§ 5. Formal calculations in Lie groups

Let $f, g$ be two formal power series with coefficients in $K$ in the same indeterminates, let $f_i$ (resp. $g_i$ ) be the homogeneous component of $f$ (resp. $g$ ) of degree $i$ . We shall write

f \equiv g \mod \deg p

f_i = g_i

for

i < p

In this §, G denotes a Lie group germ of finite dimension $n$ ; the base field $K$ is assumed to be of characteristic zero. We identify once and for all, by means of a chart, an open neighbourhood of $e$ in G with an open neighbourhood U of 0 in $K^n$ , so that $e$ is identified with 0. For $x, y$ in U and $n \in \mathbf{Z}$ , $x.y$ denotes the product of $x$ and $y$ and $x^{[m]}$ the m-th power of $x$ in G (when they are defined). The coordinates of $x \in U$ are denoted by $x_1, x_2, \ldots, x_n$ .

1. The coefficients $c_{\alpha \beta \gamma}$

Let $\Omega$ be the set of $(x, y) \in U \times U$ such that $x.y$ is defined and belongs to U. Then $\Omega$ is open in $U \times U$ and the mapping $(x, y) \mapsto x.y$ of $\Omega$ into U is analytic. The coordinates $z_1, \ldots, z_n$ of $z = x.y$ therefore admit expansions as integral series about the origin in the powers of $x_1, \ldots, x_n, y_1, \ldots, y_n$ . Therefore, there exist well defined constants $c_{\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_n, \gamma_1, \ldots, \gamma_n} \in K$ , such that

z_1^{\gamma_1} \cdots z_n^{\gamma_n} = \sum_{\alpha_1, \ldots, \beta_n \in \mathbf{N}} c_{\alpha_1, \ldots, \alpha_n, \beta_1, \ldots, \beta_n, \gamma_1, \ldots, \gamma_n} x_1^{\alpha_1} \cdots x_n^{\alpha_n} y_1^{\beta_1} \cdots y_n^{\beta_n}

for

\gamma_1, \ldots, \gamma_n

\mathbf{N}

. Adopting the conventions of Differentiable and Analytic Manifolds, R, we shall write these formulae more briefly:

(x.y)^{\gamma} = \sum_{\alpha, \beta \in \mathbf{N}^n} c_{\alpha, \beta, \gamma} x^{\alpha} y^{\beta} \quad (\gamma \in \mathbf{N}^n).

Since

x.0 = 0.x = x

for

x \in U

c_{\alpha, 0, \gamma} = c_{0, \alpha, \gamma} = \delta_{\alpha \gamma}

where

\delta_{\alpha \gamma}

is the Kronecker index. In particular, writing henceforth

k

instead of

\varepsilon_k

for

k = 1, \ldots, n

(x.y)_k = x_k + y_k + \sum_{|\alpha| \geq 1, |\beta| \geq 1} c_{\alpha, \beta, k} x^{\alpha} y^{\beta}.

Writing

c_{\alpha \beta} = (c_{\alpha \beta 1}, c_{\alpha \beta 2}, \ldots, c_{\alpha \beta n}) \in K^n

, it then follows that

x.y = x + y + \sum_{|\alpha| \geq 1, |\beta| \geq 1} c_{\alpha \beta} x^{\alpha} y^{\beta}.

On the right hand side of (5), we consider the homogeneous component of degree 2:

B(x, y) = \sum_{i,j=1}^n c_{ij} x_i y_j

so that $(x, y) \mapsto B(x, y)$ is a bilinear mapping of $K^n \times K^n$ into $K^n$ . Then

(6)

x.y \equiv x + y + B(x, y) \mod \deg 3.

