The odd-slice theorem for the rank-one supertannakian carrier

Date: 6 September 2026

Mode: ATTACK

Status: exact restricted theorem and a global implication; precursor to the later Kac–Grassmann candidate resolution, not by itself a resolution of Coulembier–Sherman Question 5.16

External review: none

Priority status: unassessed

1. Result in one line

Let $A$ be a finite-dimensional associative superalgebra in the minimal rank-one representation/distribution category generated by an odd operator $Q$, with

$$Z=Q^2,\qquad [Q,Z]=0.$$

If $A$ has no nonzero nilpotent $Q$-stable two-sided ideal, then $A$ is a product of simple algebra objects and is absolutely flat in the rank-one category.

The proof does not classify algebra presentations. It uses the ordinary Jacobson radical $J$ as one intrinsic object and shows first that

$$\boxed{J^2=0.}$$

Every non-semisimple ordinary factor is then forced into the exact odd-slice normal form

$$A_{\rm sl}\cong B\widehat\otimes\Lambda(\theta), \qquad Q\theta=1, \qquad Qb=\theta D(b),$$

where $B$ is a finite-dimensional semisimple superalgebra, $D$ is an even derivation of $B$, and the hat records the super tensor product. The identity $Q\theta=1$ contracts every equivariant module in the exterior direction and forces flatness.

There is also a separate global simplification: in every characteristic-zero supertannakian category, internal $\underline{\operatorname{Ext}}^1$-vanishing for all finitely presented modules already forces the internal Jacobson radical to vanish. Thus only the implication

$$\text{internal semisimplicity}\Longrightarrow\text{absolute flatness}$$

remains to be transferred from rank one to a general supergroup.

2. Exact setting

Work over an algebraically closed field $k$ of characteristic zero. Let $A=A_{\bar0}\oplus A_{\bar1}$ be a finite-dimensional unital associative superalgebra and let $Q$ be an odd superderivation:

$$Q(ab)=Q(a)b+(-1)^{|a|}aQ(b).$$

Assume that $Z=Q^2$ is the even infinitesimal action associated with the rank-one relation $q^2=z$. In the integrated case, subobjects are rational subrepresentations of the connected rank-one supergroup. In the distribution formulation, subobjects are the graded subspaces stable under $Q$. The argument below only needs the following two facts:

1. a $Q$-stable subspace is stable under $Z=Q^2$; and
2. the ordinary Jacobson radical is $Z$-stable.

The second fact is automatic. An algebraic even action preserves the Jacobson radical. Equivalently, in characteristic zero every derivation of a finite-dimensional algebra preserves its Jacobson radical. The radical is graded because the parity automorphism preserves it.

For this packet, internally semisimple means that there is no nonzero nilpotent graded two-sided ideal stable under the rank-one action. This agrees with the internal Jacobson-radical condition for finite algebra objects by Coulembier–Sherman, Lemma 3.11.

3. Radical-collapse lemma

Let $J=\operatorname{Jac}(A)$ be the ordinary Jacobson radical.

Lemma 3.1

For every $p\ge1$,

$$Q(J^p)\subseteq J^{p-1}.$$

Proof. Apply the super Leibniz rule to a product of $p$ elements of $J$. Every summand retains $p-1$ radical factors, hence belongs to the two-sided ideal $J^{p-1}$. $\square$

This gives an odd degree $-1$ derivation on the associated graded algebra,

$$d:\operatorname{gr}_J^pA\longrightarrow \operatorname{gr}_J^{p-1}A.$$

Because $Q^2=Z$ preserves every $J^p$, its degree $-2$ symbol vanishes:

$$d^2=0.$$

Thus the first radical symbol always lies on the square-zero side, even when $Q^2\ne0$ on $A$. The important point is that internal semisimplicity forces a much stronger collapse.

