Shank–Wehlau–Broer conjecture on modular invariant rings

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\title{A Counterexample to the Degree-Bound Conjecture\ for Transfer Quotients of Modular Invariant Rings} \author{} \date{}

\begin{document} \maketitle

\begin{abstract} We give an explicit counterexample to the conjecture that, for a finite group $G$ in characteristic $p>0$, the transfer quotient

$$\Bbbk[V]^G\Big/\sum_{Q<P}\Tr_Q^G\bigl(\Bbbk[V]^Q\bigr),$$

where $P$ is a Sylow $p$-subgroup, is generated in degrees at most $|G|$. Over $\mathbb F_2$, take $G=P=C_2\times C_2$ and

$$V=\bigl(\rad(\mathbb F_2P)^{\oplus 5}\bigr)^*.$$

We prove that the corresponding transfer quotient has a nonzero algebra indecomposable in degree $5$, although $|G|=4$. In fact, the squarefree multidegree $(1,1,1,1,1)$ contains a two-dimensional space of algebra indecomposables. We also exhibit an explicit $30$-monomial invariant representing a nonzero class in this space. \end{abstract}

\section{Statement of the counterexample}

Let $P=C_2\times C_2=\langle a,b\rangle$ and let $\Bbbk=\kk$. For a $\Bbbk P$-module $M$, define its transfer, or Brauer, quotient by

$$\Br_P(M) := M^P\Big/ \sum_{Q<P}\Tr_Q^P(M^Q).$$

For a graded $\Bbbk$-algebra $A$ with $A_0=\Bbbk$, write

$$Q(A):=A_+/(A_+)^2$$

for its space of algebra indecomposables.

\begin{theorem}\label{thm:counterexample} Let

$$\Lambda=\kk P, \qquad U=\rad(\Lambda), \qquad E=U_1\oplus\cdots\oplus U_5$$

with each $U_i\cong U$, and put

$$V=E^*, \qquad R=\kk[V]=\Sym(E).$$

If

$$J=\sum_{Q<P}\Tr_Q^P(R^Q), \qquad \overline R=R^P/J,$$

then

$$\dim_{\kk}Q(\overline R)_{(1,1,1,1,1)}=2.$$

In particular,

$$\beta(\overline R)\ge 5>4=|P|,$$

so the degree-bound conjecture for transfer quotients is false. \end{theorem}

The proof occupies Sections~\ref{sec:module} and~\ref{sec:multigraded}. Section~\ref{sec:explicit} gives an explicit indecomposable invariant.

\section{The three-dimensional module}\label{sec:module}

Write

$$s=a-1, \qquad t=b-1.$$

Since the characteristic is $2$,

$$\Lambda \cong \kk[s,t]/(s^2,t^2).$$

The radical

$$U=\rad(\Lambda)=(s,t)$$

has basis

$$x=st, \qquad y=t, \qquad z=s.$$

Left multiplication gives

$$\begin{array}{lll} a(x)=x, & a(y)=y+x, & a(z)=z,\\[2mm] b(x)=x, & b(y)=y, & b(z)=z+x. \end{array}$$

The key calculation is the following.

\begin{proposition}\label{prop:brauer-tensor} For every integer $n\ge 1$,

$$\dim_{\kk}\Br_P(U^{\otimes n})=\max\{n-3,0\}.$$

Thus

$$\begin{array}{c|ccccc} n&1&2&3&4&5\\ \hline \dim_{\kk}\Br_P(U^{\otimes n})&0&0&0&1&2. \end{array}$$

\end{proposition}

\begin{proof} We divide the proof into four steps.

\medskip \noindent \textbf{Step 1: tensor powers and syzygies.} The augmentation sequence is

$$0\longrightarrow U \longrightarrow \Lambda \xrightarrow{\varepsilon}\kk \longrightarrow 0.$$

Tensoring over $\kk$ with a $P$-module $M$, with diagonal $P$-action, gives

$$0\longrightarrow U\otimes M \longrightarrow \Lambda\otimes M \longrightarrow M \longrightarrow 0.$$

