Modular Traces and Brauer-Character Independence: Examples

The eigenspace decomposition for g shows any simple module is one-dimensional with one of the m distinct eigenvalues; each scalar action $g\mapsto {\lambda }^i$ is a representation since $({\lambda }^i{)}^m=1$. Its value at $g^j$ is $\widehat{{\lambda }^{ij}}={\zeta }^{ij}$. Distinct eigenvalues give nonisomorphic modules.

The determinant is ${\prod }_{0\le i<l<m}({\zeta }^l-{\zeta }^i)$. To obtain the formula, regard the determinant of $(x_i^j)$ as a polynomial in the $x_i$: it vanishes when two $x_i$ coincide, has total degree $m(m-1)\mathrm{/}2$, and the coefficient of $x_1{x}_2^2\cdots x_{m-1}^{m-1}$ is 1, as is that of the displayed product. Hence they agree. Its factors are nonzero and their reductions ${\lambda }^l-{\lambda }^i$ are nonzero, so the determinant is a unit in $\mathcal{O}$ and is nonzero in both K and k. For $m=1$ the table is $(1)$ and the product is empty, equal to 1.

Set $r=\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ and $t=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. Direct multiplication gives $r^2=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, $r^3=t^2=I$, and $trt=r^2$. The six matrices $I,r,r^2,t,rt,r^{2}t$ are distinct. Acting on the three nonzero vectors of ${\mathbb{F}}_2^2$ embeds this group in $S_3$; its size six identifies it with $S_3$.

The polynomial of r is $X^2+X+1$, with distinct nontrivial cube roots $\lambda ,{\lambda }^2$ in k. An invariant line for r must be one of its eigenlines. From $rt=t{r}^{-1}$, t sends the $\lambda$-eigenline to the ${\lambda }^{-1}$-eigenline, which is different. Hence no line is invariant under both; the two-dimensional representation is simple, and the same argument holds over every extension field.

Writing $\zeta =\widehat{\lambda }$, the identity has value 2 and r has value $\zeta +{\zeta }^2=-1$, since ${\zeta }^3=1$ and $\zeta \mathrm{\ne }1$. The trivial representation has values 1,1. Thus the determinant is $-1-2=-3$, nonzero in K and with residue 1 in k.

The p-regular permutations are the identity and the two conjugate 3-cycles, hence give exactly two classes. By independence at most two nonisomorphic simple modules can have characters on this two-dimensional space of class functions. The two already exhibited are nonisomorphic by dimension and simple, so exhaust the possibilities.

Take the rank-one lattices $L_{+}=\mathcal{O}$ and $L_{-}=\mathcal{O}$ with t acting by 1 and -1. These are representations since both scalars square to 1. Their reductions are both the trivial k-module because $-1=1$ in characteristic two. Their K-character values at t are 1 and -1, different because K has characteristic zero. The element t has order 2 and is p-singular, so this does not conflict with the domain of the Brauer character.

On $k^2$ take $J=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. Then $J^2=I$ in characteristic two, $J\mathrm{\ne }I$, and $\mathrm{tr}(J)=0=\mathrm{tr}(I)$. The p-regular part of t is 1. Thus the trace equality holds exactly as stated in the trace lemma while equality of operators fails. Both proposed conclusions have explicit counterexamples.