Formula (4) implies on the other hand that

(7)

c_{\alpha, \beta, \gamma} = 0 \quad \text{if } |\alpha| + |\beta| < |\gamma|

and that the terms of total degree $|\gamma|$ in the expansion of $z^\gamma$ are also those of $(x_1 + y_1)^{\gamma_1}(x_2 + y_2)^{\gamma_2} \ldots (x_n + y_n)^{\gamma_n} = \sum_{\alpha + \beta = \gamma} ((\alpha, \beta)) x^\alpha y^\beta$ (cf. Differentiable and Analytic Manifolds, R, Notation and conventions). Hence:

(8)

c_{\alpha, \beta, \gamma} = 0 \quad \text{if } |\alpha| + |\beta| = |\gamma| \quad \text{but } \alpha + \beta \neq \gamma

(9)

c_{\alpha, \beta, \alpha + \beta} = ((\alpha, \beta)).

The associativity of the product implies the relation

\sum_{\xi} c_{\alpha \xi n} x^\alpha \left( \sum_{\beta, \gamma} c_{\beta \gamma \xi} y^\beta z^\gamma \right) = \sum_{\xi, \gamma} c_{\xi \gamma n} \left( \sum_{\alpha, \beta} c_{\alpha \beta \xi} x^\alpha y^\beta \right) z^\gamma

for all

x, y, z

sufficiently close to 0, whence

(10)

\sum_{\xi} c_{\alpha \xi n} c_{\beta \gamma \xi} = \sum_{\xi} c_{\xi \gamma n} c_{\alpha \beta \xi} \quad (\alpha, \beta, \gamma, \eta \text{ in } \mathbf{N}^n).

The group germ G admits a commutative open subgroup germ if and only if $c_{\alpha \beta \gamma} = c_{\beta \alpha \gamma}$ for all $\alpha, \beta, \gamma$ in $\mathbf{N}^n$ .

2. Bracket in the Lie algebra

For $\alpha \in \mathbf{N}^n$ , let $e_\alpha$ be the point distribution $\frac{1}{\alpha!} \frac{\partial^\alpha}{\partial x^\alpha}$ at the origin. In particular,

e_k = e_{e_k} = \frac{\partial}{\partial x_k}.

The

e_\alpha

form a basis of the vector space

U(G)

. If

f

is an analytic function on an open neighbourhood of 0 in G and

f(x) = \sum \lambda_\alpha x^\alpha

is its expansion as an integral series about the origin, then

\langle e_\alpha, f \rangle = \lambda_\alpha.

In particular,

\langle e_\alpha, x^\gamma \rangle = \delta_{\alpha \gamma}.

Hence

\langle e_\alpha * e_\beta, x^\gamma \rangle = \langle e_\alpha \otimes e_\beta, (x.y)^\gamma \rangle \\ = \langle e_\alpha \otimes e_\beta, \sum_{\alpha', \beta'} c_{\alpha'\beta'\gamma} x^{\alpha'} y^{\beta'} \rangle \\ = \sum_{\alpha', \beta'} c_{\alpha'\beta'\gamma} \langle e_\alpha, x^{\alpha'} \rangle \langle e_\beta, y^{\beta'} \rangle = c_{\alpha\beta\gamma}

and hence

e_\alpha * e_\beta = \sum_\gamma c_{\alpha\beta\gamma} e_\gamma.

(Formula (10) then expresses associativity on

U(G)

.) In particular, since

L(G)

is stable under the bracket.

[e_i, e_j] = \sum_k (c_{ijk} - c_{jik}) e_k.

The constants of structure of

L(G)

relative to the basis

(e_1, \ldots, e_n)

are therefore the

c_{ijk} - c_{jik}

. In other words, canonically identifying

L(G)

with

K^n

, we obtain:

[x, y] = B(x, y) - B(y, x).

Proposition 1

(i)

x^{[-1]} \equiv -x + B(x, x) \mod \deg 3

(ii)

x.y.x^{[-1]} \equiv y + [x, y] \mod \deg 3

(iii)

y^{[-1]}.x.y \equiv x + [x, y] \mod \deg 3

(iv)

x^{[-1]}.y^{[-1]}.x.y \equiv [x, y] \mod \deg 3

(v)

x.y.x^{[-1]}.y^{[-1]} \equiv [x, y] \mod \deg 3.

