Character table

Orthogonality relations

The space of complex-valued class functions of a finite group G has a natural inner product:

${\displaystyle ⟨\alpha ,\beta ⟩:=\frac{1}{\left|G\right|}\sum _{g\in G}\alpha (g)\overline{\beta (g)}}$

where ${\displaystyle \overline{\beta (g)}}$ denotes the complex conjugate of the value of ${\displaystyle \beta }$ on ${\displaystyle g}$. With respect to this inner product, the irreducible characters form an orthonormal basis for the space of class functions, and this yields the orthogonality relation for the rows of the character table:

${\displaystyle ⟨{\chi }_i,{\chi }_j⟩=\{\begin{array}{ll}0& \text{ if }i\ne j,\\ 1& \text{ if }i=j.\end{array}}$

For ${\displaystyle g,h\in G}$ the orthogonality relation for columns is as follows:

${\displaystyle \sum _{{\chi }_i}{\chi }_i(g)\overline{{\chi }_i(h)}=\{\begin{array}{ll}\left|C_G(g)\right|,& \text{ if }g,h\text{ are conjugate}\\ 0& \text{ otherwise.}\end{array}}$

where the sum is over all of the irreducible characters ${\displaystyle {\chi }_i}$ of G and the symbol ${\displaystyle \left|C_G(g)\right|}$ denotes the order of the centralizer of ${\displaystyle g}$.

For an arbitrary character ${\displaystyle {\chi }_i}$, it is irreducible if and only if ${\displaystyle ⟨{\chi }_i,{\chi }_i⟩=1}$.

The orthogonality relations can aid many computations including:

• Decomposing an unknown character as a linear combination of irreducible characters, i.e. # of copies of irreducible representation V_i in ${\displaystyle V=⟨\chi ,{\chi }_i⟩}$.
• Constructing the complete character table when only some of the irreducible characters are known.
• Finding the orders of the centralizers of representatives of the conjugacy classes of a group.
• Finding the order of the group, ${\displaystyle \left|G\right|=\left|Cl(g)\right|\ast \sum _{{\chi }_i}{\chi }_i(g)\overline{{\chi }_i(g)}}$, for any g in G.

If the irreducible representation V is non-trivial, then ${\displaystyle \sum _g\chi (g)=0.}$

More specifically, consider the regular representation which is the permutation obtained from a finite group G acting on (the free vector space spanned by) itself. The characters of this representation are ${\displaystyle \chi (e)=\left|G\right|}$ and ${\displaystyle \chi (g)=0}$ for ${\displaystyle g}$ not the identity. Then given an irreducible representation ${\displaystyle V_i}$,

${\displaystyle ⟨{\chi }_{\text{reg}},{\chi }_i⟩=\frac{1}{\left|G\right|}\sum _{g\in G}{\chi }_i(g)\overline{{\chi }_{\text{reg}}(g)}=\frac{1}{\left|G\right|}{\chi }_i(1)\overline{{\chi }_{\text{reg}}(1)}=\dim V_i}$.

Then decomposing the regular representations as a sum of irreducible representations of G, we get ${\displaystyle V_{\text{reg}}=⨁V_i^{\dim V_i}}$, from which we conclude

${\displaystyle |G|=\dim V_{\text{reg}}=\sum (\dim V_i{)}^2}$

over all irreducible representations ${\displaystyle V_i}$. This sum can help narrow down the dimensions of the irreducible representations in a character table. For example, if the group has order 10 and 4 conjugacy classes (for instance, the dihedral group of order 10) then the only way to express the order of the group as a sum of four squares is ${\displaystyle 10=1^2+1^2+2^2+2^2}$, so we know the dimensions of all the irreducible representations.

Properties

Complex conjugation acts on the character table: since the complex conjugate of a representation is again a representation, the same is true for characters, and thus a character that takes on non-real complex values has a conjugate character.

