Arveson-Douglas Essential Normality Conjecture for Homogeneous Quotient Modules

Arveson-Douglas Essential Normality Conjecture for Homogeneous Quotient Modules

Definitions and mathematical statement

The Drury--Arveson space $H_d^2$ is the reproducing-kernel Hilbert space on the unit ball of ${\mathbb{C}}^d$ with kernel \[ K(z,w)=\frac1{1-\langle z,w\rangle}. \] For a proper homogeneous ideal $I\subset{\mathbb{C}}[z_1,\ldots,z_d]$, let $[I]$ be its Hilbert-space closure and $Q_I=H_d^2\ominus[I]$. Let $S_j=P_{Q_I}M_{z_j}|_{Q_I}$ be the compressed coordinate multiplier. For $p>0$, a compact operator $T$ lies in the Schatten class $\mathcal S^p$ when $\sum_ns_n(T)^p<\infty$, where $s_n(T)$ are its singular values.

The assertion is \[ [S_i^*,S_j]\in\mathcal S^p \quad\text{for all }i,j\text{ and every }p>\dim_{{\mathbb{C}}} Z(I). \] Here $Z(I)$ is the affine zero variety. All finite $d$ and all proper homogeneous ideals, including nonradical ones, are included; radical or smooth-variety cases alone do not give the full result.

Formulation and concept references: [S1].

Short English statement

Do homogeneous quotient modules of the Drury--Arveson space have commutators with the predicted Schatten-class decay, even for nonradical ideals?

Research attempt: the compression defect and a nonradical family

Status. The assertion for an arbitrary homogeneous ideal is not proved. A complete argument is given for coordinate power ideals, including arbitrarily thick nonradical examples. The calculation isolates a sufficient degreewise estimate for a general ideal and identifies exactly where replacing an ideal by its radical loses information.

Source audit. The Main Theorem on page 4 of [S1] treats the vanishing module of a homogeneous variety smooth away from the origin. Its introduction explicitly distinguishes the geometric, radical-ideal version from the full ideal formulation. The family below includes nonradical ideals, but belongs to the elementary monomial setting already discussed in that introduction; no novelty is claimed. This source check does not turn the geometric theorem into a theorem for every nonradical ideal or every singular projective variety.

1. Grading and the uncompressed calculation

Let $H_k$ denote the homogeneous polynomials of total degree $k$. Expanding the reproducing kernel shows that \[ e_\alpha=\sqrt{\frac{|\alpha|!}{\alpha!}}z^\alpha, \qquad \alpha\in\mathbb N^d, \] is an orthonormal basis. Consequently, with $k=|\alpha|$, \[ M_j e_\alpha=\sqrt{\frac{\alpha_j+1}{k+1}}e_{\alpha+e_j}, \qquad M_i^*e_\alpha=\sqrt{\frac{\alpha_i}{k}}e_{\alpha-e_i}\quad(k\geq1), \tag{1} \] where the second expression is zero when $\alpha_i=0$, and every backward shift kills the constant vector.

For $i\ne j$, direct composition gives \[ [M_i^*,M_j]e_\alpha =-\frac{\sqrt{\alpha_i(\alpha_j+1)}}{k(k+1)} e_{\alpha-e_i+e_j}. \tag{2} \] For $i=j$ it gives \[ [M_i^*,M_i]e_\alpha=\frac{k-\alpha_i}{k(k+1)}e_\alpha. \tag{3} \] The maps of indices in $2$ are injective on their nonzero domain. Hence their operator norm is the largest absolute coefficient, rather than a sum of coefficients. Since $\sqrt{\alpha_i(\alpha_j+1)}\leq k+1$, $2$--$3$ imply \[ \|[M_i^*,M_j]|_{H_k}\|\leq k^{-1}\qquad(k\geq1). \tag{4} \] The degree-zero block is finite dimensional and will never affect a Schatten threshold.

For homogeneous $I$, put $Q_k=Q_I\cap H_k$ and $J_k=[I]\cap H_k$. Each finite-dimensional $I\cap H_k$ is closed; orthogonality of the grading then gives \[ Q_I=\bigoplus_{k\geq0}Q_k,\qquad H_k=Q_k\oplus J_k. \] Invariance of $[I]$ under every $M_j$ implies coinvariance of $Q_I$ under $M_j^*$. This last fact is essential: forward multiplication can leave $Q_I$, whereas backward multiplication cannot.

