formal integral curve

Consider an analytic coordinate neighborhood $U$ at a real analytic $n$-manifold; coordinates are $x^1,\ldots,x^n:U\to\mathbf{R}$ and at a distinguished point $e$ all coordinates are $0$. Then an arbitrary vector field around $e$ can be represented locally as

$X = (a^i + a^i_j x^j + a^{i}_{j k} x^j x^k + a^i_{j k l} x^j x^k x^l + \ldots)\frac{\partial}{\partial x^i}$

where the sum (from $1$ to $n$) over repeated indices is understood. We look for an integral curve $\gamma = \gamma_X:I\to U$, where $\gamma(0) = e$, $I$ is some interval containing $0$, the components are given by power series

$\gamma^i(t) = \sum_{s = 1}^n \frac{P^i_s}{s!} t^s = a^i t + \frac{1}{2!}P^i_2 t^2 + \frac{1}{3!} P^i_3 t^3 +\ldots$

where $P^i_s\in\mathbb{R}$ and $P^i_1 = a^i$. Then $\frac{d}{d t}\gamma^i = X^i$ for all $t$ in some interval around $0$ iff for all integers $s\geq 1$

$\frac{d^s \gamma^i}{(d t)^s}|_{t = 0} = \frac{d^{s-1} X}{(d t)^{s-1}}|_{t = 0}$

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

$P^i_k = \sum_{r = 1}^k \sum_{m_1+m_2+\ldots+m_r = s - 1, m_i\geq 1} \frac{(s-1)!}{m_1!\cdots m_R!}
\sum_{j_1,\ldots,j_r = 1}^n a^i_{j_1 \ldots j_r} P^{j_1}_{m_1}\cdots P^{j_r}_{m_r}$

or in low order as

$P^i_s = a^i_j P^j_{s-1} + a^i_{j } \sum_{l=1}^{s-2}\binom{n-1}{m} P^j_m P^k_{s-1-l}
+ a^i_{j k l} \sum_{l_i\geq 1, l_1+l+2+l_3 = s-1}\frac{(s-1)!}{m_1! m_2! m_3!}
P^j_{m_1} P^k_{m_2} P^{l}_{m_3}$

and therefore

$P^i_2 = a^i_j a^j$

$P^i_3 = a^i_j a^j_k a^k + 2 a^i_{j k} a^j a^k$

$P^i_4 = a^i_j P^j_3 + a^i_{j k} (3 P^j_2 P^k_1 + 3 P^j_1 P^k_2) + a^i_{j k l} ( 6 P^j_1 P^k_1 P^l_1)$

that is

$P^i_4 = a^i_j a^j_k a^k_l a^l + 2 a^i_j a^j_{k l} a^k a^l +
3 a^i_{j k} a^j_l a^l a^k + 3 a^i_{j k} a^k_l a^l a^j + 6 a^i_{j k l} a^j a^k a^l$

Now suppose in the similar vain that we are given $n$ analytic vector fields $X_\alpha$, $\alpha = 1,\ldots,n$ which are linearly independent in each point in $U$ and $\gamma_{X_\alpha}$ are their analytic integral curves around $e$. By the linear independence they give a trivialization of the frame bundle around $e$ 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 $T_e U$ into $U$ by a map

$exp : c^\alpha\frac{\partial}{\partial x^\alpha}|_e\mapsto \gamma_{c^\alpha X_\alpha}(1)$

Thus the formulas for $\sum_s (P^i_s)_X$ where $X = \sum_\alpha c^\alpha X_\alpha$ give the coordinates in the exponential chart in terms of $c_\alpha$ and the coefficients $(a^i_{j_1\ldots j_r})_\alpha$ for $X_\alpha$.

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 $\sum^r P^\bullet_r$, 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 $e$ 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

$X_\alpha = \sum_i \phi_\alpha^i(x) \frac{\partial}{\partial x^i}_e$

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

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

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

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 $\phi^i_j = \phi^i_j(x^1,\ldots,x^n)$ and the evaluation at $0$, hence obtaining

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

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.