A Kac–Grassmann reduction for supertannakian semisimplicity

Date: 6 September 2026

Mode: ATTACK

Status: superseded candidate packet; the publication audit repaired Lemma 4.1 and clarified the separability transfer. Use FPRD-supertannakian-publication-audit-2026-09-06.md and the final paper for the authoritative proof.

External review: none

Priority status: unassessed

1. Candidate theorem

Let $\mathscr C$ be a supertannakian category over an algebraically closed field $k$ of characteristic zero, and let $A\in\mathscr C$ be a finite associative algebra object, not necessarily commutative.

Then the six conditions in Coulembier–Sherman, Theorem 3.17, are equivalent:

1. $A$ is absolutely flat;
2. $\underline{\operatorname{Ext}}^1_A(M,-)=0$ for every finitely presented left $A$-module $M$;
3. $A$ is a finite product of simple algebra objects;
4. $A$ is semisimple in the internal Jacobson-radical sense;
5. $A$ has no nonzero nilpotent two-sided ideal object; and
6. the category of finitely presented $A$-bimodules is rigid.

The new implication is

$$(5)\Longrightarrow(1).$$

There is also a short direct proof of $(2)\Rightarrow(5)$. Together with Coulembier–Sherman’s established implications, these two arrows close their diagram.

The proof is structural rather than classificatory. After passage to the connected supergroup, a minimal invariant ideal is differentiably simple. Kac’s 1977 theorem identifies its underlying superalgebra as

$$S\widehat\otimes\Lambda(V),$$

where $S$ is simple and $\Lambda(V)$ is a Grassmann algebra. The Grassmann factor is intrinsic: it is the supercenter. It is a simple commutative algebra object, so Coulembier–Sherman’s commutative theorem makes it absolutely flat. Separability of the simple factor $S$ then lifts flatness to the whole algebra.

2. Reduction to an algebraic supergroup

By characteristic-zero supertannakian duality, and by the same finite-generation reduction used in Coulembier–Sherman, Section 5.5, it suffices to work in

$$\mathscr C=\operatorname{Rep}^f G$$

for an algebraic supergroup $G$. Write $H=G^\circ$. Its component group $G/H$ is finite etale in characteristic zero.

The action of $G$ on $A$ is by algebra automorphisms. Infinitesimally, $\mathfrak h=\operatorname{Lie}H$ acts by superderivations:

$$x(ab)=x(a)b+(-1)^{|x||a|}a\,x(b).$$

For a connected algebraic supergroup in characteristic zero, a graded subspace of a rational module is $H$-stable exactly when it is stable under the distribution action; on finite-dimensional modules this is generated by the Lie-superalgebra action together with the connected even group. In particular, an ideal stable under every superderivation is $H$-stable.

3. Passing from $G$ to the connected component

Assume condition (5): $A$ has no nonzero nilpotent $G$-stable two-sided ideal.

Lemma 3.1

There is no nonzero nilpotent $H$-stable two-sided ideal in $A$.

Proof. Let $I$ be such an ideal. Every nilpotent ideal lies in the ordinary Jacobson radical $J(A)$. Choose representatives for the finite component group $G/H$, and form

$$I_G=\sum_{g\in G/H}gI.$$

The ordinary Jacobson radical is characteristic, so every summand lies in $J(A)$. Hence $I_G\subseteq J(A)$ is nilpotent. It is nonzero and $G$-stable, contradicting (5). $\square$

This small step is important: no assertion that a sum of arbitrary nilpotent ideals is nilpotent is being used. All summands lie in the one nilpotent ideal $J(A)$.

4. Minimal invariant ideals and Kac’s theorem

Choose a minimal nonzero $H$-stable graded two-sided ideal $I\subseteq A$. Its square is again $H$-stable. Minimality gives

$$I^2=0\quad\text{or}\quad I^2=I.$$

Lemma 3.1 excludes the first case, so $I^2=I$.

Lemma 4.1

The superalgebra $I$ is differentiably simple in Kac’s sense.

Proof. Let $0\ne J\triangleleft I$ be invariant under every superderivation of $I$. In particular, it is invariant under the restrictions of $\operatorname{Lie}H$, so connectedness and rationality make it $H$-stable. The candidate version originally invoked minimality at this point, but $J$ need not be an ideal of $A$. Instead form the ambient saturation $AJA$. It is a nonzero $H$-stable two-sided ideal of $A$ contained in $I$, so minimality gives $AJA=I$. Since $IA=AI=I$ and $I^2=I$,

$$I=I^3=I(AJA)I=IJI\subseteq J.$$

Thus $J=I$. Since $I^2\ne0$, this is exactly differentiable simplicity. $\square$

Kac’s Proposition 5.1.1 now applies:

$$I\cong S\widehat\otimes\Lambda(V),$$

where $S$ is a finite-dimensional simple associative superalgebra and $\Lambda(V)$ is a finite Grassmann superalgebra. The isomorphism need not be $H$-equivariant; only its algebraic consequences are used.

