What are the irreducible representations of point groups and how do I find them?

[Asteropaeus]

As part of physical chemistry I am reading up group theory for molecular symmetries.

I realize the way chemistry textbooks treat this must be very different from what mathematicians do.

So I want to know how I take a point group, find the matrix operations and get the character table.For an C2 rotation, in terms of coordniates, x changes into -x, y into -y and z stays the same.
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1 \end{array}

But my textbook has an example where they have something similar to d-orbitals, with lobes of different electron densities, on a molecule. It expresses everything not in terms of coordinates, but objects. So d orbitals change sign when the directions the lobes point into change. And the two d orbitals not on the central atom switch.
So it gives the matrix like this
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 0 & -1 \\
0 & -1 & 0 \end{array}

And that operating on (Ps, Pa, Pb) gives (-Ps, -Pa, -Pb) where Ps is central, Pa and Pb are terminal. It seems it only talks about the positive lobe of that p orbital.

And then later introduce the same character table for C2v.

It is so confusing what the basis is, and how to find an irreducible representation.
Especially so when the character table doesn't talk about x, y, z but about abstract rotations and d orbitals, apparently a certain way in which two axes are mixed.

All youtube videos I checked, the lecturer explains this stuff and then says, "You don't need to think about what, A1, A2, B1, E and the basis are or what is really happening. You just learn to do this operation, then later on you will find it very useful."

Something is wrong in my understanding, and I can't figure out where it is. I keep going in loops. I am stumped. I don't even know if I asked the correct question.

 

[DrDu]

You are discussing how functions transform under symmetry operations. In general $T f(r)=f(T^{-1}r)$. Now in case of the atomic orbitals, you can expand the rotated functions as linear combinations of the original functions, i.e. if $T f_j(r)=\sum_i a_i f_i(r)$, that is, within this set of functions $f_i$, the operation T can be represented as a matrix.

 

[DrDu]

You are discussing how functions transform under symmetry operations. In general $T f(r)=f(T^{-1}r)$. Now in case of the atomic orbitals, you can expand the rotated functions as linear combinations of the original functions, i.e. if $T f_i(r)=\sum_i a_{ij} f_j(r)$, that is, within this set of functions $f_i$, the operation T can be represented as a matrix with elements $a_{ij}$

 

[Stephen Tashi]

I realize the way chemistry textbooks treat this must be very different from what mathematicians do.

I don't know the chemistry, but I think you're correct !

The approach of considering "objects" is what a mathematician would call a "group acting on a set" or a "group action". For example if you consider a molecule with a C2 symmetry, you can focus on the idea that a 180 degree rotation about a certain axis takes certain atoms to the location of other atoms of the same element. But a C2 rotation can also be viewed as an operation that takes an arbitrary point (x,y,z) in space to another point in space, and the point (x,y,z) may or may not be at the exact location of an atom. So we have the same "group", but two different "group actions".

If you are considering the action of a group on a finite set, you think about operations that permute the elements of the set. If we picking a definite order to write the elements of the set, then all ways of rearranging the set can be represented by matrices whose elements are 0's or 1's. For example, if the set is $S = \{s_1,s_2\}$ then we think of the elements as having the (trivial) coordinate system $s_1 = (1,0), \ s_2 = (0,1)$ and the matrix that interchanges the elements (represented as column vectors) is $\begin{pmatrix} 0 \ 1 \\ 1, 0 \end{pmatrix}$.
For example $\begin{pmatrix} 0 \ 1 \\ 1, 0 \end{pmatrix} \begin{pmatrix} 1\\0 \end{pmatrix} = \begin{pmatrix}0\\1\end{pmatrix}$ represents mapping $s_1$ to $s_2$.

If you consider C2 acting on all points in 3 dimensional space then some points with a positive coordinate are going to be mapped to a point with a negative coordinate, so the matrices representing that group action will need to have some negative entries.

From a mathematical point of view, a set of matrices "represents" a group and the same group can have many different representations. For example, if you think of C2 as a 180 degree rotation about a given axis and you pick a coordinate system where the z-axis is that given axis then the matrices representing it may look nice (e.g. contain only 1's, 0's and -1's). But if you use a different coordinate system where the axis of rotation doesn't coincide with your choice of x,y or z-axes then the matrices representing that same group can get more complicated.

 

[Asteropaeus]

I looked back at my textbook and it indeed says that the matrix that is a representative of a certain symmetry operation can have different form according to the basis.

I guess I get that now. Different sources are using different ways to define the basis.

 

[DrDu]

One of the books I liked best on the subject is: Sternberg, Group theory and physics
Sternberg is a mathematician, but the book deals with both point symmetry group of molecules and the representations of the symmetric group, both tremendously important in physical chemistry.

 

[my2cts]

A
So I want to know how I take a point group, find the matrix operations and get the character table.

The irreps of point groups should be easy to find as well as the characters.
The matrix representations depend on the basis.
In the case of C2 and d orbitals obviously there are only 2 irreps, even and odd.
Orbitals zx, zy are odd and the others are even.
Search for "irreducible representations of c2" or any other point group and you first result is
http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=202&option=4