The $6\times6$ equality case of matrix spaces with rank-two commutators

Abstract: Let $\mathcal V\subseteq M_6(\mathbb C)$ be a $17$-dimensional linear subspace such that $$\operatorname{rank}[S,T]\leq2 \quad(S,T\in\mathcal V).$$ We prove that $\mathcal V$, or its transpose, is conjugate to the algebra $ \left{ \begin{pmatrix} A&B&C\ 0&λI_2&D\ 0&0&λI_2 \end{pmatrix}: A,B,C,D\in M_2(\mathbb C),\ λ\in\mathbb C \right}. $ Consequently, the corresponding closed algebraic locus in $\operatorname{Gr}(17,M_6(\mathbb C))$ is the disjoint union of two nonsingular irreducible components, each isomorphic to $\operatorname{Fl}(2,4;6)$. We also prove that the Zariski tangent space at $\mathcal A$ of the corresponding closed algebraic locus is equal to the tangent space to the conjugacy orbit of $\mathcal A$.

• The paper proves that every 17-dimensional subspace of $M_6(\mathbb C)$ with commutators of rank at most 2 is conjugate to the algebra $\mathcal A$ or its transpose.
• The method uses a tangent-space computation and the Zariski tangent space which is leveraged to resolve the geometry of an entire locus.
• The $\mathcal A$ is the significant algebra.

Context and main result

Omladić, Radjavi, and Šivic proved that a linear subspace $\mathcal V\subseteq M_n(\mathbb C)$ with $\operatorname{rank}[S,T]\le k$ for all $S,T\in\mathcal V$ has dimension at most

$nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$

and conjectured a classification of the subspaces attaining this bound [ORS, Linear Algebra Appl. 676 (2023)]. Their conjecture was settled for $k=1$, for $k=n-1$, and for arbitrary $k$ under the assumption that $\mathcal V$ is an algebra. This paper resolves the first open nontrivial case of the full conjecture: $(n,k)=(6,2)$, where the bound is $17$. The extremal space is the algebra

$\operatorname{rank}[S,T]\le k$0

of dimension $\operatorname{rank}[S,T]\le k$1, whose commutators are supported entirely in the upper-left $\operatorname{rank}[S,T]\le k$2 block. The main theorem states that every $\operatorname{rank}[S,T]\le k$3-dimensional $\operatorname{rank}[S,T]\le k$4 with all pairwise commutators of rank at most $\operatorname{rank}[S,T]\le k$5 satisfies $\operatorname{rank}[S,T]\le k$6 or $\operatorname{rank}[S,T]\le k$7 for some $\operatorname{rank}[S,T]\le k$8. In other words, the only equality cases in dimension $\operatorname{rank}[S,T]\le k$9 are $S,T\in\mathcal V$0 and its transpose, up to conjugacy.

The proof strategy has three parts: identify the conjugacy orbit of $S,T\in\mathcal V$1 with a flag variety; compute the Zariski tangent space to the relevant algebraic locus at $S,T\in\mathcal V$2 and show it coincides with the tangent space to that orbit; then use the Borel fixed-point theorem together with an upper-triangular fixed-point classification due to Omladić–Radjavi–Šivic to exclude any further components.

The algebraic locus and the conjugacy orbit

Let

$S,T\in\mathcal V$3

The paper gives a direct affine-chart proof that $S,T\in\mathcal V$4 is closed: on each standard chart, the vanishing of every $S,T\in\mathcal V$5 minor of $S,T\in\mathcal V$6 for all coefficient vectors $S,T\in\mathcal V$7 is equivalent to the vanishing of finitely many regular functions in the chart parameter $S,T\in\mathcal V$8. Hence $S,T\in\mathcal V$9 is a projective algebraic set.

Writing $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$0 as three coordinate planes, the radical $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$1 of $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$2 is strictly upper triangular in blocks, with $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$3 equal to the single $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$4 block. The key intrinsic identity is

$nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$5

Every element of the normalizer must preserve these two subspaces, and conversely any matrix stabilizing the flag $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$6 normalizes $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$7. Consequently

$nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$8

so $nk+\left\lfloor\frac{(n-k)^2}{4}\right\rfloor+1,$9 is nonsingular, projective, irreducible, closed, and $k=1$0-dimensional. Note that the identification of the orbit uses the radical-square flag, which is intrinsic; this is what later allows the two transpose orbits to be distinguished.

Tangent-space computation

The technical core of the paper is the equality $k=1$1, which implies $k=1$2. The argument proceeds in the affine chart of subspaces complementary to a carefully chosen complement $k=1$3 (lower-triangular-type matrices with a trace-zero condition), so tangent vectors are linear maps $k=1$4 decomposed into five component maps $k=1$5.