Theorem 3.2 (radical collapse)

If $A$ has no nonzero nilpotent $Q$-stable two-sided ideal, then

$$J^2=0.$$

Proof. Suppose $J^2\ne0$, and form

$$I=J^2+Q(J^2).$$

The Leibniz rule shows that $I$ is a two-sided ideal. For example, for homogeneous $a\in A$ and $x\in J^2$, both $aQx$ and $Qx a$ can be rewritten as an element of $Q(J^2)$ plus an element of $J^2$. Moreover,

$$Q(I)\subseteq Q(J^2)+Q^2(J^2) \subseteq Q(J^2)+J^2=I.$$

Lemma 3.1 gives $Q(J^2)\subseteq J$, so $0\ne I\subseteq J$. Hence $I$ is a nonzero nilpotent invariant ideal, a contradiction. $\square$

This argument is uniform: it does not depend on the nilpotence order of $Q$, the Jordan type of $Z$, or the dimension of $A$.

4. The canonical radical-to-head map

Set $S=A/J$. Since $J^2=0$, define

$$d_1:J\longrightarrow S, \qquad d_1(j)=Qj\pmod J.$$

For homogeneous $a\in A$ and $j\in J$,

$$d_1(aj)=(-1)^{|a|}\bar a\,d_1(j), \qquad d_1(ja)=d_1(j)\bar a.$$

Thus $T=\operatorname{im}d_1$ is a graded two-sided ideal of the semisimple algebra $S$.

Lemma 4.1

The map $d_1$ is injective.

Proof. Its kernel is

$$K=\{j\in J:Qj\in J\}.$$

The two displayed bimodule identities show that $K$ is a two-sided ideal. If $j\in K$, then $Qj\in J$ and $Q(Qj)=Zj\in J$, so $Qj\in K$. Hence $K$ is invariant. Since $K\subseteq J$ and $J^2=0$, it is nilpotent. Internal semisimplicity forces $K=0$. $\square$

The image $T$ measures exactly the part of the ordinary semisimple head paired with the radical by the odd action.

5. Canonical product split

By the graded Wedderburn–Malcev theorem, choose a semisimple graded subalgebra $B\subset A$ mapping isomorphically to $S$, so $A=B\oplus J$ as a vector space. Decompose

$$S=T\times U$$

and write $B_T,B_U$ for the corresponding factors of $B$.

If $u\in B_U$ and $j\in J$, then the bimodule identities give $d_1(uj)=d_1(ju)=0$. Injectivity of $d_1$ yields

$$B_UJ=JB_U=0.$$

Consequently

$$A=(B_T\oplus J)\times B_U.$$

The first factor equals $J+QJ$, so it is $Q$-stable. Its annihilator $B_U$ is also $Q$-stable. We call the factors $A_{\rm sl}$ and $A_{\rm ss}$, respectively. The ordinary radical of $A_{\rm ss}$ is zero. It remains to understand $A_{\rm sl}$, where

$$d_1:J\xrightarrow{\sim}B_T.$$

6. Exact odd slice

Let $e$ be the identity of $A_{\rm sl}$, and let $\theta=d_1^{-1}(e)$. It is odd. Write

$$Q\theta=e+r, \qquad r\in J.$$

Since $A_{\rm sl}$ is invariant, $Qe=0$. Since $Z$ preserves $J$,

$$Qr=Q^2\theta=Z\theta\in J.$$

Therefore $d_1(r)=0$, and injectivity gives $r=0$. We have proved

$$\boxed{Q\theta=e.}$$

Also $\theta^2=0$ because $J^2=0$. For homogeneous $a\in A_{\rm sl}$, the two bimodule identities and injectivity give

$$\theta a=(-1)^{|a|}a\theta.$$

Thus $\theta$ is supercentral and $J=\theta B_T$.

The initially chosen semisimple complement need not be preserved by $Q$. Write, for $b\in B_T$,

$$Qb=\alpha(b)+\theta\beta(b),$$

where $\alpha$ is an odd derivation and $\beta$ is an even derivation of $B_T$. The graph change

$$\phi(b)=b-\theta\alpha(b)$$

is a graded algebra embedding because $\theta^2=0$ and $\theta$ is supercentral. Its image is another semisimple complement, and direct calculation gives

$$Q\phi(b)=\theta\phi\bigl((\beta+\alpha^2)b\bigr).$$

Renaming $\phi(B_T)$ as $B$ and $D=\beta+\alpha^2$, we obtain the promised normal form

$$A_{\rm sl}\cong B\widehat\otimes\Lambda(\theta), \qquad Q\theta=1, \qquad Qb=\theta D(b), \qquad Q^2b=D(b).$$

This is the exact structural translation sought by the HUNT. The nonzero odd square is retained in the even derivation $D$; it is not discarded by passing to the square-zero associated graded.

