# Thread: Character Tables and Basis Functions: An Opinion Question

1. ## Character Tables and Basis Functions: An Opinion Question

This is not a question about how to do a problem. I know how to build a character table and how to find basis functions for a representation.

I am working out of a text written by one of my former Purdue professors. In it he gives an example of the construction of the character table for D3. His approach is to pick basis functions for a representation of the group and calculate the characters. My approach is to calculate the characters using the various theorems, which is simply a Math problem. Then I can decide which basis functions I want to use and can construct the representation from that.

I figure both approaches come out to be about the same amount of work. I was wondering if anyone had an opinion on which approach they liked better (or perhaps one I haven't thought of.) Thanks!

-Dan

2. Originally Posted by topsquark
This is not a question about how to do a problem. I know how to build a character table and how to find basis functions for a representation.

I am working out of a text written by one of my former Purdue professors. In it he gives an example of the construction of the character table for D3. His approach is to pick basis functions for a representation of the group and calculate the characters. My approach is to calculate the characters using the various theorems, which is simply a Math problem. Then I can decide which basis functions I want to use and can construct the representation from that.

I figure both approaches come out to be about the same amount of work. I was wondering if anyone had an opinion on which approach they liked better (or perhaps one I haven't thought of.) Thanks!

-Dan
I mean, this is a very vague question. For most groups of small order getting the character table can usually be done without finding any of the actual irreps. For example here I found the character table for $Q$ without finding any of the irreps (of course the fact that I only used that it was a non-abelian group of order eight shows that every non-abelian group of order eight has the same character table--a case of a more general theorem that says there are always two non-isomorphic groups of order $p^3$ where $p$ is prime which have the same character table). If you are forced to actually find the irreps/irreducible characters this usually becomes a game of putting them together from smaller groups. For example, if you are given an abelian group $G$ you can find a decomposition of $G$ as the product of cyclic groups $\mathbb{Z}_{a_1}\times\cdots\times\mathbb{Z}_{a_n}$ (which you know is possible by the structure theorem) then use the fact that every irrep of $G$ is just the tensor product of irreps of $\mathbb{Z}_{a_1},\cdots\mathbb{Z}_{a_n}$ which are easy to find. Or, maybe it happens that your group $G$ is the semidirect product of two nice groups (usually $G=A\rtimes_\varphi H$ where $A$ is abelian) and you can use the irreps of $A,H$ to get the irreps of $G$. Also, it is a common technique to induce the characters from smaller subgroups so that we can get a lot of irreps (if we do it right) for $S_n$ by inducing them from $S_{n-1}$. But, for some groups like $S_n$ there is an almost systematic way of finding the irreps (viz. books on algebraic combinatorics)