My character tables for point group $T$(Schönflies-notation but easily convertible into other point group notations) tell me that the rotation around the $z$-axis, $R_z$ (the $z$-direction corresponding to the highest order ($n=3$) axis of rotation (see comments section $\alpha$ in the tables)), transforms like the irrep named $\mathrm{T}$ (Mulliken notation simply refering to a 3-dim irrep). How do I see this? I am running in trouble, since there are multiple "versions" of each of the non-trivial group elements (e.g. 8 different $C_3$ type rotations and 3 $C_2$ rotations and $R_z$ is apparently not an "eigenoperation" of all of them. I understand $$ R_z = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ if one $C_3$ is aligned along $z$.


For clarity the question is: Why does $R_z$ transform like the irrep $T$ under point group T (unlucky coincidence in names)?

For example the irrep $T$ has the character entry $0$ for all eight threefold rotation operations ($C_3$). But when one uses the one $C_3$ which is collinear with $z$ one yields $C_3 R_z = 1 R_z$, which in my eyes seems to be in disagreement with any $3\times 3$ matrix operation with trace $0$.

  • $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$ – Qmechanic Aug 29 '18 at 17:26
  • $\begingroup$ ^ or maybe Chemistry? I don't think I've ever seen this notation in the context of physics. $\endgroup$ – Mike Aug 29 '18 at 17:26
  • $\begingroup$ @Qmechanic I fear not, I have tried to discuss a similar problem various times there and for me its virtually impossible to explain the meaning of "X transforms like irrep Y" to a mathematician. It seems that terminology (and in part idea) is very alien to them. That is why I am here. $\endgroup$ – Rudi_Birnbaum Aug 29 '18 at 17:29
  • $\begingroup$ @Mike Yes, that's a suggestion, but then I do not see the special "chemistry" in here. I would opt for cross posting there after being unsuccessful here, if thats fine with people here. $\endgroup$ – Rudi_Birnbaum Aug 29 '18 at 17:30
  • $\begingroup$ I think that this is on-topic here. But, that said, it isn't particularly clear to me what you're actually asking. $\endgroup$ – Emilio Pisanty Aug 29 '18 at 18:47

It's a long time since I learned about point groups. But I'll have a go.

I believe that we are talking about the matrix representation in the basis $(x,y,z)$, in which case they are the familiar looking $3\times 3$ rotation matrices. But not the set of all $3\times 3$ rotation matrices! There are (as you said) 8 symmetry operations of the $C_3$ type, which are basically clockwise and anticlockwise rotations of $120^\circ$ about the four three-fold axes of an object with tetrahedral symmetry. Each of them can be represented by a $3\times 3$ rotation matrix, and the ones about the $z$-axis take the form given in your question. It is just conventional to orient the system so that $z$ lies along this high-symmetry axis. It is not true that a general $R_z$ with arbitrary $\theta$ corresponds to the $T$ irreducible representation: just the rotation matrices corresponding to the specific angles and axes of the symmetry operations.

The key point is that the character of the operation is given by the trace of the matrix representative in each case. For the $C_3^+$ and $C_3^-$ operations about $z$, $\theta=\pm120^\circ=\pm2\pi/3$, $\cos\theta=-\frac{1}{2}$ and the trace of the matrix in your question can be seen to be zero, which is what appears in the character tables. The same is true for all the other matrices representing the other $C_3^\pm$ operations about other axes: although they have a more complicated form in general, they have this property in common: the trace of a rotation matrix is always $1+2\cos\theta$ where $\theta$ is the overall angle of rotation. (In a similar way, for the $C_2$ operations for which the rotation matrices correspond to $\theta=180^\circ=\pi$, $\cos\theta=-1$, the character as calculated from the trace, is $1+2\times(-1)=-1$, and this is the number appearing in the character tables.)

