Similarity Classes of Nilpotents

We compute the closure of the similarity classes of nilpotents of \(M_n(\mathbb C)\). This is the Gerstenhaber–Hesselink theorem.

Theorem 1 (Gerstenhaber–Hesselink). Let \(m\in\operatorname{GL}_n(\mathbb C)\) be a nilpotent matrix. Then the closure of \[\left\{gmg^{-1}:g\in\operatorname{GL}_n(\mathbb C)\right\}\] is \[\left\{m’\in M_n(\mathbb C):\operatorname{rank}(m’)^k\le\operatorname{rank}m^k\text{ for all }k\ge0\right\}.\]

Throughout, we will work with the complex analytic topology on \(M_n(\mathbb C)\) for simplicity, but the reader familiar with the Zariski topology is encouraged to rewrite the arguments accordingly to work with arbitrary algebraically closed fields. Let’s begin by explaining why the target set should be closed.

Lemma 2. Fix a nonnegative integer \(r\). Then the subset \[\{g\in M_n(\mathbb C):\operatorname {rank}g\ge r\}\] is open.

Proof. Observe that \(\operatorname{rank}g\ge r\) if and only if the columns of \(g\) feature \(r\) linearly independent vectors. By a row-reduction, this is in turn equivalent to asserting that there is an \((r\times r)\)-minor (coming from these linearly independent vectors) which is invertible. Thus, we may encode the condition \(\operatorname{rank}g\ge r\) as asserting that the determinant of some \((r\times r)\)-minor of \(g\) is nonzero. The determinant of each minor is a polynomial in \(g\), so we conclude that the given subset is a union of open subsets and hence open (because the nonzero locus of a continuous function is open). \(\blacksquare\)

Proposition 3. Fix a nonnegative integer \(r\). Then the closure of \[\{m\in M_n(\mathbb C):\operatorname {rank}m=r\}\] is \[\{m\in M_n(\mathbb C):\operatorname {rank}m\le r\}.\]

Proof. By the lemma, we see that the second set is open, so it only remains to show density. Choose some \(m\) of rank at most \(r\), and we will show that any open neighborhood of \(m\) contains an element of rank exactly \(r\). By conjugating \(m\), we may assume that \(m\) is in Jordan normal form. By rearranging the rows and columns (notably breaking the Jordan normal form), we may assume that the first \(n-s\) columns all vanish, where \(s\) is the rank of \(m\). (The same form of \(m\) could be achieved by a judicious upper-triangularization of \(m\), where the point is that our basis should begin with a basis with the kernel of \(m\).)

We are given that \(s\le r\), so it is true that the first \(n-r\) columns all vanish. However, \(m\) also features \(s-r\) more vanishing columns. Well, any open neighborhood \(U\) of \(m\) contains a small open neighborhood where we can modify each individual coefficient of \(m\), so we simply define \(m’\) to agree with \(m\) except along the diagonal, where we adjust \(s-r\) of the vanishing entries to a very small complex number. Then \(m’\in U\) and \(\operatorname{rank}m’=r\), as required. \(\blacksquare\)

This now upgrades to show that the target set in Theorem 1 is in fact closed. The rest of the proof of Theorem 1 amounts to combinatorics with Jordan blocks. To see where this comes from, note that a nilpotent matrix \(m\in M_n(\mathbb C)\) admits a Jordan normal form, which can be encoded into a partition of \(n\): letting \(J_d\) be the Jordan block of size \(d\) with all eigenvalues equal to \(0\), we see that \[m=J_{d_1}\oplus\cdots\oplus J_{d_r},\] for some partition \((d_1,\ldots,d_r)\) of \(n\). Accordingly, we make the following provisional definition.

Definition 4. Given a nilpotent matrix \(m\in M_n(\mathbb C)\), we define the Jordan partition of \(m\) to be the partition \((d_1,\ldots,d_r)\) of \(n\), where the Jordan normal form of \(m\) is \[J_{d_1}\oplus\cdots\oplus J_{d_r}.\] By convention, a partition has \(d_1\ge\cdots\ge d_r\).

Remark 5. The theory of Jordan normal forms implies that two nilpotent matrices are similar if and only if they have the same Jordan partition.

It will be useful to note that the Jordan partition of \(m\) can be used to read off some ranks of \(m\).