Certain properties of the group G can be deduced from its character table:

• The order of G is given by the sum of the squares of the entries of the first column (the degrees of the irreducible characters). More generally, the sum of the squares of the absolute values of the entries in any column gives the order of the centralizer of an element of the corresponding conjugacy class.
• All normal subgroups of G (and thus whether or not G is simple) can be recognised from its character table. The kernel of a character χ is the set of elements g in G for which χ(g) = χ(1); this is a normal subgroup of G. Each normal subgroup of G is the intersection of the kernels of some of the irreducible characters of G.
• The number of irreducible representations of G equals the number of conjugacy classes that G has.
• The commutator subgroup of G is the intersection of the kernels of the linear characters of G.
• If G is finite, then since the character table is square and has as many rows as conjugacy classes, it follows that G is abelian iff each conjugacy class has size 1 iff the character table of G is ${\displaystyle |G|\times |G|}$ iff each irreducible character is linear.
• It follows, using some results of Richard Brauer from modular representation theory, that the prime divisors of the orders of the elements of each conjugacy class of a finite group can be deduced from its character table (an observation of Graham Higman).

The character table does not in general determine the group up to isomorphism: for example, the quaternion group and the dihedral group of order 8 have the same character table. Brauer asked whether the character table, together with the knowledge of how the powers of elements of its conjugacy classes are distributed, determines a finite group up to isomorphism. In 1964, this was answered in the negative by E. C. Dade.

The linear representations of G are themselves a group under the tensor product, since the tensor product of 1-dimensional vector spaces is again 1-dimensional. That is, if ${\displaystyle {\rho }_1:G\to V_1}$ and ${\displaystyle {\rho }_2:G\to V_2}$ are linear representations, then ${\displaystyle {\rho }_1\otimes {\rho }_2(g)=({\rho }_1(g)\otimes {\rho }_2(g))}$ defines a new linear representation. This gives rise to a group of linear characters, called the character group under the operation ${\displaystyle [{\chi }_1\ast {\chi }_2](g)={\chi }_1(g){\chi }_2(g)}$. This group is connected to Dirichlet characters and Fourier analysis.

Outer automorphisms

The outer automorphism group acts on the character table by permuting columns (conjugacy classes) and accordingly rows, which gives another symmetry to the table. For example, abelian groups have the outer automorphism ${\displaystyle g\mapsto g^{-1}}$, which is non-trivial except for elementary abelian 2-groups, and outer because abelian groups are precisely those for which conjugation (inner automorphisms) acts trivially. In the example of ${\displaystyle C_3}$ above, this map sends ${\displaystyle u\mapsto u^2,u^2\mapsto u,}$ and accordingly switches ${\displaystyle {\chi }_1}$ and ${\displaystyle {\chi }_2}$ (switching their values of ${\displaystyle \omega }$ and ${\displaystyle {\omega }^2}$). Note that this particular automorphism (negative in abelian groups) agrees with complex conjugation.

Formally, if ${\displaystyle \varphi :G\to G}$ is an automorphism of G and ${\displaystyle \rho :G\to \mathrm{GL}}$ is a representation, then ${\displaystyle {\rho }^{\varphi }:=g\mapsto \rho (\varphi (g))}$ is a representation. If ${\displaystyle \varphi ={\varphi }_a}$ is an inner automorphism (conjugation by some element a), then it acts trivially on representations, because representations are class functions (conjugation does not change their value). Thus a given class of outer automorphisms, it acts on the characters – because inner automorphisms act trivially, the action of the automorphism group ${\displaystyle \mathrm{A}\mathrm{u}\mathrm{t}}$ descends to the quotient ${\displaystyle \mathrm{O}\mathrm{u}\mathrm{t}}$.

This relation can be used both ways: given an outer automorphism, one can produce new representations (if the representation is not equal on conjugacy classes that are interchanged by the outer automorphism), and conversely, one can restrict possible outer automorphisms based on the character table.

Finding the vibrational modes of a water molecule using character table

To find the total number of vibrational modes of a water molecule, the irreducible representation Γ_irreducible needs to calculate from the character table of a water molecule first.

Finding Γreducible from the Character Table of H²O molecule

Water (${\displaystyle {\text{H}}_2^{}\text{O}}$) molecule falls under the point group ${\displaystyle C_{2v}}$. Below is the character table of ${\displaystyle C_{2v}}$ point group, which is also the character table for a water molecule.

In here, the first row describes the possible symmetry operations of this point group and the first column represents the Mulliken symbols. The fifth and sixth columns are functions of the axis variables.

