In a multiplicative group , the commutator of two group elements is defined as (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 * (or more precisely, step ), if all iterated commutators of order or higher necessarily vanish. For instance, a group is nilpotent of order if and only if it is abelian, and it is nilpotent of order if and only if for all (i.e. all commutator elements are central), and so forth. A good example of an -step nilpotent group is the group of upper-triangular unipotent matrices (i.e. matrices with 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 is the vector space of real polynomials of one variable of degree at most , then there are two natural affine actions on . Firstly, every polynomial in gives rise to an “vertical” shift . Secondly, every gives rise to a “horizontal” shift . The group generated by these two shifts is a nilpotent group of step ; this reflects the well-known fact that a polynomial of degree vanishes once one differentiates more than 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 of generators. Using abstract algebra, one can then construct the *free nilpotent group* of step , defined as the group generated by the subject to the relations that all commutators of order involving the generators are trivial. This is the universal object in the category of nilpotent groups of step with marked elements . In other words, given any other -step nilpotent group with marked elements , there is a unique homomorphism from the free nilpotent group to that maps each to for . 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 . For instance, when , is simply the free abelian group generated by , and so every element of can be described uniquely as

for some integers , with the obvious group law. Indeed, to obtain existence of this representation, one starts with any representation of in terms of the generators , and then uses the abelian property to push the factors to the far left, followed by the factors, and so forth. To show uniqueness, we observe that the group of formal abelian products is already a -step nilpotent group with marked elements , and so there must be a homomorphism from the free group to . Since distinguishes all the products from each other, the free group must also.

It is only slightly more tricky to describe the free nilpotent group of step . Using the identities

(where is the conjugate of by ) we see that whenever , one can push a positive or negative power of past a positive or negative power of , at the cost of creating a positive or negative power of , or one of its conjugates. Meanwhile, in a -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 of has a representation of the form

for some integers for and for . Note that we don’t need to consider commutators for , since

and

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

forms a -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 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*

(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 -step nilpotent group, it turns out that for representations of the form

one has uniqueness but not existence (e.g. even in the simplest case , there is no place in this representation for, say, or ), 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 case, but the case looks more fearsome still, especially now that the quadruple commutators split into several distinct-looking species such as and which are nevertheless still related to each other by identities such as (3). While one can eventually disentangle this mess for any fixed and by a finite amount of combinatorial computation, it is not immediately obvious how to give an explicit description of uniformly in and .

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 . 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.

** — 1. Generalisation — **

The first step is to generalise the concept of a free nilpotent group to one where the generators have different “degrees”. Define a *graded sequence* to be a finite ordered sequence of formal group elements , indexed by a finite, totally ordered set , where each is assigned a positive integer , which we call the *degree* of . We then define the degree of any formal iterated commutator of the by declaring the degree of to be the sum of the degrees of and . Thus for instance has degree . (The ordering on is not presently important, but will become useful for the polycyclic representation; note that such ordering has already appeared implicitly in (1) and (2).)

Define the *free -step nilpotent group* generated by a graded sequence to be the group generated by the , subject to the constraint that any iterated commutator of the of degree greater than is trivial. Thus the free group corresponds to the case when all the are assigned a degree of .

Note that any element of a graded sequence of degree greater than is automatically trivial (we view it as a -fold commutator of itself) and so can be automatically discarded from that sequence.

We will recursively define the free -step nilpotent group of some graded sequence in terms of simpler sequences, which have fewer low-degree terms at the expense of introducing higher-degree terms, though as mentioned earlier there is no need to introduce terms of degree larger than . Eventually this process exhausts the sequence, and at that point the free nilpotent group will be completely described.

** — 2. Shift — **

It is convenient to introduce the iterated commutators for by declaring and , thus for instance .

Definition 1 (Shift)Let be an integer, let be a non-empty graded sequence, and let be the minimal element of . We define the (degree )shiftof by defining to be formed from by removing , and then adding at the end of all commutators of degree at most , where and . For sake of concreteness we order these commutators lexicographically, so that if , or if and . (These commutators are also considered to be larger than any element of ). We give each a degree of , and define the group element to be .

