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Rank varieties and an exact higher Frobenius functor

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Lesson 6 produced, for a finite symmetric tensor category \(\mathcal C\) with projective generator \(P\), a least rank \(r\) for which the \(kE_r\)-module \(M=\operatorname{Hom}(P,P^{\otimes p^r})\) is not projective, while its restrictions to all proper subgroups are. This lesson chooses the embedding \(\iota\colon E_r\to U(k)\) so that \(R_\iota(\operatorname{St}_{r-1})\otimes M\) is still not projective. By the restricted tilting theorem this module is the restriction of a tilting module, so \(\Theta\) does not kill it, hence does not kill \(M\), and the higher Frobenius functor \(\Phi\) is exact (Theorem 5.2). The choice is made with Carlson's rank varieties, which detect projectivity and turn tensor products into intersections. The support of \(M\) consists of points with coordinates linearly independent over \(\mathbb F_p\), the support of the restricted Steinberg module is cut out by explicit equations in the embedding parameters, and a Moore determinant argument solves those equations.

Prerequisites. Higher Frobenius functors and Norm maps and a first nonprojective tensor power; Rational representations of SLâ‚‚ and tilting modules, Sections 1, 2 and 4.

1. Rank varieties

Let \(E=E_r=(\mathbb Z/p)^r\) with generators \(g_1,\dots,g_r\). For \(\alpha=(\alpha_1,\dots,\alpha_r)\in k^r\) put \[ t_\alpha=\sum_{i=1}^r\alpha_i(g_i-1)\in kE . \] The elements \(g_i-1\) commute and have \(p\)-th power zero, so \(t_\alpha^p=\sum_i\alpha_i^p(g_i-1)^p=0\), and \(t\mapsto t_\alpha\) makes every \(kE\)-module a module over \(k[t]/(t^p)\). A module over \(k[t]/(t^p)\) is a direct sum of Jordan blocks of sizes at most \(p\), and it is free exactly when all blocks have size \(p\).

Definition 1.1. The rank variety of a finite-dimensional \(kE\)-module \(D\) is \[ \begin{aligned} &\operatorname{Supp}(D)=\bigl\{[\alpha]\in\mathbb P^{r-1}(k):\\ &\qquad D\text{ is not free over }k[t]/(t^p)\\ &\qquad\text{via }t\mapsto t_\alpha\bigr\}. \end{aligned} \] Replacing \(\alpha\) by a nonzero multiple replaces \(t_\alpha\) by a multiple, which does not change freeness, so the condition depends only on \([\alpha]\).

Theorem 1.2 (named result N6, part 2 [Ben, Corollary 2.25.8, Theorems 2.25.7 and 2.26.9]). For finite-dimensional \(kE\)-modules \(D,D'\):

  1. (Dade's lemma.) \(\operatorname{Supp}(D)=\varnothing\) if and only if \(D\) is projective.
  2. (Tensor product theorem.) \(\operatorname{Supp}(D\otimes D')\)\(=\operatorname{Supp}(D)\cap\operatorname{Supp}(D')\).

In Benson's notes, (1) is Dade's lemma for the cyclic shifted subgroups generated by \(1+t_\alpha\), and (2) follows from the identification of the rank variety with the cohomological support variety (Theorem 2.25.7) and the tensor product formula for the latter (Theorem 2.26.9(v)). Both statements hold for every prime, including \(p=2\).

The following computation is used twice.

Lemma 1.3. Let \(h(z)\in zk[z]\) and let \(k[z]/(z^p)\) be a module over \(k[t]/(t^p)\) through \(t\mapsto h(z)\). It is free if the coefficient of \(z\) in \(h\) is nonzero, and not free otherwise.

Proof. If \(h(z)=cz+\dots\) with \(c\ne0\), then \(t\mapsto h(z)\) is an isomorphism \(k[t]/(t^p)\to k[z]/(z^p)\), since its inverse can be computed degree by degree, so the module is free of rank one. If \(c=0\), then \(h(z)\in(z^2)\) and \(h(z)^{p-1}\in(z^{2p-2})=0\), as \(2p-2\ge p\); so \(t^{p-1}\) acts by zero, which is impossible on a free module of rank one. \(\square\)

2. The support of the least nonprojective module

Lemma 2.1 ([OpenAI-208, Lemma 6.1]). Let \(M\) be a nonprojective finite-dimensional \(kE\)-module whose restriction to every proper subgroup of \(E\) is projective. Then every point \([\alpha]\) of \(\operatorname{Supp}(M)\) has coordinates \(\alpha_1,\dots,\alpha_r\) that are linearly independent over \(\mathbb F_p\).