Remark 6. The kernel of \(J_d\) has dimension \(1\), and more generally, the kernel of \(J_d^k\) has dimension \(\min\{k,d\}\). Thus, if \(m\) has Jordan partition \((d_1,\ldots,d_r)\), then \[\dim\ker m^k=\sum_{i=1}^{j}\min\{k,d_i\},\] so \[\dim\ker m^{k+1}-\dim\ker m^k=\left|\{i:d_i>k\}\right|.\] Equivalently, \[\operatorname{rank}m^{k}-\operatorname{rank}m^{k+1}=\left|\{i:d_i>k\}\right|.\] For example, this tells us that one can read the Jordan partition off of the numbers \(\dim\ker m^k\) or the numbers \(\operatorname{rank}m^k\).

Now, for nilpotent \(m\), we are interested in computing the closure of the set \[\operatorname{GL}_n(\mathbb C)\cdot m=\left\{gmg^{-1}:g\in\operatorname{GL}_n(\mathbb C)\right\}.\] Of course, the set \(\operatorname{GL}_n(\mathbb C)\cdot m\) is uniquely determined by the Jordan normal form of \(m\) and hence the Jordan partition of \(m\), but the closure may pick up other nilpotent elements. Let’s see some examples of this of roughly increasing power.

Example 7. Consider a nilpotent matrix \(m\) with Jordan partition \((2)\). Then we claim that zero is in the closure of \(\operatorname{GL}_2(\mathbb C)\cdot m\). Indeed, the family of matrices \[\begin{bmatrix}0 & a \\ & 0\end{bmatrix}\] can be seen to have Jordan partition \((2)\) whenever \(a\ne0\) by computing the ranks of the powers. However, the closure of this family includes the specialization \(a=0\), which is the zero matrix.

Example 8. Consider a nilpotent matrix \(m\) with Jordan partition \((n)\), meaning that the Jordan normal form of \(m\) has a single block. Then we claim that the closure of \(\operatorname{GL}_n(\mathbb C)\cdot m\) contains all nilpotents! Indeed, consider the family of matrices \[\begin{bmatrix} 0 & a_1 \\ & 0 & a_2 \\ & & \ddots & \ddots \\ & & & 0 & a_{n-1} \\ & & & & 0\end{bmatrix}.\] By computing the ranks of the powers of this matrix, we see that this matrix has Jordan partition \((n)\) as long as all the \(a_i\)s are nonzero. Thus, the entire family of matrices (even with some of the \(a_i\)s equal to \(0\)) is in the desired closure.

For example, we can specialize a select family of the \(a_i\)s to vanish to produce any Jordan normal form, up to scalars just above the diagonal. However, adjusting these nonzero scalars just above the diagonal does not change the Jordan partition because the ranks of the powers does not change.

Example 9. Consider the nilpotent matrix \(m\) with Jordan partition \((3,1)\). Then we claim that the closure of \(\operatorname{GL}_4(\mathbb C)\cdot m\) contains a nilpotent matrix with Jordan partition \((2,2)\). Indeed, consider the family of matrices \[m_a=\begin{bmatrix}0 & 1 \\ & 0 & a \\ & & 0 \\ & & 1 & 0 \end{bmatrix}.\] If \(a\ne0\), then \(m_a\) has Jordan partition \((3,1)\) by giving \(\mathbb C^4\) the basis \(\left\{e_3,m_ae_3,m_a^2e_3\right\}\cup\{e_4\}\). Thus, the specialization \(a=0\) lives in our closure, but this of course gives a nilpotent matrix with Jordan partition \((2,2)\).

Example 10. Consider a nilpotent matrix \(m\) with Jordan partition \((d_1,d_2)\) with \(d_1\ge d_2+2\). Then we claim that the closure of \(\operatorname{GL}_n(\mathbb C)\cdot m\) contains a nilpotent matrix with Jordan partition \((d_1-1,d_2+1)\). Indeed, consider the family of matrices \[m_a=\begin{bmatrix}J_{d_1-1} & aE_{d_11} \\ & J_{d_2+1}\end{bmatrix}.\] (Here, \(E_{d_11}\) is the elementary matrix which is nonzero everywhere except for a \(1\) in row \(d_1\) and column \(1\).) As long as \(a\ne0\), we can give \(\mathbb C^n\) the basis \[\left\{m_a^ke_{d_1}:0\le k\le d_1-1\right\}\cup\left\{e_{d_1+k}:1\le k\le d_2-1\right\}\] to see that \(m_a\) has Jordan partition \((d_1,d_2)\). However, this means that the specialization \(a=0\) must live in the closure, and this specialization has Jordan partition \((d_1-1,d_2+1)\).

