Explicit-generator conjecture for squared Schubert varieties

A counterexample to Friedman--Rosana--Sturmfels, Conjecture 6.1

Exact source statement

In Hannah Friedman, Andrea Rosana, and Bernd Sturmfels, Distance Optimization in the Grassmannian of Lines, arXiv:2601.22843v1, Section 6, the authors define, for $1\leq i<j\leq n$,

$$\mathcal S_{ij}=\operatorname{Gr}(2,n)\cap \mathcal L_{ij}, \qquad \mathcal L_{ij}=\{X=[x_{rs}]:x_{rs}=0\text{ unless }i\leq r, \ j\leq s\},$$

and let $\mathcal S_{ij}^2$ be the projective image closure under $X\mapsto P=X^2$. Immediately before Conjecture 6.1, $P_{rs}$ is defined to be the $2\times n$ submatrix of the symmetric matrix $P$ consisting of rows $r,s$. Conjecture 6.1 asserts that the prime ideal of $\mathcal S_{ij}^2$ is generated by

1. the entries of $2P^2-\operatorname{tr}(P)P$;
2. the $3\times3$ minors of $P$; and
3. the $2\times2$ minors of $P_{rs}$ for $r<i$ or $s<j$.

There is no radical or saturation in the statement.

Universal extraneous projective locus

Work over $\mathbb C$, as in the paper, and let

$$z=(1,\sqrt{-1},0,\ldots,0)^T,\qquad P_z=zz^T.$$

Then $P_z\neq0$, so $[P_z]$ is a genuine point of $\mathbb P(\operatorname{Sym}^2\mathbb C^n)$. Since

$$z^Tz=1+(\sqrt{-1})^2=0, \qquad \operatorname{tr}(P_z)=z^Tz=0, \qquad P_z^2=z(z^Tz)z^T=0,$$

every entry of $2P_z^2-\operatorname{tr}(P_z)P_z$ vanishes. Also $\operatorname{rank}(P_z)=1$, so every $3\times3$ minor of $P_z$ vanishes and every $2\times2$ minor of every two-row submatrix $(P_z)_{rs}$ vanishes. Thus $[P_z]$ belongs to the projective zero set of the proposed generators for every pair $(i,j)$.

More invariantly, the proposed zero set always contains the Veronese image

$$\nu_2\bigl(\{[z]\in\mathbb P^{n-1}:z^Tz=0\}\bigr),$$

the traceless symmetric rank-one locus.

Contradiction for the point Schubert variety

Take $(i,j)=(n-1,n)$ (for instance $(n,i,j)=(3,2,3)$). By the definition above, a skew-symmetric matrix in $\mathcal S_{n-1,n}$ has only the entry $x_{n-1,n}$ and its negative possibly nonzero. Hence $\mathcal S_{n-1,n}$ is one point, and so is its square image:

$$\mathcal S_{n-1,n}^2 =\bigl\{[\operatorname{diag}(0,\ldots,0,1,1)]\bigr\}.$$

(The harmless overall sign depends on the convention for $X^2$.) The point $[P_z]$ is not this point: its $(1,1)$-entry is nonzero, whereas the displayed point has zero $(1,1)$-entry when $n\geq3$. Therefore the proposed generators have a projective zero outside $\mathcal S_{n-1,n}^2$. They do not generate its prime ideal, even set-theoretically.

For the smallest concrete instance $(n,i,j)=(3,2,3)$, the extraneous matrix is

$$P_z= \begin{bmatrix} 1&\sqrt{-1}&0\\ \sqrt{-1}&-1&0\\ 0&0&0 \end{bmatrix},$$

while the square image is the single point represented by $\operatorname{diag}(0,1,1)$.

In this smallest instance one can classify the entire proposed zero set. On the chart $t=\operatorname{tr}(P)\neq0$, the matrix equation makes $2P/t$ a symmetric rank-two idempotent; the two-row conditions force its first row and column to vanish, leaving precisely the point $[\operatorname{diag}(0,1,1)]$. On $t=0$, the matrix equation says $P^2=0$. A $3\times3$ symmetric square-zero matrix has rank at most one, because $\operatorname{im}(P)\subseteq\ker(P)$. Every nonzero rank-one symmetric matrix is projectively $zz^T$, and square-zero is equivalent to $z^Tz=0$. Hence the proposed projective zero set is exactly

$$\bigl\{[\operatorname{diag}(0,1,1)]\bigr\} \;\sqcup\; \nu_2\bigl(\{[z]\in\mathbb P^2:z^Tz=0\}\bigr).$$

It is reducible (a point disjoint from the displayed Veronese curve), so even the radical of the proposed generator ideal is not prime.

There is also an immediate graded obstruction. All listed proposed generators have degree $2$ or $3$, so their homogeneous ideal has zero degree-one part. The ideal of the displayed point has many nonzero linear forms. The two ideals therefore cannot be equal.

In fact, the same graded obstruction disproves the claim for every $i>1$. The first $i-1$ rows and columns of every $X\in\mathcal S_{ij}$ are zero. Therefore the first $i-1$ rows and columns of $P=X^2$ are zero, so the prime ideal of $\mathcal S_{ij}^2$ contains all linear forms $p_{ab}$ with $a<i$ (or, symmetrically, $b<i$). The ideal generated by the listed quadrics and cubics has no nonzero linear forms.

