The characters of the identity operation, *E*, reveal the degeneracy of the orbitals of each symmetry species.

Thus in a C_{3v} molecule any *a*_{1} or *a*_{2} orbital is non-degenerate. Any doubly degenerate pair of orbitals in a C_{3v} molecule must be labeled *e*, as the only symmetry species with a character greater than one is E. (For example, the p_{x} and p_{y} orbitals on the central nitrogen atom of ammonia are doubly degenerate, and together belong to the E irreducible representation, so are labeled *e*. Note that this could have been predicted from the C_{3v} character table, which told us that the x and y functions were required jointly to compose a function that transformed as E.)

Since there are no characters greater than two in the column under *E*, we can immediately say that no orbitals in a C_{3v} molecule can be triply degenerate (or greater). That is, the character table of a point group immediately gives the maximum possible degeneracy of the orbitals of a molecule in that group.

The characters in rows labeled with an A or B symmetry species indicate the behaviour of an orbital under the corresponding symmetry operation.

As in the example on the previous page, a 1 indicates that the orbital is unchanged by the operation, while a -1 indicates that the operation causes it to change sign.

This means that it is relatively simple to identify the symmetry labels of non-degenerate orbitals – we need merely to consider the behaviour of the orbital under each operation of the group, and then match this behaviour to the correct row of 1’s and -1’s in the character table.

For rows labeled with E or T symmetry species (which refer to the behaviour of doubly or triply degenerate orbitals respectively) the characters in the row are the **sums** of the characters which give the behaviour of the individual orbitals. Thus, since the character of any orbital under the identity is 1, the character under *E* __must__ give the degeneracy of the symmetry species, as each orbital has contributed 1 to the overall character.

As a further example, if we consider a doubly degenerate pair of orbitals, if there is a symmetry operation that leaves one member of the pair unchanged but changes the sign of the other, then the overall character is reported as 1 + (-1) = 0.

Note that the above discussion has been limited to assigning the symmetry species of an individual orbital (by consideration of its behaviour under each operation and then comparison of this to the rows of characters in a character table). It is also possible to classify linear combinations of orbitals on atoms that are related by symmetry operations of the molecule.

For example, the three hydrogen atoms in ammonia are related by the C_{3} rotation of the molecule, and by the three mirror planes. Thus we may classify the symmetry of a linear combination of the three 1s orbitals of these atoms. We write this as considering the combination φ_{1}, where φ_{1} = φ_{A} + φ_{B} + φ_{C} with A, B and C being the three hydrogen atoms in the molecule, and φ_{X} being the 1s orbital on atom X. The notation signifies that the linear combination under consideration has an equal contribution from each of the three 1s orbitals, and it may thus be visualised as three spheres of equal size and the same sign at the corners of an equilateral triangle (the same spatial arrangement as the three hydrogen atoms in ammonia).

Note the C_{3} axis runs perpendicular to the plane of this triangle, and passes through its centre, while the mirror planes are also perpendicular to the plane of the triangle but each bisect one side of it. i.e. the symmetry elements are in exactly the same place relative to the 1s orbitals as they are in the ammonia molecule.

We can see that this combination remains unchanged under a C_{3} rotation or any of the reflections of the point group, so the characters of this linear combination of orbitals are 1 under all the operations of the C_{3v} point group. From the character table, it can be seen that this makes the orbital combination of A_{1} symmetry. This orbital combination may thus only contribute to *a*_{1} molecular orbitals in an NH_{3} molecule.