Suzuki exceptional characters with $\epsilon = -1$ I have a question about Suzuki's theory of exceptional characters of finite groups. If you are familiar with this theory, then I'm just asking: can we always choose $\epsilon=1$?
If not, here is a simpler version of the question:
Suppose we have a subgroup $H \leq G$ and a set $Y$ of irreducible characters of $H$ such that whenever $\psi_1$ and $\psi_2$ are differences of characters from $Y$, then the scalar products $(\psi_1,\psi_2)_H = (\psi_1^G,\psi_2^G)_G$ are equal.
For example, $G$ could be a Frobenius group with Frobenius complement $H$, and $Y$ could be all the irreducible characters of $H$ with a fixed degree.
If $|Y| \geq 3$, then we can prove that there is a unique sign $\epsilon \in \{ 1, -1\}$ and unique irreducible characters $\chi_i$ of $G$ so that $\phi_i^G - \phi_j^G = \epsilon( \chi_i -\chi_j)$ for all $\phi_i,\phi_j \in Y$.
(The case $|Y|=1$ is silly since $\phi_i^G-\phi_i^G=0$, and the case $|Y|=2$ loses uniqueness, because you can always switch $\chi_i$ and $\chi_j$ by flipping the sign $\epsilon$; in particular in both of these cases we can choose $\epsilon=1$.)
In other words, to each irreducible character $\phi_i$ of $H$ in $Y$, there is some irreducible character $\chi_i$ of $G$ so that we either always have $\phi_i^G - \phi_j^G = \chi_i - \chi_j$ (this is $\epsilon=1$) or we always have $\phi_i^G - \phi_j^G = \chi_j - \chi_i$ (this is $\epsilon=-1$).
In all the examples I've checked (a somewhat small number admittedly, this is a lot of setup to check), $\epsilon=1$. Does the other case actually occur?
 A: Let $G = \newcommand{\PSL}{\operatorname{PSL}}\PSL(2,8)$. Let $A$ be the non-identity elements of a Sylow 3-subgroup, and let $H=N_G(A)\cong D_{18}$ be its normalizer. Partition $\newcommand{\Irr}{\operatorname{Irr}}\Irr(H)$ into equivalence classes of those irreducible characters that agree outside of $A$. We get three equivalence classes, $\{ 1 \} \subset \hat H$, $\{-1\} \subset \hat H$, and $Y=\{ \phi_i \in \Irr(H) : \deg(\phi_i) = 2 \}$. Similarly we find a set $X=\{ \chi_i \in \Irr(G) : \deg(\chi_i) = 7 \}$ that agrees outside of the $G$-normal closure of $A$.
Since $A$ is a TI-set, Suzuki's theory (or a direct calculation) shows that induction to $G$ is an isometry from the span of the differences of characters in $Y$ onto the span of differences of characters in $X$. However, direct calculation shows that $\epsilon=-1$ here: $\phi_i^G - \phi_j^G = \chi_j - \chi_i$ for every $1 \leq i,j \leq 4$.
Here is the character table of $G$ on the left (with $H\setminus A$ pressed to the left, followed by $A$, followed by $G \setminus H$) and the induced characters in a second copy of the table on the right. You can see how clearly the $\epsilon=-1$ shows up with all the negated character values: $\chi_i(a) = - \phi_i(a) = - \phi_i^G(a)$.


It is somewhat neat that this same group also supports $\epsilon=1$ for the bottom half of the table. Here we take $\hat A$ to be the non-identity elements of a Sylow 7-subgroup, and otherwise do the obvious thing: $\hat H = N_G(\hat A) \cong D_{14}$, $\hat Y = \{ \hat\phi_i \in \Irr(\hat H) : \deg(\hat\phi_i) = 2 \}$ is the only non-trivial collection of irreducible characters of $\hat H$ agreeing outside of $\hat A$, and $\hat X = \{ \hat\chi_i \in \Irr(G) : \deg(\hat \chi_i) = 9 \}$ is the only non-trivial collection of irreducible characters of $G$ agreeing outside the normal closure of $\hat A$. Here the value of $\hat \epsilon=1$ is very clear as the characters simply agree on $\hat A$.
$\PSL(2,2^n)$ is a good group to look at because of its Suzuki nature, but actually this thing happens in $\PSL(2,q)$ for $q \in \{ 13,17,19\}$ as well. My TI-set finder is still finding all TI-sets, so is still prohibitively slow for many groups, and so I'm not sure what other sorts of groups might work. I put a few more groups up at my homepage.
