Problem: Let $t \to P_t$ be a one parameter subgroup $\mathbb{C}^* \to \text{Gl}_{n}(\mathbb{C})$.

Let $X$ be a $n \times n$ nilpotent matrix.

I want to show that if $lim_{t \to 0} P_t^{-1}XP_t = Y$ then $\text{rank}(X) \geq \text{rank}(Y)$.

I have checked that this is true if $X$ is in the Jordan canonical form and the one parameter subgroups are in diagonal form. Firstly in this case we can assume that $X$ is indecomposable(hence rank is n-1) and so only one jordan block(since the actions are blockwise due to the nature of $P_t$), then if $P_t(i, i) = t^{i}$, then the $Y$ as defined above has $t^{i_k +i_{k+1}}$ on the super diagonal entries and $0$ elsewhere.

So, the rank is at most $n-1$.

But for an arbitrary nilpotent matrix I don't know how to show.

We can make one simplification: if $Z$ is a nilpotent matrix then $Z = A^{-1}XA$ where $X$ is in Jordan canonical form, but then we cannot assume that $P_t$ are diagonal and so on...

So, either we prove that given $X$ in Jordan Canonical form and $P_t$ in arbitrary form..

or $X$ in arbitrary form and $P_t$ in special form that the rank cannot increase on taking limits.

I am interested in this question since combined with

  1. fact that any point in the Zariski closure of a conjugacy class in the affine variety of these nilpotent matrices is captured by a one parameter subgroup

  2. If $Y$ is in closure of $X$ then, $Y^2$ is in closure of $X^2$ and so on....

  3. Partition(there is one partition corresponding ot every class) of $X$ majorizes partition of $Y$ iff $rank^{i}(X) \geq rank^{i}(Y)$

we would have shown that if $Y$ is in the closure of the conjugacy class of $X$ then the partition corresponding to $X$ majorizes that of $Y$.

I have shown the converse using one parameter subgroups and block matrix identitities here


1 Answer 1


This has nothing to do with one-parameter subgroup. In general, a sequence of (possibly rectangular) constant-rank matrices may converge only to a matrix of equal or smaller rank.

Let $Y=\lim_{i\to\infty}X_i$, where each $\operatorname{rank}(X_i)=r$ for each $i$ and $\operatorname{rank}(Y)=s$. If $Y=0$, there is nothing to prove. Suppose $Y\ne0$. Perform an economic singular value decomposition $Y=U\Sigma V^\ast$ where $V$ has $s$ orthonormal columns. Let $\mathcal V$ be the column space of $V$. Then $\|Yv\|_2\ge\sigma_s(Y)>0$ for all unit vectors $v\in\mathcal V$.

If $r<s$, then $n<(n-r)+s$. Hence $\ker(X_i)\cap\mathcal V\ne0$ and there exists some unit vector $u_i\in \ker(X_i)\cap\mathcal V$. Since the intersection of the unit sphere and $\mathcal V$ is compact, $\{u_i\}$ has a subsequence $\{u_{i_k}\}_{k\in\mathbb N}$ that converges to some unit vector $v\in\mathcal V$. But then $Yv=\lim_{k\to\infty}X_{i_k}u_{i_k}=0$, which contradicts our previous observation that $\|Yv\|_2>0$. Hence we must have $r\ge s$.

Remark. By considering an appropriate subsequence, the above result is equivalent to the following: if $X_1,X_2,\ldots$ have possibly different ranks but they converge to $Y$, then $\operatorname{rank}(Y)\le\liminf_{i\in\mathbb N}\operatorname{rank}(X_i)$.

  • $\begingroup$ Thank you! I was looking to show this for all points in the Zariski closure(which is a priori bigger than analytic closure), but due to the fact that constructible sets have same Zariski and Analytic closure this will do(or even by invoking that every limit point is captured by one parameter subgroups and hence is in the naalytic cloure) $\endgroup$ Commented Jun 8, 2022 at 4:53

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