# Doesn't completion of a representation ring preserve its indecomposables?

For $G = PSU(3)$, it is known that $\dim I(G;\mathbb Q) / I(G;\mathbb Q)^2 = 3$, while $H^{**}(BG;\mathbb Q)$ is obviously a power series ring in two indeterminates since $G$ has rank 2. This would would be fine except that there is an elementary argument the natural map should induce a bijection on spaces of indecomposable elements and $3\neq2$.

What goes wrong?

Here's the argument. Let $T$ be compact torus, $R(T)$ its representation ring, and $\widehat R(T)$ the completion with respect to the augmentation ideal $I(T)$. One has maps $$R(T) \longrightarrow \widehat R(T) \overset\sim\longrightarrow K^*(BT) \overset{\textrm{ch}}\longrightarrow H^{**}(BT;\mathbb Q),$$ where the double-star insists we view the cohomology ring as the direct product of the $H^n$. If we tensor $R(T)$ with $\mathbb Q$ beforehand, we get $$R(T;\mathbb Q) \longrightarrow \widehat R(T;\mathbb Q) \longrightarrow H^{**}(BT;\mathbb Q).$$ The last map sends a one-dimensional representation $t \in R(T)$ to $e^u$ for $u = c_1(ET \times_T \mathbb C_t)$, and hence is an isomorphism, as both rings are power series rings on $\dim T$ indeterminates and $\log t$ is a power series in $t-1$. Since $t-1 \mapsto e^u - 1 = u + u^2/2 + \cdots$, the map also sends $\hat I(T;\mathbb Q)$ to the augmentation ideal $H^{\geq 1}(BT;\mathbb Q)$ of $H^{**}(BG;\mathbb Q)$.

Now consider a compact, connected Lie group $G$ with maximal torus $T$. If we make the identification $BT = EG / T$, then $BT$ admits a right action of the Weyl group $W$ of $G$. The maps are equivariant with respect to the action of $w \in W$ since $t w$ is sent to $\exp c_1(ET \times_T \mathbb C_{t w})$ and $ET \times_T \mathbb C_{t w}$ is the pullback of $ET \times_T \mathbb C_{t}$ under $w\colon BT \to BT$. This equivariance and the standard isomorphisms $$R(T)^W = R(G),$$ $$\widehat R(T)^W = \widehat R(G),$$ $$H^{**}(BT;\mathbb Q)^W = H^{**}(BG;\mathbb Q)$$ then show $\widehat R(G;\mathbb Q) \to H^{**}(BG;\mathbb Q)$ is also an isomorphism preserving the augmentation ideal. Then, since $R(G)$ is Noetherian, the maps $$R(G;\mathbb Q) \longrightarrow \widehat R(G;\mathbb Q) \overset\sim\longrightarrow H^{**}(BG;\mathbb Q)$$ induces isomorphisms $$\frac{I(G;\mathbb Q)\phantom{2}}{I(G;\mathbb Q)^2} \overset\sim\longrightarrow \frac{\hat I(G;\mathbb Q)\phantom{2}}{\hat I(G;\mathbb Q)^2} \overset\sim\longrightarrow \frac{H^{\geq 1}(BG;\mathbb Q)\phantom{2}}{H^{\geq 1}(BG;\mathbb Q)^2}$$ of modules of indecomposables.

• Currently there is a call to blacklist the 'completion' tag (see this meta discussion: meta.mathoverflow.net/questions/3227/…), so I removed it. Commented May 2, 2017 at 10:28
• OK. Sorry if I stepped into a beehive. I was just trying to think of relevant tags.
– jdc
Commented May 2, 2017 at 10:43
• You didn't step into a beehive; it was my bad. At the moment the completion (as opposed to completeness) tag is not being considered as problematic. I've rolled back to the prior version. Commented May 3, 2017 at 11:49
• To me everything seems fine, except that I haven't checked dim $I/I^2= 3$, and I would naturally suspect the mistake to be there... are you sure about it? Commented May 4, 2017 at 10:09

In fact $\text{dim }I(G)/I(G)^2=2$. Let $x=[V_{3L_1}]$, $y=[V_{2L_1+L_2}]$ and $z=[V_{3L_1+3L_2}]$. Then \begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[x, y, z]/(y^3-y^2-xz-2y(x+z)-x-y-z).\end{eqnarray}

Note that $\text{dim }V_{3L_1}=\text{dim }V_{3L_1+3L_2}=10$ and $\text{dim }V_{2L_1+L_2}=8$. If $\overline{x}=x-10$, $\overline{y}=y-8$, $\overline{z}=z-10$, then \begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[\overline{x}, \overline{y}, \overline{z}]/(\overline{y}^3+23\overline{y}^2-2(\overline{x}+\overline{z})\overline{y}-\overline{x}\overline{z}-27(\overline{x}+\overline{z})+135\overline{y})).\end{eqnarray} Note that $I(G)=(\overline{x}, \overline{y}, \overline{z})$. It follows that \begin{eqnarray}I(G; \mathbb{Q})/I(G; \mathbb{Q})^2=\text{span}_\mathbb{Q}\{\overline{x}, \overline{y}, \overline{z}\}/\text{span}_\mathbb{Q}\{-27(\overline{x}+\overline{z})+135\overline{y}\}\end{eqnarray} which is of dimension 2.

Remark: If we let \begin{cases}x&=X+1-2Y\\ y&=Y-1\\ z&=Z+1-2Y\end{cases} we will get the more compact description \begin{eqnarray}R(PSU(3))\cong\mathbb{Z}[X, Y, Z]/(Y^3-XZ)\end{eqnarray} (see Lemma 7.1 of this paper).

Added: The following is the SAGE code which verifies the relation $y^3-y^2-xz-2y(x+z)-x-y-z=0$. A2(a, b) means the irreducible representation $V_{(a+b)L_1+bL_2}$.

sage: A2=WeylCharacterRing("A2", style="coroots")
sage: x=A2(3, 0)
sage: y=A2(1, 1)
sage: z=A2(0, 3)
sage: y^3-y^2-x*z-2*y*(x+z)-x-y-z
0