# How to select bases for Matrix representation of a point group?

1. Aug 6, 2014

### hokhani

To represent operations of a point group by matrix we need to choose basis for this representation. What is the criteria for doing that? How to realize that how many bases are necessary for a matrix representation and how to select them? Or could you please give me an elementary reference to find my answer simply?

2. Aug 6, 2014

### hokhani

About matrix representation of a point group

In a point group representation of a molecule, is it necessary to write the bases as row vectors? If yes, does the matrix operate to the bases from the left or in other words do we have to consider two sequent operations,say AB, from the left (in which B acts after A)?

This question arises to me from the statement below (taken from http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Symmetry/Matrix_representations_of_groups):

"Note: We have written the vectors representing our basis as row vectors. This is important. If we had written them as column vectors, the corresponding transformation matrices would be the transposes of the matrices above, and would not reproduce the group multiplication table (try it as an exercise if you need to convince yourself)."

Another question:
Why don't we make representation matrices by simply sandwiching an operator between the bases (what that we do in quantum mechanics for matrix representation of an operator)?

Last edited by a moderator: Aug 7, 2014
3. Aug 6, 2014

### Fredrik

Staff Emeritus
Let V be a vector space. The set of invertible linear operators on V is a group. It's denoted by GL(V).

A representation of a group G is a group homomorphism from G into GL(V), for some vector space V. In other words, it's a function $\Gamma:G\to\operatorname{GL}(V)$ such that $\Gamma(xy)=\Gamma(x)\Gamma(y)$ for all $x,y\in G$.

To define a representation $\Gamma:G\to\operatorname{GL}(V)$, you must specify what operator $\Gamma(x)$ is for all $x\in G$. So let $x\in G$ be arbitrary. How do you specify what operator $\Gamma(x)$ is? The same way we do it for any other function. To define a function f, we must specify a domain X, and then say what f(x) is for all x in X. The domain of $\Gamma(x)$ is V (by definition of GL(V)). So to define $\Gamma(x)$, you only need to specify what $\Gamma(x)v$ is for all $v\in V$.

However, since $\Gamma(x)$ is a linear operator, it's sufficient to specify $\Gamma(x)v$ for all v in a basis for V. It doesn't matter which basis. The reason is this: If $\{e_i\}$ is a basis for V, then for all $v\in V$, we have $\Gamma(x)v=\Gamma(x)\big(\sum_i v_i e_i\big) =\sum_i v_i \Gamma(x)e_i$. So the value of $\Gamma(x)v$ is determined by the values of the $\Gamma(x)e_i$.

Edit: I realized that I didn't explain a very important detail: If V is an n-dimensional vector space, then every basis defines a one-to-one correspondence between the set of linear operators on V and the set of n×n matrices, and also between GL(V) and the set of invertible n×n matrices. So a map into GL(V) can also (by a mild abuse of notation) be viewed as a map into the set of invertible n×n matrices. If you want to understand this sort of thing, I suggest that you take a look at the FAQ post about the relationship between linear operators and matrices. https://www.physicsforums.com/showthread.php?t=694922 [Broken]

Last edited by a moderator: May 6, 2017
4. Aug 7, 2014

### Staff: Mentor

The OP had two threads on this. Since they were both different I have merged them into one.

5. Aug 7, 2014

### Fredrik

Staff Emeritus
The questions are closely related, so it's probably better this way, with both in the same thread. You can also consider moving them to the linear & abstract algebra forum and fixing the broken link (move the closing url tag to the left of the right parenthesis).

I find it difficult to answer this because I don't understand what's going on on that page. It's horribly formatted, and this makes it hard to read. The symbol that's displayed as a white question mark on a black background is probably supposed to be something else. Also, the product of a 1×4 matrix and a 3×4 matrix is undefined.

This is a good way to find the matrix corresponding to an operator, i.e. the numbers $\Gamma(x)_{ij}$ associated with the operator $\Gamma(x)$ (and some specific ordered basis). I still don't understand what's going on on that page, but if you're trying to define a representation of a group, what you need to do is to specify the operator that corresponds to the group element, i.e. specify $\Gamma(x)$ for each $x$ in the group. Because of the 1-to-1 correspondence between GL(V) and invertible matrices, you can do this by specifying the matrix components $\Gamma(x)_{ij}$ instead of the operator $\Gamma(x)$.

Last edited: Aug 7, 2014