In their classic paper outlining the Jahn–Teller theorem (Proc. R. Soc. Lond. A 1937, 161 (905), 220–235), the authors wrote "where $[\phi^2]$ denotes the representation of the symmetrical product of $\phi$ with itself". Reading this I'm thinking about a regular product of representation (scalar product with each characters).

But in the next paragraph there is this equality for the $D_\mathrm{4h}$ group:

$$[\mathrm{E_g^2}] = [\mathrm{E_u^2}] = \mathrm{A_{1g} + B_{1g} + B_{2g}}$$

Obviously something is wrong, as the direct product is a 4-dimensional representation, but the right-hand side is 3-dimensional.

I would like to add that there is a reference of Weyl 1928 unfortunately this article is in German and I cannot find anything in it.

What is the correct way to carry out this direct product?

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    $\begingroup$ In a (typical) direct product table there are some things in square brackets and some not in square brackets. What you want are those which aren't in square brackets. See e.g. chemistry.meta.stackexchange.com/questions/3435 (let the page load for a minute or so, then click on the "Desktop" link for "direct product tables" - the footnote there has more detail). $\endgroup$
    – orthocresol
    Feb 27, 2019 at 19:45
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    $\begingroup$ @Tom When you take the direct product of two degenerate irreducible representations, the result can subdivided into a symmetric and antisymmetric part. To obtain the symmetric part, you can use the formula from this question. $\endgroup$
    – Tyberius
    Feb 27, 2019 at 19:57

1 Answer 1


In your case you need the equations for the characters as there are symmetric and antisymmetric parts in the product. Your question gives just the symmetric part and the antisymmetric part is $A_2$ normally written as $[A_2]$ to indicate this.

The characters are found using $\chi^+ =\left([\chi(R)]^2+ \chi(R^2) \right)/2$ for the symmetrical part and $\chi^- =\left([\chi(R)]^2- \chi(R^2) \right)/2$ for the antisymmetric part. Using the point group table you need to work out the square of all the operations for $E_g$ this gives $\chi(R^2)$ (e.g. $(C_2)^2 \to E$) and then square the operations which gives $[\chi(R)]^2$, i.e. $[E]^2\to 4$. From the list of characters you then identify the symmetry species, either directly by inspection (as for the antisymmetric in this case), or by forming the irreps from the reducible representation.


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