The middle term is free as a $\kk P$-module. Indeed, the map

$$g\otimes m\longmapsto g\otimes g^{-1}m$$

identifies the diagonal action on $\Lambda\otimes M$ with the action on the first tensor factor alone. By Schanuel's lemma,

$$U\otimes M\simeq_{\mathrm{st}}\Omega(M),$$

and hence, by iteration,

$$U^{\otimes n}\simeq_{\mathrm{st}}\Omega^n(\kk).$$

The Brauer quotient of a free module is zero. In fact,

$$\Lambda^P=\kk(1+a+b+ab)=\kk st$$

and

$$st=\Tr_1^P(1).$$

Therefore $\Br_P(\Lambda)=0$, and additivity gives the same conclusion for all finite free modules. Stable isomorphism consequently implies

$$\Br_P(U^{\otimes n})\cong \Br_P(\Omega^n(\kk)).$$

\medskip \noindent \textbf{Step 2: an explicit free resolution.} For $n\ge 0$, let

$$F_n=\Lambda^{n+1}.$$

For $n\ge 1$, define

$$d_n:F_n\longrightarrow F_{n-1}$$

by

$$d_n(c_0,\ldots,c_n) = (sc_0+tc_1,\ sc_1+tc_2,\ \ldots,\ sc_{n-1}+tc_n),$$

and set $d_0=\varepsilon$. Since $s^2=t^2=0$ and $st=ts$, one has $d_{n-1}d_n=0$.

We claim that this complex is exact. Let

$$m=(m_0,\ldots,m_{n-1})\in\ker d_{n-1}.$$

Writing each $m_j$ in the basis $1,s,t,st$ and comparing coefficients shows that there are unique scalars

$$\beta_0,\ldots,\beta_n, \delta_0,\ldots,\delta_{n-1}\in\kk$$

such that \begin{equation}\label{eq:kernel-form} m_j=\beta_js+\beta_{j+1}t+\delta_jst \qquad(0\le j\le n-1). \end{equation} Conversely, every vector of this form is in the image of $d_n$: take

$$c_j=\beta_j+\delta_jt\quad(0\le j<n), \qquad c_n=\beta_n.$$

Then

$$sc_j+tc_{j+1} =\beta_js+\beta_{j+1}t+\delta_jst.$$

Hence the complex is exact and

$$\Omega^n(\kk)=\ker d_{n-1}\subseteq \Lambda^n.$$

\medskip \noindent \textbf{Step 3: fixed points.} For $0\le j\le n-1$, let $w_j\in\Lambda^n$ be the vector whose $j$th coordinate is $st$ and whose other coordinates are zero. From \eqref{eq:kernel-form},

$$sm_j=\beta_{j+1}st, \qquad tm_j=\beta_jst.$$

Thus $m$ is fixed by both $a=1+s$ and $b=1+t$ if and only if every $\beta_j$ is zero. Therefore

$$\Omega^n(\kk)^P =\Span_{\kk}\{w_0,\ldots,w_{n-1}\},$$

so

$$\dim_{\kk}\Omega^n(\kk)^P=n.$$

\medskip \noindent \textbf{Step 4: transfers from the three maximal subgroups.} The maximal subgroups are

$$Q_a=\langle a\rangle, \qquad Q_b=\langle b\rangle, \qquad Q_c=\langle ab\rangle.$$

Transfer from the trivial subgroup factors through every maximal subgroup, so these three subgroups suffice.

For $Q_a$, choose coset representatives $1,b$. Then

$$\Tr_{Q_a}^P=1+b=t, \qquad \Omega^n(\kk)^{Q_a}=\ker s.$$

The condition $sm=0$ is equivalent to

$$\beta_1=\cdots=\beta_n=0,$$

and then $tm=\beta_0w_0$. Hence

$$\Tr_{Q_a}^P\bigl(\Omega^n(\kk)^{Q_a}\bigr)=\kk w_0.$$

Similarly,

$$\Tr_{Q_b}^P=1+a=s, \qquad \Tr_{Q_b}^P\bigl(\Omega^n(\kk)^{Q_b}\bigr)=\kk w_{n-1}.$$