In particular, $I$ has an identity $e$. Since $I$ is a two-sided ideal of $A$, this identity is a central idempotent of $A$. Indeed, for $a\in A$, both $ae$ and $ea$ lie in $I$, and

$$ae=eae=ea.$$

The identity is unique, so the $H$-stability of $I$ implies that $e$ is $H$-fixed. Therefore

$$A=I\times(1-e)A$$

as $H$-algebras. Induction on dimension gives an algebra product

$$A=A_1\times\cdots\times A_t$$

in which every $A_i$ is $H$-simple and, as an underlying superalgebra,

$$A_i\cong S_i\widehat\otimes\Lambda(V_i).$$

The finite component group permutes these factors. Taking its orbits gives a product of $G$-simple algebra objects. Thus the argument already proves

$$(5)\Longrightarrow(3).$$

This factorization uses no list of supergroups or algebra laws.

5. The Grassmann factor is intrinsic

For a superalgebra $B$, write

$$Z_s(B)=\{z\in B:zb=(-1)^{|z||b|}bz\text{ for homogeneous }b\in B\}$$

for its supercenter.

Lemma 5.1

If $S$ is a finite-dimensional simple associative superalgebra over algebraically closed $k$ of characteristic zero, then

$$Z_s(S)=k.$$

Proof. A nonzero homogeneous supercentral element generates a nonzero graded ideal, hence is invertible. An odd supercentral element squares to zero in characteristic different from two, so cannot be invertible. The even supercenter is therefore a finite-dimensional commutative division algebra over the algebraically closed field $k$, and is $k$. $\square$

Expanding an element coefficientwise in $\Lambda(V_i)$, Lemma 5.1 gives

$$C_i:=Z_s(A_i)=1\widehat\otimes\Lambda(V_i).$$

So the Grassmann factor is not a choice made by Kac’s presentation. It is the canonical supercenter and is preserved by the $H$-action.

Lemma 5.2

The commutative $H$-algebra $C_i$ is simple.

Proof. If $0\ne L\subsetneq C_i$ were an $H$-stable ideal, then $A_iL$ would be a nonzero proper $H$-stable two-sided ideal of $A_i$. This contradicts $H$-simplicity. $\square$

Coulembier–Sherman’s Theorem 6.4 and Corollary 6.6 apply to this canonical commutative algebra. They imply that every simple commutative algebra in a characteristic-zero supertannakian category is absolutely flat. Consequently,

$$C_i\text{ is absolutely flat in }\operatorname{Rep}H.$$

This is the point at which the open noncommutative problem is reduced exactly to the already-settled commutative theorem.

6. Separability lifts flatness

The simple superalgebra $S_i$ is separable in $\operatorname{sVec}$. One way to see this is the graded Artin–Wedderburn classification: over an algebraically closed field of characteristic zero, the simple associative superalgebras are of matrix or queer type, and both have an even bimodule section of multiplication. Equivalently, they are the central simple superalgebras over $k$.

Tensoring such a section with $C_i$ shows that the extension

$$C_i\longrightarrow A_i$$

is separable after forgetting the $H$-action. Equivariance of the chosen section is not required: exactness in $\operatorname{Rep}H$ is detected by the exact faithful forgetful functor to $\operatorname{sVec}$.

We record the standard transfer explicitly.

Lemma 6.1 (separable transfer)

Let $C\to B$ be a separable extension of finite superalgebras. If a right $B$-module $M$, restricted to $C$, is flat, then $M$ is flat over $B$.

Proof. A separability section makes

$$M\widehat\otimes_B N$$

a natural direct summand of

$$M\widehat\otimes_C N$$

for every left $B$-module $N$. Thus the first tensor functor is a natural retract of the second. A retract of an exact additive functor is exact. $\square$

Let $M$ be any right $A_i$-module in $\operatorname{Rep}H$. For an equivariant short exact sequence of left $A_i$-modules, restriction to $C_i$ remains exact, so $M\otimes_{C_i}-$ is exact internally. After applying the exact faithful forgetful functor, the nonequivariant separability section makes $M\otimes_{A_i}-$ a natural retract of that exact functor. Hence it is exact, and faithful forgetting reflects the required internal exactness. No claim about arbitrary nonequivariant $C_i$-module sequences is used.

Each $A_i$ is therefore absolutely flat in $\operatorname{Rep}H$. Products are absolutely flat, and restriction

$$\operatorname{Rep}G\longrightarrow\operatorname{Rep}H$$

is exact, faithful, and preserves relative tensor products. Hence $A$ is absolutely flat in $\operatorname{Rep}G$. This proves

$$\boxed{(5)\Longrightarrow(1).}$$

7. The Ext arrow closes directly

Lemma 7.1

For a finite algebra object in any supertannakian category,

$$(2)\Longrightarrow(5).$$

Proof. Let $R\ne0$ be a nilpotent two-sided ideal object and set $M=A/R$. The module $M$ is finite, hence finitely presented. Condition (2), applied to the extension

$$0\longrightarrow R\longrightarrow A\longrightarrow A/R\longrightarrow0$$

gives an epimorphism of internal-Hom objects

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

Apply the exact faithful fiber functor to $\operatorname{sVec}$. Internal $A$-linear Hom is the equalizer imposing the action relation, so its underlying vector space is the full space of underlying $A$-linear maps. The identity of the underlying module $M$ therefore lifts to an underlying $A$-linear section $s:M\to A$. No claim that this section is equivariant is needed.

