is now actually a function of s, and that the general Lax equation can be decomposed into a collection of Lax equations that are of a type simpler, so as to be solvable, and that this can be organized by successive polynomial approximations. TRANSFORMATIONS AND INVARIANTS Obviously for arbitrary A A → S^{-1}A S leaves the spectrum of A invariant for any nonsingular S, therefore trace and determinant are invariant. If S = U is unitary, then U^{-1}= U^{†}so A → U^{†}A U leaves the spectrum invariant, but not so for nonsingular and nonunitary S where the mapping is A → S^{†}A S We have Tr(S^{†}A S) = Tr(S S^{†}A) X= Tr( A ) Since Det( S^{†}) = Det^{*}( S ) Det( S^{†}A S ) = Det( S^{†}) Det( A ) Det( S ) = |Det( S )|² Det( A ) If S is special ( Det( S ) = 1 ), so then Det( S^{†}) = 1, Det( S^{†}A S ) = Det( A )

1. See also The Toda Equation 2. 3.

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