I know how to do it for just about every other point group, but the $D_{\infty \mathrm h}$ and $C_{\infty \mathrm v}$ character tables aren't as straightforward. In particular, I'm interested in the vibrational modes of carbon dioxide, $\ce{CO2}$.

$$\begin{array}{c|cccc|cc} \hline C_{\infty\mathrm{v}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & & \\ \hline \mathrm{A_1} \equiv \Sigma^+ & 1 & 1 & \cdots & 1 & z & x^2 + y^2, z^2 \\ \mathrm{A_2} \equiv \Sigma^- & 1 & 1 & \cdots & -1 & R_z & \\ \mathrm{E_1} \equiv \Pi & 2 & 2 \cos\phi & \cdots & 0 & (x,y), (R_x,R_y) & (xz,yz) \\ \mathrm{E_2} \equiv \Delta & 2 & 2 \cos 2\phi & \cdots & 0 & & (x^2-y^2,xy) \\ \mathrm{E_3} \equiv \Phi & 2 & 2 \cos 3\phi & \cdots & 0 & & \\ \vdots & \vdots & \vdots & \ddots & \vdots & & \\ \hline \end{array}$$


$$\small \begin{array}{c|cccccccc|cc} \hline D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \mathrm{A_{1g}} \equiv \Sigma^+_{\mathrm{g}} & 1 & 1 & \cdots & 1 & 1 & 1 & \cdots & 1 & & x^2 + y^2, z^2 \\ \mathrm{A_{2g}} \equiv \Sigma^-_{\mathrm{g}} & 1 & 1 & \cdots & -1 & 1 & 1 & \cdots & -1 & R_z & \\ \mathrm{E_{1g}} \equiv \Pi_{\mathrm{g}} & 2 & 2\cos\phi & \cdots & 0 & 2 & -2\cos\phi & \cdots & 0 & (R_x,R_y) & (xz,yz) \\ \mathrm{E_{2g}} \equiv \Delta_{\mathrm{g}} & 2 & 2\cos 2\phi & \cdots & 0 & 2 & 2\cos 2\phi & \cdots & 0 & & (x^2-y^2,xy) \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & & \\ \mathrm{A_{1u}} \equiv \Sigma^+_{\mathrm{u}} & 1 & 1 & \cdots & 1 & -1 & -1 & \cdots & -1 & z & \\ \mathrm{A_{2u}} \equiv \Sigma^-_{\mathrm{u}} & 1 & 1 & \cdots & -1 & -1 & -1 & \cdots & 1 & & \\ \mathrm{E_{1u}} \equiv \Pi_{\mathrm{u}} & 2 & 2\cos\phi & \cdots & 0 & -2 & 2\cos\phi & \cdots & 0 & (x,y) & \\ \mathrm{E_{2u}} \equiv \Delta_{\mathrm{u}} & 2 & 2\cos 2\phi & \cdots & 0 & -2 & -2\cos 2\phi & \cdots & 0 & & \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & & \\ \hline \end{array}$$

  • $\begingroup$ Perhaps not the answer you are looking for, but the most straightforward way is just to learn what they are instead of figuring them out. It is quite logical, there are two stretching modes (symmetric $\Sigma_g^+$ + antisymmetric $\Sigma_u^+$) and two degenerate bending modes $\Pi_u$. You can visualise them here: chemtube3d.com/vibrationsCO2.htm $\endgroup$ Mar 20, 2017 at 18:23
  • 1
    $\begingroup$ @orthocresol True, but that only gives an answer for the case where it is $D_{\infty h}$ with 3 atoms. This is the case that we see most often, but it is useful to have a general way to derive it rather than memorizing cases. $\endgroup$
    – Tyberius
    Mar 20, 2017 at 22:15

4 Answers 4


As Tyberius noted, the projection formula does not work for infinite order groups (this is because the Hermitian form on characters is defined to be G-invariant by averaging over all elements in a group. That is, it works because it is possible to hit all the elements of the group in some order).

How can one work in the infinite dimensional groups then? One property that still holds for these particular infinite groups (though not in general 1) is Maschke's Theorem, that every representation is the direct sum of irreducible representations. So we can still determine what irreps make up a rep by adding them up.

