Character Values for Alternating Groups of degree $\geq 7$

Question

Is it true that in each row and column of the character table of alternating groups with degree $\geq 7$ there are at most two complex values? Any reference will be highly appreciated.

Answer 1

As Geoff thought, the answer is contained in James and Kerber (it's Theorem 2.5.13 in "The Representation Theory of the Symmetric Group", Encyclopedia of Mathematics and its Applications vol. 16, 1981).

Each row or column of the character table contains at most one pair of irrationalities.

If $\lambda$ is a self-conjugate partition (and so corresponds to an irreducible character of $S_n$ that splits on restriction to $A_n$), then the only cycle type of elements where the corresponding character values are irrational is the one with cycle lengths equal to the lengths of the hooks centred on the main diagonal of the Young diagram (notice that these are all odd and distinct, so correspond to a conjugacy class of $S_n$ that splits in $A_n$).

For example, for the partition $(6,6,6,6,6,6)$ this cycle type would be $[11,9,7,5,3,1]$.

Answer 2

Answer amended due to useful comment of Peter Mueller ( so should be unaccepted, as there may be a better one). In a 1994 Journal of Algebra paper, John Thompson and I determined the extension of $\mathbb{Q}$ generated by the values of its irreducible characters. For large enough $n$, this can contain an arbitrarily large number of complex quadratic extensions of $\mathbb{Q}$. However, as Peter Mueller points out, this does not, of itself, answer the question posed here. A key Lemma used in the above paper was a Lemma of James and Kerber ( in Encyclopedia of Mathematics) determines the extension $\mathbb{Q}[\{\chi(g): \chi \in {\rm Irr}(G) \}].$ This is always (at most) a quadratic extension of $\mathbb{Q}$. However, it would take a detailed examination of the proof of that Lemma of James and Kerber to see whether there can be more that two irreducible characters $\chi$ with $\chi(g)$ non-real.

Later edit: Jeremy Rickard explains in his answer exactly what the Lemma of James and Kerber says about the question.

Answer 3

This answer essentially summarizes information from the other answers, hopefully, making the whole picture clear. For each self-conjugate partition $\lambda$ (i.e., $\lambda=\lambda'$) of $n$, the irreducible character $\chi_\lambda$ of $S_n$ is a sum of two irreducible representations of $A_n$: $\chi_\lambda = \chi_\lambda^+ + \chi_\lambda^-$. Also, for each partition with distinct odd parts $\mu=(2m_1+1,\dotsc,2m_d+1)$, where $m_1>\dotsb >m_d$ the conjugacy class $C_\mu$ of elements with cycle type $\mu$ in $S_n$ splits into two $A_n$-conjugacy classes, represented by elements $w_\mu^+$ and $w_\mu^-$.

There is a bijective correspondence $\lambda\mapsto \phi(\lambda)$ from the set of self-conjugate partitions onto the set of partitions with distinct odd parts. The image of $\lambda$ is $\mu=(2m_1+1,\dotsc,2m_d+1)$ if the number of boxes in the Young diagram of $\lambda$ strictly to the right (or strictly below) a box of the form $(i,i)$ is $m_i$.

Theorem. All character values of $A_n$ are real except possibly those of the form $\chi_{\lambda}^\pm(w_\mu^\pm)$, where $\mu=\phi(\lambda)=(2m_1+1,\dotsc,2m_d+1)$. The character value $\chi_{\lambda^\pm}(w_\mu^\pm)$ is not real if and only if $\sum_{i=1}^d m_i$ is odd.

It follows that, in each row and column of the character table there is at most one pair of irrationalities. The only rows with irrationalities are those corresponding to $\lambda^\pm$ where $\lambda$ is a self-conjugate partitions such that the number of off-diagonal boxes in its Young diagram is odd. The only columns with irrationalities are of the classes of elements with cycle decomposition $(2m_1+1,\dotsc,2m_d+1)$, where $m_1>\dotsb > m_d$, and $\sum_{i=1}^d m_d$ is odd.

The only alternating groups $A_n$ with real character table are for the values $n=2,5,6,10,14$.

James and Kerber discuss the characters of alternating groups, but do not give any interesting examples. Frobenius (Georg Ferdinand Frobenius. 1901. ‘Über die Charaktere der alternirenden Gruppe.’ [On the characters of alternating groups.] S’ber Akad. Wiss. Berlin, 303–15) has computations for $A_8$. My book has many examples and a more detailed explanation in Chapters 4 and 5.

Answer 4

As a related question, one can ask how many of the irreducible characters have only real values. It is known that this is the same as the number of real conjugacy classes, where $[x]$ is said to be real iff it is the same as $[x^{-1}]$. Let $G$ be the alternating group on $n$ letters, let $S$ denote the corresponding symmetric group, and let $\epsilon\colon S\to\mathbb{Z}/2$ be the map with kernel $G$. If $x\in G$ then there is an obvious way (starting from a disjoint cycle decomposition) to construct $y\in S$ with $yxy^{-1}=x^{-1}$. If $m_i$ is the number of $i$-cycles in $x$ then it works out that $$ \epsilon(y) = \sum_im_i\lfloor i/2\rfloor = \sum_j (m_{4j+2}+m_{4j+3}). $$ If $\epsilon(y)=0$ then clearly $[x]$ is real. If $\epsilon(y)=1$ then $[x]$ is real iff there is an odd permutation $z$ that commutes with $x$ (so we can replace $y$ by $yz$). I think that this holds iff there is an even $i$ with $m_i>0$, or an odd $i$ with $m_i>1$. Thus, we see that $[x]$ is non-real iff it is a product of cycles of distinct odd lengths, an odd number of which are congruent to $3$ mod $4$. There will be many classes like this when $n$ is large, so there will be many non-real irreducible characters. It seems unlikely to me that each such character has only two non-real values, but I do not see a proof. Anyway, I hope that this analysis may at least shed some light on the original question.