Functions:

• ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ are related to translational movement and IR active bands.
• ${\displaystyle R_x}$, ${\displaystyle R_y}$ and ${\displaystyle R_z}$ are related to rotation about respective axis.
• Quadratic functions (such as ${\displaystyle x^2+y^2}$, ${\displaystyle x^2-y^2}$, ${\displaystyle x^2}$, ${\displaystyle y^2}$,${\displaystyle z^2}$, ${\displaystyle xy}$, ${\displaystyle yz}$,${\displaystyle zx}$) are related to Raman active bands.

When determining the characters for a representation, assign ${\displaystyle 1}$ if it remains unchanged, ${\displaystyle 0}$ if it moved, and ${\displaystyle -1}$ if it reversed its direction. A simple way to determine the characters for the reducible representation ${\displaystyle {\mathrm{\Gamma }}_{\text{reducible}}}$, is to multiply the "number of unshifted atom(s)" with "contribution per atom" along each of three axes (${\displaystyle x,y,z}$) when a symmetry operation is carried out.

Unless otherwise stated, for the identity operation ${\displaystyle E}$, "contribution per unshifted atom" for each atom is always ${\displaystyle 3}$, as none of the atom(s) change their position during this operation. For any reflective symmetry operation ${\displaystyle \sigma }$, "contribution per atom" is always ${\displaystyle 1}$, as for any reflection, an atom remains unchanged along with two axes and reverse its direction along with the other axis. For the inverse symmetry operation ${\displaystyle i}$, "contribution per unshifted atom" is always ${\displaystyle -3}$, as each of three axes of an atom reverse its direction during this operation. An easy way to calculate "contribution per unshifted atom" for ${\displaystyle C_n}$ and ${\displaystyle S_n}$ symmetry operation is to use below formulas

${\displaystyle C_n=2\cos \theta +1}$

${\displaystyle S_n=2\cos \theta -1}$

where, ${\displaystyle \theta =\frac{360}{n}}$

A simplified version of above statements is summarized in the table below

Character of ${\displaystyle {\mathrm{\Gamma }}_{\text{reducible}}}$ for any symmetry operation ${\displaystyle =}$ Number of unshifted atom(s) during this operation ${\displaystyle \times }$ Contribution per unshifted atom along each of three axes

Calculating the irreducible representation Γirreducible from the reducible representation Γreducible along with the character table

From the above discussion, a new character table for a water molecule (${\displaystyle C_{2v}}$ point group) can be written as

Using the new character table including ${\displaystyle {\mathrm{\Gamma }}_{\text{red}}}$, the reducible representation for all motion of the ${\displaystyle {\text{H}}_2^{}\text{O}}$ molecule can be reduced using below formula

${\displaystyle N=\frac{1}{h}\sum _x(X_i^x\times X_r^x\times n^x)}$

where,

${\displaystyle h=}$ order of the group,

${\displaystyle X_i^x=}$ character of the ${\displaystyle {\mathrm{\Gamma }}_{\text{reducible}}}$ for a particular class,

${\displaystyle X_r^x=}$ character from the reducible representation for a particular class,

${\displaystyle n^x=}$ the number of operations in the class

So,

${\displaystyle N_{A_1}=\frac{1}{4}[(9\times 1\times 1)+((-1)\times 1\times 1)+(3\times 1\times 1)+(1\times 1\times 1)]=3}$

${\displaystyle N_{A_2}=\frac{1}{4}[(9\times 1\times 1+((-1)\times 1\times 1)+(3\times (-1)\times 1)+(1\times (-1)\times 1)]=1}$

${\displaystyle N_{B_1}=\frac{1}{4}[(9\times 1\times 1)+((-1)\times (-1)\times 1)+(3\times 1\times 1)+(1\times (-1)\times 1)]=3}$

${\displaystyle N_{B_2}=\frac{1}{4}[(9\times 1\times 1)+((-1)\times (-1)\times 1)+(3\times (-1)\times 1)+(1\times 1\times 1)]=2}$

So, the reduced representation for all motions of water molecule will be

${\displaystyle {\mathrm{\Gamma }}_{\text{irreducible}}=3A_1+A_2+3B_1+2B_2}$

Translational motion for water molecule

Translational motion will correspond with the reducible representations in the character table, which have ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ function