First we need to find the representation we wish to reduce. I find that the easiest way to do this is to start with just the atoms (instead of the vectors) and then include the vector contribution afterwards. For the sake of a new example, I will do the linear $\ce{I4^{2-}}$ anion instead of $\ce{CO2}$.

$$\small \begin{array}{c|cccccccc} \hline D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \Gamma_\text{atoms} & 4 & 4 & \cdots & 4 & 0 & 0 & \cdots & 0 &\\ \end{array}$$

Now, we multiply by the vector's contributions to the character. $$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

These come from taking the trace of the matrix representations of each symmetry operation. For example, a rotation about the z-axis by an angle $\theta$ is given by: $$\left[\begin{array}{ccc}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{array}\right]$$

$$\small \begin{array}{c|cccccccc} \hline D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\ \end{array}$$

Now, we are only interested in the vibrational modes so we must subtract off the translational and rotational degrees of freedom. We can check the character table for these (they are the ones which correspond to $z$, $(x,y)$ and $(R_x, R_y)$, $R_z$ doesn't count because we define the molecular axis to be the $z$-axis. $$\small \begin{array}{c|cccccccc} \hline D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\ -\Gamma_\text{rot} & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&0 \\ -\Gamma_\text{trans} & 3 & 1+2\cos\theta&\cdots& 1 & -3 & -1+2\cos\theta&\cdots&-1 \\ \hline\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\ \end{array}$$

So now we have the actual representation of the vibrations. How do we go about decomposing it? We want to eventually subtract off all of the irreps until we have the zero vector so we consider what effect each irrep has and try to move our remaining rep closer to zero. The first thing to note is that only $\Pi_g$ and $\Pi_u$ irreps have $2\cos\theta$. $\Pi_u$ adds $2\cos\theta$ under both proper and improper rotations, while $\Pi_g$ adds to proper and subtracts from improper. We note that in our rep, we only have $\cos\theta$ terms under proper rotations and so we need to subtract off enough $\Pi_u$'s until there are the same number of $\cos\theta$ (with different signs) under proper and improper rotations, at which point we should subtract off $\Pi_g$s until there are no $\cos\theta$ terms.

$$\small \begin{array}{c|cccccccc} \hline D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\ -\Pi_u & 2 & 2\cos\theta&\cdots&0 & -2 & 2\cos\theta&\cdots&1 \\ -\Pi_g & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&1 \\ \hline\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\ \end{array}$$

Similar to the $\Pi_g/\Pi_u$, the $\Sigma_g^+$ shift adds 1 to all positions, while $\Sigma_u^+$ adds 1 to all positions before $i$ and subtracts from all after. We subtract enough $\Sigma_g^+$ until all the positions before and after $i$ have equal magnitude. $$\small \begin{array}{c|cccccccc} \hline D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\ -2\Sigma_g^+ & 2 & 2 &\cdots&2 & 2 & 2&\cdots&2 \\ -\Sigma_u^+ & 1 & 1 &\cdots&1 & -1 & -1&\cdots&-1 \\ \hline \Gamma& 0 & 0 &\cdots&0 & 0 & 0&\cdots&0 \end{array}$$

Thus, we have found that the vibrational modes of $\ce{I_4^{2-}}$ to be $2\Sigma_g^+ +\Sigma_u^+ + \Pi_u + \Pi_g$.

We can check this with the formulas given in Group Theory And Chemistry by David M. Bishop as noted by Feodoran. One can also follow Tyberius's suggestion and work in $D_{2h}$ which will give the decomposition $2A_g+B_{1u}+B_{2u}+B_{3u}+B_{2g}+B_{3g}$.

It is important to note that correlation from the $D_{2h}$ point group to $D_{\infty h}$ works fine for vibrations because you cannot have terms higher than $\Pi$ and so there is no ambiguity. For things like electronic states, though, in which direct products must be decomposed, you will run into some problems. If the $D_{2h}$ result includes $A_g+B_{1g}$ does that correlate with a $\Sigma_g^+ +\Sigma_g^-$ or a $\Delta_g$ state in $D_{\infty h}$? Working in the parent infinite group may be more difficult without the reduction formula, but these problems are avoided.

The reason going to lower symmetry works is because all characters under the infinitely many proper and improper rotations are the same for each irrep (namely $2\cos n\theta$) and so we can represent the behavior of all of them by the behavior of one of them. Hence, we pick $\theta=\pi$ and ignore all the others and this produces the $D_{2h}$ point group (the two improper rotation axes are equivalent to $i$ and the two proper axes are equivalent to one $C_2$ about $z$). This is origin of the ambiguity. In $D_{\infty h}$, $\Sigma_g^+ + \Sigma_g^-$ are equal to $\Delta_g$ at $\theta=\pi$, but not in general.