Why standard repairs do not change the conclusion

Let $J_{ij}$ be the homogeneous ideal generated by the three listed families and let $\mathfrak m$ be the irrelevant ideal. The point $[P_z]$ remains in $V(J_{ij}:\mathfrak m^\infty)$. Indeed, if $f\in J_{ij}:\mathfrak m^k$, choose a homogeneous coordinate $h$ with $h(P_z)\neq0$. Then $h^kf\in J_{ij}$, whence $h(P_z)^k f(P_z)=0$ and $f(P_z)=0$. Thus ordinary projective saturation cannot remove this genuine projective locus. Passing to the radical cannot remove it either, because radicalization preserves the zero set.

Saturation by the special equation $\operatorname{tr}(P)$ would remove the displayed extraneous points on the trace-zero boundary, but no such saturation appears in Conjecture 6.1. It would be a different statement and would still require a separate proof that the resulting saturated ideal is the desired projective image ideal.

Consequently, Conjecture 6.1 as literally stated in the cited paper is false.

Convention audit

The obstruction is unaffected by the possible global sign convention $P=X^2$ versus $P=-X^2$, because all varieties are projective. It is also not an issue of working over the real numbers: the paper places the prime ideal in $\mathbb P(\operatorname{Sym}^2\mathbb C^n)$, and, even if one started over $\mathbb R$, equality of the stated polynomial ideals would survive extension of scalars to $\mathbb C$, where the counterexample exists.

If one instead normalizes to affine projection matrices by imposing $\operatorname{tr}(P)=2$, then the trace-zero matrix $P_z$ is outside that affine chart and homogeneity is lost. But Conjecture 6.1 explicitly concerns the homogeneous prime ideal of the projective square image, not merely equations on this affine chart. Taking the projective closure restores trace-zero boundary points, and the listed equations alone do not select the correct Schubert-dependent part of that boundary.

The precise saturation repair

There is a clean corrected statement. Let $J_{ij}$ be the ideal generated by the three families in Conjecture 6.1 and put $t=\operatorname{tr}(P)$. Then

$$I(\mathcal S_{ij}^2)=J_{ij}:t^\infty.$$

Here is a proof, included to isolate exactly what goes wrong in the published formulation.

After localizing at $t$, put $E=2P/t$. The quadratic matrix equation becomes

$$E^2=E,\qquad E^T=E,\qquad \operatorname{tr}(E)=2.$$

Thus $E$ is the symmetric idempotent onto a nondegenerate two-plane $U\subset\mathbb C^n$. Conversely every nondegenerate two-plane has a unique symmetric idempotent with image $U$. Scheme-theoretically, the symmetric idempotents of trace two form a smooth irreducible scheme: orthogonal conjugation is transitive, and at $E_0=\operatorname{diag}(1,1,0,\ldots,0)$ the linearized equation $HE_0+E_0H=H$, for symmetric $H$, forces the two diagonal blocks of $H$ to vanish and leaves the $2\times(n-2)$ off-diagonal block free. This is precisely the smooth open subset of $\operatorname{Gr}(2,n)$ where the ambient symmetric form restricts nondegenerately. In particular, the $3\times3$ minors are redundant after localizing at $t$.

Let $x=(x_{ab})_{a<b}$ be the Pluecker vector of $U$, and set

$$Q(x)=\sum_{a<b}x_{ab}^2.$$

Nondegeneracy of $U$ is equivalent to $Q(x)\neq0$. The second exterior power of the orthogonal projector is the rank-one projector onto $\bigwedge^2U$:

$$\bigwedge^2E=\frac{xx^T}{Q(x)}.$$

Its $((r,s),(a,b))$-entry is exactly the $2\times2$ minor of $E$ using rows $r,s$ and columns $a,b$. Therefore the last family of generators, pulled back to the Grassmannian chart $Q(x)\neq0$, is

$$x_{rs}x_{ab}\quad(a<b), \qquad\text{for every }(r,s)\text{ with }r<i\text{ or }s<j.$$

They cut out the Schubert variety scheme-theoretically on this open chart. To see the scheme assertion without choosing a square root of $Q$, use the degree-zero coordinate ring of the homogeneous localization at $Q$. Every even-degree homogeneous element of the Pluecker ideal $\langle x_f:f\text{ forbidden}\rangle$ is a sum of terms $x_fx_gH_{fg}$: write each positive-degree coefficient of $x_f$ as a sum $\sum_gx_gH_{fg}$. After dividing by a power of $Q$, this says that the contracted ideal in the degree-zero localization is generated by the functions $x_fx_g/Q$, exactly the displayed row-minor functions. (The identity $x_fQ=\sum_gx_g(x_fx_g)$ also makes the set-theoretic implication transparent.) It follows that

$$(J_{ij})_t=I(\mathcal S_{ij}^2)_t.$$

The right-hand ideal is prime and does not contain $t$, since the Schubert variety has nondegenerate real points. Contracting the localized equality back to the homogeneous coordinate ring gives

$$J_{ij}:t^\infty =I(\mathcal S_{ij}^2):t^\infty =I(\mathcal S_{ij}^2),$$

as claimed. Thus the proposed equations are correct on the trace-nonzero chart, but their unsaturated homogeneous ideal always retains too much of the trace-zero boundary.