As only the reducible representations ${\displaystyle B_1}$, ${\displaystyle B_2}$ and ${\displaystyle A_1}$ correspond to the ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ function,

${\displaystyle {\mathrm{\Gamma }}_{\text{translational}}=A_1+B_1+B_2}$

Rotational motion for water molecule

Rotational motion will correspond with the reducible representations in the character table, which have ${\displaystyle R_x}$, ${\displaystyle R_y}$ and ${\displaystyle R_z}$ function

As only the reducible representations ${\displaystyle B_2}$, ${\displaystyle B_1}$ and ${\displaystyle A_2}$ correspond to the ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$ function,

${\displaystyle {\mathrm{\Gamma }}_{\text{rotational}}=A_2+B_1+B_2}$

Total vibrational modes for water molecule

Total vibrational mode, ${\displaystyle {\mathrm{\Gamma }}_{\text{vibrational}}={\mathrm{\Gamma }}_{\text{irreducible}}-{\mathrm{\Gamma }}_{\text{translational}}-{\mathrm{\Gamma }}_{\text{rotational}}}$

${\displaystyle =(3A_1+A_2+3B_1+2B_2)-(A_1+B_1+B_2)-(A_2+B_1+B_2)}$

${\displaystyle =2A_1+B_1}$

So, total ${\displaystyle 2+1=3}$ vibrational modes are possible for water molecules and two of them are symmetric vibrational modes (as ${\displaystyle 2A_1}$) and the other vibrational mode is antisymmetric (as ${\displaystyle 1B_1}$)

Checking whether the water molecule is IR active or Raman active

There is some rules to be IR active or Raman active for a particular mode.

• If there is a ${\displaystyle x}$, ${\displaystyle y}$ or ${\displaystyle z}$ for any irreducible representation, then the mode is IR active
• If there is a quadratic functions such as ${\displaystyle x^2+y^2}$, ${\displaystyle x^2-y^2}$, ${\displaystyle x^2}$, ${\displaystyle y^2}$,${\displaystyle z^2}$, ${\displaystyle xy}$, ${\displaystyle yz}$ or ${\displaystyle xz}$ for any irreducible representation, then the mode is Raman active
• If there is no ${\displaystyle x}$, ${\displaystyle y}$, ${\displaystyle z}$ nor quadratic functions for any irreducible representation, then the mode is neither IR active nor Raman active

As the vibrational modes for water molecule ${\displaystyle {\mathrm{\Gamma }}_{\text{vibrational}}}$ contains both ${\displaystyle x}$, ${\displaystyle y}$ or ${\displaystyle z}$ and quadratic functions, it has both the IR active vibrational modes and Raman active vibrational modes.

Similar rules will apply for rest of the irreducible representations ${\displaystyle {\mathrm{\Gamma }}_{\text{irreducible}},{\mathrm{\Gamma }}_{\text{translational}},{\mathrm{\Gamma }}_{\text{rotational}}}$

Finding the vibrational modes of the ethylene molecule using character table

Ethylene is a member of the D2h point group, which has eight Mulliken symbols in the first column. Besides, the ethylene molecule contains six atoms, each with an x, y, and z axis. So, the molecule has a total of 18 axes.

For vibrational modes of the molecule, it is necessary to calculate the irreducible representation Γ_irreducible. Also, the irreducible representation is related with the reducible representation.

Here is another method to calculate the representation calculation. It is necessary to find the change of x, y and z axes. If the atom changes the place after the operation, there is no contribution to the Γ_reducible. If the atom keeps the same place after the operation, then check the axis, if the axis keeps same direction, the contribution to the Γ_reducible.is 1; if the axis reverses to the opposite direction, the contribution to the Γ_reducible.is -1; if the axis rotates at a certain angle θ, the contribution is cos θ. After calculating all axes of all atoms, there is the value of the reducible representation Γ_reducible for this operation. In this case, ethylene is the D_2h point group with eight symmetry operations in the first line, each operation provides the different Γ_reducible.

E: Identity Symmetry. All atoms remain in their original positions, so they all have the same x, y, and z axes. The 18 axes remain in the same position, each contributing one to the reducible. The reducible number for E is 18.

C₂(x), C₂(y): As the molecule rotates along the x or y axis, each atom moves and contributes zero to the reducible. The overall Γ_reducible for C2(x) and C2(y) are 0.

