Dixmier Conjecture

Dixmier Conjecture

Definitions and mathematical statement

For a characteristic-zero field $K$, the first Weyl algebra is \[ A_1(K)=K\langle x,\partial\rangle/(\partial x-x\partial-1), \] the associative algebra of polynomial differential operators in one variable. The selected Dixmier conjecture asks whether every unital $K$-algebra homomorphism $\phi:A_1(K)\to A_1(K)$ is bijective, and hence an automorphism. An endomorphism is specified by a pair $P=\phi(x)$, $Q=\phi(\partial)$ satisfying $QP-PQ=1$; the assertion is that such a pair generates all of $A_1(K)$ and yields an invertible substitution. This entry is specifically the first Weyl algebra in characteristic zero. The proof-claim title in the supplied bibliography is not treated as proof of this assertion.

Formulation and concept references: [S1].

Short English statement

Must every endomorphism of the one-variable Weyl algebra over a characteristic-zero field be invertible?

Research attempt

Status: unresolved by this attempt; an external proof claim is present. The supplied source is not merely a paper on a related variant. Zheglov's Theorem 1.1 in arXiv:2410.06959v5 claims exactly the characteristic-zero first-Weyl-algebra assertion. The inspected version history on 27 September 2026 lists v5, dated 19 January 2026, and describes expanded arguments and simplified final calculations. The abstract and introductory theorem were checked against the selected statement. This document does not independently certify its long normal-form argument, nor does it claim to have located an error in it. The inherited open-status metadata, the catalogue's proof-claim qualification, and the local unresolved status are retained separately from that external assertion.

The independent work below proves injectivity of every endomorphism, completely solves several coordinate cases, and isolates the obstruction in a degree-reduction approach. It keeps the field arbitrary of characteristic zero and does not substitute a result about a higher Weyl algebra, a completion, or positive characteristic.

1. Normal ordering and a faithful model.

Write $D=\partial$. Repeatedly using $Dx=xD+1$ expresses every element as \[ T=\sum_{j=0}^m f_j(x)D^j,\qquad f_j\in K[x]. \] These expressions are unique. To prove this, act on $K[x]$, with $x$ multiplication and $D$ formal differentiation. Suppose a normally ordered expression acts as zero, and take its largest nonzero index $m$. Commuting it with multiplication by $x$ exactly $m$ times gives multiplication by $m!f_m(x)$, since \[ [D^j,x]=jD^{j-1}. \] A zero operator has zero iterated commutator. Multiplication by a nonzero polynomial is not zero, and $m!\ne0$ in characteristic zero. This contradiction proves uniqueness, descending on $m$ if necessary. It also proves faithfulness of the differential-operator model and that the normally ordered monomials $x^iD^j$ form a basis.

Induction on $j$ gives the useful exact product formula \[ D^j f(x)=\sum_{r=0}^j\binom jr f^{(r)}(x)D^{j-r}. \tag{336.1} \] The inductive step differentiates each coefficient and also increases the remaining power of $D$; Pascal's identity combines the two contributions. This formula is the basis of the exact finite diagnostic, rather than treating $x$ and $D$ as commuting variables during multiplication.

2. A complete proof of simplicity and automatic injectivity.

Let $I$ be a nonzero two-sided ideal. Choose a nonzero element of $I$ of minimum $D$-order. If that order is positive, its commutator with $x$ is nonzero by normal ordering and has order one less, a contradiction. Thus $I$ contains a nonzero polynomial in $x$. Among such polynomials choose one of minimum degree. If its degree is positive, the commutator with $D$ is its nonzero derivative, of smaller degree, again a contradiction. Hence $I$ contains a nonzero scalar and therefore contains one.

The Weyl algebra is consequently simple. The kernel of a unital endomorphism is a two-sided ideal and cannot contain one, so it is zero. Thus every map in the question is already injective. The entire missing assertion is surjectivity: one must recover $x$ and $D$ as noncommutative polynomials in the given $P,Q$.

This is not a finite-dimensional vector-space problem. Injective linear self-maps of infinite-dimensional spaces need not be onto. Even the commutative ring $K[x]$ has an injective nonsurjective endomorphism $x\mapsto x^2$. Simplicity establishes one half of the desired statement but supplies no bound on how to express the original generators in the image.

The same commutator calculation identifies a centralizer needed below: \[ C_{A_1}(x)=K[x]. \tag{336.2} \] Indeed, $[T,x]=\sum_{j\ge1}j f_j(x)D^{j-1}$ vanishes only if every positive-order coefficient vanishes. Characteristic zero is again essential.

3. Complete solution when one image lies in the polynomial coordinate ring.

Suppose $P=f(x)\in K[x]$ and $[Q,P]=1$. The polynomial $f$ cannot be constant. Write $Q=\sum_{j=0}^m a_j(x)D^j$ with $a_m\ne0$. Formula $336.1$ shows that for $m\ge1$ the leading coefficient of $[Q,f]$, of $D$-order $m-1$, is \[ m a_m(x)f'(x). \] It is nonzero. Order zero for $Q$ would give commutator zero, whereas order greater than one would give a commutator of positive order. Therefore $m=1$ and \[ a_1(x)f'(x)=1. \] The only units of $K[x]$ are nonzero constants, so $f=ax+b$ for $a\ne0$, and \[ P=ax+b,\qquad Q=a^{-1}D+g(x). \tag{336.3} \] Conversely every such pair has the required commutator. Its inverse coordinates are explicit: \[ x=(P-b)/a,\qquad D=a\left(Q-g\left((P-b)/a\right)\right). \tag{336.4} \] Thus the endomorphism is onto in this complete subcase. Interchanging the two Weyl coordinates, with the necessary sign change, gives the analogous case when an image is a polynomial in $D$.