Having now identified the origin of the ambiguity, one can tune how fast these ambiguities occur though. The problem occurs when the $\theta$ chosen to represent all $\theta$ has the property $1=\cos n\theta$ where $n$ represents the period of the ambiguity. So for $D_{2h}$, $\theta=\pi$ and $1=\cos2\theta$ and the period of the ambiguity is 2. That is, every other term will be ambiguously identical ($\Sigma, \Delta, \Gamma, \ldots$ are the same and $\Pi,\Phi,\ldots$ are the same). By dropping to $D_{4h}$ instead, $\theta=\pi/2$ and $1=\cos4\theta$. The ambiguity period is longer and we can distinguish $\Delta$ from $\Sigma$ by correlation from the $D_{4h}$ point group. We can thus select the point group to descend to based on how much resolving power we expect to need in the parent point group (though we should stick to even numbers to maintain inversion symmetry for $D_{\infty h}$). Of course we could also stay in the parent point group and never have a problem, but also have no reduction formula.


Instead of trying to use symmetry tables it is possible to calculate the vibrational normal modes directly using the equations of motion, rather as would be done for a double pendulum for example.

It is normally assumed that the vibrational motion of the atoms is harmonic, i.e. Hook's law applies, and we additionally assume that the only force constants are those between adjacent atoms; this is equivalent to assuming that the valence bond model describes the bonding. A molecular orbital model would consider forces constants between one atom and every other atom.

The starting point is to calculate the potential energy and, from this, the forces on the atoms. As carbon dioxide is symmetrical, there is only one force constant, k, but as the masses are different it is found that the mass-weighted force constants are needed in the calculation and these are different for each type of atom.

Stretching vibrations

The normal modes for a simple molecule can easily be sketched and as the molecule is linear there are $3N − 5$ or four modes in total, two are stretches and the other two are the degenerate bending motion in the plane of the page. One where the carbon atom moves down and the two oxygen atoms move up and then vice versa. The other bending normal mode is the similar motion but perpendicular to the plane of the page. Only the stretching modes are considered here.


Placing a vector $s_1,\,s_2,\,s_3$ on each atom and pointing in the same direction and along the long axis of the molecule, the potential energy V is the sum of terms for the stretching of each bond; therefore,

$$V(s_1,s_2,s_3)=\frac{k}{2}(s_1-s_2)^2+\frac{k}{2}(s_2-s_3)^2 $$

These equations are just for the stretching motion, a second calculation is needed with vectors perpendicular to the molecular axis for the bending modes.

The forces are the negative derivatives with respect to the displacements

$$-\frac{dV}{ds_1}=-k(s_1-s_2); \quad -\frac{dV}{ds_2}=k(s_1-s_2)-k(s_2-s_3);\quad -\frac{dV}{ds_3}=k(s_2-s_3)$$

With f as the force, placing these equations into matrix form gives

$$\begin{bmatrix} f_1\\ f_2\\f_3 \end{bmatrix} = \begin{bmatrix} -k & k & 0 \\ k & -2k & k\\ 0 & k & -k \end{bmatrix} \begin{bmatrix} s_1\\ s_2\\s_3 \end{bmatrix}$$

but, because the masses are different, we must change to mass weighted force constants, using the formula $K_{i,j}=k_{i,j}/\sqrt{m_im_j}$ where i and j are the atom indices; for example $K_{1,2} = k\sqrt{ m_Om_C}$, is the mass-weighted force constant between atoms 1 and 2, if $m_O$ is the oxygen mass and $m_C$ that of the carbon. The matrix of force constants becomes

$$\boldsymbol{K} = \begin{bmatrix} \displaystyle\frac{-k}{m_O} & \displaystyle\frac{k}{\sqrt{m_Om_C}} & 0 \\ \displaystyle\frac{k}{\sqrt{m_Om_C}} & \displaystyle\frac{-2k}{m_C} & \displaystyle\frac{k}{\sqrt{m_Om_C}}\\ 0 & \displaystyle\frac{k}{\sqrt{m_Om_C}} &\displaystyle \frac{-k}{m_O} \end{bmatrix}$$

which is solved as a secular determinant with eigenvalues $\lambda$,

$$ \lambda_1=-k/m_O; \quad \lambda_2= -k\frac{2m_O+m_C}{m_Om_C}; \quad \lambda_3 = 0 $$

The frequency of each vibration is $\omega^2 = −\lambda$ so the square of the normal mode frequencies are

$$\omega_1^2=k/m_O; \quad \omega_2^2 = k\frac{2m_O+m_C}{m_Om_C} ; \quad \omega_3=0$$

Using the measured stretching vibrational frequencies for $\ce{CO2}$, for example, which are 1337 & 2349 cm$^{-1}$, the force constants are 1680 & 1418 N/m respectively. Note that the frequency $\omega = 2\pi\nu c$ with $\nu$ in cm$^{-1}$.

