0
$\begingroup$

First I would like to mention that I do not know that should I post this question here or in the math community, but since my background is in physics and this kind of question is usually asked by people in the physics community I decided to post it here.

My question is how to construct terms (to be used in Lagrangian or free energy, for example) which are invariant under the effect of an arbitrary group(not the rotation group necessarily)?

To be more specific, consider a group $G$ with the trivial irreducible representation $\Gamma_0$ and another irreducible representation $\Gamma$ where $\text{dim}(\Gamma)\ge 2$ with basis vectors $\{|\Gamma,e_i\rangle\}$. Therefore an arbitrary vector $|A\rangle$ in $\Gamma$ can be written as: $$ |A\rangle = \sum_i A_i |\Gamma,e_i \rangle$$ Then we know that under the effect of an element of $G$ which we call $g$ the components of $|A\rangle$ transform as: $$ A'_i=\sum_j D^{\Gamma}_{ij}(g) A_j $$ Now I want to build terms with arbitrary orders of $A_i$ like the quadratic terms $\sum_{ij} \lambda_{ij}A_iA_j$ or the quartic terms $\sum_{ijkl} \beta_{ijkl}A_iA_jA_kA_l$ which are invariant under the action of all the elements of $G$. There seems to be a way by using the tensor product representations $\Gamma \otimes \Gamma$ or $\Gamma \otimes \Gamma \otimes \Gamma \otimes \Gamma$ if decomposing these (usually) reducible representations to the direct sum of the irreducible representations of $G$, for example: $$ \Gamma\otimes \Gamma = \sum_i \oplus \Gamma_i$$ We can use the character table of $G$ to find which irreducible reps (and each one how many times) are included in the direct sum above. So we can check whether $\Gamma_0$ is included in the direct sum or not. Now for example if we want to construct a quadratic invariant form from $|A\rangle$, we construct the tensor product: $$|A\rangle \otimes |A\rangle = \sum_{ij} A_i A_j (|\Gamma,e_i\rangle \otimes |\Gamma,e_j \rangle)$$ and try to write the basis vectors $|\Gamma,e_i\rangle \otimes |\Gamma,e_j \rangle$ of the tensor product space in terms of the basis vectors of the direct product space and then the coefficient of $|\Gamma_0,e'\rangle$ will give us the desired invariant form which will be quadratic in $A_i$ for sure. Basically, this is exactly like the addition of angular momenta and finding the Clebsch-Gordon coefficients but for a general case. But I cannot find any general method for finding these coefficients for arbitrary $G$.

Especially I need to find such terms for point groups. Obviously, any rotational invariant term will be invariant under $G$ in this case. But there can be other terms which are not invariant under every rotation whilst are $G$ invariant.

$\endgroup$
  • 1
    $\begingroup$ You just gave a general method -- character tables. There are also faster methods for specific groups. What else do you want to know, specifically? $\endgroup$ – knzhou May 15 at 11:31
1
$\begingroup$

The book "Group Theory: Birdtracks, Lie's, and Exceptional Groups" by Predrag Cvitanović is built around the construction of invariants. It is quite different from the usual approaches and is probably exactly what you want.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.