Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 138 / 4

An integral form of a group representation $V$ is a $G$-stable finite free $\mathcal{O}$-submodule $W\subset V$ with $K{\otimes }_{\mathcal{O}}W\cong V$. Such a form exists: take a basis lattice $L$ and replace it by ${\sum }_{g\in G}gL$, which is finite and torsion-free and therefore free over the discrete valuation ring $\mathcal{O}$.

Choose ordinary simple modules $V_i$ with forms $W_i$, and modular simple modules $S_j$ with projective covers $P_j$. With $\pi$ a uniformizer, put ${\overline{W}}_i=W_i/\pi W_i$. The decomposition matrix and Cartan matrix of a group algebra have entriesThe first numbers do not depend on the integral form: the Brauer character of its reduction is the ordinary character restricted to p-regular elements, and part 2(a) determines all composition multiplicities from this restriction.

For the assertion in (i), set $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$. It is a submodule of the finite free module ${\mathrm{Hom}}_{\mathcal{O}}(W,W^{\prime })$, hence is finite free over $\mathcal{O}$. There is a natural injective mapFor any homomorphism $f$ in the target, clearing the finitely many denominators of its values on an $\mathcal{O}$-basis of $W$ gives ${\pi }^{N}f(W)\subset W^{\prime }$. Thus ${\pi }^{N}f\in I$, and the map is surjective. Therefore $\boxed{I\text{ is an }\mathcal{O}\text{-form of }{\mathrm{Hom}}_{KG}(M,M^{\prime })}$, regarding this Hom space as a vector space. The original PDF supplies part (i), which is missing from the supplied TeX.

Write $\mathfrak{p}=\pi \mathcal{O}$. Certainly multiplication by $\pi$ sends every $\mathcal{O}G$-homomorphism $W\to W^{\prime }$ to one with image in $\pi W^{\prime }$. Conversely, if $f:W\to \pi W^{\prime }$ is such a homomorphism, define $h(w)={\pi }^{-1}f(w)\in W^{\prime }$. This is well defined because $W^{\prime }$ is torsion-free. Cancellation of $\pi$ shows that $h$ is $\mathcal{O}$-linear and commutes with the $G$-action. Hence $f=\pi h$, andThis step in reduction of Hom from a projective group-algebra lattice does not itself require projectivity.

Put $I={\mathrm{Hom}}_{\mathcal{O}G}(W,W^{\prime })$. Since $W$ is a projective module, its Hom functor is exact. Applied to $0\to \pi W^{\prime }\to W^{\prime }\to W^{\prime }/\pi W^{\prime }\to 0$, it givesThe kernel is $\pi I$ by (ii). Every map to $W^{\prime }/\pi W^{\prime }$ kills $\pi W$ and factors uniquely through $W/\pi W$; this gives the second isomorphism. ThereforeBy (i), $I$ is finite free with rank ${\dim }_K{\mathrm{Hom}}_{KG}(M,M^{\prime })$. Its reduction has dimension equal to that rank. Thus reduction of Hom from a projective group-algebra lattice yieldsThe original PDF has the group-algebra subscripts used here; the supplied TeX drops or corrupts several of them.

Completeness permits idempotent lifting from $kG$ to $\mathcal{O}G$. In particular, each projective cover $P_j$ lifts to a finite projective $\mathcal{O}G$-lattice ${\tilde{P}}_j$; one can lift an idempotent presenting it as a summand of a finite free module. Set $Q_j=K{\otimes }_{\mathcal{O}}{\tilde{P}}_j$.

For a simple $S_{\ell }$, every map $P_j\to S_{\ell }$ factors through its simple head $S_j$. Splitting and Schur lemma give ${\dim }_k{\mathrm{Hom}}_{kG}(P_j,S_{\ell })={\delta }_{j\ell }$. Exactness of this Hom functor along a composition series therefore givesPart (iii) identifies this with ${\dim }_K{\mathrm{Hom}}_{KG}(Q_j,V_i)$. By Maschke's theorem and splitting, $KG$ is split semisimple, soThe reductions of any two integral forms of the same ordinary module have identical Brauer characters, hence identical composition multiplicities. We may thus reduce a direct-sum form for the displayed decomposition instead of ${\tilde{P}}_j$. In the modular representation ring this givesComparing the simple basis coefficients proves $c_{\ell j}={\sum }_i{d}_{i\ell }{d}_{ij}$, or $\boxed{C=D^{T}D}$.