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Preliminaries: "Group Embedding" and "Compound Characters"

I'm sorry for not knowing how to say "group embedding" in the mathmatical language, but specifically I want to obtain the character table of $S_4$ from that of $A_4$ by group embedding.

First, let me introduce some general basics of embedding.

I assume there are two groups $G, H$ with $H$ being a subgroup of $G$ ($H \leq G$) and denote irreducible representations of $G$ and $H$ by $\mathscr{R}^{[\alpha]}$ and $\mathscr{T}^{[a]}$, respectively. The matrices which correspond to these reps. are denoted by $\mathcal{M}^{[\alpha]}, \mathcal{N}^{[a]}$ with their characters $\Xi^{[\alpha]}, \chi^{[a]}$.

Since $H \leq G$, if we restrict $\mathcal{M}^{[\alpha]}$ to $H$ the result matrices form the representation of $H$. This representation is not irreducible in general and can be decomposed into irreducible representations of $H$ as \begin{align} {\cal M}^{[\alpha]}(h) = \sum_{a} {f^{\alpha}}_{a} {\cal N}^{[a]}(k). \end{align} Here, ${f^{\alpha}}_{a}$ are called "embedding coefficients" and positive integers. Then our goal is to determine all these embedding coeffcients.

The characters $\Xi_i$ of ${\scr R}$ and $\chi_{i_k}$ of ${\scr T}$ on a conjugacy class of $G: C_i$ are related as \begin{align} \Xi_i^{[\alpha]} = \sum_{a} {f^{\alpha}}_{a} \chi_{i_k}^{[a]} \end{align} with $\Gamma_{i_k}$ being conjugacy classes of $H$ which are gathered into $C_i$. Note that this relation does not depend on the choice of $i_k$ and we assume $C_i$ contains elements of $H$.

In general, for characters of irreps. $\chi$ there are two orhogonal relations \begin{align} \langle \chi^{[\alpha]}, \chi^{[\beta]} \rangle \equiv \frac{1}{|G|} \sum_{i} n_i \chi^{[\alpha]}_i \bar{\chi}^{[\beta]}_i &= \delta^{\alpha \beta}, \\ \sum_{\alpha} \chi^{[\alpha]}_i \bar{\chi}^{[\alpha]}_j &= \frac{|G|}{n_i} \delta_{ij}, \end{align} where the indices $i$ denote conjugacy classes of the group and $n_i$ is the number of elements of the $i$-th class.

From these orthogonalities, we have \begin{align} \sum_{\alpha} f^{\alpha}_{a} \Xi^{[\alpha]}_i = \frac{|G|}{|H|} \sum_{i_k} \frac{n_{i_m}}{n_i} \chi^{[\alpha]}_{i_k}. \end{align} On the other hand, if there is a class $C_v$ which does not contain element of $H$, we can see \begin{align} \sum_{\alpha} f^{\alpha}_{a} \Xi^{[\alpha]}_v = 0. \end{align}

This quantities $\sum_{\alpha} f^{\alpha}_{a} \Xi^{[\alpha]}_i$ are called "compound character" in the reference.

Example: $A_4 \subset S_4$

More concretely, let me consider an example embedding $A_4$ into $S_4$. $S_4$ have five classes $C_{1, 2, 3, 4, 5}$ with # of its elements are $1, 6, 3, 8, 6$ and $\Gamma_1 \subset C_1, \Gamma_4 \subset C_3, \Gamma_{2}, \Gamma_{3} \subset C_4$. Then the compound characters on each class are given by \begin{align} \sum_{\alpha} {f^{\alpha}}_{a} \Xi_1^{[\alpha]} &= 2 \chi_1^{[a]},\\ \sum_{\alpha} {f^{\alpha}}_{a} \Xi_3^{[\alpha]} &= 2 \chi_4^{[a]},\\ \sum_{\alpha} {f^{\alpha}}_{a} \Xi_4^{[\alpha]} &= \chi_2^{[a]} + \chi_3^{[a]},\\ \sum_{\alpha} {f^{\alpha}}_{a} \Xi_{2, 5}^{[\alpha]}&= 0. \end{align} Recall that the indices $a$ runs over the irreducible representations of $A_4$ (${\bf 1}, {\bf 1_1},{{\bf \bar{1}_1}},{\bf 3}$), we can construct the following table from the character table of $A_4$(the following table is cited from the reference).

\begin{array}[ccccccc] ~~ & C_1 & C_2 & C_3 & C_4 & C_5& {\rm Length}& \\\hline {\bf 1} & 2 & 0 & 2 & 2 & 0 & 2 & \\ {\bf 1_1}& 2 & 0 & 2 & -1 & 0 & 1& \\ {\bf \bar{1}_1} & 2 & 0 & 2 & -1 & 0 & 1 & \\ {\bf 3} & 6 & 0 & -2 & 0 & 0 & 2 \end{array}

Here, I evaluated the compound characters $\sum_{\alpha} f^{\alpha}_{a} \Xi^{[\alpha]}_i$ for each class (indices $i$) and each irreps. of $H$ (indices $a$). Since the evaluation can be done for each irrep. of $A_4$ independently, we can list their values as above.

