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In a multiplicative group ${G}$, the commutator of two group elements ${g, h}$ is defined as ${[g,h] := g^{-1}h^{-1}gh}$ (other conventions are also in use, though they are largely equivalent for the purposes of this discussion). A group is said to be nilpotent of step ${s}$ (or more precisely, step ${\leq s}$), if all iterated commutators of order ${s+1}$ or higher necessarily vanish. For instance, a group is nilpotent of order ${1}$ if and only if it is abelian, and it is nilpotent of order ${2}$ if and only if ${[[g_1,g_2],g_3]=id}$ for all ${g_1,g_2,g_3}$ (i.e. all commutator elements ${[g_1,g_2]}$ are central), and so forth. A good example of an ${s}$-step nilpotent group is the group of ${s+1 \times s+1}$ upper-triangular unipotent matrices (i.e. matrices with ${1}$s on the diagonal and zero below the diagonal), and taking values in some ring (e.g. reals, integers, complex numbers, etc.).

Another important example of nilpotent groups arise from operations on polynomials. For instance, if ${V_{\leq s}}$ is the vector space of real polynomials of one variable of degree at most ${s}$, then there are two natural affine actions on ${V_{\leq s}}$. Firstly, every polynomial ${Q}$ in ${V_{\leq s}}$ gives rise to an “vertical” shift ${P \mapsto P+Q}$. Secondly, every ${h \in {\bf R}}$ gives rise to a “horizontal” shift ${P \mapsto P(\cdot+h)}$. The group generated by these two shifts is a nilpotent group of step ${\leq s}$; this reflects the well-known fact that a polynomial of degree ${\leq s}$ vanishes once one differentiates more than ${s}$ times. Because of this link between nilpotentcy and polynomials, one can view nilpotent algebra as a generalisation of polynomial algebra.

Suppose one has a finite number ${g_1,\ldots,g_n}$ of generators. Using abstract algebra, one can then construct the free nilpotent group ${{\mathcal F}_{\leq s}(g_1,\ldots,g_n)}$ of step ${\leq s}$, defined as the group generated by the ${g_1,\ldots,g_n}$ subject to the relations that all commutators of order ${s+1}$ involving the generators are trivial. This is the universal object in the category of nilpotent groups of step ${\leq s}$ with ${n}$ marked elements ${g_1,\ldots,g_n}$. In other words, given any other ${\leq s}$-step nilpotent group ${G'}$ with ${n}$ marked elements ${g'_1,\ldots,g'_n}$, there is a unique homomorphism from the free nilpotent group to ${G'}$ that maps each ${g_j}$ to ${g'_j}$ for ${1 \leq j \leq n}$. In particular, the free nilpotent group is well-defined up to isomorphism in this category.

In many applications, one wants to have a more concrete description of the free nilpotent group, so that one can perform computations more easily (and in particular, be able to tell when two words in the group are equal or not). This is easy for small values of ${s}$. For instance, when ${s=1}$, ${{\mathcal F}_{\leq 1}(g_1,\ldots,g_n)}$ is simply the free abelian group generated by ${g_1,\ldots,g_n}$, and so every element ${g}$ of ${{\mathcal F}_{\leq 1}(g_1,\ldots,g_n)}$ can be described uniquely as

$\displaystyle g = \prod_{j=1}^n g_j^{m_j} := g_1^{m_1} \ldots g_n^{m_n} \ \ \ \ \ (1)$

for some integers ${m_1,\ldots,m_n}$, with the obvious group law. Indeed, to obtain existence of this representation, one starts with any representation of ${g}$ in terms of the generators ${g_1,\ldots,g_n}$, and then uses the abelian property to push the ${g_1}$ factors to the far left, followed by the ${g_2}$ factors, and so forth. To show uniqueness, we observe that the group ${G}$ of formal abelian products ${\{ g_1^{m_1} \ldots g_n^{m_n}: m_1,\ldots,m_n \in {\bf Z} \} \equiv {\bf Z}^k}$ is already a ${\leq 1}$-step nilpotent group with marked elements ${g_1,\ldots,g_n}$, and so there must be a homomorphism from the free group to ${G}$. Since ${G}$ distinguishes all the products ${g_1^{m_1} \ldots g_n^{m_n}}$ from each other, the free group must also.