2. An exact formula for what compression adds

Let $P=P_{Q_I}$ and define the leakage maps \[ A_{j,k}=(1-P)M_j|_{Q_k}:Q_k\longrightarrow J_{k+1}. \] Using $S_j=PM_j|_{Q_I}$ and inserting $P+(1-P)$ yields \[ [S_i^*,S_j]|_{Q_k} =P[M_i^*,M_j]|_{Q_k}-A_{i,k}^*A_{j,k}. \tag{5} \] Indeed, $S_i^*S_j=PM_i^*PM_j|_Q$, while $S_jS_i^*=PM_jM_i^*|_Q$ because $M_i^*Q\subset Q$. The omitted term in the first product is precisely $PM_i^*(1-P)M_j|_Q=A_i^*A_j$. In particular its sign in $5$ is negative.

This formula explains why the estimate for the ambient shift does not automatically settle the quotient problem. The ambient commutator is already $O(k^{-1})$, but multiplication followed by projection onto the ideal introduces a new term. Boundedness of coordinate multiplication gives only $\|A_{j,k}\|\leq1$, which has no decay in $k$.

Here is a useful sufficient criterion, proved without any operator-ideal interpolation. Suppose $r\geq1$ and \[ \dim Q_k\leq C(k+1)^{r-1},\qquad \|A_{j,k}\|\leq C_j(k+1)^{-1/2}. \tag{6} \] Then $[S_i^*,S_j]\in\mathcal S^p$ for all $p>r$. To see this, $4$--$6$ bound each degree-$k$ block by $C'(k+1)^{-1}$. The blocks preserve degree and are orthogonal in both domain and range. Truncating to degrees at most $N$ is finite rank, and the norm of the remaining direct sum tends to zero. Thus the operator is compact. Each block has at most $\dim Q_k$ singular values, so for every $p>0$, \[ \sum_{\ell}s_\ell([S_i^*,S_j])^p \leq C''\sum_{k\geq0}(k+1)^{r-1-p}<\infty \quad\text{if }p>r. \tag{7} \] This argument works also when $0<p<1$; it does not use a triangle inequality for a Schatten quasi-norm. All summation is over the singular values of orthogonal finite-dimensional blocks.

For a general homogeneous ideal, the expected rank exponent in $6$ is the usual Hilbert-function exponent, $r=\dim_{\mathbb C}Z(I)$. The conditional criterion itself only assumes the displayed rank bound; the proof of the concrete family below obtains that bound by direct counting and does not require a general Hilbert-polynomial theorem. The unproved general step in this approach is the leakage estimate, not the summation in $7$.

3. Complete verification for coordinate thickenings

Fix $0\leq r\leq d$ and positive integers $a_{r+1},\ldots,a_d$. Consider \[ I=(z_{r+1}^{a_{r+1}},\ldots,z_d^{a_d}), \tag{8} \] with the empty list interpreted as $I=(0)$ when $r=d$. Its zero variety is the coordinate $r$-plane. If any exponent is greater than one, $8$ is generally nonradical: for example $z_j\notin I$ but $z_j^{a_j}\in I$ when $a_j>1$.

The standard monomials in the quotient are exactly those satisfying \[ 0\leq\alpha_j<a_j\quad(j>r), \tag{9} \] with no restriction on the first $r$ coordinates. Orthogonality of the monomials proves that these $e_\alpha$ form an orthonormal basis of $Q_I$. In particular, there is no choice of a possibly ill-conditioned algebraic basis hidden in this identification.

If $j\leq r$, multiplying a standard monomial by $z_j$ preserves $9$, so $A_{j,k}=0$. If $j>r$, leakage occurs precisely on the boundary layer $\alpha_j=a_j-1$. On that layer $1$ gives \[ A_{j,k}e_\alpha=\sqrt{\frac{a_j}{k+1}}e_{\alpha+e_j}. \] These output monomials are orthogonal, and on the other layers the leakage is zero. Therefore \[ \|A_{j,k}\|\leq\sqrt{a_j}\,(k+1)^{-1/2}. \tag{10} \] If the layer is empty in small degrees, the norm is simply zero; $10$ remains valid.

