An infinite family disproving the Prasad–Ram invariant-flag formula

Source. Gaurav Bhatnagar and Amritanshu Prasad, A bibasic double sum extension of a q-binomial theorem arising out of subspace enumeration, Conjecture 5, equation (2.3). The distinct scalar-matrix specialization, Conjecture 1, is proved in that paper and is not disputed here.

The proposed formula fails for every prime power $q$ and every positive integer $m\neq3$. Indeed, every operator with irreducible characteristic polynomial of degree $3m$ is a counterexample. The smallest counterexample has $q=2$ and $m=1$. A second example with $q=2$ and $m=2$ avoids negative powers entirely: the conjectured integer is $168$, whereas the actual integer is $336$.

1. The source definitions

Let $V=\mathbb F_q^{3m}$ and $T\in\operatorname{End}_{\mathbb F_q}(V)$. The quantity on the left side of the conjecture is the number of splitting subspaces

$$\sigma^T_{(m,m,m)} = \#\left\{ W\leq V: \dim W=m,\quad V=W\oplus T(W)\oplus T^2(W) \right\}. \tag{1}$$

For nonnegative integers $j_1,j_2,j_3$ with sum $3m$, the quantity on the right side is the number of invariant flags

$$Y^T_{(j_1,j_2,j_3)} = \#\left\{ 0=W_0\subseteq W_1\subseteq W_2\subseteq W_3=V: T(W_i)\subseteq W_i,\quad \dim(W_i/W_{i-1})=j_i \right\}. \tag{2}$$

Zero increments are allowed, as required by the source's sum over all triples of nonnegative integers. Put

$$g(j_1,j_2,j_3) = \begin{cases} 2j_2+j_3,&j_2+2j_3\equiv2\pmod3, \\ j_2+2j_3,&j_2+2j_3\equiv0,1\pmod3. \end{cases} \tag{3}$$

For a primitive complex cube root of unity $\omega$, the source conjectures

$$\sigma^T_{(m,m,m)} = q^{m^2-m} \sum_{j_1+j_2+j_3=3m} q^{\frac13\left( \sum_{i=1}^3\binom{j_i}{2}-g(j_1,j_2,j_3) \right)} Y^T_{(j_1,j_2,j_3)} \omega^{j_2+2j_3}. \tag{4}$$

The left side counts splitting subspaces, not invariant chains; invariant chains occur only on the right.

2. Irreducible operators leave exactly three flags

Choose any monic irreducible polynomial of degree $3m$ over $\mathbb F_q$, and let $T$ be its companion operator. Equivalently, identify

$$V=\mathbb F_{q^{3m}}, \qquad T(v)=\alpha v, \qquad \mathbb F_q(\alpha)=\mathbb F_{q^{3m}}. \tag{5}$$

If a nonzero subspace $U\leq V$ is invariant under $T$, then it is invariant under every polynomial in $T$. Hence it is invariant under multiplication by every element of $\mathbb F_q[\alpha]=\mathbb F_{q^{3m}}$. Any nonzero $u\in U$ then gives

$$V=\mathbb F_{q^{3m}}u\subseteq U,$$

so $U=V$. Therefore the only invariant subspaces are $0$ and $V$.

Consequently, the only nonzero flag counts are

$$Y^T_{(3m,0,0)} = Y^T_{(0,3m,0)} = Y^T_{(0,0,3m)} =1. \tag{6}$$

Write

$$C=\binom{3m}{2}=\frac{3m(3m-1)}2.$$

For the three triples in (6), respectively, the values of $g$ are $0$, $3m$, and $6m$, and all three powers of $\omega$ equal one. Thus the proposed right-hand side becomes

$$\begin{aligned} R(q,m) &= q^{m^2-m} \left( q^{C/3} +q^{(C-3m)/3} +q^{(C-6m)/3} \right) \\ &= q^{(5m^2-7m)/2} \left(q^{2m}+q^m+1\right). \end{aligned} \tag{7}$$

All exponents in the first line are integers. For $m\geq2$, all three source summands have nonnegative exponents.

