Let $N \in \mathfrak{M}_n(\mathbb{C})$ nilpotent, such that there exists $X \in \mathfrak M_n(\mathbb{C})$ with $X^2=N$ (take for instance $n>2$ and $N(1,n)=1$; $N(i,j)=0$ otherwise).

Denote by $\mathcal{S}_N$ the set of $X \in \mathfrak M_n(\mathbb{C})$ such that $X^2=N$. Is $\mathcal{S}_N$ connected or path-connected ? What happens when we change $2$ by $3,4,\ldots $?


This is a memorial to an incorrect solution that used to be here. Unfortunately, I can't delete it since it was accepted.

  • $\begingroup$ The first statement about closure and refining partitions is not clear, can you elaborate? What about the following 6x6 matrix: the upper left and the lower right quarters are standard 3x3 Jordan blocks, and the upper right quarter has $\epsilon$ in its center (everything else is 0). I believe it has Jordan type (4,2) but converges to (3,3) as $\epsilon$ goes to 0. $\endgroup$ – Sergei Ivanov May 17 '10 at 9:28
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    $\begingroup$ The relevant order on partitions is not refinement, but dominance (en.wikipedia.org/wiki/Dominance_order). Thus two elements of $S_N$ can be in the same connected component only if the corresponding partitions can be "linked", using dominance, via partitions whose "square" is the partition of N. It is not clear to me that this is also sufficient: you are only allowed to conjugate elements of $S_N$ by matrices commuting with N, so even if the "poset connectedness" condition holds, you may not be able to conclude that the various elements of $S_N$ are in the same connected component. $\endgroup$ – damiano May 17 '10 at 10:44
  • $\begingroup$ Sergei -- I don't know about that particular example, but yes, I was too hasty with the first statement. $\endgroup$ – algori May 17 '10 at 11:50

Edit: This is just half an answer: I can only show that the sets matrices with $X^2=N$ and fixed Jordan type are path connected.

Every nilpotent matrix is conjugate to a nilpotent matrix in Jordan form, which is unique up to permutation of Jordam blocks. So we have a bijection

$$\mathrm{Nilp}_n(\mathbb C)/\mathrm{conjugation} \quad \cong \quad \mbox{integer partitions of }n$$

associating with a conjugacy class of a nilpotent matrix $X$ the sizes of its Jordan blocks $(a_1, \ldots, a_r)$ which sum up to $n$. The max of the $a_i$'s is the nilpotency-degree of $X$. To the class of $X^2$ is associated the partition $$(\lfloor (a_1 +1)/2 \rfloor, \lfloor a_1/2\rfloor ,\lfloor (a_2 +1)/2 \rfloor ,\lfloor a_2/2\rfloor, \ldots, \lfloor a_r/2 \rfloor)$$

From here we can derive a necessary and sufficient condition on a nilpotent matrix to be a square.

Now fix your preferred nilpotent matrix $N$. Let $X$ be a matrix with $X^2=N$ and Jordan type $(a_1, \ldots, a_r)$. Conjugating the whole setup, we may assume $X$ is in Jordan form.

Let $Y$ be a matrix with $Y^2=X^2 = N$ having the same Jordan type as $X$, and let us construct a path from $X$ to $Y$. Since $X$ and $Y$ have the same Jordan type, there exists an invertible matrix $S$ with $Y=SXS^{-1}$. Because $Y^2=X^2=N$ the matrix $S$ commutes with $N$. It is enough to construct a path from the identity matrix to $S$ in the set $\mathcal C_N$ of invertible matrices that commute with $N$.

I claim $\mathcal C_N$ is path connected (for just any $N$). Indeed, the set $[N]$ of all commutators of $N$ is linear subspace of the vector space $\mathrm M_n(\mathbb C)$. The determinant, as a function on $[N]$ is a polynomial function which is not identically zero since the identity matrix belongs to $[N]$. Thus, the zero set of the determinant is Zariski closed, so $\mathcal C_N$ is Zariski open in $[N]$. Any Zariski-open in a complex vector space is path connected.

What remains is to connect different Jordan types. We certainly can connect $(5,2)$ with $(5,1,1)$ by the 1 in the $2\times 2$--block with $t$ and vary $t$ from $1$ to zero. The problem that remains is to connect, say, type $(4,2)$ with type $(3,3)$ as pointed out in the comments.

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    $\begingroup$ How do you connect Jordan type (3,3) to (4,2)? They have the same square. $\endgroup$ – Sergei Ivanov May 16 '10 at 11:45
  • $\begingroup$ Jordan type $(3,3)$ has square $(2,1,2,1) = (2,2,1,1)$ and Jordan type $(4,2)$ has square $(3,1,1,1)$. Is that not so? $\endgroup$ – Xandi Tuni May 16 '10 at 11:57
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    $\begingroup$ No, the square of Jordan type (4) is (2,2). $\endgroup$ – Sergei Ivanov May 16 '10 at 12:02
  • $\begingroup$ Oops, my fault. I will look if I can fix that. $\endgroup$ – Xandi Tuni May 16 '10 at 12:11

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