Brauer correspondence for SL2(Fp) in defining characteristic

Counting a nonzero first column and then the $p$ second columns with determinant one gives $\mathrm{\mid }G\mathrm{\mid }=p(p^2-1)$, so $D$ is Sylow of order $p$. Its common fixed line is $kY$. A normalizer must preserve that line, hence is lower triangular. Conversely the lower triangular matrices normalize $D$ by direct conjugation. Thus $N=D⋊T$, where $T=\{\mathrm{diag}(a,a^{-1}):a\in {\mathbf{F}}_p^{\times }\}$. Such a diagonal conjugates $c$ to $a^{-2}c$, so $C_N(D)=D\times ⟨-I⟩$.

Put $n=i-1<p$. On the basis $X^j{Y}^{n-j}$, the operator $u-1$, for $u$ with $c=1$, lowers the highest $X$-degree by one with leading coefficient $j\mathrm{\ne }0$ when $j>0$. Its successive powers on $X^n$ therefore give a triangular basis. It is one Jordan block of size $i$, with kernel $k{Y}^n$. Every nonzero invariant subspace contains a nonzero kernel vector, by applying a maximal nonvanishing power of this nilpotent operator. The upper unipotent acts in the reverse way, and its powers of $h-1$ on $Y^n$ span the whole space. Hence every nonzero $G$-submodule is all of $V_i$: these $p$ modules are simple and have distinct dimensions. This also covers $n=0$.

The $p$-regular matrices in $G$ are exactly the semisimple ones. Indeed finite order prime to $p$ gives a square-free annihilating polynomial; conversely a semisimple matrix has eigenvalues in ${\mathbf{F}}_{p^2}^{\times }$ and hence order prime to $p$. For every trace $t\mathrm{\ne }\pm 2$, the polynomial $Z^2-tZ+1$ has distinct roots, and its companion matrix gives one ${\mathrm{G}\mathrm{L}}_2({\mathbf{F}}_p)$ class. It gives one $G$ class too: the determinant map from its centralizer is onto. In the split case this follows from diagonal matrices. In the nonsplit case the centralizer is ${\mathbf{F}}_{p^2}^{\times }$ acting by multiplication, with determinant $x^2-d{y}^2$ for a nonsquare $d$. For each $c\mathrm{\ne }0$, the sets of squares and $c+d$ times squares, meaning $\{x^2\}$ and $\{c+d{y}^2\}$, each have $(p+1)\mathrm{/}2$ elements and intersect; thus $x^2-d{y}^2=c$ has a solution. Multiplying a conjugator by a centralizer element adjusts its determinant to one. Trace $\pm 2$ gives only the semisimple matrices $\pm I$. There are exactly $p$ regular classes. F4 proves that the list in step 1.2 exhausts all simple modules.

The subspaces spanned by $Y^n,X{Y}^{n-1},\dots ,X^j{Y}^{n-j}$ are $N$-stable and give a full composition flag. Since $u-1$ is a single Jordan block, its invariant subspaces are precisely these: viewing the module as $k[z]\mathrm{/}(z^i)$, submodules are ideals $(z^j)$. Thus restriction to $N$ is indecomposable and has a unique head, the line represented by $X^n$. Define $U_j$ on $N$ by trivial $D$ action and $\mathrm{diag}(a,a^{-1})$ acting by $a^j$. The head is $U_n$, and the factors from head down are $U_n,U_{n-2},\dots ,U_{-n}$. The element $-I$ acts throughout by $(-1{)}^n$.

By F7 every central block idempotent of $kN$ lies in $k{C}_N(D)=kD\otimes k⟨-I⟩$. This commutative algebra is $kD\times kD$, whose only primitive idempotents are $e_{+}=(1+(-I))\mathrm{/}2$ and $e_{-}=(1-(-I))\mathrm{/}2$, by F5. These are central in $kN$, so they are exactly its two block idempotents. Each has nonzero Brauer projection at $D$ (all its support centralizes $D$); since $D$ is Sylow, F8 gives defect $D$ for both.

For either $L=G$ or $N$, every finite $kL$-module is relatively $D$-projective: the relative trace of $[L:D{]}^{-1}\mathrm{i}\mathrm{d}$ is the identity, so F6 applies. Also $kD\cong k[z]\mathrm{/}(z^p)$, with $z=u-1$. Every finite projective module over this algebra is free. To see this, lift a basis of $P\mathrm{/}zP$ to obtain a surjection $A^r\to P$, since its cokernel $C=zC$ vanishes by $z^p=0$. Split this surjection; its kernel $K$ has $K\mathrm{/}zK=0$ by dimensions and thus is zero by the same nilpotence argument. Consequently the single Jordan module of dimension $i<p$ is not projective on $D$, and cannot be projective on $L$, since restriction preserves finite free modules and their summands. F10 and F11 give it vertex $D$, the only alternative inside $D$ being $1$, which would imply projectivity by F6's finite counit. This applies to both $V_i$ and its restriction. For $i=p$ the restriction to $D$ is regular free; the split counit from relative $D$-projectivity makes $V_p$ a summand of a free induced module, hence projective.

F1 gives exactly two global blocks with defect $D$; every positive defect is conjugate to $D$ since its order divides the $p$-part $p$ of $\mathrm{\mid }G\mathrm{\mid }$. For $i<p$, step 3.1 and F2 identify the block of $V_i$ with induction of the local block containing its restriction. Step 2.2 identifies this as $e_{+}$ when $i$ is odd and $e_{-}$ when $i$ is even. Both sets occur, since $1,2<p$ or $p=3$ with these same two indices. Under A1, F12 identifies the indecomposable vertex-$D$ restriction itself as the Green correspondent. AC is used only in this invocation; the trace, sign and block calculations did not use it.

By F9 the projective simple $V_p$ is alone in a defect-zero block. The exhaustive list in step 2.1 and the partition in F3 leave no further blocks: every nonzero finite block has a simple quotient by a proper left ideal of largest dimension. Thus the two positive-defect blocks and this Steinberg block are all the blocks. The endpoint $i=1$ was included in the constant-polynomial calculation, and $i=p$ is precisely the projective exception. The hypothesis $p>2$ ensures the two sign idempotents exist and are distinct; the assertion does not extend this calculation to $p=2$.