Sorry for writing such a long sentences, but my question concerns the "Length" in the above table. According to the reference, the length of a compound character is the number of irreducible character which is contained in it.

The reference states the fomula giving the length is \begin{align} {\rm Length} = \sum_i \frac{n_i}{n} \left( \sum_{\alpha} f^{\alpha}_a \bar{\Xi}_i^{[\alpha]}\right)^2. ​ \end{align}

For example, please see the first line of the table. The values of $\sum_{\alpha} f^{\alpha}_{{\bf 1}} \Xi^{[\alpha]}_i$ are listed there, and the reference reads since the length is 2, this line consists of $\Xi^{[{\bf 1}]} + \Xi^{[{\bf 1'}]}$ with ${\bf 1}$ being the trivial rep. of $S_4$ and \begin{align} \Xi^{[{\bf 1'}]} = \{ 1, -1, 1, 1, -1\} (= \{2, 0, 2, 2, 0, 2\} - \{1, 1, 1, 1, 1, 1\}). \end{align}

My question

How we determine the length of a compound character? How can we derive the formula of the length? Actually, the reference does not determine the embedding coefficient directly, but find the decomposition at first by using lengths.

My thoughts

I suspect that the formula is a typo of \begin{align} {\rm Length(?)} &= \sum_i \frac{n_i}{n} \left| \sum_{\alpha} f^{\alpha}_a \Xi_i^{[\alpha]}\right|^2\\ &= \sum_i \frac{n_i}{n} \sum_{\alpha, \beta} f^{\alpha}_a f^{\beta}_a \Xi_i^{[\alpha]} \bar{Xi}_i^{[\beta]}\\ &= \sum_{\alpha, \beta} f^{\alpha}_a f^{\beta}_a \langle \Xi^\alpha, \Xi^\beta \rangle \\ &= \sum_{\alpha, \beta} f^{\alpha}_a f^{\beta}_a \delta^{\alpha \beta} = \sum_{\alpha} ({f^{\alpha}}_a)^2, \end{align} because if so this gives us $\sum_{\alpha} ({f^{\alpha}}_a)^2$ for each $a$ by using the orthogonality. From this corrected formula, we can obtain the same lengths as above. However, $\sum_{\alpha} ({f^{\alpha}}_a)^2$ is only a sum of squared multiplicities, not a sum of multiplicities. For the case of length $2$, this is uniquely decomposed into $1^2 + 1^2$, so it is not a problem but for the case of more complicated table, we cannot decompose the sum of squared multiplicities uniquely.

More precisely, my questions are:

  1. Is the above formula is actually typo?
  2. If so, the length is just a sum of squared multiplicities. How can we determine the decomposition completely?
  3. For general cases, should I determine the embedding coefficients at first?

Sorry for very long texts, but I could not find the word "compound character" in mathematical textbooks and I had to introduce it at first.

References

Ramond, Pierre, Group theory. A physicist’s survey., Cambridge: Cambridge University Press (ISBN 978-0-521-89603-0/hbk)(2010). ZBL1205.20001.

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    $\begingroup$ It may be a bit difficult to get answers if you don't make your post a bit shorter? Can you try to summarize it a bit? (For example, I could explain to you how to compute the character table of $S_4$ by inducing characters from $A_4$, and if you're stuck with that, maybe a shorter post can help see what the problem is.) $\endgroup$
    – Pedro Tamaroff
    Jun 5 at 13:59
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    $\begingroup$ One odd notation thing: this question uses $\chi_i$ to mean what I would usually write as $\chi(g_i)$ where $g_i$ is in the $i$th conjugacy class $C_i$. $\endgroup$ Jun 5 at 17:53
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    $\begingroup$ I think the short version of the question is "How do you solve for a,b,c given only a^2 + b^2 + c^2?" which sounds like a terrible question, so Keyspire is asking "Surely there is a typo?" but sadly no, that really is a thing $\endgroup$ Jun 5 at 18:26
  • $\begingroup$ @PedroTamaroff I'm really sorry, but I could not help writing such long sentences because I don't understand details of, the discussion which is not written in the mathematical language. However, I will try to do so, thank you for your kind advice. $\endgroup$
    – Keyspire
    Jun 5 at 19:03
  • $\begingroup$ @JackSchmidt Thank you very much for following my questions! I have some comments, so let me reply to your great answer below. $\endgroup$
    – Keyspire
    Jun 5 at 19:05
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I believe you are asking a detailed question where most of the time we only answer if it is easy to do.