(In (i),

x^{[-1]}

of course represents the expansion of the function

x \mapsto x^{[-1]}

as an integral series about the origin; the other formulae can be interpreted analogously.) Let

g_1

and

g_2

be the homogeneous components of

x^{[-1]}

of degrees 1 and 2. Then

0 = x.x^{[-1]} \\ \equiv x + g_1(x) + B(x, g_1(x)) \mod \deg 2 \quad \text{(by (6))} \\ \equiv x + g_1(x) \mod \deg 2

and hence

g_1(x) = -x

. Then

0 = x.x^{[-1]} \\ \equiv x + (-x + g_2(x)) + B(x, -x + g_2(x)) \mod \deg 3 \\ \equiv g_2(x) - B(x, x) \mod \deg 3

and hence $g_2(x) = B(x, x)$ . This proves (i). Then, using (i),

x.y.x^{[-1]} \equiv (x + y + B(x, y)).(-x + B(x, x)) \quad \text{mod deg 3}

\equiv x + y + B(x, y) + (-x + B(x, x)) + B(x + y, -x) \quad \text{mod deg 3}

\equiv y + B(x, y) - B(y, x) \quad \text{mod deg 3}

\equiv y + [x, y] \quad \text{mod deg 3 (by (13))}

whence (ii). The proof of (iii) is analogous. Combining (i) and (iii), we obtain

x^{[-1]},y^{[-1]}.x.y \equiv (-x + B(x, x)).(x + [x, y]) \quad \text{mod deg 3}

\equiv -x + B(x, x) + x + [x, y] + B(-x, x) \quad \text{mod deg 3}

\equiv [x, y] \quad \text{mod deg 3}

whence (iv). The proof of (v) is analogous.

3. Powers

Consider j points of G:

x(1) = (x(1)_1, x(1)_2, \ldots, x(1)_n)

x(2) = (x(2)_1, x(2)_2, \ldots, x(2)_n)

\ldots \ldots \ldots \ldots \ldots \ldots \ldots

x(j) = (x(j)_1, x(j)_2, \ldots, x(j)_n).

The mapping

(x(1), x(2), \ldots, x(j)) \mapsto x(1).x(2)\ldots x(j)

admits an expansion as an integral series about the origin:

x(1).x(2)\ldots x(j) = \sum_{\alpha(1), \alpha(2), \ldots, \alpha(j) \in \mathbf{N}^n} a_{\alpha(1), \ldots, \alpha(j)} x(1)^{\alpha(1)} \ldots x(j)^{\alpha(j)}

where the $a_{\alpha(1), \ldots, \alpha(j)}$ are elements of $\mathbf{K}^n$ . We write, for $j = 0, 1, 2, \ldots$ ,

\psi_j(x) = \sum_{\alpha(1) \neq 0, \ldots, \alpha(j) \neq 0} a_{\alpha(1), \ldots, \alpha(j)} x^{\alpha(1) + \ldots + \alpha(j)}

where the right hand side is a convergent integral series in the variable $x \in \mathbf{K}^n$ . This series is obtained by suppressing in (14) the terms in which one of the variables $x(1), \ldots, x(j)$ does not occur explicitly and then writing $x(1) = x(2) = \cdots = x(j) = x$ .

If $t \in \mathbf{K}$ , all the t-th power mappings of G have the same expansion as an integral series about the origin, since any two of them coincide on a neighbourhood of 0. We denote this expansion as an integral series by $x^{[t]}$ .

Proposition 2

(i) $\psi_j \equiv 0 \mod \deg j$ . (ii) If $t \in \mathbf{K}$ , then

x^{[t]} = \sum_{i=1}^8 \binom{t}{i} \psi_i(x)

*where the formal power series on the right is meaningful because of (i).

(We write

\binom{t}{i} = \frac{t(t-1)\ldots(t-i+1)}{i!}

for all $t \in \mathbf{K}$ .)

Assertion (i) is obvious from the definition of the $\psi_j$ .

We prove (ii) for $t$ an integer $\geq 0$ . By (14),

x^{[t]} = \sum_{\alpha(1), \ldots, \alpha(t) \in \mathbf{N}^n} a_{\alpha(1), \ldots, \alpha(t)} x^{\alpha(1) + \ldots + \alpha(t)}.