7. Flatness from the slice

Let $M$ be any equivariant left module over a slice factor and let $Q_M$ be its odd action. Write $L_\theta$ for multiplication by $\theta$. The module Leibniz rule and $Q\theta=1$ give

$$Q_ML_\theta+L_\theta Q_M=1_M, \qquad L_\theta^2=0.$$

It follows immediately that

$$\ker L_\theta=\operatorname{im}L_\theta.$$

Choose a vector-space complement to $\ker L_\theta$. Multiplication by $\theta$ identifies that complement with the kernel, so $M$ is free as a module over $\Lambda(\theta)$. Since $B$ is a finite-dimensional semisimple, hence separable, superalgebra, a $B\widehat\otimes\Lambda(\theta)$-module that is projective over $\Lambda(\theta)$ is projective over the full algebra: the separability idempotent splits the induction/action map.

Therefore every equivariant module over a slice factor is flat. Modules over the ordinary semisimple factors are already flat after forgetting symmetry. The exact faithful fiber functor preserves relative tensor products and reflects exactness, so the product algebra $A$ is absolutely flat internally.

8. Product of internal simple factors

Decompose $B$ into its graded-simple factors. Every derivation kills the central idempotents, so $D$ preserves those factors. If $e_i$ is one such idempotent, then

$$\theta_i=e_i\theta, \qquad Q\theta_i=e_i.$$

The corresponding factor $B_i\widehat\otimes\Lambda(\theta_i)$ has only three graded two-sided ideals: $0$, $\theta_iB_i$, and the whole algebra. The middle ideal is not invariant because $Q(\theta_iB_i)$ projects onto $B_i$. Hence every slice factor is simple as an internal algebra. The ordinary semisimple factor similarly decomposes into invariant simple factors.

We have proved:

Rank-one odd-slice theorem

For a finite algebra object in the connected rank-one category generated by $q$ with $q^2=z$, internal semisimplicity implies both absolute flatness and decomposition as a product of simple algebra objects. Together with the general implications of Coulembier–Sherman Theorem 3.17, these conditions are equivalent in the rank-one category once the Ext implication below is included.

9. A global Ext implication

The following argument does not use rank one.

Proposition 9.1

Let $\mathscr C$ be a characteristic-zero supertannakian category with an exact faithful tensor functor $\omega:\mathscr C\to\operatorname{sVec}$, and let $A\in\mathscr C$ be a finite algebra object. If

$$\underline{\operatorname{Ext}}_A^1(M,-)=0$$

for every finitely presented left $A$-module $M$, then $\mathcal J_{\mathscr C}(A)=0$.

Proof. Let $R=\mathcal J_{\mathscr C}(A)$. It is a nilpotent internal ideal. If $R\ne0$, set $M=A/R$. This is a finite, hence finitely presented, internal $A$-module. Apply $\underline{\operatorname{Hom}}_A(M,-)$ to

$$0\longrightarrow R\longrightarrow A\longrightarrow M\longrightarrow0.$$

The assumed Ext vanishing makes

$$\underline{\operatorname{Hom}}_A(M,A) \longrightarrow \underline{\operatorname{Hom}}_A(M,M)$$

an epimorphism. After applying $\omega$, this is a surjective map between the full spaces of underlying $\omega(A)$-linear maps: exact tensor functors preserve the equalizer defining internal $A$-linear Hom. Hence the identity of $\omega(M)$ lifts to an underlying $\omega(A)$-linear splitting $s:\omega(M)\to\omega(A)$.

Such a splitting cannot exist. If $e=s(1+R)$, then $e\equiv1\pmod R$ and $e^2=e$. Because $R$ is nilpotent, $e$ is invertible, so the idempotent $e$ equals $1$. But for $r\in R$, A-linearity gives $0=s(r+R)=re=r$. Thus $R=0$, a contradiction. $\square$

This removes the Ext-to-semisimplicity direction from the unresolved part of Question 5.16. The remaining global problem is the transfer of the rank-one odd-slice theorem to arbitrary supergroups.