For $r\geq1$, let $\beta=(\alpha_{r+1},\ldots,\alpha_d)$ and write $|\beta|$ for its sum. Directly counting the unrestricted coordinates gives \[ \dim Q_k= \sum_{\substack{0\leq\beta_j<a_j\\ |\beta|\leq k}} {k-|\beta|+r-1\choose r-1} \leq \left(\prod_{j>r}a_j\right){k+r-1\choose r-1}. \tag{11} \] The stars-and-bars formula used here follows by inserting $r-1$ separators among $k-|\beta|$ identical objects. The last binomial is bounded by a constant times $(k+1)^{r-1}$. Equations $10$--$11$ establish $6$, and hence prove exactly the requested conclusion $p>\dim Z(I)=r$ for every ideal $8$.

If $r=0$, all coordinates have bounded exponent in the quotient, so its dimension is $\prod_j a_j$. Every commutator is then finite rank and belongs to every $\mathcal S^p$, $p>0$, as required. This handles the dimension-zero endpoint separately instead of inserting $r=0$ into an inappropriate infinite-degree estimate.

For example, in four variables the nonradical ideal \[ I=(z_3^2,z_4^3) \] has variety dimension two and at most $6(k+1)$ standard monomials of degree $k$. The only leakage maps are in coordinates three and four, with squared norms at most $2/(k+1)$ and $3/(k+1)$. Formula $5$ then gives summability for all $p>2$, independently of the ambient dimension four. This is an actual test of the refined dimension in the statement, not just the weaker threshold $p>d$.

4. What the exponent does and does not say

The threshold in this family is not merely an artifact of counting, at least when $r\geq2$. Restrict to the subfamily $I=(z_{r+1},\ldots,z_d)$, so the quotient is $H_r^2$. By $3$, on degree $k$ the eigenvalues of $[S_1^*,S_1]$ are \[ \frac{k-\alpha_1}{k(k+1)}. \] For all standard monomials with $\alpha_1\leq k/2$, these eigenvalues are at least $1/[2(k+1)]$. There are at least $c_r k^{r-1}$ such monomials for $k$ sufficiently large. For $r=2$ this follows by choosing $\alpha_1$ from $0$ to $\lfloor k/2\rfloor$. For $r>2$, each such choice has ${k-\alpha_1+r-2\choose r-2}\geq c'_r k^{r-2}$ completions. Consequently \[ \sum_\ell s_\ell([S_1^*,S_1])^r \geq c\sum_{k\geq k_0} k^{r-1}(k+1)^{-r}=\infty. \tag{12} \] Thus one cannot generally replace $p>r$ by $p\geq r$. When $r=1$, the unthickened quotient is the one-variable Hardy shift, whose self-commutator is rank one. That exceptional endpoint does not invalidate $12$, which was explicitly proved only for $r\geq2$.

Replacing $8$ by its radical retains the same variety but changes the Hilbert space. For instance $I=(z_2^2)\subset\mathbb C[z_1,z_2]$ has two infinite layers $\alpha_2=0,1$, whereas $\sqrt I=(z_2)$ has one. The difference is infinite dimensional, despite the equality of zero sets. It therefore cannot be dismissed as a finite-rank perturbation without an additional operator argument. Our proof controls the second layer through its shift weights; it does not remove that layer algebraically.

Likewise, a general linear change of polynomial variables need not be unitary on Drury--Arveson space. Coordinate orthogonality in $1$ is preserved by unitary changes of variables, but not by arbitrary badly conditioned linear maps in a way that makes $10$ automatic. The explicit calculation should not be extended to a new ideal simply by saying that generators can be put in normal form.

5. Verification and the first remaining gap

The accompanying finite script enumerates standard monomials for $(z_3^2,z_4^3)$, checks $11$ against enumeration, and builds the degree blocks of $5$ independently from the compressed shift formulas. It verifies the compression sign and the leakage norm on bounded degrees. Those calculations check indexing and algebra, not the infinite Schatten sum; $7$, $10$, and $11$ prove that sum.

The approach now has a concrete stopping point. For arbitrary homogeneous $I$, show that the compression defect has sufficient decay and multiplicity control. A bound $\|A_{j,k}\|=O(k^{-1/2})$ together with the Hilbert-function rank estimate would suffice, but has not been established here. A weaker estimate might still suffice if the large singular values occupy sufficiently small subspaces; merely knowing the dimension of $Z(I)$ does not prove either assertion. The coordinate-power theorem, the endpoint obstruction, and formula $5$ are complete partial results. The full conjecture remains unresolved by this attempt.

Sources

• S1 Englis and Eschmeier, Geometric Arveson-Douglas conjecture. https://users.math.cas.cz/~englis/85.pdf. Formulation source.