Actual images and local fibers in the second commutator layer

Introduction

Throughout, $K$ is algebraically closed of characteristic zero. We write $M_n=M_n(K)$ and $\mathfrak{sl}_n=\mathfrak{sl}_n(K)$. For a polynomial $f$ in $m$ original variables, let

be its set of single values, and write $S_f=\operatorname{Val}_n(f)$. No linear span or Zariski closure is taken in this notation. An ordinary multilinear polynomial uses each of its $m$ distinct original variables exactly once in every word. It is proper if substituting $1$ for any variable gives zero in the free algebra, equivalently if it is invariant under adding a scalar to any input. Let $J$ denote the two-sided commutator ideal of the unital free associative algebra.

The L’vov–Kaplansky conjecture predicts that the actual image of an ordinary multilinear polynomial on $M_n$ is one of $\{0\}$, $K I_n$, $\mathfrak{sl}_n$, and $M_n$ (see [2]). We studied the quartic case by repeated-input restrictions in [5]. Tang gives a quartic classification in [9 Theorem 1.1]. In [6 Theorem A], we classify the images in degrees four through six, starting with proper sextics.

Here we fix the second commutator layer and allow arbitrary degree. For $n \ge3$ we determine the images and construct local fibers for proper multilinear polynomials in $J^2 \setminus J^3$.

These assertions hold for $f+R$ for every proper ordinary multilinear $R\in J^3$ in the same variables. The trace and the input $a$ are taken for the complete polynomial $f+R$.

The proof uses complete restrictions to two active variables. § 6 treats a nonzero restriction with one common passive matrix. If all such restrictions vanish, $\operatorname{Id}(UT_3)=J^3$ (see [3 Theorem 2.1]) gives a nonzero pair of symbols with independent diagonal passive matrices. § 7 treats that pair in an étale chart of the original source. The alternating polynomial $s_4$ belongs to this second case.

The two local conditions in Theorem 1.1 hold at the same original input. If the restriction has nonzero trace, its full differential is surjective at a rank-one square-zero input. If $h\ne0$ and its restriction is zero, the first nonzero term in the transverse expansion of $h$ gives a bilinear test on the two full active matrices. Theorems 6.2 and 7.2 give a nilpotent input at which this test is nonzero and the traceless differential along the restriction is surjective. Solving $g=N$ in those tangent directions leaves the transverse variables unchanged. The selected term therefore survives on the exact fiber $g^{-1}(N)$ (see Section 6.4 and Proposition 7.14). A correction in $J^3$ can change this term even though it vanishes on the restriction. The input is selected after the correction is included in $h$. The local-image criterion of § 2 then gives the stated sets of values.

§ 3 gives coordinates for the proper second quotient and the vanishing lemma for restrictions with few active variables. § 4 constructs the one-order operators. § 5 gives the size-three two-symbol construction and its traceful extension to larger sizes. § 6 and § 7 treat the common-background and independent-diagonal restrictions, respectively. § 8 formulates the remaining local-fiber questions and relates them to the higher-degree results in [6, 7, 8].

Local fibers and actual images

A single local condition for arbitrary degree

For noncentral-valued $f$ define the target vector space

Interior and closure in the next theorem use the Zariski topology on $V_f$.

Consequently $LK$, for this field class and all degrees and sizes, is equivalent to the middle condition for every noncentral-valued $f$. Moreover, $N\in S_f$ already holds for every such $f$; the issue is coverage of a neighborhood, not merely the existence of a nilpotent value.

Proof. First note that $S_f$ is invariant under conjugation and multiplication by nonzero scalars, and contains zero. If $\operatorname{tr} f \ne0$, then $K I_n \subseteq S_f$. To see this, choose Weyl matrices $U,V$ with $UV=qVU$, $U^n=V^n=I$, where $q$ is a primitive $n$th root. The $U^aV^b$ form a basis, and only its degree-zero element has nonzero trace. Expand a nonzero-trace evaluation multilinearly in that basis. Some basis tuple has nonzero output trace. All words at this tuple have the same total $(\mathbb{Z}/n\mathbb{Z})^2$ degree, so their sum is a multiple of one basis element. It must be a nonzero multiple of $I$. Scaling one input gives all scalars. If $\operatorname{tr} f \equiv0$, the only scalar in $V_f$ is zero, already in $S_f$.

Here is the elementary degeneration underlying the result. Every nonscalar $W$ has a similar representative $\left(\begin{smallmatrix} u & v \\ w & Z \end{smallmatrix}\right)$ with $v \ne0$. For otherwise every row covector would be a left eigenvector, forcing $W$ to be scalar. The polynomial curve

has nonzero rank-one square-zero limit $R(0)$, conjugate to $N$. If $W\in\mathfrak{sl}_n$, the whole curve lies in $\mathfrak{sl}_n$. Suppose an open neighborhood $O$ of $N$ in $V_f$ lies in $S_f$. After a fixed conjugation the curve passes through $N$. Its inverse image of $O$ is an open subset of $\mathbb{A}^1$ containing zero, hence contains some nonzero $t$. Conjugation and scalar invariance now give $W\in S_f$. The scalar discussion completes $S_f=V_f$. The reverse implication and the closure formulation are immediate from the definition of interior. Central-valued polynomials have image either zero or $K I_n$ by multilinearity, so the asserted equivalence with LK follows.

Finally, a nonscalar value has, after conjugation, a nonzero off-diagonal entry. Expand its inputs in matrix units. Some matrix-unit tuple gives an output with a nonzero off-diagonal entry. Regard these units as directed edges. Any nonzero word is an Euler trail using the same edges. An open trail has uniquely determined initial and terminal vertices, so every nonzero word in this tuple is a multiple of the same off-diagonal unit. The output is therefore a nonzero multiple of $E_{ij}$, $i\ne j$. Scaling and conjugation give $N$ itself. □

This argument is independent of degree, properness, and a chosen coefficient family. We use it as a characterization and make no priority claim for the local reformulation. The proof also applies to a restricted evaluation domain whenever its image has the stated conjugation and scalar invariance, and all the required scalar targets have separately been obtained.

The first condition implies the second, but neither is claimed necessary for polynomial surjectivity. In particular, generic full rank only proves dominance; it does not locate an open image neighborhood at $N$. The distinction is real even for invariant constructible cones: $\{0\}\cup\{A:\det A\ne0\}$ is dense, contains every scalar, and omits $N$. It is an illustration of the logical issue, not a claimed polynomial image.

The rank-one attainment used in Theorem 2.1 is already in Kanel-Belov–Malev–Rowen: see Lemma 5 on p. 6 of arXiv:1005.0191v3, the author version cited in [15], and its explicit restatement for arbitrary $n$ in Lemma 2 on p. 2 of arXiv:1306.4389v3 [16]. These locators refer to the specified preprint versions.

Higher-order trace on a smooth traceless fiber