Three ingredients are combined:

Determinantal tangent condition. At a rank-$k=1$6 point $k=1$7 of the determinantal variety $k=1$8, the tangent space consists of those $k=1$9 with $k=n-1$0. The paper proves this via an explicit normal form and inspection of minors involving both pivots. Applied along curves through $k=n-1$1, it yields, for each test commutator $k=n-1$2 of rank exactly $k=n-1$3,

$k=n-1$4

Test commutators. Since $k=n-1$5 spans $k=n-1$6, restrictions derived from invertible test matrices extend linearly. Testing pairs such as $k=n-1$7 forces the components $k=n-1$8 to match, entry by entry, the explicit differential of the conjugation action:

$k=n-1$9

with $k$0, $k$1, and all remaining components vanishing except possibly $k$2. After subtracting $k$3 — a genuine tangent vector to the orbit — one is left with a residual map supported on the $k$4-summand.

Functional identity. The surviving diagonal-block data satisfy

$k$5

The paper solves this functional equation completely: writing $k$6 reduces it to $k$7, and an explicit computation against the generators $k$8 of $k$9 shows $\mathcal V$0 is scalar. The trace condition $\mathcal V$1 then forces the scalar functional to vanish, so $\mathcal V$2, $\mathcal V$3 for some $\mathcal V$4. The residual tangent vector is therefore $\mathcal V$5, which lies in $\mathcal V$6. This closes the reverse inclusion.

An important consequence follows immediately: since $\mathcal V$7 and both have dimension $\mathcal V$8, the point $\mathcal V$9 is nonsingular, lies on a unique irreducible component, and that component equals $(n,k)=(6,2)$0. The same holds at every point of $(n,k)=(6,2)$1 and of $(n,k)=(6,2)$2 (transposition is an algebraic automorphism of $(n,k)=(6,2)$3 because $(n,k)=(6,2)$4).

Global classification via the Borel fixed-point theorem

Two facts combine to determine all of $(n,k)=(6,2)$5. First, each irreducible component $(n,k)=(6,2)$6 is stable under the conjugation action of $(n,k)=(6,2)$7, by irreducibility of the action map's image closure. Second, $(n,k)=(6,2)$8 is projective and stable under the connected solvable group $(n,k)=(6,2)$9 of invertible upper-triangular matrices, so the Borel fixed-point theorem supplies a $17$0-fixed point $17$1. Such a point satisfies both hypotheses of the specialized Omladić–Radjavi–Šivic fixed-point theorem, hence lies in $17$2. But no other irreducible component meets either orbit, so $17$3 itself is that orbit. Therefore

$17$4

This is the structural strengthening over the earlier work: rather than classifying individual extremal spaces, the entire equality locus in $17$5 is identified.

Distinctness of the two components

It remains to rule out $17$6. Conjugate algebras are isomorphic, so an isomorphism invariant distinguishing $17$7 from $17$8 suffices. Both have radical cube zero and semisimple quotient $17$9; the ordered pair of dimensions

$\operatorname{rank}[S,T]\le k$00

is preserved by isomorphisms, where $\operatorname{rank}[S,T]\le k$01 is the idempotent of the noncommutative simple summand. For $\operatorname{rank}[S,T]\le k$02 the radical square is the $\operatorname{rank}[S,T]\le k$03 block, giving $\operatorname{rank}[S,T]\le k$04; for the transpose, $\operatorname{rank}[S,T]\le k$05. The two orbits are consequently distinct, disjoint, nonsingular, projective, and irreducible, each isomorphic to $\operatorname{rank}[S,T]\le k$06.

Limitations and scope

The result is specific to $\operatorname{rank}[S,T]\le k$07; the general equality-case classification of Omladić–Radjavi–Šivic remains open for other pairs. Two external inputs are load-bearing: the closedness argument relies on the standard affine-chart framework, and the global step depends on [Proposition 11] of Omladić–Radjavi–Šivic for $\operatorname{rank}[S,T]\le k$08-stable subspaces — without that fixed-point classification the tangent-space analysis alone would not yield the full locus. The tangent-space computation exploits special features of the $\operatorname{rank}[S,T]\le k$09 block structure (e.g., the solvability of the functional equation via $\operatorname{rank}[S,T]\le k$10 representation-theoretic constraints); whether analogous computations succeed at larger $\operatorname{rank}[S,T]\le k$11 is not addressed. The paper also notes that the manuscript was developed with extensive use of ChatGPT, including drafting and verification, with the author responsible for correctness.

Conclusion

The paper settles the $\operatorname{rank}[S,T]\le k$12 equality case of the Omladić–Radjavi–Šivic dimension-bound conjecture, proving that every $\operatorname{rank}[S,T]\le k$13-dimensional subspace of $\operatorname{rank}[S,T]\le k$14 with commutators of rank at most $\operatorname{rank}[S,T]\le k$15 is conjugate to $\operatorname{rank}[S,T]\le k$16 or to its transpose. Methodologically, it demonstrates that a Zariski tangent-space computation — here reduced to solving an explicit functional equation on $\operatorname{rank}[S,T]\le k$17 — can pin down an entire Grassmannian locus, yielding the clean geometric statement that the locus is the disjoint union of two copies of $\operatorname{rank}[S,T]\le k$18. The natural next question raised by the technique is whether the same orbit-plus-tangent strategy extends to the remaining open cases of the general conjecture.