The previous example gives us quite a bit of power. We are now ready to reveal the answer.

Definition 11. We define the dominance order on partitions as follows: for two partitions \(d=(d_1,d_2,\ldots)\) and \(d’=(d_1’,d_2’,\ldots)\) of \(n\), we say \(d\trianglerighteq d’\) if and only if \[(d_1+\cdots+d_k)\ge(d’_1+\cdots+d’_k)\] for each \(k\). Implicitly, we are appending infinitely many \(0\)s to the end of the partitions to allow this to make sense.

Remark 12. The dominance ordering is reflexive, antisymmetric, and transitive. In other words, \(\trianglerighteq\) defines a partial ordering on partitions.

Lemma 13. The dominance order is generated by the relations \(d\trianglerighteq d\) and \[(d_1,d_2,\ldots,d_i,\ldots,d_j,\ldots)\trianglerighteq(d_1,d_2,\ldots,d_i-1,\ldots,d_j+1,\ldots)\] provided that the right-hand side is a partition (still ordered in decreasing order).

Proof. Certainly these relations are included in the dominance order because the partial sums on the left are larger than the ones on the right. It remains to show that these relations generate the dominance order, so suppose we have \(d\trianglerighteq d’\) for two partitions \(d\) and \(d’\) of \(n\). By inducting on \(n\), we may assume that \(d\) and \(d’\) have no nonzero entries in common; otherwise, we may throw out the entries in common and reduce the value of \(n\).

For example, we now must have \(d_1>d_1’\). We will explain how to use the given relation to make the difference \(d_1-d_1’\) smaller, which will complete the proof because we can then inductiely apply this process. Because the sum of all the coefficients must be equal, there is an index \(i\) where \(d_{i-1}-d_i>d_{i-1}’-d_i’\ge0\). Thus, we may replace \(d\) with the partition given by \(d_1\mapsto d_1-1\) and \(d_i\mapsto d_i+1\). This new partition \(d\) has decreased the value of \(d_1-d_1’\), as required. \(\blacksquare\)

We are now ready to prove our theorem.

Proof of Theorem 1. Our proof is more or less taken from Joyce O’Halloran. In fact, we will show that the following are equivalent for a nilpotent matrix \(m\) and arbitrary matrix \(m’\).

1. \(m’\) is in the closure of \(\operatorname{GL}_n(\mathbb C)\cdot m\).
2. The matrix \(m’\) is nilpotent, and the Jordan partition of \(m’\) is dominated by the Jordan partition of \(m\).
3. We have \(\operatorname{rank}(m’)^k\le\operatorname{rank}m^k\) for all \(k\ge0\).

The equivalence of 1 and 3 then shows that the closure of \(\operatorname{GL}_n(\mathbb C)\cdot m\) equals the required set.

It remains to show our equivalences. Quickly, we note that both 1 and 3 still imply that \(m’\) is nilpotent: for 1, we see that the closure should be cut out by the equation \(x^n=0\), forcing \((m’)^n=0\); for 3, we see that the inequality \(\operatorname{rank}(m’)^n\le\operatorname{rank}m^n\) forces \((m’)^n=0\). Thus, throughout, we may let \(d\) and \(d’\) be the Jordan partitions of \(m\) and \(m’\), respectively.

• Note that 1 implies 3 because Lemma 2 shows that the conditions in 3 cut out a closed set, and this closed set contains \(\operatorname{GL}_n(\mathbb C)\cdot m\).
• To see that 3 implies 2, we note that 3 is equivalent to having \(\dim\ker(m’)^k\ge\dim\ker m^k\) for all \(k\). Remark 6 then explains that this is equivalent to having \[\sum_{i=1}^{r’}\min\{k,d_i’\}\ge\sum_{i=1}^{r}\min\{k,d_i\}\] for all \(k\). Notably, subtracting the left-hand sum from \(n=d_1’+\cdots+d_{r’}’\) gives \[\sum_{d_{i}’>k}(d_i’-k).\] Applying similar logic to the right-hand side shows that the partial sums of \(d’\) (namely, the partial sums of elements exceeding \(k\)) are bounded above by the partial sums of \(d\), so the result follows.
• Lastly, to see that 2 implies 1, we note that it is enough to handle the generating dominance relations of Lemma 13, which is covered by Example 10 (after adding in extra Jordan blocks). \(\blacksquare\)