Proof. Let \(c\colon E\to\mathbb F_p\) be a nonzero homomorphism, \(c_i=c(g_i)\), and \(H=\ker c\), a proper subgroup. Put \(N=\operatorname{Ind}_H^Ek=k[E/H]\). Choose \(\gamma\in E/H\) with \(c(\gamma)=1\); then \(k[E/H]=k[\gamma]/(\gamma^p-1)=k[z]/(z^p)\) with \(\gamma=1+z\), and \(g_i\) acts by \((1+z)^{c_i}\), with \(c_i\) represented by an integer in \(\{0,\dots,p-1\}\). So \(t_\alpha\) acts by multiplication by \[ h_\alpha(z)=\sum_i\alpha_i\bigl((1+z)^{c_i}-1\bigr), \] whose coefficient of \(z\) is \(\sum_ic_i\alpha_i\). By Lemma 1.3, \[ \operatorname{Supp}(N)=\Bigl\{[\alpha]:\sum_ic_i\alpha_i=0\Bigr\}. \] By the tensor identity (Lesson 5, Section 1), \(M\otimes N\cong\operatorname{Ind}_H^E(\operatorname{Res}^E_HM)\), which is projective because \(\operatorname{Res}^E_HM\) is. By Theorem 1.2, \(\operatorname{Supp}(M)\cap\operatorname{Supp}(N)=\varnothing\): the hyperplane \(\sum_ic_i\alpha_i=0\) misses \(\operatorname{Supp}(M)\). This holds for every nonzero \(c\in\mathbb F_p^r\), which is the claim. \(\square\)

3. The support of a restricted Steinberg module

For \(\lambda=(\lambda_1,\dots,\lambda_r)\in k^r\) let \(\iota_\lambda\colon E\to U(k)\) send \(g_i\) to \(u(\lambda_i)\). It is injective exactly when \(\lambda_1,\dots,\lambda_r\) are linearly independent over \(\mathbb F_p\), since \(\iota_\lambda(\sum_ic_ig_i)=u(\sum_ic_i\lambda_i)\). Write \(R_\lambda\) for restriction along \(\iota_\lambda\).

Lemma 3.1 ([OpenAI-208, Lemma 6.2]). Let \(\lambda_1,\dots,\lambda_r\) be linearly independent over \(\mathbb F_p\). Then \[ \begin{aligned} &\operatorname{Supp}\bigl(R_\lambda(\operatorname{St}_{r-1})\bigr)\\ &\quad=\Bigl\{[\alpha]:\sum_i\alpha_i\lambda_i^{p^j}=0\\ &\qquad\qquad\text{for }0\le j\le r-2\Bigr\}. \end{aligned} \]

Proof. On \(\operatorname{St}_1=\operatorname{Sym}^{p-1}(V)\) let \(N\) be the operator \(e_1^ae_2^b\mapsto b\,e_1^{a+1}e_2^{b-1}\). Then \[ u(a)=\sum_{b=0}^{p-1}\frac{a^bN^b}{b!}=\exp(aN), \] because \(N^b/b!\) sends \(e_1^ie_2^m\) to \(\binom mbe_1^{i+b}e_2^{m-b}\), which is the term of degree \(b\) in \(e_1^i(e_2+ae_1)^m\); here \(m\le p-1\), so \(b!\) is invertible. The operator \(N\) has a single Jordan block of size \(p\), since \(N^{p-1}e_2^{p-1}=(p-1)!\,e_1^{p-1}\ne0\). So \(\operatorname{St}_1\cong k[N]/(N^p)\), and \(t_\alpha\) acts by \[ \begin{aligned} &\sum_i\alpha_i\bigl(\exp(\lambda_iN)-1\bigr)\\ &\quad=\Bigl(\sum_i\alpha_i\lambda_i\Bigr)N\\ &\qquad+(\text{terms in }N^2,N^3,\dots). \end{aligned} \] By Lemma 1.3, \(R_\lambda(\operatorname{St}_1)\) is free exactly when \(\sum_i\alpha_i\lambda_i\ne0\). On the twist \(\operatorname{St}_1^{(j)}\), the element \(u(a)\) acts by \(\exp(a^{p^j}N)\) (Lemma 1.2(3) of Lesson 2), so the same computation replaces \(\lambda_i\) by \(\lambda_i^{p^j}\). By formula (4.1) of Lesson 2, \(\operatorname{St}_{r-1}\cong\bigotimes_{j=0}^{r-2}\operatorname{St}_1^{(j)}\), and Theorem 1.2(2) turns the tensor product into the intersection of the hyperplanes. For \(r=1\), \(\operatorname{St}_0=k\) is not free over \(k[t]/(t^p)\), so its support is all of \(\mathbb P^0(k)\), which agrees with the empty system of equations. \(\square\)

