Brauer Characters and Decomposition Matrices - Examples

If $g\in P$ is $p$-regular, then $\mathrm{\mid }g\mathrm{\mid }$ divides the order of the $p$-group $P$ and is also coprime to $p$, so $\mathrm{\mid }g\mathrm{\mid }=1$. Thus $P^0=\{1\}$.

By [L2], the only simple $kP$-module is the trivial one, so the only irreducible Brauer character is the trivial character. Then [L1] is consistent with step 1.1 because the class-function space on the one-point set $P^0$ is one-dimensional.

For any finite-dimensional $kP$-module $V$, the Brauer character is therefore determined by ${\phi }_V(1)={\dim }_{k}V$.

In $S_3$, the conjugacy classes are represented by $1$, a transposition $(12)$ of order $2$, and a $3$-cycle $(123)$ of order $3$.

By [F1], the $2$-regular elements are those of order $1$ or $3$, so they form the classes of $1$ and $(123)$. Likewise the $3$-regular elements are those of order $1$ or $2$, so they form the classes of $1$ and $(12)$.

Hence a Brauer character of $S_3$ in characteristic $2$ or $3$ is determined by the two values on the corresponding classes, in agreement with [L1].

By [L1], only the values at $1$ and $(123)$ matter. The ordinary trivial and sign characters both restrict to $(1,1)$ on those classes, because a $3$-cycle is even. The ordinary standard character restricts to $(2,-1)$.

In characteristic $2$, the trivial module gives the Brauer character ${\phi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}}=(1,1)$, while the $2$-dimensional simple module gives ${\phi }_{\mathrm{s}\mathrm{t}\mathrm{d}}=(2,-1)$. Therefore [F1] writes the restricted ordinary characters as $1^0={\phi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}},{\mathrm{sgn}}^0={\phi }_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{v}},{\mathrm{std}}^0={\phi }_{\mathrm{s}\mathrm{t}\mathrm{d}}\mathrm{.}$

Reading off the coefficients yields the displayed decomposition matrix.

By [L2], $D^{\mathsf{T}}=\left(\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\end{array}\right)\mathrm{.}$

Multiplying gives $D^{\mathsf{T}}D=\left(\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\mathrm{.}$

Therefore [L1] yields the displayed Cartan matrix.

In the displayed block-ordered matrix, the last diagonal sector is the $1\times 1$ block $(1)$.

That sector therefore corresponds to a block containing exactly one ordinary irreducible character and exactly one irreducible Brauer character, namely the standard row and the $2$-dimensional Brauer column. This contrasts with [L1], where having one Brauer character does not force a unique ordinary one.

So such one-row one-column blocks do occur.

The element $u$ has order $p$, so it is $p$-singular. Its action matrix still has ordinary trace $2$.

By [F1], the Brauer character of this module is not defined at $u$, because $u\mathrm{\notin }{C}_p^0$. So the trace from step 1.1 cannot be a Brauer-character value.

This directly refutes the statement [L1].