C₂(z): The molecule rotates along the z axis, with only two carbon atoms remaining in the same position. The x and y axes of each carbon atom reverse to the opposite place, but z axis keeps the same direction, contributing negative one of each atom. The overall Γ_reducible is -2.

i: The molecule is inverse through the center. Since all atoms move places, the overall Γ_reducible for i is 0.

σ(xy): The molecule flips across the xy plane. The overall Γ_reducible for σ(xy) is 0, as all atoms move places.

σ(xz): The molecule flips across the xz plane, but two carbon atoms remain in the same place. The x and z axes remain unchanged, each contributing to a single reducible number. However, the y axis reverses and contributes to negative one Γ_reducible. So, each carbon contributes one Γ_reducible, the overall Γ_reducible is 2.

σ(yz): It is different from other operations. All six atoms maintain their original positions. The y and z axes remain the same, but the x axis reverses, resulting in one Γ_reducible for each atom. The total Γ_reducible is 6.

New character table for ethylene ${\displaystyle {\mathrm{\Gamma }}_{\text{red}}}$

The next step is to calculate the irreducible presentation based on the reducible presentation. Here is the calculation.

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width:88.01ex; height:5.176ex;" alt="{\displaystyle N_{A_{g}}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times 1)+(6\times 1\times 1)]=3}"/></span><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt259" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{B_1g} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times 1)+0+0+(2\\times 1\\times (-1))+(6\\times 1\\times (-1))] = 1"}}' id="mwAzM"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{B_{1}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))]=1}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>B</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>1</mn> </mrow> </msub> <mi>g</mi> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>1</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{B_{1}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))]=1}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/ed207e4f8225ec37333901aed4faf63c74e5d5a4" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:96.075ex; height:5.176ex;" alt="{\displaystyle N_{B_{1}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))]=1}"/></span><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt260" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{B_2g} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times (-1))+0+0+(2\\times 1\\times 1)+(6\\times 1\\times (-1))] = 2"}}' id="mwAzQ"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{B_{2}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))]=2}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>B</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>2</mn> </mrow> </msub> <mi>g</mi> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>2</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{B_{2}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))]=2}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/a5dc8ff4f406c2b7038b339913835b845bb0d926" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:96.075ex; height:5.176ex;" alt="{\displaystyle N_{B_{2}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))]=2}"/></span><span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt261" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{B_3g} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times (-1))+0+0+(2\\times 1\\times (-1))+(6\\times 1\\times 1)] = 3"}}' id="mwAzU"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{B_{3}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)]=3}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>B</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msub> <mi>g</mi> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>3</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{B_{3}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)]=3}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/2ce828c07f46c64fe311cfde0017806250eafcd2" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:96.075ex; height:5.176ex;" alt="{\displaystyle N_{B_{3}g}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)]=3}"/></span>

<span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt262" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{A_u} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times 1)+0+0+(2\\times 1\\times (-1))+(6\\times 1\\times (-1))] = 1"}}' id="mwAzc"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{A_{u}}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))]=1}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>A</mi> <mrow class="MJX-TeXAtom-ORD"> <mi>u</mi> </mrow> </msub> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>1</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{A_{u}}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))]=1}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/fa8afff5d5879edba883e9e396a197bd0eee5681" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:95.367ex; height:5.176ex;" alt="{\displaystyle N_{A_{u}}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times (-1))+(6\times 1\times (-1))]=1}"/></span>

<span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt263" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{B_1u} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times 1)+0+0+(2\\times 1\\times 1)+(6\\times 1\\times 1)] = 3"}}' id="mwAzk"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{B_{1}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times 1)+(6\times 1\times 1)]=3}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>B</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>1</mn> </mrow> </msub> <mi>u</mi> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>3</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{B_{1}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times 1)+(6\times 1\times 1)]=3}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c825ffe8987c9ec11b890f71cde9aa3ed88e0bb0" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:88.992ex; height:5.176ex;" alt="{\displaystyle N_{B_{1}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times 1)+0+0+(2\times 1\times 1)+(6\times 1\times 1)]=3}"/></span>