Example 1If , and the graded sequence consists entirely of elements of degree , then the shift of this sequence is given bywhere have degree , and , have degree , and , , etc.

The key lemma is then

Lemma 2 (Recursive description of free group)Let be an integer, let be a non-empty graded sequence, and let be the minimal element of . Let be the shift of of . Then is generated by and , and furthermore the latter group is a normal subgroup of that does not contain . In other words, we have a semi-direct product representationwith being identified with and the action of being given by the conjugation action of . In particular, every element in can be uniquely expressed as , where .

*Proof:* The argument will be motivated by the identities

(note that the products terminate in finite time due to nilpotency). In particular, we of course have

We now form a semi-direct product

To do this, we need a conjugation action of on . Inspired by the identities (4), (5), we define this action on generators by mapping

Note that this map and its inverse preserve the degree of the generators, and so by the definition of the weighted free group this defines an automorphism on . By construction, we have the counterpart of (6):

The group is generated by together with the for . Giving the degree of , one checks by hand (using plenty of commutator identities, as well as (9)) that any iterated commutator of generators with total degree greater than in is trivial. Thus there is a homomorphism from to which maps to and to for all . Comparing (6) with (9) with, we see that this homomorphism also maps to .

On the other hand, there is an obvious homomorphism from to that maps to , and from the compatibility of (4), (5) with (7), (8) we may extend this homomorphism to a homomorphism from to which also maps to . By checking this homomorphism on generators, we see that this map inverts the previous homomorphism from to , and so the two groups are isomorphic, as required.

We can now iterate this. Observe that every time one shifts a non-empty graded sequence, one removes one element (the minimal element ) but replaces it with zero or more elements of higher degree. Iterating this process, we eventually run out of elements of degree one, then degree two, and so forth, until the sequence becomes completely empty. We glue together all the elements encountered this way and refer to the full sequence as the *completion* of the original sequence . As a corollary of the above lemma we thus have

Corollary 3 (Explicit description of free nilpotent group)Let be an integer, and let be a graded sequence. Then every element of can be represented uniquely aswhere is an integer, and is the completion of .

Example 2We continue with the sequence from Example 1, with . We already saw that shifting once yielded the sequenceAnother shift gives

and shifting again gives

At this point, all remaining terms in the sequence have degree at least two, and further shifting simply removes the first element without adding any new elements. Thus the completion is

and every element of can be uniquely expressed as

In a recent paper of Leibman, a related argument was used to expand bracket polynomials (a generalisation of ordinary polynomials in which the integer part operation is introduced) of degree in several variables into a canonical basis , where is the same completion of that was encountered here. This was used to show a close connection between such bracket polynomials and nilpotent groups (or more precisely, nilsequences).

## 21 comments

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22 December, 2009 at 8:46 am

Dylan ThurstonIn Example 2, aren’t you missing ?

[Oops! Corrected, thanks – T.]22 December, 2009 at 7:27 pm

AnonymousI’m probably just confused, but I would’ve guessed there would be 3+3+8 terms in the completion.

13 May, 2013 at 1:37 pm

Sean EberhardIs also missing?

[Corrected, thanks – T.]22 December, 2009 at 9:19 am

AnonymousIn a 2-step nilpotent groups, when one pushes g_j pass g_i, one may need to use one of the conjugates of the commutator [g_i,g_j]. But in the representation (2), there are only integers powers of the commutators and (non-trivial) conjugates of them. I don’t see why. Is it something obvious that I don’t see?

22 December, 2009 at 3:06 pm

Terence TaoIn a 2-step nilpotent group, all commutators are central, and are thus their own conjugates.

22 December, 2009 at 9:22 am

AnonymousOops, I meant no nontrivial conjugates of the commutators in the above post.

By the way, is there anyway that one can edit one’s own posts (like the reviews at Amazon.com)?

