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Chapter 19: Problem 1

A function \(f: G \rightarrow \mathbb{R}\) is said to be an aralytic function on \(G\) if it can be expanded as a Taylor serics at any point \(g \in G\). Show that if \(X\) is a left-invariant vector ficld and \(f\) is an analytic function on \(G\) then $$ f(g \exp t X)=\left(c^{t x} f\right)(g) $$ where, for any vector ficld \(Y\), we define $$ \mathrm{e}^{\gamma} f=\boldsymbol{f}+Y \boldsymbol{f}+\frac{1}{21} Y^{2} f+\frac{1}{3 !} Y^{3} f+\cdots=\sum_{i=0}^{\infty} \frac{Y^{n}}{n !} f $$ The operator \(Y^{\pi}\) is defined inductively by \(Y^{n} f=Y\left(Y^{n-1} f\right)\).

Short Answer

Expert verified
The formula \(f(g \exp t X)=\left(c^{t X} f\right)(g)\) holds for an analytic function \(f\) and a left-invariant vector field \(X\). The crucial concept applied here is the notion of a Taylor series expansion for the analytic function, left invariance for the vector field and the operator definition given for this case.

Step by step solution

01

- Recognize the Problem

We have to find out if the given formula \(f(g \exp t X)=\left(c^{t X} f\right)(g)\) holds for an analytic function \(f\) and left-invariant vector field \(X\). As the first step of a solution, we should understand the context of analytic functions and left-invariant vector fields.
02

- Apply the Definition of an Analytic Function

An analytic function on \(G\) can be expanded as a Taylor series at any point \(g \in G\). That is, \(f(g) = \sum_{n=0}^{\infty} a_n(g - g_0)^n\), where \(g_0\) is the point around which the Taylor series is expanded.
03

- Define the Left-Invariant Vector Field

A left-invariant vector field \(X\) is such that for any transform \(h: G \rightarrow G\), and for \(g\) in \(G\), we have \(X(h(g)) = h(X(g))\), that is, \(X\) is invariant under the application of the transformation \(h\).
04

- Calculate Using the Operator Definition

Now, we can rewrite \(f(g \exp t X)\) as \(f(g) \exp(t X)\), based on the given operator definition for exponential. So, \(f(g) \exp(t X) = \sum_{i=0}^{\infty} \frac{(t X)^{n}}{n!} f(g)\). Because \(X\) is left-invariant, it commutes with \(f(g)\), so the equation simplifies to \(f(g) \sum_{i=0}^{\infty} \frac{(t X)^{n}}{n!}\).
05

- Compare the Sides of the Equation

The right-hand side of the original equation is \((c^{t X} f)(g) = \sum_{i=0}^{\infty} \frac{(t X)^{n}}{n !} f(g)\). This is exactly what we've got on the left-hand side during previous steps. So, the given formula \(f(g \exp t X)=\left(c^{t X} f\right)(g)\) is proven.

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