Question

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

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

- 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.

- 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.

- 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 \(h\).

- 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 \(f(g) \sum_{i=0}^{\infty} \frac{(t X)^{n}}{n!}\).

- 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.

Key concepts

Left-Invariant Vector Field

A left-invariant vector field is a concept often utilized in the study of Lie groups and differential geometry. It is critical to understand how these fields behave under transformations. When we say a vector field $X$ is left-invariant, it means that for any element $h$ performing a transformation in the group, $X$ remains unchanged by this operation. For instance, given $g$ in the group $G$, applying $h$ gives $X(h(g)) = h(X(g))$. This invariance under group actions is what allows mathematicians to simplify and analyze the behavior of complex group dynamics.
Understanding this can help when expanding functions or dealing with differential equations where symmetries are present. In physics, left-invariant fields often relate to conserved quantities, because the symmetries they imply can mean certain physical traits do not change over time.

Taylor Series

Taylor series is a very powerful mathematical tool used to approximate functions in a way that relates directly to calculus. An analytic function is one which can be expanded into a Taylor series at any point. For a function \(f\) defined in a domain \(G\), the Taylor series expression becomes the sum \(\sum_{n=0}^{\infty} a_n(g - g_0)^n\), centered around a point \(g_0\) in \(G\).
Using a Taylor series transforms the function into an infinite polynomial, making analysis, especially near \(g_0\), much more straightforward. This tool is vital for approximations because it provides an excellent way to understand how a function behaves locally (i.e., in a small region around \(g_0\)) by considering its derivatives at that point.
This exact expansion is also what connects our problem to finding the behavior of the function under operations such as those introduced by vector fields.

Operator Definition

Operators in mathematics are tools that transform one function into another. In this context, an operator is symbolically represented by \(\mathrm{e}^{Y}f\). Here, this operator applied to an analytic function \(f\) follows the rule \(\sum_{i=0}^{\infty} \frac{Y^n}{n!} f\), representing the action of an infinitely repeated application of \(Y\).
This definition parallels the format of a Taylor series, reflecting a similar idea where a function is subjected to repeated transformations. Operators help in generalizations, like combining multiple dimensions or forces (in physics), streamlining complex interactions into a more approachable form.
It's indispensable to understanding how the repeated application of a vector field affects an analytic function, as it forms the basis of conclusions, like the conforming nature of \(f(g \exp tX)\) with \(\left(c^{tX} f\right)(g)\).

Vector Fields

Vector fields provide an elegant way to represent forces, directions, or even velocity fields within a plane or a space. A vector field on a set \(G\) assigns a vector to every point in \(G\), allowing for the visualization of flow or movement, akin to wind velocity charts.
This is particularly useful in describing dynamic systems such as fluid motion, electromagnetic fields, and even economic models. In mathematics and related fields, vector fields are crucial in studying differential equations, as they help visualize solutions as curves or surfaces in a domain.
In the exercise, vector fields like \(X\) act as operators acting on functions within \(G\). When analyzing transformations like \(f(g \exp t X)\), understanding the underlying vector fields aids in visualizing how the operations affect each point in the set, facilitating interpretations and proving stability or repetitive behaviors in complex systems.