25 December, 2009 at 3:26 pm

AnonymousMerry Christmas, Terence Tao!

25 December, 2009 at 8:32 pm

AnonymousA nilpotent group of step s is a very neat concept!

Do we know the nilpotent subgroup of step 2 of GL(n,\mathbb{K})? I could find them for n=2,3, but after that it seems more difficult.

3 March, 2011 at 8:13 am

Anonymouswhen you define a free nilpotent group, do you not have to make all s step commutators trivial rather than just the commutators involving the generators?

3 March, 2011 at 11:42 am

Terence TaoIt turns out that either approach gives an equivalent group. (Once one makes all s-step commutators trivial, one can show inductively that the i^th group in the lower central series is generated by all the iterated commutators of generators of step i or greater, and in particular that vanishes.)

17 March, 2012 at 9:12 pm

254A, addendum: Some notes on nilprogressions « What’s new[…] also be profitably viewed as a special type of polycylic group; this is the perspective taken in this previous blog post. Last, but not least, one can view nilpotent groups from a combinatorial group theory perspective, […]

31 May, 2012 at 5:46 am

AlexeyThank you for the article. Not completely clear to me that is isomorphic to a subgroup of , i only see that it maps onto the subgroup. I’ll need to think more to understand.

31 May, 2012 at 6:03 am

Terence Tao~~From the universality properties, there is a map from the latter group to the former which is the identity on the with but sends the with to the identity. By using the universality properties again one can show that this map is a right inverse of the obvious embedding of the former subgroup into the latter, which gives the injectivity.~~[Argument incorrect, see comments below – T.]31 May, 2012 at 3:50 pm

AlexeyThis particular argument false: if $s=2$, $A = (a, b)$, $A’ = (b, [b,a])$, then if $a$ is mapped to $1$, so is $[a,b]$. The group ${\mathcal F}_{\le2}(A’)$ is indeed isomorphic to a subgroup of ${\mathcal F}_{\le2}(A)$ and isomorphic to $\mathbb Z^2$, but it is not a retract of ${\mathcal F}_{\le2}(A)$.

1 June, 2012 at 9:11 am

Terence TaoAh, you’re right, of course. One has to rearrange the argument by constructing the semi-direct product by abstract nonsense first, rather than by embedding first into the free nilpotent group, and only establishing the isomorphism between the two groups afterwards. I’ve rewritten the argument to reflect this.

1 June, 2012 at 10:46 am

AlexeyThanks! I will read the new version.

4 June, 2012 at 1:59 am

AlexeyI do not find formula (5), i find instead:

[Corrected, thanks – T.]14 December, 2013 at 3:39 am

Martin SeysenA description of free nilpotent groups of step s is also given in section 11 (Basic commutators) of the book ‘The Theory of Groups’ by Marshall Hall, Macmillan Company, 12th printing, 1973. Equivalence of that description with your description is not obvious to me.

Btw., do you know if a presentation of such a group in terms of commutators of shape […[[g_1,g_2], g_3], …, g_n] exists? (Of course, some of the g_i may be the same).

11 November, 2014 at 2:49 am

Alexander Shamov>

> Note that this map and its inverse preserve the degree of the generators

I’m not sure I understand what this means. The map that you described doesn’t even take generators to generators…

11 November, 2014 at 7:46 am

Alexander ShamovIt seems that the key lemma here is , where is the submonoid generated by the generators of degree , and this can be checked by hand.

26 November, 2018 at 7:00 am

DSAlexander Shamov you are right, I had the same problem in that place. This is the heart of the proof so it’s a shame the argument is not expanded. A key fact useful here is $[xy,z]=y^{-1}x^{-1}x^z y^z=[x,z]^y[y,z]$

In short, “commutator are bilinear up to conjugation”.

Thus when applying the homomorphism defined above to a relation, which is iterated commutators of generators, you get iterated commutators of \emph{products} of generators, but by bilinearity it’s a product of conjugates of commutators of generators, which are themselves relations, so the map is well-defined.