The normalised eigenvector matrix x is, after some rearranging and with the total mass as $M=2m_O+m_C$

$$\boldsymbol {x}= \frac{1}{\sqrt{2M}} \begin{bmatrix} -\sqrt{M} & \sqrt{m_C} & \sqrt{2m_O} \\ 0 & -2\sqrt{m_O} & \sqrt{2m_C}\\ \sqrt{M}& \sqrt{m_C} & \sqrt{2m_O} \end{bmatrix}$$

Notice that the eigenvectors do not depend on the force constants; this matrix is used to produce the geometry that is related to the symmetry of the vibrations, and cannot depend on the value of the force constants. The normal modes depend only on the geometry because the ‘springs’ connecting the atoms can only vibrate in certain patterns governed by the geometry or symmetry of the molecule; the frequency and size of extension depend on the force constants.

The normal mode coordinates are calculated using $\boldsymbol{Q = x^Tq}$ where $\boldsymbol{x}^T$ is the transpose of the eigenvector matrix and q the vector of mass weighted coordinates with $q=s\sqrt{m}$.

$$\begin{bmatrix} Q_1\\Q_2\\Q_3 \end{bmatrix} = \begin{bmatrix} -\sqrt{M} & 0 & \sqrt{M} \\ \sqrt{m_C} & -2\sqrt{m_O} & \sqrt{m_C}\\ \sqrt{2m_O}& \sqrt{2m_C} & \sqrt{2m_O} \end{bmatrix} \begin{bmatrix} s_1\sqrt{m_O}\\s_2\sqrt{m_C}\\s_3\sqrt{m_O} \end{bmatrix}$$

and the individual modes are

$$Q_1=\sqrt{m_O/2}(-s_1+s_3); \quad Q_2 = \sqrt{\frac{m_om_C}{2M}}(s_1-2s_2+s_3);\\ \quad Q_3=(m_Os_1+m_Cs_2+m_Os_3)/\sqrt{M}$$

The symmetry of the normal mode can be seen from the displacements s. The last, $Q_3$, is just a translation of the molecule as all the s vectors point in the same direction and has frequency $\omega=0$. $Q_1$ would correspond to a symmetric stretch and therefore $Q_2$ to the assymetric stretch, both O atoms move in one direction and the carbom moves the opposite way to keep the centre of gravity fixed in space.

We can choose any Q to be zero to find the s displacements

(i) Suppose we choose $Q_1 = Q_2 = 0$, then all the atom displacements are the same, $s_1 = s_2 = s_3$ and are $Q_3/\sqrt{M}$. As each atom is moving in the same direction this cannot be a vibrational normal mode as the bonds are neither stretched nor compressed, but represents a translation, and clearly, this corresponds to the zero frequency eigenvalue $\omega_3 = 0$. Note that if $Q_3 = 0$ this must mean that the centre of mass does not change, no translation occurs, and so we expect to produce normal modes with this condition.

(ii) If $Q_1=Q_3=0$ then $\displaystyle s_1=s_3= \frac{m_C}{2m_OM}Q_2$ and $\displaystyle s_2= - \frac{2m_O}{2m_OM}Q_2$ In this case this is the asymmetric stretch with the oxygen atoms moving in the same direction and opposed to that of the carbon, and this is $\omega_2$. The relative motion is $s_1 = s_3 = 0.115Q_2$ to $s_2 = −0.308Q_2$ so the carbon atom moves further than the oxygen atoms do, which is not surprising, because the centre of mass has to be held constant and the C atom has to compensate for the motion of two O atoms.

(iii) Finally, if $Q_2 = Q_3 = 0$ then the equations are solved when $s_2 = 0$ and $s_1 = −s_3$, which is the symmetric stretch $\omega_1$ with displacements $\pm Q_1/\sqrt{2m_O}$ or $0.176Q_1$.