Finally,

$$ab-1=(1+s)(1+t)-1=s+t+st=:u.$$

For $m$ as in \eqref{eq:kernel-form},

$$um_j=(\beta_j+\beta_{j+1})st.$$

Thus $um=0$ exactly when

$$\beta_0=\beta_1=\cdots=\beta_n.$$

Choosing $1,a$ as coset representatives gives

$$\Tr_{Q_c}^P=1+a=s,$$

so

$$\Tr_{Q_c}^P\bigl(\Omega^n(\kk)^{Q_c}\bigr) =\kk(w_0+\cdots+w_{n-1}).$$

The full proper-subgroup transfer subspace is therefore

$$\Span_{\kk} \bigl\{ w_0, w_{n-1}, w_0+\cdots+w_{n-1} \bigr\}.$$

Its dimension is $1$ for $n=1$, $2$ for $n=2$, and $3$ for $n\ge 3$. Since the fixed-point space has dimension $n$, we obtain

$$\dim_{\kk}\Br_P(\Omega^n(\kk))=\max\{n-3,0\}.$$

Together with Step~1, this proves the proposition. \end{proof}

\section{The multigraded transfer quotient}\label{sec:multigraded}

Take five copies $U_1,\ldots,U_5$ of $U$ and write their bases as

$$x_i,y_i,z_i \qquad(1\le i\le 5).$$

Then

$$R=\Sym(U_1\oplus\cdots\oplus U_5) =\kk[x_i,y_i,z_i\mid 1\le i\le 5],$$

with action

$$\begin{aligned} a(x_i)&=x_i,& a(y_i)&=y_i+x_i,& a(z_i)&=z_i,\\ b(x_i)&=x_i,& b(y_i)&=y_i,& b(z_i)&=z_i+x_i. \end{aligned}$$

Give $R$ its $\mathbb N^5$-grading by declaring

$$\deg(x_i)=\deg(y_i)=\deg(z_i)=e_i,$$

where $e_i$ is the $i$th standard basis vector. The $P$-action, fixed-point spaces, and all transfer maps preserve this multigrading.

For a subset $S\subseteq\{1,\ldots,5\}$, let $\one_S\in\{0,1\}^5$ be its indicator vector. The multihomogeneous component of degree $\one_S$ is naturally a $P$-module

$$R_{\one_S} \cong \bigotimes_{i\in S}U_i \cong U^{\otimes |S|}.$$

It follows degree by degree that \begin{equation}\label{eq:component-brauer} \overline R_{\one_S} \cong \Br_P(U^{\otimes |S|}). \end{equation} By Proposition~\ref{prop:brauer-tensor},

$$\overline R_{\one_S}=0 \qquad\text{for }|S|=1\text{ or }2,$$

whereas

$$\dim_{\kk}\overline R_{(1,1,1,1,1)}=2.$$

We now show that every element of this two-dimensional component is an algebra indecomposable. Consider

$$(\overline R_+)^2_{(1,1,1,1,1)}.$$

A product of two positive multihomogeneous elements contributing to this component has multidegrees $\alpha,\gamma\in\mathbb N^5$ satisfying

$$\alpha+\gamma=(1,1,1,1,1).$$

Thus the supports of $\alpha$ and $\gamma$ form a nontrivial partition

$$\{1,\ldots,5\}=S\sqcup S^c.$$

One of the two sets has cardinality at most $2$. By \eqref{eq:component-brauer}, the corresponding factor lies in a zero component of $\overline R$. Hence every such product is zero and

$$(\overline R_+)^2_{(1,1,1,1,1)}=0.$$

Therefore

$$Q(\overline R)_{(1,1,1,1,1)} =\overline R_{(1,1,1,1,1)}$$

has dimension $2$. Its elements have total degree $5$, so

$$\beta(\overline R)\ge 5>4=|P|.$$

This proves Theorem~\ref{thm:counterexample}.

\section{An explicit indecomposable invariant}\label{sec:explicit}

The dimension argument above already disproves the conjecture. We now give an explicit representative of a nonzero degree-$5$ indecomposable class.

For a word $w=w_1w_2w_3w_4w_5$ in the letters $x,y,z$, define

$$m_w=(w_1)_1(w_2)_2(w_3)_3(w_4)_4(w_5)_5,$$

where $(x)_i=x_i$, $(y)_i=y_i$, and $(z)_i=z_i$. Let $\rho$ be cyclic left rotation of five-letter words, and set

$$C(w)=\sum_{r=0}^{4}m_{\rho^r(w)}.$$

Define \begin{equation}\label{eq:explicit-F} \begin{aligned} F={}&C(xxyyz)+C(xxzyy)+C(xyyyz)\ &+C(xyyzy)+C(xyzyy)+C(xzyyy). \end{aligned} \end{equation} This is a sum of $30$ distinct squarefree multihomogeneous monomials of total degree $5$.