<span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt264" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{B_2u} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times (-1))+0+0+(2\\times 1\\times (-1))+(6\\times 1\\times 1)] = 3"}}' id="mwAzs"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{B_{2}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)]=3}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>B</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>2</mn> </mrow> </msub> <mi>u</mi> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>3</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{B_{2}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)]=3}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/c2dbd0c253fdf715590b1c9103f855dce8d0ebd7" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:96.226ex; height:5.176ex;" alt="{\displaystyle N_{B_{2}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times (-1))+(6\times 1\times 1)]=3}"/></span>

<span class="mwe-math-element mwe-math-element-inline" typeof="mw:Extension/math" about="#mwt265" data-mw='{"name":"math","attrs":{},"body":{"extsrc":"N_{B_3u} = \\frac{1}{8}[(18\\times 1\\times 1)+0+0+((-2)\\times 1\\times (-1))+0+0+(2\\times 1\\times 1)+(6\\times 1\\times (-1))] = 2"}}' id="mwAz0"><span class="mwe-math-mathml-a11y" style="display: none;"><math xmlns="http://www.w3.org/1998/Math/MathML" class="mathjax_ignore" alttext="{\displaystyle N_{B_{3}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))]=2}"> <semantics> <mrow class="MJX-TeXAtom-ORD"> <mstyle displaystyle="true" scriptlevel="0"> <msub> <mi>N</mi> <mrow class="MJX-TeXAtom-ORD"> <msub> <mi>B</mi> <mrow class="MJX-TeXAtom-ORD"> <mn>3</mn> </mrow> </msub> <mi>u</mi> </mrow> </msub> <mo>=</mo> <mrow class="MJX-TeXAtom-ORD"> <mfrac> <mn>1</mn> <mn>8</mn> </mfrac> </mrow> <mo stretchy="false">[</mo> <mo stretchy="false">(</mo> <mn>18</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>2</mn> <mo stretchy="false">)</mo> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mn>0</mn> <mo>+</mo> <mo stretchy="false">(</mo> <mn>2</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo>+</mo> <mo stretchy="false">(</mo> <mn>6</mn> <mo>×<!-- × --></mo> <mn>1</mn> <mo>×<!-- × --></mo> <mo stretchy="false">(</mo> <mo>−<!-- − --></mo> <mn>1</mn> <mo stretchy="false">)</mo> <mo stretchy="false">)</mo> <mo stretchy="false">]</mo> <mo>=</mo> <mn>2</mn> </mstyle> </mrow> <annotation encoding="application/x-tex">{\displaystyle N_{B_{3}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))]=2}</annotation> </semantics> </math></span><img src="https://wikimedia.org/api/rest_v1/media/math/render/svg/d323cbec6b846ebb4fd2d1852bdaa74b9b5af0e4" class="mwe-math-fallback-image-inline mw-invert skin-invert" aria-hidden="true" style="vertical-align: -1.838ex; width:96.226ex; height:5.176ex;" alt="{\displaystyle N_{B_{3}u}={\frac {1}{8}}[(18\times 1\times 1)+0+0+((-2)\times 1\times (-1))+0+0+(2\times 1\times 1)+(6\times 1\times (-1))]=2}"/></span>

Γ_irreducible = 3A_g+1B_1g+2B_2g+3B_3g+1A_u+3B_1u+3B_2u+2B_3u

Translational motion has x, y and z functions in “linear functions, roatations”. So, Γ_trans = 1B_1u+1B_2u+1B_3u

Rotational motion has R_x, R_y and R_z functions in “linear functions, roatations”. So, Γ_rot = 1B_1g+1B_2g+1B_3g

Vibrational motio: Γ_vib = Γ_irreducible-Γ_trans-Γ_rot = 3A_g+1B_2g+2B_3g+1A_u+2B_1u+2B_2u+1B_3u

The final step is to determine which vibrations are IR or Raman active. This means that the symmetry operation can be detected using the infrared or Raman spectrum.

First, for IR to work, they must have x, y, and z functions in "linear functions, rotations". In Γ_vib, only 2B_1u+2B_2u+1B_3u are IR active.

To be Raman active, "quadratic functions" must include x², y², z²,xy, xz, yz, x²+y² or x²-y² functions. In Γ_vib, only 3A_g+1B_2g+2B_3g are Raman active.