Zoran Skoda
formal integral curve

Consider an analytic coordinate neighborhood UU at a real analytic nn-manifold; coordinates are x 1,,x n:URx^1,\ldots,x^n:U\to\mathbf{R} and at a distinguished point ee all coordinates are 00. Then an arbitrary vector field around ee can be represented locally as

X=(a i+a j ix j+a jk ix jx k+a jkl ix jx kx l+)x i X = (a^i + a^i_j x^j + a^{i}_{j k} x^j x^k + a^i_{j k l} x^j x^k x^l + \ldots)\frac{\partial}{\partial x^i}

where the sum (from 11 to nn) over repeated indices is understood. We look for an integral curve γ=γ X:IU\gamma = \gamma_X:I\to U, where γ(0)=e\gamma(0) = e, II is some interval containing 00, the components are given by power series

γ i(t)= s=1 nP s is!t s=a it+12!P 2 it 2+13!P 3 it 3+ \gamma^i(t) = \sum_{s = 1}^n \frac{P^i_s}{s!} t^s = a^i t + \frac{1}{2!}P^i_2 t^2 + \frac{1}{3!} P^i_3 t^3 +\ldots

where P s iP^i_s\in\mathbb{R} and P 1 i=a iP^i_1 = a^i. Then ddtγ i=X i\frac{d}{d t}\gamma^i = X^i for all tt in some interval around 00 iff for all integers s1s\geq 1

d sγ i(dt) s| t=0=d s1X(dt) s1| t=0 \frac{d^s \gamma^i}{(d t)^s}|_{t = 0} = \frac{d^{s-1} X}{(d t)^{s-1}}|_{t = 0}

The left hand side is just P s iP^i_s hence for k=1k = 1 we get P 1 i=a iP^i_1 = a^i and for s2s\geq 2 these equalities read as

P k i= r=1 k m 1+m 2++m r=s1,m i1(s1)!m 1!m R! j 1,,j r=1 na j 1j r iP m 1 j 1P m r j r P^i_k = \sum_{r = 1}^k \sum_{m_1+m_2+\ldots+m_r = s - 1, m_i\geq 1} \frac{(s-1)!}{m_1!\cdots m_R!} \sum_{j_1,\ldots,j_r = 1}^n a^i_{j_1 \ldots j_r} P^{j_1}_{m_1}\cdots P^{j_r}_{m_r}

or in low order as

P s i=a j iP s1 j+a j i l=1 s2(n1m)P m jP s1l k+a jkl i l i1,l 1+l+2+l 3=s1(s1)!m 1!m 2!m 3!P m 1 jP m 2 kP m 3 l P^i_s = a^i_j P^j_{s-1} + a^i_{j } \sum_{l=1}^{s-2}\binom{n-1}{m} P^j_m P^k_{s-1-l} + a^i_{j k l} \sum_{l_i\geq 1, l_1+l+2+l_3 = s-1}\frac{(s-1)!}{m_1! m_2! m_3!} P^j_{m_1} P^k_{m_2} P^{l}_{m_3}

and therefore

P 2 i=a j ia j P^i_2 = a^i_j a^j
P 3 i=a j ia k ja k+2a jk ia ja k P^i_3 = a^i_j a^j_k a^k + 2 a^i_{j k} a^j a^k
P 4 i=a j iP 3 j+a jk i(3P 2 jP 1 k+3P 1 jP 2 k)+a jkl i(6P 1 jP 1 kP 1 l) P^i_4 = a^i_j P^j_3 + a^i_{j k} (3 P^j_2 P^k_1 + 3 P^j_1 P^k_2) + a^i_{j k l} ( 6 P^j_1 P^k_1 P^l_1)

that is

P 4 i=a j ia k ja l ka l+2a j ia kl ja ka l+3a jk ia l ja la k+3a jk ia l ka la j+6a jkl ia ja ka l P^i_4 = a^i_j a^j_k a^k_l a^l + 2 a^i_j a^j_{k l} a^k a^l + 3 a^i_{j k} a^j_l a^l a^k + 3 a^i_{j k} a^k_l a^l a^j + 6 a^i_{j k l} a^j a^k a^l