The proof says more than that a one-variable polynomial map must be invertible. It uses the order of the other differential operator to force the degree of $f$ to be one. Replacing the commutator by a formal derivative without controlling those orders would omit the main step.

4. Every nonconstant affine-linear image is also settled.

Suppose $P=ax+bD+c$ with $(a,b)\ne(0,0)$. Choose $u,v\in K$ with $av-bu=1$, and put $R=ux+vD$. Then \[ [R,P]=1. \] The affine-linear change from $(x,D)$ to $(P,R)$ is invertible because its linear determinant is one. Hence it is an automorphism of the Weyl algebra; its inverse is obtained by the ordinary inverse matrix, with the scalar translation removed.

Now $[Q-R,P]=0$. Transporting $336.2$ through the affine-linear automorphism gives $C_{A_1}(P)=K[P]$. Consequently \[ Q=R+h(P) \] for some polynomial $h$, and $P,Q$ generate $P,R$, hence all of $A_1$. This proves the selected assertion whenever either image has Bernstein degree one. There is no need to assume $K$ algebraically closed to choose $u,v$.

More generally, the two triangular substitutions \[ (x,D)\longmapsto(x,D+f(x)),\qquad (x,D)\longmapsto(x+g(D),D) \tag{336.5} \] are automorphisms, with inverses obtained by subtracting the same polynomial. Any finite composition with affine-linear determinant-one substitutions is therefore an explicit positive family. Verifying that a particular pair belongs to this generated family is different from proving that every commutator-one pair does.

5. What the associated graded calculation actually proves.

Give $x,D$ total degree one. Normal ordering shows that the associated graded algebra is the commutative polynomial ring $K[X,\Xi]$. If $P,Q$ have total degrees $a,b$, their commutator has degree at most $a+b-2$. The homogeneous component in that degree is \[ \frac{\partial q}{\partial\Xi}\frac{\partial p}{\partial X} -\frac{\partial q}{\partial X}\frac{\partial p}{\partial\Xi}, \tag{336.6} \] where $p,q$ are their leading symbols. One verifies the formula first on normally ordered monomials using the $r=1$ term of $336.1$, then extends bilinearly. If cancellation makes the degree smaller, the displayed expression is zero.

For $[Q,P]=1$ and $a+b>2$, the expression in $336.6$ must vanish. It does not equal one: the constant commutator lies below that leading degree. Thus one cannot simply assert that the two leading symbols form a polynomial map of Jacobian one and then invert it.

There is already a triangular automorphism illustrating the loss: \[ P=x,\qquad Q=D+x^2. \] Here $[Q,P]=1$, but the leading symbols are $X$ and $X^2$, which are algebraically dependent. Their Poisson bracket is zero, exactly as $336.6$ predicts. Surjectivity is restored by the lower-order operation $Q-P^2=D$. A prospective degree-reduction proof must find an analogous simplifying operation for every possible leading configuration and prove that a well-founded complexity decreases. The existence of this operation in the example does not establish its existence in general.

Nor does conjugating operators inside a larger completion automatically give an inverse inside $A_1(K)$. Formal series may have infinitely many terms, and rational or negative-order differential operators may lie outside the polynomial Weyl algebra. A normal-form argument must prove the required return to the original algebra, rather than use invertibility in a larger ring as a surrogate for polynomial generation. This is a requirement on a proposed route, not an assertion that [S1] fails to meet it.

6. A finite-matrix shortcut is impossible in characteristic zero.

There are no finite-dimensional matrices $P,Q$ over $K$ satisfying $QP-PQ=I$. Taking traces gives zero on the left and the positive matrix size on the right, nonzero in characteristic zero. Thus exact finite-matrix searches cannot enumerate the pairs in the problem. Truncating differentiation and multiplication to polynomials of degree at most $N$ changes their commutator at the boundary: multiplication exits that subspace. Any numerical experiment ignoring the boundary has silently changed the algebra.

The offline diagnostic instead uses sparse dictionaries of normally ordered monomials and $336.1$ over exact rational coefficients. It checks the triangular and affine families, verifies explicit inverse compositions, and compares the leading-symbol commutator formula on finite examples. It also verifies the boundary discrepancy in a truncated polynomial model. These are finite identity checks, not a search over all endomorphisms or a verification of [S1].

Outcome and first gap.

Every unital endomorphism is proved injective, and all pairs with a polynomial-coordinate or affine-linear image are explicitly inverted. The degree calculation locates, rather than removes, the main obstruction: the leading symbols can be dependent, and a uniform terminating simplification inside the original algebra has not been established. The external v5 manuscript claims a full resolution; this independent attempt supplies only the stated subcases and does not turn either its title or the injectivity argument into an unqualified solution claim.

Sources

• S1 Alexander Zheglov, The Conjecture of Dixmier for the first Weyl algebra is true, arXiv:2410.06959v5, 2026. https://arxiv.org/abs/2410.06959. Formulation source.