It is only slightly more tricky to describe the free nilpotent group ${{\mathcal F}_{\leq 2}(g_1,\ldots,g_n)}$ of step ${\leq 2}$. Using the identities

$\displaystyle gh = hg [g,h]; \quad gh^{-1} = ([g,h]^{-1})^{g^{-1}} h^{-1} g; \quad g^{-1} h = h [g,h]^{-1} g^{-1}; \quad g^{-1} h^{-1} := [g,h] g^{-1} h^{-1}$

(where ${g^h := h^{-1} g h}$ is the conjugate of ${g}$ by ${h}$) we see that whenever ${1 \leq i < j \leq n}$, one can push a positive or negative power of ${g_i}$ past a positive or negative power of ${g_j}$, at the cost of creating a positive or negative power of ${[g_i,g_j]}$, or one of its conjugates. Meanwhile, in a ${\leq 2}$-step nilpotent group, all the commutators are central, and one can pull all the commutators out of a word and collect them as in the abelian case. Doing all this, we see that every element ${g}$ of ${{\mathcal F}_{\leq 2}(g_1,\ldots,g_n)}$ has a representation of the form

$\displaystyle g = (\prod_{j=1}^n g_j^{m_j}) (\prod_{1 \leq i < j \leq n} [g_i,g_j]^{m_{[i,j]}}) \ \ \ \ \ (2)$

for some integers ${m_j}$ for ${1 \leq j \leq n}$ and ${m_{[i,j]}}$ for ${1 \leq i < j \leq n}$. Note that we don’t need to consider commutators ${[g_i,g_j]}$ for ${i \geq j}$, since

$\displaystyle [g_i,g_i] = id$

and

$\displaystyle [g_i,g_j] = [g_j,g_i]^{-1}.$

It is possible to show also that this representation is unique, by repeating the previous argument, i.e. by showing that the set of formal products

$\displaystyle G := \{ (\prod_{j=1}^k g_j^{m_j}) (\prod_{1 \leq i < j \leq n} [g_i,g_j]^{m_{[i,j]}}): m_j, m_{[i,j]} \in {\bf Z} \}$

forms a ${\leq 2}$-step nilpotent group, after using the above rules to define the group operations. This can be done, but verifying the group axioms (particularly the associative law) for ${G}$ is unpleasantly tedious.

Once one sees this, one rapidly loses an appetite for trying to obtain a similar explicit description for free nilpotent groups for higher step, especially once one starts seeing that higher commutators obey some non-obvious identities such as the Hall-Witt identity

$\displaystyle [[g, h^{-1}], k]^h\cdot[[h, k^{-1}], g]^k\cdot[[k, g^{-1}], h]^g = 1 \ \ \ \ \ (3)$

(a nonlinear version of the Jacobi identity in the theory of Lie algebras), which make one less certain as to the existence or uniqueness of various proposed generalisations of the representations (1) or (2). For instance, in the free ${\leq 3}$-step nilpotent group, it turns out that for representations of the form

$\displaystyle g = (\prod_{j=1}^n g_j^{m_j}) (\prod_{1 \leq i < j \leq n} [g_i,g_j]^{m_{[i,j]}}) (\prod_{1 \leq i < j < k \leq n} [[g_i,g_j],g_k]^{n_{[[i,j],k]}})$

one has uniqueness but not existence (e.g. even in the simplest case ${n=3}$, there is no place in this representation for, say, ${[[g_1,g_3],g_2]}$ or ${[[g_1,g_2],g_2]}$), but if one tries to insert more triple commutators into the representation to make up for this, one has to be careful not to lose uniqueness due to identities such as (3). One can paste these in by ad hoc means in the ${s=3}$ case, but the ${s=4}$ case looks more fearsome still, especially now that the quadruple commutators split into several distinct-looking species such as ${[[g_i,g_j],[g_k,g_l]]}$ and ${[[[g_i,g_j],g_k],g_l]}$ which are nevertheless still related to each other by identities such as (3). While one can eventually disentangle this mess for any fixed ${n}$ and ${s}$ by a finite amount of combinatorial computation, it is not immediately obvious how to give an explicit description of ${{\mathcal F}_{\leq s}(g_1,\ldots,g_n)}$ uniformly in ${n}$ and ${s}$.

Nevertheless, it turns out that one can give a reasonably tractable description of this group if one takes a polycyclic perspective rather than a nilpotent one – i.e. one views the free nilpotent group as a tower of group extensions of the trivial group by the cyclic group ${{\bf Z}}$. This seems to be a fairly standard observation in group theory – I found it in this book of Magnus, Karrass, and Solitar, via this paper of Leibman – but seems not to be so widely known outside of that field, so I wanted to record it here.