For $\alpha = (\alpha(1), \ldots, \alpha(t)) \in (\mathbf{N}^n)^t$ , let $\sigma(\alpha)$ denote the set of $j \in \{1, 2, \ldots, t\}$ such that $\alpha(j) \neq 0$ . If, in the sum (17), we group together the terms for which $\sigma(\alpha)$ is the same, then

x^{[t]} = \sum_{\sigma \subset \{1, t\}} h_{t, \sigma}(x)

where

h_{t, \sigma}(x) = \sum_{\sigma(\alpha) = \sigma} a_{\alpha(1), \ldots, \alpha(t)} x^{\alpha(1) + \ldots + \alpha(t)}.

Let $\sigma = \{j_1, j_2, \ldots, j_q\}$ with $j_1 < j_2 < \ldots < j_q$ . In (14) (where $j$ is replaced by $t$ ), we substitute 0 for $x(k)$ for $k \notin \sigma$ ; as 0 is the identity element of $G$ , we obtain the expansion of $x(j_1) \cdot x(j_2) \ldots x(j_q)$ as an integral series about the origin:

x(j_1) \cdot x(j_2) \ldots x(j_q) = \sum_{\sigma(\alpha) \subset \sigma} a_{\alpha(1), \ldots, \alpha(t)} x(j_1)^{\alpha(j_1)} x(j_2)^{\alpha(j_2)} \ldots x(j_q)^{\alpha(j_q)}

and hence, by the definition of $\psi_q$ :

\psi_q(x) = \sum_{\sigma(\alpha) = \sigma} a_{\alpha(1), \ldots, \alpha(t)} x^{\alpha(j_1) + \ldots + \alpha(j_q)}.

By (19) and (20), we see that $h_{t, \sigma}(x) = \psi_{\mathrm{card}\,\sigma}(x)$ . Then (18) implies

x^{[t]} = \sum_{i=0}^t \binom{t}{i} \psi_i(x) = \sum_{i=0}^\infty \binom{t}{i} \psi_i(x).

Then we write $x^{[t]'} = \sum_{i=0}^\infty \binom{t}{i} \psi_i(x)$ for all $t \in \mathbf{K}$ . In the integral series $x^{[t]}$ and $x^{[t]'}$ , each coefficient is a polynomial function of $t$ . For this is obvious for $x^{[t]'}$ . As far as $x^{[t]}$ is concerned, it suffices to prove that, for all $u \in \mathrm{U}(G)$ , the image of $u$ under $x \mapsto x^{[t]}$ is a polynomial function of $t$ . Now, for $u \in \mathrm{U}^m(G)$ , this image is $t^m u$ (§ 4, no. 3, Proposition 7 (iv)).

As $x^{[t]} = x^{[t]'}$ for $t$ an integer $\geq 0$ , it then follows that $x^{[t]} = x^{[t]'}$ for all $t \in \mathbf{K}$ .

Remark

(1) We write condition (ii) of Proposition 2 for t an integer $\geqslant 0$ :

\begin{align*} 0 &= \psi_0(x) \\ x &= \psi_0(x) + \psi_1(x) \\ x^{[2]} &= \psi_0(x) + 2\psi_1(x) + \psi_2(x) \\ &\ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{align*}

These formulae suffice to determine the $\psi_i$ .

(2) We see that $\psi_0(x) = 0, \psi_1(x) = x, \psi_2(x) = x^{[2]} - 2x$ ,

x^{[-1]} = \sum_{i=1}^\infty (-1)^i \psi_i(x).

(3) The above expression for $\psi_2$ and formula (6) prove that

\psi_2(x) \equiv \mathrm{B}(x, x) \mod \deg 3.

Using Proposition 2, (i) and (ii), we see that

x^{[t]} \equiv tx + \binom{t}{2} \mathrm{B}(x, x) \mod \deg 3.