4. Solving the support equations

Lemma 4.1 ([OpenAI-208, Lemma 6.3]). Let \(\alpha_1,\dots,\alpha_r\in k\) be linearly independent over \(\mathbb F_p\). There are \(\lambda_1,\dots,\lambda_r\in k\), linearly independent over \(\mathbb F_p\), with \[ \sum_{i=1}^r\alpha_i\lambda_i^{p^j}=0\qquad(0\le j\le r-2). \]

Proof. For \(r=1\) there are no equations; take \(\lambda_1=1\). Let \(r\ge2\). Since \(k\) is perfect, put \(\beta_i=\alpha_i^{p^{-(r-2)}}\); these are again linearly independent over \(\mathbb F_p\), because \(x\mapsto x^{p^{-(r-2)}}\) is an additive injection commuting with multiplication by elements of \(\mathbb F_p\). Let \(B\) be the \((r-1)\times r\) matrix with entries \(\beta_i^{p^s}\), \(0\le s\le r-2\), \(1\le i\le r\).

The rows of \(B\) are linearly independent. A relation \(\sum_sa_s\beta_i^{p^s}=0\) for all \(i\) gives the additive polynomial \(f(X)=\sum_{s=0}^{r-2}a_sX^{p^s}\), which vanishes on the \(\mathbb F_p\)-span of the \(\beta_i\), a set of \(p^r\) elements. As \(\deg f\le p^{r-2}<p^r\), \(f=0\).

The row space of \(B\) contains no nonzero vector of \(\mathbb F_p^r\). Suppose \(c_i=f(\beta_i)\in\mathbb F_p\) for all \(i\), with \(f\) as above. The polynomial \(f^p-f\) is additive, of degree at most \(p^{r-1}\), and vanishes at each \(\beta_i\) because \(c_i^p=c_i\), hence on their span of \(p^r\) elements. So \(f^p=f\) as polynomials; comparing degrees, \(f\) is constant, and \(f=0\) because it has no constant term. So \(c=0\).

The kernel of \(B\) is a line; let \(\lambda\) span it. If \(c\in\mathbb F_p^r\) satisfies \(\sum_ic_i\lambda_i=0\), then \(c\) lies in the annihilator of \(\ker B\), which is the row space of \(B\), so \(c=0\). Hence the \(\lambda_i\) are linearly independent over \(\mathbb F_p\). The row of \(B\) with \(s=r-2-j\) is \((\alpha_i^{p^{-j}})_i\), and the equation \(\sum_i\alpha_i^{p^{-j}}\lambda_i=0\), raised to the \(p^j\)-th power, gives \(\sum_i\alpha_i\lambda_i^{p^j}=0\). \(\square\)

Proposition 4.2 ([OpenAI-208, Proposition 6.4]). Let \(M\) be a nonprojective finite-dimensional \(kE\)-module whose restriction to every proper subgroup is projective. There is an injective homomorphism \(\iota_\lambda\colon E\to U(k)\) such that \(R_\lambda(\operatorname{St}_{r-1})\otimes M\) is not projective.

Proof. By Theorem 1.2(1) there is a point \([\alpha]\in\operatorname{Supp}(M)\). By Lemma 2.1 its coordinates are linearly independent over \(\mathbb F_p\). Lemma 4.1 gives \(\lambda\) with linearly independent coordinates, so \(\iota_\lambda\) is injective, and by Lemma 3.1 the point \([\alpha]\) lies in \(\operatorname{Supp}(R_\lambda(\operatorname{St}_{r-1}))\). By Theorem 1.2(2), \([\alpha]\in\operatorname{Supp}(R_\lambda(\operatorname{St}_{r-1})\otimes M)\), which is therefore not projective by Theorem 1.2(1). \(\square\)

5. An exact higher Frobenius functor

Theorem 5.1 ([OpenAI-208, Theorem 6.5]). Let \(\mathcal C\) be a finite symmetric tensor category over \(k\). There are \(r\ge1\) and an injective homomorphism \(E_r\to U(k)\) for which the higher Frobenius functor of Lesson 5, \[ \begin{gathered} \Phi\colon\mathcal C\to\mathcal C\boxtimes\operatorname{Ver}_{p^r}(k),\\ \Phi(X)=\Theta_{\mathcal C}(X^{\otimes p^r}), \end{gathered} \] is exact. It is an additive symmetric tensor functor with \(\Phi(af)=a^{p^r}\Phi(f)\), and for every projective \(Q\) \[ \begin{aligned} &(\operatorname{Hom}(Q,-)\boxtimes\mathrm{Id})(\Phi(X))\\ &\quad\cong\Theta\bigl(\operatorname{Hom}(Q,X^{\otimes p^r})\bigr). \end{aligned} \]

