Triviality of SK_0(Lambda) for Lambda an order in a group algebra over a $p$-adic field

Question

CR refers to Methods of Representation Theory by Charles Curtis and Irving Reiner.

Let $F$ be a finite extension of $\mathbb{Q}_p$ with valuation ring $\mathcal{O}_F$. Let $G$ be a finite group and let $A$ be the group algebra $F[G]$. Let $\Lambda$ be an $\mathcal{O}_F$-order in $A$. Let $K_0(\Lambda)$ (resp. $K_0(A)$) denote the Grothendieck group of the category of finitely generated projective left $\Lambda$-modules (resp. $A$-modules).

Let $SK_0(\Lambda)=\ker \varphi$ where $\varphi: K_0(\Lambda) \longrightarrow K_0(A)$ is the map given by $\varphi([M])=[F \otimes_{\mathcal{O}_F} M]$ (this definition is taken from CR vol 2, top of page 222.)

Is $SK_0(\Lambda)$ always trivial? I know that this is true in the following cases:

• $\Lambda=\mathcal{O}_{F}[G]$ - see CR vol 1, Theorem 32.1
• more generally, when $\Lambda$ satisfies the conditions of CR vol 1, Theorem 32.5
• $\Lambda$ is a maximal order - see CR vol 1, Theorem 26.24iii
• $\Lambda$ is commutative - see CR vol 1, Proposition 35.7

Can anyone provide (a reference to) a proof of the general case or counterexample?

Answer 1

No, the kernel of the map $\varphi: K_0(\Lambda)\longrightarrow K_0(A)$ is not necessarily trivial for all orders $\Lambda$ in $A$. In fact, it is only trivial for all orders in $A$ if $A$ is a direct sum of fields and skew fields.

Assume the Wedderburn decomposition of $A$ has the matrix algebra $M_n(D)$ as a simple component (with $n > 1$ and $D$ a skew field). Let $\Delta$ be the maximal order in $D$ and let $\Pi$ be its maximal ideal. Then define an order $$ \Lambda_1 := \left[ \begin{array}{ccccc} \Delta & \Pi & \Pi& \cdots & \Pi \newline \Delta & \Delta & \Pi & \cdots & \Pi \newline \vdots & \vdots & \ddots & & \vdots \newline \Delta &\Delta &\Delta &\cdots & \Delta \end{array}\right] \subset M_n(D) $$ Note that $\Lambda_1$ has $n$ non-isomorphic projective lattices (corresponding to the rows or, if you prefer left modules, the columns ). All of these become isomorphic to $D^{n}$ upon tensoring with the ground field. Thus the map $K_0(\Lambda_1)\cong \mathbb Z^n\longrightarrow K_0(A)\cong \mathbb Z$ has non-trivial kernel. Now if $A$ decomposes as $A \cong M_n(D)\oplus B$, then just choose an any order $\Lambda_2 \subset B$ and you get a counterexample $\Lambda=\Lambda_1\oplus \Lambda_2 \subset A$.

Note that the choice of $\Lambda_1$ above is arbitrary. Any order in $M_n(D)$ with at least two non-isomorphic projective modules would do the trick.