Answer to the first question

Using group theorist's notation so I don't get lost:

$$[\theta,\phi] = \dfrac{1}{|G|} \sum_{g \in G} \theta(g) \overline{\phi(g)}$$

We have the nice relation: if there are non-negative integers $d_\chi$ so that

$$\theta = \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi} \cdot \chi$$

Then we have the nice projection / use dot products to get coefficients:

$$[\theta,\chi] = d_\chi$$

and the standard sort of pythagorean thing

$$[\theta,\theta] = \left[ \theta, \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi} \cdot \chi \right] = \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi} \cdot [ \theta, \chi ] = \sum_{\chi \in \operatorname{Irr}(G)} d_{\chi}^2$$

So, I think this means 1. No, there is no typo in the formula (or at least there is a similar formula with no typo, and it does have squares)

Answer to the second question

So how do you determine the $d_{\chi}$ from the sum of their squares? Well, this is pretty hard in general. It is like finding the angle inside a circle from its radius ... you can't in general. However, we have assumed the $d_{\chi}$ are non-negative integers, and as you have mentioned, often there are not many possibilities.

  • $1 = 1$
  • $2 = 1^2 + 1^2$
  • $3 = 1^2 + 1^2 + 1^2$
  • $4 = 1^2 + 1^2 + 1^2 + 1^2$ or $4 = 2^2$
  • $5 = 1^2$ plus one of the fours

Unless the number is small, you don't have much hope. However, the number is often small, so this trick gets used a lot.

A better version not only computes $[\theta,\theta]$, but also $[\theta,\phi]$ and looks for all the possible $d_\chi$ for $\theta$ and $\phi$ that would create that matrix. GAP has a command for this, OrthogonalEmbeddings and it is super-fun when it works (as in, gives you only a single solution -- usually it gives you many solutions).

Sorry on the third

I'm not sure about your third question. You might find those numbers early or late in the process. I think of them as "decomposition numbers", but that might not be the right word in this case. Another word is "multiplicities of the restriction".

Sometimes that is a good idea. In a recent question I asked, I need 7 positive integers that add up to 7 and that include 7 copies of "1". So I guess that was easy? But the next part needs some positive integers that add up to 14. I don't know how many (maybe I know it is 1, 7, or 14?).

I think the case you are studying is called spin and often the questions you ask are very successfully answered. A more general version is clifford theory and is also very successful. The even more general version is just induced characters. I am not having much luck personally there, but I am hoping my question will be Clifford theory once I figure out the next step.

Vocabulary

compound character is "reducible character"

group embedding is fine, or you can say "subgroup" if the way the two groups are related is clear. "fusion" is the name of how the conjugacy classes between the two groups match up (the way you added Class 2 and Class 3 of A4 to get Class 2 of S4, and the way you got 0 for Class 2 and Class 5 of S4).

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  • $\begingroup$ Then, the above formula is $[\theta, \theta]$, I think. Is it correct? If so, we multiply $\theta$ and its conjugate in the summation. The quoted formula $\sum_i \frac{n_i}{n} (\sum_{\alpha} {f^{\alpha}}_a \bar{\Xi}^{\alpha})^2$ does only contain $\bar{\Xi}$ only, so it is not a standard inner product of characters. If we change the second power to the second power of the absolute value of $\sum_{\alpha} {f^{\alpha}}_a \bar{\Xi}^{\alpha}$, we obtain the inner product as you mentioned. Hence, I still think this is a typo but I am not confident. Is it OK? $\endgroup$
    – Keyspire
    Jun 5 at 19:11
  • $\begingroup$ Sorry for not mentioning, but the $()^2$ is just the square, not an inner product. So $\sum_i \frac{n_i}{n} (\sum_{\alpha} {f^{\alpha}}_a \bar{\Xi}^{\alpha})^2 = \sum_i \frac{n_i}{n} \sum_{\alpha, \beta} {f^{\alpha}}_a {f^{\beta}} _a \bar{\Xi}^{\alpha} \bar{\Xi}^{\beta}$ and sadly it is the multiplication of two $\bar{\Xi}$. I think this cannot be the inner product since in general $\Xi$ is complex. $\endgroup$
    – Keyspire
    Jun 5 at 19:28
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    $\begingroup$ Yes, I think you are right. When you use it in this case, all the numbers are real. But yes, in general I think the length formula in your question is missing a complex conjugate. Probably instead of length, we should call it $\operatorname{Length}^2$, since it is the sum of squares of the "coordinates". $\endgroup$ Jun 5 at 19:59
  • $\begingroup$ I am so relieved to find it is actually a typo. Though I've read through out your great answer and I'd like to comment to other parts of your answer, let me do it several hours later since it is late night in my region. But now, I just want to say thank you. $\endgroup$
    – Keyspire
    Jun 5 at 20:03
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    $\begingroup$ $[\theta,\chi]$ is the coefficient, yes. I think usually this is hard to do very early, because we are trying to find $\chi$ (we don't know the irreducibles yet). But if you know $\theta = \chi^{[a]} + \chi^{[b]}$ (like when $2=1^2+1^2$), then as soon as you know one of the $\chi^{[a]}$ you can find the other $\chi^{[b]} = \theta - \chi^{[a]}$ by subtracting.$\qquad$ $\endgroup$ Jun 6 at 19:30

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