3. The established splitting-subspace count gives a different answer

The splitting-subspace theorem of Eric Chen and Dennis Tseng, The Splitting Subspace Conjecture, Finite Fields and Their Applications 24 (2013), 15–28, Corollary 3.4, states that whenever $\mathbb F_q(\alpha)=\mathbb F_{q^{md}}$, the number of $m$-dimensional subspaces satisfying

$$\mathbb F_{q^{md}} = W\oplus\alpha W\oplus\cdots\oplus\alpha^{d-1}W$$

is

$$\frac{q^{md}-1}{q^m-1}\,q^{m(m-1)(d-1)}. \tag{8}$$

Applying this previously established theorem with $d=3$ to (5), the actual left side of (4) is

$$\sigma^T_{(m,m,m)} = \left(q^{2m}+q^m+1\right)q^{2m(m-1)}. \tag{9}$$

Comparing (9) with (7) gives the exact discrepancy

$$\boxed{ \frac{\sigma^T_{(m,m,m)}}{R(q,m)} =q^{m(3-m)/2}. } \tag{10}$$

Since $q>1$ and $m\geq1$, equality holds for this family precisely when $m=3$. Therefore every irreducible operator of degree $3m$ disproves (4) whenever $m\neq3$. Irreducible polynomials exist in every positive degree over every finite field, so the counterexamples cover every prime power and every such dimension.

4. The smallest counterexample needs no external enumeration theorem

Take $q=2$, $m=1$, and the irreducible polynomial $X^3+X+1$. In the basis $1,\alpha,\alpha^2$, multiplication by $\alpha$ is

$$T= \begin{pmatrix} 0&0&1 \\ 1&0&1 \\ 0&1&0 \end{pmatrix}. \tag{11}$$

For every nonzero $v\in\mathbb F_2^3$, the vectors

$$v,\quad T(v),\quad T^2(v)$$

are linearly independent: any dependence would give a nonzero annihilating polynomial for $v$ of degree at most two, contradicting the irreducibility of $X^3+X+1$. Every one-dimensional subspace therefore splits. Since $\mathbb F_2^3$ has seven one-dimensional subspaces,

$$\sigma^T_{(1,1,1)}=7. \tag{12}$$

By (6), the only invariant flags have increment triples

$$(3,0,0),\qquad(0,3,0),\qquad(0,0,3).$$

Their respective contributions to the source formula are

$$2,\qquad1,\qquad\frac12.$$

Hence (4) predicts

$$R(2,1)=2+1+\frac12=\frac72\neq7. \tag{13}$$

More generally, the same elementary cyclic-vector argument works for every prime power when $m=1$, giving

$$\sigma^T_{(1,1,1)}=q^2+q+1, \qquad R(q,1)=q+1+q^{-1}. \tag{14}$$

5. An integral counterexample eliminates negative-exponent concerns

Take instead $q=2$, $m=2$, and the irreducible polynomial

$$X^6+X+1\in\mathbb F_2[X]. \tag{15}$$

The three flag contributions in (7) are all positive integers:

$$R(2,2)=2^7+2^5+2^3=168. \tag{16}$$

On the other hand, (9) gives

$$\sigma^T_{(2,2,2)} =(2^4+2^2+1)2^4 =336. \tag{17}$$

This value can also be checked without invoking (8): there are exactly

$$\begin{bmatrix}6\\2\end{bmatrix}_2 =651$$

two-dimensional subspaces of $\mathbb F_2^6$, and direct Gaussian elimination shows that exactly $336$ satisfy

$$\mathbb F_2^6=W\oplus T(W)\oplus T^2(W). \tag{18}$$

Thus the failure is substantive even when every term of the conjectural formula is an integer. The scalar specialization proved in the source remains correct; what fails is its proposed extension to arbitrary operators.

Conclusion: Source Conjecture 5 is DISPROVED for every prime power $q$ and every positive dimension parameter $m\neq3$.