# Math Help - Symmetry "decomposition" of functions

1. ## Symmetry "decomposition" of functions

First, sorry for the incorrect terminology, I'm not sure what else to call this.

I know it is possible to use the group $C_2$ to show that a function f(x) breaks down into the sum of an even and an odd function. Since I can't find my notes or reference on the subject my questions are
1) How do you do this using $C_2$
and
2) How do you generalize this process? As an example, how can you do this with, say, $D_3$?

Thanks!
-Dan

2. Thank you to any who have looked at this problem.

I finally found the text reference. I'll show how to do the $C_2$ problem. The $D_3$ problem can be done similarly, but I have yet to manage to construct any useful basis functions for it.

The relevant equation to find functions with symmetries under a particular irreducible group representation is
$\sum_R \chi ^{(\mu)*}(R)O_R \psi^{( \mu )} = \frac{g}{n_{\mu}} \psi ^{(\mu)}$
where the sum is over the group elements, $\chi(R)$ is the character of the group element R, $O_R$ is the operator (corresponding to the group element R) that acts in function space, $\psi$ is a function, g is the order of the group, and n is the dimension of the representation under study. The superscript $\mu$ indicates that we are working in the $\mu$th representation.

So applying this to the A representation of $C_2$ we get
$1 \cdot O_E f(x) + 1 \cdot O_{C_2} f(x) = \frac{2}{1}f(x)$

$f(x) + f(-x) = 2f(x)$

$f(x) = f(-x)$

So the A representation leads to basis functions that are even.

Applying this to the B representation of $C_2$ we get
$1 \cdot O_E f(x) + (-1) \cdot O_{C_2} f(x) = \frac{2}{1}f(x)$

$f(x) - f(-x) = 2f(x)$

$f(x) = -f(-x)$

So the B representation leads to basis functions that are odd.

Thus any function can be represented by the sum of the basis functions over all representations. In this case that means
$f(x) = f_A(x) + f_B(x)$
ie. any function may be written in terms of the sum of an even function and an odd function.

This process can be applied to any group, though the interpretation of the basis functions can be a bit hairy for certain group symmetries.

-Dan