(4) Let $\psi_{p,m}(x)$ and $h_{t,m}(x)$ denote the homogeneous components of $\psi_p(x)$ and $x^{[t]}$ of degree m. Then $\psi_{p,m} = 0$ for $m < p$ . On the other hand, Proposition 2 (ii) gives

h_{t,m}(x) = \sum_{p \leq m} \frac{t(t-1)\ldots(t-p+1)}{p!} \psi_{p,m}(x)

that is

h_{t,m}(x) = \sum_{1 \leq r \leq m} t^r \phi_{r,m}(x)

where the

\phi_{r,m}

are homogeneous polynomial mappings of degree m of

\mathbf{K}^n

into

\mathbf{K}^n

. In particular, by (23),

\phi_{1,m}(x) = \sum_{p \leq m} \frac{(-1)^{p-1}}{p} \psi_{p,m}(x)

\phi_{m,m}(x) = \frac{1}{m!} \psi_{m,m}(x).

(5) If $\mathbf{K}$ is of characteristic $> 0$ , the results of nos. 1 and 2 remain true, provided $e_\alpha$ , in no. 2, is defined by $\langle e_\alpha, \sum_\beta \lambda_\beta x^\beta \rangle = \lambda_\alpha$ . In no. 3, if the $\psi_j$ are defined as above, the argument again proves that $x^{[t]} = \sum_{i=1}^\infty \binom{t}{i} \psi_i(x)$ if $t \in \mathbf{N}$ .

4. Exponential

Let $E(x)$ be the expansion of an exponential mapping of $G$ as an integral series about 0. Let $L(x)$ be the expansion of the inverse mapping of an injective exponential mapping of $G$ as an integral series about 0. Since the tangent mapping at 0 to any exponential mapping is the identity of $L(G)$ , $E(x) \equiv x \mod \deg 2$ and $L(x) \equiv x \mod \deg 2$ . Since $E(L(x)) = L(E(x))$ for $x$ sufficiently close to 0, the formal power series $E$ and $L$ are such that $E(L(X)) = L(E(X)) = X$ . An analogous argument shows that

E(tX) = (E(X))^{[t]}, \quad L(X^{[t]}) = tL(X)

for

t \in K

Proposition 3

(27)

L = \sum_{p=1}^{\infty} \frac{(-1)^{p-1}}{p} \psi_p

(28)

E = \sum_{p=1}^{\infty} \frac{1}{p!} \psi_{p,p}

(Recall that

\psi_{p,p}

is the homogeneous component of

\psi_p

of degree

p

or, by (24),

E(tx) = (E(x))^{[t]}

The two sides are formal power series in $t$ and $x$ . Equating the terms of first degree in $t$ , we obtain

x = \sum_{m \geq 0} \phi_{1,m}(E(x)).

Now the inversion of a system of formal power series, when it is possible, is possible in only one way (Algebra, Chapter IV, § 6, Corollary to Proposition 8). Then

L(x) = \sum_{m \geq 0} \phi_{1,m}(x) \quad \text{by (29)}

= \sum_{p,m} \frac{(-1)^p}{p} \psi_{p,m}(x) \quad \text{by (25)}

= \sum_p \frac{(-1)^p}{p} \psi_p(x),

whence (i). Similarly, for

t \neq 0

L(tx) = tL((tx)^{[t^{-1}]})

= tL\left( \sum_{m \geq 0} \sum_{1 \leq r \leq m} t^{m-r} \phi_{r,m}(x) \right) \quad \text{by (24)}.

Equating the terms of first degree in t, we obtain

x = L \left( \sum_{m \geq 0} \phi_{m,m}(x) \right)

whence

E(x) = \sum_{m \geq 0} \phi_{m,m}(x)

= \sum_{m \geq 0} \frac{1}{m!} \psi_{m,m}(x) \quad \text{by (26).}

Proposition 4

For the chart of G used to be canonical, it is necessary and sufficient that $\psi_j = 0$ for $j \geq 2$ .

This is sufficient by Proposition 3. Suppose that the chart is canonical and that $\psi_i = 0$ for $2 \leq i < n$ . Then $nx = x^{[n]} = \sum_{i=0}^n \binom{n}{i} \psi_i(x) = nx + \psi_n(x)$ , whence $\psi_n = 0$ . Hence $\psi_j = 0$ for $j \geq 2$ by induction on $j$ .

Exercises

See the exercises for § 5.