# 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. – Todd Trimble May 2 '17 at 10:28
• OK. Sorry if I stepped into a beehive. I was just trying to think of relevant tags. – jdc May 2 '17 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. – Todd Trimble May 3 '17 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? – Pierre May 4 '17 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