There is a very general method with which to do these types of calculations called the GFG matrix method or some similar name equivalent to this. As this is a matrix method it is easy to calculate equations using symbolic algebra programmes such as SymPy or to do the calculation numerically using for example Python. (Both SymPy & Python are free to use).

The secular determinant to be solved is

$$ | \boldsymbol{GFG} -\omega^2\boldsymbol{I} |=0$$

where I is a unit diagonal matrix, $\omega$ the normal mode frequencies and the matrices are G is a matrix zero everywhere except on the diagonal where it has values $1/\sqrt{m}$, i.e.

$$\boldsymbol{G} = diag \left [ 1/\sqrt{m_i}\right ] $$

and the F matrix is that of the force constants; $$\boldsymbol{F} = \begin{bmatrix} k_{11} & k_{12} & k_{13} & \cdots\\ k_{21} & k_{22} &\cdots & \cdots \\ \vdots & \vdots & & k_{nn} \end{bmatrix}$$

and the product is just the mass weighted matrix of force constants

$$\boldsymbol{GFG} = \begin{bmatrix} k_{11}/m_1 & k_{12}/\sqrt{m_1m_2} & k_{13}/\sqrt{m_1m_3} & \cdots\\ k_{21}/\sqrt{m_1m_2} & k_{22}/m_2 &\cdots & \cdots \\ \vdots & \vdots & & k_{nn}/\sqrt{m_n} \end{bmatrix}$$

The eigenvectors x can be used to produce the normal mode displacements as $\boldsymbol{Q=x^Tq}$ where $\boldsymbol{q=G^{-1}s}$ so that each element is $q_i= s_i\sqrt{m_i}$ and the coordinate displacements via, $\boldsymbol{ s =GxQ}$.

Bending vibrations

In the case of bending vibrations in a molecule with masses $m_{1,2,3}$, such as HCN, the potential is written as

$$V= k_br_1r_2(\delta\theta)^2/2$$

where the bond lengths are $r_{1,2}$ and the bending force constant $k_b=k_\theta r_1r_2$ which keeps its units in N/m. For small angle bends $\displaystyle \delta\theta = \frac{(s_1 - s_2)}{r_1} - \frac{(s_3 - s_2)}{r_2}$ (where vector s is perpendicular to the internuclear axis just as s was along the axis) and the mass weighted K matrix of force constants becomes

$$K = \begin{bmatrix} \displaystyle \frac{k_3}{m_1}\left(\frac{r_2}{r_1}\right) & \displaystyle\frac{-k_3}{\sqrt{m_1m_2}}\left(1+\frac{r_2}{r_1}\right) & \displaystyle \frac{k_3}{\sqrt{m_1m_3}} \\ \displaystyle\frac{-k_3}{\sqrt{m_1m_2}}\left(1+\frac{r_2}{r_1}\right) & \displaystyle\frac{k_3}{m_2}(2+\frac{r_2}{r_1}+\frac{r_1}{r_2}) & \displaystyle \frac{-k_3}{\sqrt{m_2m_3}}\left(1+\frac{r_1}{r_2}\right)\\ \displaystyle \frac{k_3}{\sqrt{m_1m_3}} & \displaystyle\frac{-k_3}{\sqrt{m_2m_3}}\left(1+\frac{r_1}{r_2}\right) & \displaystyle\frac{k_3}{m_3}\left(\frac{r_1}{r_2} \right) \end{bmatrix}$$

which means that the ratio of bond lengths must be known if they are different from 1.

The algebraic eigenvalues are immensely complex so it is necessary to calculate them numerically except for the case of $D_{\infty h}$ point group. In this case the force constants are zero, or $2k_3(2/m_2 +1/m_1)$ where $m_2$ is the central atom.

In the case of $\mathrm{CO}_2$, $m_1 = 16$, $m_2 = 12$ and the bond length $r=0.116$ nm. The only non zero (and doubly degenerate) frequency is $\omega^2 =2k_3(2/m_2 +1/m_1)$ . In $\mathrm{CO}_2$ the bending vibration has a frequency of $667$ cm$^{-1}$ and then $k_3 = 57$ N/m.

(source. The arguments presented here follow closely that in Beddard 'Applying Maths in the Chemical & Biomolecular Sciences, an example based approach' publ OUP, and where there is a more detailed description. Herzberg gives several examples (Vol II, chapter 2) and Wilson, Decius & Cross 'Molecular Vibrations describe the method in great detail.)