10. A genuine nonzero-square control

Let $B=M_2(k)$, let $N=E_{12}$, and set

$$D=\operatorname{ad}N.$$

The derivation $D$ is nonzero and nilpotent. On $A=B\widehat\otimes\Lambda(\theta)$, define

$$Q\theta=1, \qquad Qb=\theta[N,b].$$

Then

$$Q^2|_B=\operatorname{ad}N\ne0, \qquad Q^6=0.$$

The ordinary radical $\theta B$ is nonzero, but it is not invariant because $Q(\theta b)=b$. Since $B$ is simple, $A$ is internally simple. The module contraction above proves internal absolute flatness. This is a positive control in the genuinely non-square-zero unipotent rank-one category, not the previous exterior-translation boundary with $Q^2=0$.

11. Why the global detection gate remains open

Restriction to one odd line does not preserve internal semisimplicity. The smallest exact warning is

$$A=\Lambda(\theta_1,\theta_2), \qquad Q_i=\partial_{\theta_i}.$$

Under the full two-dimensional odd translation action, $A$ is internally simple: repeated contractions take every nonzero invariant ideal to the unit. For any nonzero line $q=aQ_1+bQ_2$, however,

$$\eta=b\theta_1-a\theta_2, \qquad q\eta=0.$$

The proper nilpotent ideal $(\eta)$ is stable under that line. Thus every rank-one restriction has a new radical even though the global algebra is simple and absolutely flat.

This is not a counterexample to Question 5.16, and it does not show that a global failure is invisible on all lines. It does rule out the naive transfer

$$A\text{ globally semisimple} \Longrightarrow A|_{\mathfrak h_q}\text{ semisimple}.$$

The next attack must work at the level of a specific Tor class and separate the radical created by restriction from the global obstruction. Merely checking whether the restricted algebra is semisimple gives false positives on every line in this control.

This warning is consistent with characteristic-zero support theory: standard Duflo–Serganova/associated-variety detectors are organized by square-zero odd directions. They do not supply the required non-square-zero radical transfer theorem.

12. Scope, dependencies, and novelty boundary

• The exact open question, general implication diagram, internal radical theorem, and solved ambient cases are due to Kevin Coulembier and Alexander Sherman, Absolutely flat algebras in tensor categories, especially Theorem 3.17, Lemma 3.11, Theorem 5.18, and Question 5.16.
• The finite tensor-category equivalence used in their partial theorem is due to Coulembier, Stroiński, and Zorman, Simple algebras and exact module categories.
• The interpretation of tangent Lie elements as distributions and invariant derivations is part of the framework developed by Benson and Pevtsova, Group schemes and their Lie algebras over a symmetric tensor category.
• Odd-line reduction is influenced by Duflo–Serganova theory. The standard associated-variety detector uses square-zero odd elements; see Drupieski and Kujawa’s support-theory survey.
• Graded Wedderburn–Malcev splitting, separability of finite-dimensional semisimple algebras in characteristic zero, and the exterior-module contraction criterion are classical.
• Victor Kac’s 1977 differentiably-simple-superalgebra theorem gives the global normal form $S\widehat\otimes\Lambda(V)$. It was not used in the rank-one derivation above, but it is the decisive input in the subsequent candidate resolution. Kac explicitly traces the method to Richard Block’s differentiably-simple-ring work.

The radical-collapse, exact-slice, and global Ext arguments above have not been externally reviewed. A targeted source search did not locate this exact rank-one formulation, but that is not a priority claim. They should be treated as a Lab theorem packet pending expert audit.

13. Superseding global route

The line-by-line detection question above was the next question at the time this packet was written. The subsequent attack found a stronger route that does not restrict a globally semisimple algebra to individual odd lines. Instead it decomposes by minimal invariant ideals, applies Kac’s intrinsic normal form, identifies the Grassmann factor with the canonical supercenter, and invokes the already-established commutative absolute-flatness theorem.

That argument is recorded in FPRD-supertannakian-candidate-resolution-2026-09-06.md. The two-translation example in Section 11 remains a valid warning against the abandoned naive restriction argument.