Proof. Let \(P\) be a projective generator and \(r\) the least rank of Proposition 4.1 of Lesson 6, so that \(M=\operatorname{Hom}(P,P^{\otimes p^r})\) is not projective over \(kE_r\) and is projective on every proper subgroup. Proposition 4.2 gives an embedding \(\iota_\lambda\) for which \(R_\lambda(\operatorname{St}_{r-1})\otimes M\) is not projective. By the restricted tilting theorem (Theorem 4.1 of Lesson 4) there is a tilting module \(T\) with \(R_\lambda(\operatorname{St}_{r-1})\otimes M\cong R_\lambda(T)\). Lemma 3.1(1) of Lesson 5 gives \(\Theta(R_\lambda(T))\ne0\), and since \(\Theta\) is a tensor functor, \[ \begin{aligned} &\Theta\bigl(R_\lambda(\operatorname{St}_{r-1})\bigr)\otimes\Theta(M)\\ &\quad\cong\Theta\bigl(R_\lambda(T)\bigr)\ne0 . \end{aligned} \] So \(\Theta(M)\ne0\), and Proposition 5.1 of Lesson 5, with \(Q=P\), shows that \(\Phi\) is exact. The remaining properties are Proposition 4.4 of Lesson 5. \(\square\)

All steps include \(p=2\) and \(r=1\).

6. Exercises

6.1. Let \(r=2\) and let \(H=\langle g_1\rangle\). Compute \(\operatorname{Supp}(\operatorname{Ind}_H^Ek)\) directly from Definition 1.1.

6.2. Show that \(\operatorname{Supp}(k)=\mathbb P^{r-1}(k)\) for the trivial module, and deduce from Theorem 1.2(2) that \(\operatorname{Supp}(D\otimes D^*)=\operatorname{Supp}(D)\) when \(\operatorname{Supp}(D^*)=\operatorname{Supp}(D)\).

6.3. Let \(r=2\). Given \(\alpha_1,\alpha_2\) linearly independent over \(\mathbb F_p\), show that \(\lambda=(\alpha_2,-\alpha_1)\) satisfies the conclusion of Lemma 4.1.

6.4. Let \(p=2\) and \(r=2\). Compute the action of \(t_\alpha\) on \(R_\lambda(V)=R_\lambda(\operatorname{St}_1)\) and verify Lemma 3.1 directly.

7. Solutions

6.1. Here \(c(g_1)=0\), \(c(g_2)=1\) defines \(H\), and \(\operatorname{Ind}_H^Ek=k[z]/(z^p)\) with \(g_1\) acting trivially and \(g_2\) by \(1+z\). So \(t_\alpha\) acts by \(\alpha_2z\). By Lemma 1.3 the module is free exactly when \(\alpha_2\ne0\), and \(\operatorname{Supp}=\{[1:0]\}\).

6.2. Each \(g_i-1\) acts by zero on \(k\), so \(t_\alpha\) acts by zero, and a module of dimension \(1<p\) is not free over \(k[t]/(t^p)\). Hence every point lies in \(\operatorname{Supp}(k)\). By Theorem 1.2(2), \(\operatorname{Supp}(D\otimes D^*)\)\(=\operatorname{Supp}(D)\cap\operatorname{Supp}(D^*)\)\(=\operatorname{Supp}(D)\) under the assumption.

6.3. With \(r=2\) there is one equation, \(\alpha_1\lambda_1+\alpha_2\lambda_2=0\), and \(\alpha_1\alpha_2+\alpha_2(-\alpha_1)=0\). If \(c_1\alpha_2-c_2\alpha_1=0\) with \(c_1,c_2\in\mathbb F_p\), the independence of \(\alpha_1,\alpha_2\) gives \(c_1=c_2=0\), so \(\lambda_1,\lambda_2\) are independent.

6.4. On \(V\), \(u(\lambda_i)-1\) is the matrix \(\lambda_ie_{12}\), where \(e_{12}\) sends \(e_2\) to \(e_1\) and \(e_1\) to \(0\). So \(t_\alpha\) acts by \((\alpha_1\lambda_1+\alpha_2\lambda_2)e_{12}\). The module \(V\) of dimension \(2=p\) is free over \(k[t]/(t^2)\) exactly when this is nonzero. So \(\operatorname{Supp}(R_\lambda(V))\)\(=\{[\alpha]:\alpha_1\lambda_1+\alpha_2\lambda_2=0\}\), as Lemma 3.1 states for \(r=2\).

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