# 6. Active transformations [Draft]

Some prerequisites: Differential geometry. E.g. a course “897602 Geometrical Methods of Physics” at IF.

Drawings as you read would help.

1. Introduction

[For later…]

2. Set up and coordinate maps

Let ${M}$ be an ${m-}$dimensional manifold, and let ${(U,x)}$ be a coordinate chart. This means that ${U}$ is an open region in ${M,}$ and ${x}$ is a bijection

$\displaystyle x:U\rightarrow x(U)\subseteq {\mathbb R}^m. \ \ \ \ \ (1)$

It is convenient to use the maps ${x^{\mu}}$ defined via

$\displaystyle x = (x^1, x^2, \cdots, x^m). \ \ \ \ \ (2)$

That is, for ${{\mu} = 1,2,\cdots, m,}$

$\displaystyle x^{\mu}: U\rightarrow x^{\mu}(U)\subseteq {\mathbb R}. \ \ \ \ \ (3)$

In most physics (and perhaps also mathematics) textbooks, ${x^{\mu}}$ is normally regarded as a place holder for a number. For example, we would encounter them in an equation

$\displaystyle \textrm{}x+3 = 6\textrm{''}. \ \ \ \ \ (4)$

The solution for this would be ${\textrm{}x = 3\textrm{''}.}$ For us however, writing ${x}$ or ${x^{\mu}}$ would refer to maps. So in this post, writing something like ${\textrm{}x = 3\textrm{''}}$ would be meaningless (unless agreed in advance).

We call ${x}$ as a coordinate map. For simplicity of the discussions, in this post we restrict ourselves to only one coordinate chart, and hence we use only one coordinate map. This means that although we are discussing transformations in this post, we do not make use of any coordinate transformations(!)

3. Diffeomorphisms

Let us consider a map

$\displaystyle g: M\rightarrow M \ \ \ \ \ (5)$

such that

• ${g}$ is bijective.
• ${g}$ and ${g^{-1}}$ can be differentiated infinitely many times.

In this case, the map ${g}$ is called a “diffeomorphism”.

In principle, we need to be more precise what it means by “differentiation” or “derivative” for the map ${g.}$ After all, we only know how to take a derivative on a function from ${{\mathbb R}^k}$ to ${{\mathbb R}^n.}$ But we do not know (suppose we have not yet studied differentiable manifold) how to take a derivative on a function from one manifold to another.

The idea is that, roughly speaking, we need to represent part of ${M}$ by coordinate charts, and represent ${g}$ by maps between two images of coordinate maps. If all the representatives of ${g}$ are differentiable, then ${g}$ is said to be differentiable. The same goes for differentiable infinitely many times. More details might appear in later versions……

For physicists, I would say that the concept of “differentiable infinitely many times” is related to the concept of “smoothness” (well, I think this is just the definition by mathematicians). Although I do not say much about what it means by “differentiable infinitely many times”, I think physicists have the ability to accept the concept of “smoothness” anyway even if we do not know the exact definitions. So let us simply move on.

If we collect all the diffeomorphisms on ${M,}$ and use the map composition operation ${\circ,}$ then we have a group called “diffeomorphism group” ${\textrm{Diff}(M).}$ The proof that it is a group is left as an exercise for keen readers. Let me give some hints that the identity element of ${\textrm{Diff}(M)}$ is simply the identity map, and that the inverse of ${g}$ as a group element is simply given by the inverse map ${g^{-1}.}$

So each element ${g}$ of the group ${\textrm{Diff}(M)}$ maps a point ${p\in M}$ to another point ${g(p)\in M.}$ To avoid later clutter of notation, when the context is clear, we will write ${gp}$ instead of ${g(p).}$

For a point ${p\in M,}$ we call

$\displaystyle x^{\mu}(p) \ \ \ \ \ (6)$

the ${{\mu}}$th-coordinate (in the chart ${(U,x)}$) of the point ${p.}$ Let us consider ${\xi p}$ for ${\xi\in\textrm{Diff}(M).}$ Its ${{\mu}}$th-coordinate is given by

$\displaystyle x^{\mu}(\xi p) =x^{\mu}\circ\xi\circ x^{-1}(x(p)). \ \ \ \ \ (7)$

Note that ${\tilde{\xi}^{\mu}\equiv x^{\mu}\circ\xi\circ x^{-1}}$ is the ${{\mu}}$th component of coordinate representation of ${\xi.}$ So we write

$\displaystyle x^{\mu}(\xi p)=\tilde{\xi}^{\mu}(x(p)). \ \ \ \ \ (8)$

Roughly speaking, ${x(p)\in{\mathbb R}^m}$ whereas ${\tilde{\xi}^{\mu}}$ is a map ${{\mathbb R}^m\rightarrow{\mathbb R}}$ (more precisely, it is a map ${{\mathbb R}^m\supseteq x(U)\rightarrow x^{\mu}(U)\subseteq{\mathbb R}}$).

