Zoran Skoda exponential map

This entry is about calculational aspects of the exponential map for Lie groups. For more general exponential map via connections in Riemannian geometry see $n$Lab:exponential map. Related entries in $n$Lab: invariant vector field, adjoint action, Hausdorff series, Hadamard's formula. I used the standard facts from the introductory chapters on differential geometry and Lie groups in Sigurdur Helgason, Differential geometry, Lie groups and symmetric spaces, most notably the formula for the differential of the exponential map below. We use the Einstein summation convention, skipping the sum in the notation, when summing over repeating indices, one lower and one upper, whenever not said otherwise.

Throughout, $G$ is a Lie group with tangent Lie algebra $\mathfrak{g} = \mathcal{X}^L(G)\cong T_e G \cong\mathcal{X}^R(G)$.

Given $X\in T_e G$, there is a unique 1-parametric subgroup $h = H_X: t\mapsto h(t)$ through $e$ such that $exp(t X) = h(t)$, and in particular $exp(X) = h(1)$.

If $X = X_e \in T_e G$ and $g\in G$, then the corresponding left invariant vector field $X^L$ satisfies

$$X^L_g f = (X^L f)(g) = ((L_g)_* X) f = X(f\circ L_g) = \frac{d}{d t} f(g exp(t X)) |_{t=0}$$

and, likewise, $X^R_g f = \frac{d}{d t} f(exp(t X) g)|_{t = 0}$.

If $f$ is an analytic function on $G$, for fixed $g\in G$ and $X\in T_e G$, the Taylor expansion for a function of parameter $t$ gives (Helgason formulas (6) and (11) in Sec. II.1)

$$f(g exp(t X)) = \sum_{n = 0}^\infty t^n \frac{((X^L)^n f)(g)}{n!}$$

The formula for the differential of the exponential map $T_X exp = T_{(e,X)} exp : T_X (T_e G)\to T_{exp X} G$ is

$$T_{X} exp = (d exp)_X = d (L_{exp X})_e \circ \frac{1 - exp(- ad X)}{ad X}$$

where $ad X$ on the right hand side is in fact the corresponding element in the tangent space at a vector in Lie algebra, which is itself identified with Lie algebra. Similarly we can write in terms of $ad_r X = - ad X$ and $R_{exp X}$:

$$T_{X} exp = (d exp)_X = d (R_{exp X})_e \circ \frac{1 - exp(ad X)}{- ad X}$$

Thus, if $e_1,\ldots,e_n$ is the basis of $T_e G$ with commutators $[e_i,e_j] = C^k_{i j} e_k$ and $\partial^1,\ldots, \partial^n$ are the corresponding vector fields in $\Gamma T(T_e G)$ and let $X = \sum_i X^i e_i$. Denote $\mathcal{C}^k_j = - C^k_{i j} X^i = C^k_{j i} X^i$ (this sum is just a number!). Then $(ad X)(e_j) = [X^i e_i, e_j ] = -\mathcal{C}^k_j e_k$ and $(ad X)^n(e_j) = (-\mathcal{C}^n)^k_j e_k$. Then $(T_X exp)(\partial^i_X) = (L_{exp X})_* \left(\frac{1- e^{\mathcal{C}}}{-\mathcal{C}}\right)^k_i e_k = \left(\frac{1- e^{\mathcal{C}}}{-\mathcal{C}}\right)^k_i (L_{exp X})_* e_k$. On the other hand, $e_i^L = (L_{exp_X})_* e_i$ at point $exp(X)$ by the definition of left invariant fields. Thus $e_i^L = \left(\frac{\mathcal{C}}{e^{\mathcal{C}}-1}\right)^j_i (d exp)(\partial^j)$ and, likewise, $e_i^R = \left(\frac{\mathcal{C}}{1-e^{-\mathcal{C}}}\right)^j_i (d exp)(\partial^j)$ and finally the comparison of the two gives $e_i^R = (e^{\mathcal{C}})^j_i e_j^L$.

We will call by $\mathcal{O}^{-1}$ the matrix $e^{\mathcal{C}}$ and it will play a role in the construction of a Hopf algebroid. In fact, one can consider other coordinates on a neighborhood of $e$ in $G$, not only the chart given by the exponential map; $\mathcal{O}^{-1}$ will then be a matrix of formal functions, while $e^{\mathcal{C}}$ is how it looks in the chart given by the exponential map.

The following formula, quadratic in $\mathcal{O}^{-1}$, holds:

$$C_{\mu\nu}^\gamma (\mathcal{O}^{-1})^\sigma_\gamma = C^\sigma_{\lambda\rho}(\mathcal{O}^{-1})^\lambda_\mu (\mathcal{O}^{-1})^\rho_\nu$$

In the chart given by the exponential map, $\mathcal{O}^{-1} = e^\mathcal{C}$ and the identity boils down to the sequence of identities for $n=0,1,2,\ldots$

$$C^\gamma_{\mu\nu}(\mathcal{C}^n)^\sigma_\gamma = \sum_{m=0}^n \binom{n}{m} C^\sigma_{\lambda\rho} (\mathcal{C}^m)^\lambda_\mu (\mathcal{C}^{n-m})^\rho_\nu$$

which are proved by induction on $n$, using antisymmetry $C^\mu_{\alpha\beta} = C^\mu_{\beta\alpha}$ and the Jacobi identity

$$C^\lambda_{\alpha\beta} C^\sigma_{\lambda\gamma} + C^\lambda_{\beta\gamma} C^\sigma_{\lambda\alpha} + C^\lambda_{\gamma\alpha} C^\sigma_{\lambda\beta} =0$$

as well as the Pascal triangle identity

$$\binom{n+1}{m} = \binom{n}{m} + \binom{n}{m-1}$$

Denote by $G_\epsilon$ a neighborhood of $e\in G$ such that the exponential map is diffeo from a neighborhood of zero in $\mathfrak{g}$ and $G$. The entries of $\mathcal{O}^{-1}$ are identified with certain functions on $G_\epsilon$. The above quadratic formula for $\mathcal{O}^{-1}$ is precisely the condition that the map $\mathfrak{g}\to\mathfrak{g}\otimes C^\infty(G_\epsilon)$ given on the basis

$$e_i\mapsto e_j\otimes (\mathcal{O}^{-1})^j_i$$

extends (uniquely) to a homomorphism of associative algebras $U(\mathfrak{g})\to U(\mathfrak{g})\otimes C^\infty(G_\epsilon)$. For this one first defines such a map on free generators $E_i$ of the tensor algebra $T(\mathfrak{g})$, instead of $e_i$, and then checks that the ideal generated by relations $E_i E_j - E_j E_i - [E_i, E_j]$ gives zero, what boils down to our quadratic relation.