Put $e=s(1+R)$. Then

$$e\equiv1\pmod R,\qquad e^2=e,\qquad Re=0.$$

The idempotent $1-e$ belongs to the nilpotent ideal $R$, so $1-e=0$. Thus $e=1$, and $Re=0$ forces $R=0$, a contradiction. $\square$

8. Closing the implication diagram

Coulembier–Sherman, Theorem 3.17, proves

$$(6)\Longleftrightarrow(1)\Longrightarrow(2), \qquad (3)\Longrightarrow(4)\Longleftrightarrow(5),$$

and, because $A\in\mathscr C$, also $(1)\Rightarrow(3)$.

Sections 6 and 7 add

$$(5)\Longrightarrow(1), \qquad (2)\Longrightarrow(5).$$

Every condition therefore implies every other one.

9. Independent finite-dimensional control

There is a direct module-level picture behind Section 6. Since $C_i=\Lambda(V_i)$ is a simple $H$-algebra, its homogeneous-space presentation makes the infinitesimal orbit map onto the odd tangent space. For every nonzero $\ell\in V_i$, choose an odd infinitesimal symmetry $x$ for which

$$u=x(\ell)\in\Lambda(V_i)_{\bar0}$$

has nonzero constant term, hence is a unit. On an equivariant module $M$, the Leibniz identity gives

$$\nabla_x L_\ell+L_\ell\nabla_x=L_u.$$

Therefore $L_u^{-1}\nabla_x$ contracts multiplication by $\ell$. The restriction of $M$ to every one-generator exterior subalgebra is free. The rank-variety theorem for the exterior algebra then makes finite $M$ projective over $\Lambda(V_i)$, recovering the finite part of the commutative flatness step by a different route.

This control is not needed for the formal proof, which uses Coulembier–Sherman’s stronger commutative theorem. It is retained because it checks that the reduction has the expected module-theoretic mechanism.

10. Audit obligations before publication

This ATTACK has produced a candidate resolution, not an externally checked theorem. A separate PUBLISH run must verify every bridge below against the primary literature and write out any suppressed categorical detail.

1. Connected integration. Verify the exact statement that a finite graded ideal stable under $\operatorname{Lie}H$ is $H$-stable for a connected algebraic supergroup in characteristic zero.
2. Kac scope. Check that Proposition 5.1.1 applies to possibly nonunital finite associative superalgebras with $I^2\ne0$, exactly as used for a minimal ideal.
3. Canonical center. Audit the coefficientwise proof that $Z_s(S\widehat\otimes\Lambda(V))=\Lambda(V)$, including the queer type.
4. Separable transfer. Supply an explicit even separability section for both matrix and queer simple superalgebras, or cite a source giving the super version with identical hypotheses.
5. Category reduction. Verify that the finite-generation reduction and restriction to $G^\circ$ preserve the exact relative tensor products used to test absolute flatness for all ind-modules.
6. Priority and scope. Search for earlier use of Kac’s differentiably simple normal form in the absolute-flatness problem, and check the latest version of Coulembier–Sherman before any resolution claim.

A failure at any one of these points returns the work to ATTACK. No public claim of resolving Question 5.16 should be made before this audit is complete.

11. Sources and attribution

• The open question, implication diagram, commutative absolute-flatness theorem, and supertannakian reduction are due to Kevin Coulembier and Alexander Sherman, Absolutely flat algebras in tensor categories, especially Theorem 3.17, Question 5.16, Theorem 6.4, and Corollary 6.6.
• The decisive normal form is Victor Kac, Proposition 5.1.1 in Lie Superalgebras, Advances in Mathematics 26 (1977), 8–96. Kac explicitly describes its proof as the super analogue of Block’s differentiably-simple-ring method.
• The classical method behind Kac’s proposition is Richard E. Block, Determination of the Differentiably Simple Rings with a Minimal Ideal, Annals of Mathematics 90 (1969), 433–459.
• The optional rank-variety control uses Petter Andreas Bergh and Karin Erdmann, The Avrunin–Scott theorem for quantum complete intersections, Journal of Algebra 322 (2009), 479–488, specialized to exterior algebras.

The combination of these ingredients into the proof above has not been externally reviewed. The source search found no earlier statement of this argument, but that is not a priority claim.

12. Recommended next mode

PUBLISH – conduct a hostile proof audit of the six obligations in Section 10. If all survive, prepare a self-contained resolution note and update the public claim ledger. If any fails, preserve the exact gap and return to ATTACK.