Note that the expression ${\tilde{\xi}^{\mu}(x(p))}$ resemblance the one (${\textrm{}\xi^{\mu}(x)\textrm{''},}$ or more often ${\textrm{}\xi\textrm{''}}$ is written as ${\textrm{}x'\textrm{''}}$) we often seen in physics textbooks (especially general relativity). Such expression is almost always interpreted as coordinate transformation from ${\textrm{}x^{\mu}\textrm{''}}$ to ${\textrm{}x'^{\mu}\textrm{''}.}$ So it looks to me that this point of view is the passive transformation. However, in this entire post, I am trying to discuss active transformation. The end result should be no different from the discussion involving passive transformation.

4. Transformation on scalar field

Consider a scalar function

$\displaystyle f:M\rightarrow {\mathbb R}. \ \ \ \ \ (9)$

We have learned in the previous section that each element ${g}$ of the group ${\textrm{Diff}(M)}$ is a map between points in ${M.}$ By an appropriate definition, each element ${g}$ induce a transformation on ${f.}$ That is, for each element ${g,}$ define a map (note: ${F(M)}$ means a set of smooth functions on ${M}$)

$\displaystyle {\rho}_g: F(M)\rightarrow F(M) \ \ \ \ \ (10)$

such that

$\displaystyle ({\rho}_g f)(p) = f(g^{-1}p) \ \ \ \ \ (11)$

for all ${p\in M.}$

By the above definition, ${{\rho}_g}$ also forms a group. Consider

$\displaystyle \begin{array}{rl} ({\rho}_h{\rho}_g f)(p) &= ({\rho}_g f)(h^{-1}p)\\ &= f(g^{-1}h^{-1}p)\\ &= f((hg)^{-1}p)\\ &= ({\rho}_{hg}f)(p). \end{array} \ \ \ \ \ (12)$

So we have

$\displaystyle {\rho}_h{\rho}_g = {\rho}_{hg}. \ \ \ \ \ (13)$

That is we have the closure relation. Furthermore, the multiplication rule above suggests that ${{\rho}}$ preserves the group operation and gives a one-to-one correspondence between ${g}$ and ${{\rho}_g}$. The set of all such ${{\rho}_g}$ under the map composition ${\circ}$ forms a group. Technically, we say that this group is (group) isomorphic to ${\textrm{Diff}(M).}$

Let us now consider coordinate representation. Consider

$\displaystyle \begin{array}{rl} ({\rho}_\xi {\phi})(p) &={\phi}(\xi^{-1}p)\\ &={\phi}\circ x^{-1}(x(\xi^{-1}p))\\ &={\phi}\circ x^{-1}(x^{\mu}(\xi^{-1}p)u_{\mu})\\ &=\tilde{{\phi}}((\tilde{\xi}^{-1})^{\mu} (x(p))u_{\mu}), \end{array} \ \ \ \ \ (14)$

where ${\tilde{\phi} = {\phi}\circ x^{-1}}$ is coordinate representation of ${{\phi},}$ and ${u_{\mu}}$ are such that

$\displaystyle u_1 = (1,0,0,\cdots), u_2 = (0,1,0,\cdots), etc. \ \ \ \ \ (15)$

By writing ${\widetilde{{\rho}_\xi{\phi}} = {\rho}_\xi{\phi}\circ x^{-1},}$ which is the coordinate representation of ${{\rho}_\xi{\phi},}$ the rule for transformed scalar field (14) can be rewritten as

$\displaystyle \widetilde{{\rho}_\xi{\phi}}(x^{\mu}(p)u_{\mu})=\tilde{{\phi}}((\tilde{\xi}^{-1})^{\mu} (x(p))u_{\mu}). \ \ \ \ \ (16)$

This rule resemblance (by resemblance, I mean, we simply make some renamings of the notations and abuse some others) ${\textrm{}{\phi}'(x^{\mu}) = {\phi}((\xi^{-1})^{\mu}(x))\textrm{''},}$ which is often interpreted as the result of coordinate transformation ${\textrm{}x^{\mu}\mapsto x'^{\mu} = \xi^{\mu}(x^{\nu})\textrm{''}.}$

5. Transformation on vector field

[For later…]