I actually didn't know how to do this either, but I've found how to work with these infinite symmetry point groups. You have noticed that the reduction formula would not work properly for these groups. So, what we can do to solve for $D_\mathrm {\infty h}$ is solve for the irreducible representation in a lower order symmetry group, say $D_\mathrm{2h}$, and then correlate the results to the higher order symmetry group using the tables below.

$$\begin{array}{c|cccccc} C_\mathrm{\infty v} & \mathrm{A_1 = \Sigma^+} & \mathrm{A_2 = \Sigma^-} & \mathrm{E_1 = \Pi} & \mathrm{E_2 = \Delta} & \cdots \\ \hline C_\mathrm{2v} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{B_1 + B_2} & \mathrm{A_1 + A_2} & \cdots \end{array}$$

$$\begin{array}{c|cccccc} D_\mathrm{\infty h} & \Sigma_\mathrm{g}^+ & \Sigma_\mathrm{g}^- & \Pi_\mathrm{g} & \Delta_\mathrm{g} & \cdots & \Sigma_\mathrm{u}^+ & \Sigma_\mathrm{u}^- & \Pi_\mathrm{u} & \Delta_\mathrm{u} & \cdots \\ \hline D_\mathrm{2h} & \mathrm{A_g} & \mathrm{B_{1g}} & \mathrm{B_{2g} + B_{3g}} & \mathrm{A_g + B_{1g}} & \cdots & \mathrm{B_{1u}} & \mathrm{A_u} & \mathrm{B_{2u} + B_{3u}} & \mathrm{A_u + B_{1u}} & \cdots \end{array}$$

I got these tables, and a lot of information I didn't know about point groups, from this page from the University of Massachusetts Boston.

I also found a nice walk-through on Wikipedia, which explains how to do this for $\ce{CO2}$, which is identical in terms of symmetry grouping.


Again only part of an answer. As a reference I used the book Group Theory And Chemistry by David M. Bishop. There it is stated (Chapter 9-8) that linear molecules are special cases, and they give generalized results for $N$ atomic molecules:

  • asymmetric molecules ($C_{\infty v}$) have:
    • $N-1$ longitudinal vibrations belonging to $\Sigma^+$
    • $N-2$ pairs of transverse vibrations belongig to $\Pi$
  • symmetric molecules ($D_{\infty h}$) have for even $N$:
    • $\frac{N}{2}$ longitudinal vibrations belonging to $\Sigma^+_g$
    • $\frac{N}{2}-1$ longitudinal vibrations belonging to $\Sigma^+_u$
    • $\frac{N}{2}-1$ pairs of transverse vibrations belonging to $\Pi_g$
    • $\frac{N}{2}-1$ pairs of transverse vibrations belonging to $\Pi_u$
  • symmetric molecules ($D_{\infty h}$) have for odd $N$:
    • $\frac{N-1}{2}$ longitudinal vibrations belonging to $\Sigma^+_g$
    • $\frac{N-1}{2}$ longitudinal vibrations belonging to $\Sigma^+_u$
    • $\frac{N-3}{2}$ pairs of transverse vibrations belonging to $\Pi_g$
    • $\frac{N-1}{2}$ pairs of transverse vibrations belonging to $\Pi_u$

I tried the $\ce{CO2}$ example, but didn't get very far: The representation we need to reduce is (Chapter 9-6 in the above book explains how to do this):

\begin{array}{c|cccccccc|cc} D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline \Gamma^{red} & 9 & 3+6\cos\phi & \cdots & 3 & -3 & -1+2\cos\phi & \cdots & -1 \\ \end{array}

As already stated in Tyberius' answer, the problem is to apply the reduction formula. Here we need to sum over infinitely many $C_\infty^\phi$ axes. We can do this by integration, for example in the reduction formula for the IRREP $A_{1g}$ we get the following term:

\begin{equation} 2\times 1\times\int\limits_0^{2\pi}(3+6\cos\phi)\mathrm{d}\phi = 6\pi \end{equation}

In a similar way for the improper rotational axes $S_\infty^\phi$ we get $-4\pi$. What I don't know yet is how to treat the infinit many elements of the $\sigma_v$ and $C_2$ classes.

  • $\begingroup$ The reduction formula cannot be applied to infinite dimensional groups. $\endgroup$
    – levineds
    Jul 8, 2017 at 7:02

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