One applies the matrix representing the symmetry operation to the elements in the basis. In your question, you seem to be considering applying the rotation operation ($C_3$) to the rotation matrix ($R_z$) itself, which is not how things work (by which I mean, not helpful in determining the characters appearing in character tables for the various operations in the 3-dimensional $T$ irreducible representation).

As the various commenters have suggested, I am sure that there is a broader general knowledge of point groups on Chemistry StackExchange. It is central to an understanding of molecular symmetry, and atomic orbitals in crystal fields, so there is bound to be plenty of expertise. So if this answer doesn't help, and nobody offers a better one, you should certainly try there. But hopefully this makes some kind of sense.

[Edit, following OP comments]. Several points to respond to.

Yes, the main feature of the character is that it is invariant to a change of basis (similarity transform) and hence is associated with the trace of the transformation matrices, which has that property.

I don't agree with the terminology

such an irrep (representing the highest order rotation in the group)

The set of matrices as a whole constitutes the representation (for a given basis). Subsets of those matrices will correspond to conjugate operations, i.e. operations belonging to the same class: those matrices in the same class will have the same character. Typically, they will involve the same kind of operation, but carried out with respect to symmetry elements that are (themselves) related by a symmetry operation.

This means that it is not necessarily true that similar kinds of operations carried out with respect to "different type of rotation axis" will have the same character. It all depends on the irrep, which in turn is related to the basis: the number and kind of functions that are being interconverted by the transformations. I have only been discussing this particular case of the $(x,y,z)$ basis, in which the matrices are the familiar $3\times 3$ rotation matrices (because effectively, we are rotating vectors). For the $D_{8h}$ group, there is no corresponding 3-dimensional irrep. There are various non-equivalent twofold axes (different classes), and various one- and two-dimensional irreps: in the character table, there are various different characters depending on both irrep and axis type (class). You would need to look at the matrix for a simple example of each case, to determine the character.

Coming to the icosahedral group, there is a three-dimensional $T_1$ irrep, which transforms the basis $(x,y,z)$, and as far as I can tell this fits the pattern I described above. At the Wikipedia page the character for $C_5$ is given as $2\cos\theta$ where $\theta=\pi/5$, and this is equal to $1+2\cos 2\pi/5$ (the rotation angle is $2\pi/5$). However there is another 3-dimensional irrep, $T_2$, for which the character for $C_5$ is different. So those operations are different. They are still rotations through $2\pi/5$, but the objects being transformed are not simple vectors. You would need to look into a suitable basis for this irrep, I'm not familiar enough with it. Generally, the matrices correspond to both the operation being performed, and the things being rotated. (Or, in the more general case, reflected etc).

Again, I hope that this clarifies things a bit. It's not completely straightforward. You may find a book helpful: Chemical Applications of Group Theory by FA Cotton is thorough, but there may be more up to date alternatives.

  • $\begingroup$ Thank you very much, that was exactly what I need to know (/remember)! It means that $R_z$ has to be $\in$ the group $G$ under consideration? (Which makes much sense, since then I one can define the homomorphism $G\to G/<R_z>$, and the traces of the (reduced) matrix representations of that are the characters of "$R_z$" in my tables, I suppose?). So when I go for an icosahedral group instead I'll have to check (the trace of) the matrix representation of a 5-fold rotation, right? (the rest I pretty much still knew about) $\endgroup$ – Rudi_Birnbaum Aug 30 '18 at 6:29
  • $\begingroup$ Also can one conclude that such an irrep (representing the highest order rotation in the group) must have the same entries for all group rotation elements (does it apply also to all other point group elements?) of the same type? For example in some point groups that contain "different types" of two-fold rotational axis (like $D_{8h}$) the characters in that representation must be the same for all these. (The underlying principle may be that the trace is independent upon certain operations , like basis change, rotations(?), ...). $\endgroup$ – Rudi_Birnbaum Aug 30 '18 at 6:55
  • $\begingroup$ I've edited my answer in an effort to respond to these points. Again, hopefully it will help. $\endgroup$ – user197851 Aug 30 '18 at 11:34

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