\begin{proposition}\label{prop:explicit} The polynomial $F$ lies in $R^P$, its class in $\overline R$ is nonzero, and that class is algebra indecomposable. \end{proposition}

\begin{proof} Put

$$\Delta_a=a-1, \qquad \Delta_b=b-1.$$

A direct expansion using $a(y_i)=y_i+x_i$ gives \begin{align*} &\Delta_a\bigl(C(xxyyz)+C(xxzyy)+C(xyyyz)\bigr)\ &\qquad={} C(xxxxz)+C(xxxzy)+C(xxyyz)+C(xyxyz)+C(xyyxz), \end{align*} and the same expression is obtained from \begin{align*} &\Delta_a\bigl(C(xyyzy)+C(xyzyy)+C(xzyyy)\bigr). \end{align*} The two copies cancel in characteristic $2$, so $\Delta_a(F)=0$.

Similarly, because $b$ replaces each $z_i$ by $z_i+x_i$,

$$\begin{aligned} \Delta_bC(xxyyz)&=\Delta_bC(xxzyy)=C(xxxyy),\\ \Delta_bC(xyyyz)&=\Delta_bC(xzyyy)=C(xxyyy),\\ \Delta_bC(xyyzy)&=\Delta_bC(xyzyy)=C(xyxyy). \end{aligned}$$

These terms cancel in pairs, and therefore $\Delta_b(F)=0$. Hence $F\in R^P$.

It remains to prove that $F$ is not a sum of relative transfers. Work in

$$M=R_{(1,1,1,1,1)}\cong U^{\otimes 5}.$$

For a word $w$, write $[w]$ for the coefficient-extraction functional at $m_w$. Define

$$\begin{aligned} \lambda&=[zyyyx]+[zzzyx],\\ \mu_a&=[zyyxz]+[zzzxz],\\ \mu_b&=[xyyyy]+[zzxyy],\\ \mu_c&=[zxzyy]+[zyxyy]. \end{aligned}$$

Direct evaluation on the word basis gives the identities of linear functionals \begin{equation}\label{eq:functional-identities} \begin{aligned} \lambda\circ(1+b) &=\mu_a\circ(a-1) =[zyyyz]+[zzzyz],\ \lambda\circ(1+a) &=\mu_b\circ(b-1) =\mu_c\circ(ab-1) =[zyyyy]+[zzzyy]. \end{aligned} \end{equation}

Let $Q_a=\langle a\rangle$. If $h\in M^{Q_a}$, then, using coset representatives $1,b$,

$$\begin{aligned} \lambda\bigl(\Tr_{Q_a}^P(h)\bigr) &=\lambda((1+b)h)\\ &=\mu_a((a-1)h)=0. \end{aligned}$$

If $h\in M^{\langle b\rangle}$, then

$$\lambda\bigl(\Tr_{\langle b\rangle}^P(h)\bigr) =\lambda((1+a)h) =\mu_b((b-1)h)=0.$$

Finally, if $h\in M^{\langle ab\rangle}$, choose coset representatives $1,a$ and obtain

$$\lambda\bigl(\Tr_{\langle ab\rangle}^P(h)\bigr) =\lambda((1+a)h) =\mu_c((ab-1)h)=0.$$

Transfer from the trivial subgroup factors through a maximal subgroup, so $\lambda$ vanishes on the full proper-subgroup transfer subspace of $M$.

On the other hand,

$$\lambda(F)=1.$$

Indeed, $zyyyx$ occurs once as a cyclic rotation of $xzyyy$, whereas $zzzyx$ does not occur because every monomial of $F$ contains exactly one letter $z$. Hence $F$ is not a relative transfer, so its class in $\overline R_{(1,1,1,1,1)}$ is nonzero. Section~\ref{sec:multigraded} shows that the decomposable subspace in this multidegree is zero. Thus the class of $F$ is algebra indecomposable. \end{proof}

\section{Conclusion}

For

$$\Bbbk=\mathbb F_2, \qquad G=P=C_2\times C_2, \qquad V=\bigl(\rad(\mathbb F_2G)^{\oplus 5}\bigr)^*,$$

we have

$$0\ne \overline F \in Q\left( \Bbbk[V]^G \Big/ \sum_{Q<G}\Tr_Q^G(\Bbbk[V]^Q) \right)_5.$$

Therefore

$$\boxed{ \beta\left( \Bbbk[V]^G \Big/ \sum_{Q<G}\Tr_Q^G(\Bbbk[V]^Q) \right) \ge 5>4=|G|. }$$

Thus the degree-bound conjecture is false.

\end{document}