Now suppose in the similar vain that we are given nn analytic vector fields X αX_\alpha, α=1,,n\alpha = 1,\ldots,n which are linearly independent in each point in UU and γ X α\gamma_{X_\alpha} are their analytic integral curves around ee. By the linear independence they give a trivialization of the frame bundle around ee and hence we can write a trivial connection on the frame bundle which is equivalent to a flat affine connection \nabla (by affine we mean a connection on a tangent vector bundle). To this affine connection one associates an exponential map which diffeomorphically sends some neighborhood of zero in the tangent bundle T eUT_e U into UU by a map

exp:c αx α| eγ c αX α(1) exp : c^\alpha\frac{\partial}{\partial x^\alpha}|_e\mapsto \gamma_{c^\alpha X_\alpha}(1)

Thus the formulas for s(P s i) X\sum_s (P^i_s)_X where X= αc αX αX = \sum_\alpha c^\alpha X_\alpha give the coordinates in the exponential chart in terms of c αc_\alpha and the coefficients (a j 1j r i) α(a^i_{j_1\ldots j_r})_\alpha for X αX_\alpha.

We are interested in computing the inverse function which computes the original coordinates as functions of the coordinates in the exponential chart. One should be able to do this by simply composition inverting the expressions for the vector function rP r \sum^r P^\bullet_r, say using the Lagrange inversion. But this seems to be difficult to do. It turns out that there is a neat indirect solution which holds at least in the case when the original vector fields close under the Lie bracket, that is when they are say the left invariant vector fields around the unit ee on the Lie group in a chart in which we are given the expressions for vector field but not explicitly the group multiplication. In that case we write the vector fields as

X α= iϕ α i(x)x i e X_\alpha = \sum_i \phi_\alpha^i(x) \frac{\partial}{\partial x^i}_e

and interchange in an antihomomorphic fashion x ix^i for i\partial^i and i\partial_i for x ix_i where the latter are “contravariant Weyl algebra” generators.

Then define D i=[,x i]D_i = [-,x_i] as an operators of commuting a formal series in s\partial^s with x ix_i. Then the inverse function KK is the vector function given by the formula

K i(c 1,,c n)=( s=0 ( l,jc jϕ j l()D l) sc kϕ k i())|0 K^i(c^1,\ldots,c^n) = \left(\sum_{s = 0}^\infty (\sum_{l,j} c^j \phi^l_j(\partial) D_l)^s c^k \phi^i_k(\partial)\right) |0\rangle

where the right hand side involves the left action on the Fock vacuum.

Of course, we can write the same formula using the original coordinates with ϕ j i=ϕ j i(x 1,,x n)\phi^i_j = \phi^i_j(x^1,\ldots,x^n) and the evaluation at 00, hence obtaining

K i(c 1,,c n)=( s=0 ( l,jc jϕ j l(x 1,,x n)[,x l]) sc kϕ k i(x 1,,x n))(0)=c i+(exp( jc jϕ j l[,x l])1 jc jϕ j l[,x l](c kϕ k i))(0) K^i(c^1,\ldots,c^n) = \left(\sum_{s = 0}^\infty (-\sum_{l,j} c^j \phi^l_j(x^1,\ldots,x^n) [-,\frac{\partial}{\partial x^l}])^s c^k \phi^i_k(x^1,\ldots,x^n)\right)(0) = c^i + \left(\frac{\exp(-\sum_j c^j \phi^l_j [-,\frac{\partial}{\partial x^l}])- 1}{\sum_j c^j \phi^l_j [-,\frac{\partial}{\partial x^l}]}(c^k \phi^i_k)\right)(0)

See also combinatorics in wikipedia Butcher group.

Last revised on November 13, 2018 at 17:35:03. See the history of this page for a list of all contributions to it.