Smoothness and rationality conjecture for triangularization varieties

The conjecture is false already over $k=\mathbb Q$ in dimension $n=6$. Let

$$J_2(0)=\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad M=\operatorname{diag}(J_2(0),J_2(0),0,0)\in M_6(\mathbb Z).$$

Because $M$ is already upper triangular,

$$I_6\in X_M(\mathbb Q).$$

Let $B\subseteq\mathrm{GL}_6$ be the upper-triangular Borel subgroup and

$$\pi:\mathrm{GL}_6\longrightarrow\mathrm{GL}_6/B$$

the complete-flag quotient. For $g\in\mathrm{GL}_6$,

$$g^{-1}Mg\text{ is upper triangular} \quad\Longleftrightarrow\quad \pi(g)\text{ is an $M$-stable complete flag}.$$

Therefore, with reduced structures,

$$X_M=\pi^{-1}(\mathcal B_M),$$

where

$$\mathcal B_M =\{V_\bullet:M(V_i)\subseteq V_i\text{ for every }i\}$$

is the Springer fiber of $M$.

The quotient map $\pi$ is a Zariski-locally trivial principal $B$-bundle. Consequently, if $K$ is an irreducible component of $\mathcal B_M$, then

$$Y=\pi^{-1}(K)$$

is an irreducible component of $X_M$, locally isomorphic to $K\times B$. Since $B$ is smooth,

$$Y\text{ is smooth}\quad\Longleftrightarrow\quad K\text{ is smooth}. \tag{1}$$

The nilpotent Jordan type of $M$ is

$$(2,2,1,1).$$

Fresse and Melnikov proved that all components of a type-A Springer fiber are nonsingular if and only if its Jordan partition is a hook, has two rows, has shape $(\lambda_1,\lambda_2,1)$, or equals $(2,2,2)$. See L. Fresse and A. Melnikov, On the singularity of the irreducible components of a Springer fiber in sl(n), Selecta Mathematica 16 (2010), 393–418, https://arxiv.org/abs/0905.1617.

The partition $(2,2,1,1)$ belongs to none of these classes. Therefore $\mathcal B_M$ has a singular irreducible component $K$. Since $M$ is split over $\mathbb Q$, its standard-tableau Spaltenstein strata and their component closures are defined by determinantal rank conditions over $\mathbb Q$; hence $K$ is geometrically irreducible and defined over $\mathbb Q$.

There is also an explicit rational singular point. Let

$$P=[\,e_1\ \ e_3\ \ e_5\ \ e_6\ \ e_2\ \ e_4\,].$$

The two-column singularity criterion shows that its column-filled Jordan-basis flag $\pi(P)$ belongs to every component and is singular on each singular component. See L. Fresse, Singular components of Springer fibers in the two-column case, Annales de l'Institut Fourier 59 (2009), 2429–2444, https://arxiv.org/abs/0803.2188. Therefore

$$\pi(P)\in\operatorname{Sing}(K) \quad\text{and}\quad P\in\operatorname{Sing}(Y).$$

Thus $X_M(\mathbb Q)\neq\varnothing$, but $X_M$ has an irreducible component singular at an explicit rational point. This contradicts the asserted smoothness of every component.