Commutator

A commutator is a sequence of moves that consists in doing a sequence A, then a sequence B, then the inverse of the sequence A and finally the inverse of the sequence B. As a result, only specific pieces are affected. It allows to finish the cube without disturbing pieces that are already in place.

Mathematical definition

Given a group, a commutator is an element of the form ghg'h' (also denoted [g, h]), where g and h are elements of the group with inverses g' and h'.

Cube notation is very close : A B A' B' = [A, B]

Fundamental principle

It is straightforward that A A' does nothing, as well as B B', because A' is the inverse of A and B' the inverse of B.

So the commutator A B A' B' does change something only because B A' is not the same as A' B. When B A' = A' B, or B A = A B, we say that the elements A and B commute. In this case, A B A' B' = A (B A') B' = A (A' B) B' = I where I is the identity.

For example for L and R or U and D. In this case, the commutator does nothing : L R L' R' = L L' R R' = I

So in some way, the commutator allows you to measure to which extent the moves do not commute.

Effect

The sequence A changes a set of pieces located at some locations J on the cube. The sequence B affects some locations K. If J and K have no intersection, then A and B commute, and the commutator does nothing (for example [L, R]). Otherwise, J and K have an intersection N.

Pieces that are brought into the intersection by A are NA' (apply the inverse of A to the locations N). Pieces that are brought into the intersection by B are NB'.

The pieces that are affected by the commutator [A, B] = A B A' B' are located in the union of N, NA' and NB'. In other words, pieces that are affected by a commutator are those who are at the intersection of both moves, or are brought into the intersection by A or B. Other pieces, even if they are temporarily mixed up by A and B, will be put back at their original location with A' and B'.

Trivial case

When J and K have an empty intersection, A and B commute, so the commutator does nothing.

B inplace

If NB' = N, the sequence B only moves affected pieces that are inside the intersection. It may also move pieces that are outside the intersection, but those moves will be cancelled at the end. Affected pieces will only be N and NA'. So those pieces will be in J, i.e. among pieces that are directly affected by A.

In this case, it is relevant to consider that [A, B] = (A B A') B' = [A: B] B'. First part is the conjugate of B by A, and second part is the inverse of B.

Example

Let's consider [M', U2] = M' U2 M U2

In this example, N are the top-front and top-back edges. NB' is NU2, which is exactly N, and NA' are front-bottom and front-top edges. So affected pieces are top-back, top-front, and front-bottom edges, which are directly affected by M and M'. In other words, affected pieces are in the middle slice.

The conjugated move is U2. Because only middle slice pieces are changed at the end, we can safe ignore top-left and top-right pieces. The relevant part of U2 is just a swap of top-back and top-front edges. Thus, M' U2 M swaps front-top and front-bottom edges.

So [M', U2] can be understood as : swap front-top and front-bottom edges, then swap top-back and top-front edges, which yields a 3-cycle of edges.

A inplace

This is a symmetrical case. Affected pieces will only be N and NB'. So those pieces will be in K, i.e. among pieces that are directly affected by B.

In this case, it is relevant to consider that [A, B] = A (B A' B') = A [B: A']. First part is A and second part is the conjugate of A' by B.

Examples

Let's consider the inverse of [M', U2] which is (M' U2 M U2)' = U2 M' U2 M = [U2, M'] = U2 [M': U2]. The relevant part of U2 consists in swapping top-back and top-front edges, and [M': U2] swaps top-front and front-bottom edges.

Other examples :

• A = "rotating a corner" like [R' D' R D R' D' R, U]
• A = "flipping an edge" like [R' E' R2 E2 R', U]
• A = "exchanging two corners" like [R' L' D2 R L, U]
• A = "exchanging two edges" like [M2 D2 M2, U]

Three different sets

When NA' and NB' are not included in N, there are three sets involved in the commutator :

• NA' : pieces that are brought into the intersection by A
• NB' : pieces that are brought into the intersection by B
• N : the intersection of J and K, i.e. where A and B interfere.

Note that NA' can have locations in common with N and NB' can have locations in common with N. In other words, it is possible that a sequence (A or B) does two different things in the context of a commutator :

1. moving pieces inside the intersection (from N to N)
2. bringing pieces into the intersection (from NA' \ N to N or from NB' \ N to N)

When there is only case 1, it is in fact the cases studied before. So we are left with those possible cases :

• Case 2 only : the three sets NA', NB' and N are completely separated (no overlap)
• Case 1 and 2 at the same time : the three sets overlap

No overlap

Referring to the animation below, when sets do not overlap, the commutator can be summed up by :

• A stores the content of the intersection P_N in NA, and replace it with some pieces P_A from NA'
• B stores P_A in NB, and bring some pieces P_B from NB' into the intersection
• A' retrieves P_B from the intersection and places them in NA', and brings to the intersection previously stored pieces P_N in NA
• B' retrieves P_N from the intersection and places them in NB', and finally places P_A in the intersection.

So it is a 3-cycle of (P_N, P_A, P_B) into (P_A, P_B, P_N)

We notice that NA and NB are used as temporary storage :

• The content of the intersection, P_N, is first hidden by A in its storage NA, and comes back with A', so that the only transformation applied to these pieces is B'.
• The content of NA' is brought by A, and is then hidden by B in NB, so that the only transformation applied to these pieces is A.
• The content of NB' is brought by B, and then is placed immediately in NA' by A'. So the transformation applied to these pieces is B A'.

In other words, all the commutator does is :

• P_A goes A
• P_B goes B A'
• P_N goes B'

Examples

• [R' E2 R, U'] is a 3-cycle of edges, where P_A is the back-left edge, P_B is the top-front edge and P_N is the top-right edge.
• [R' D' R, U'] is a 3-cycle of corners

Overlap

This is the most complex case, because it is like the "no overlap" case, but with an additional phenomenon.

Some pieces in the intersection N are hidden in a storage, thus protecting them from being scrambled by the interference due to the overlap.

Let's call H_B the location of pieces hidden by B:

H_B = ((NB) \ N) B'

Similarly, pieces are hidden by A:

H_A = ((HA) \ H) A'

In order to get the same results as in the "no overlap" case, we will redefine the set of pieces :

• P_A will be pieces that are brought into H_B by A, i.e. pieces that are at the beginning in H_BA'. This set may overlap with the intersection N.
• P_B will be pieces that are brought into H_A' by B, i.e. pieces that are at the beginning in H_A'B' \ N
• P_N will be pieces that are hidden by A, which are at the beggining in H_A.

We get the same result as before :

• P_A goes A
• P_B goes B A'
• P_N goes B'

Example

Let's consider [R', F]. The intersection N is the front-right column (containing the front-right-top corner, the front-right-bottom corner and the front-right edge).

H_B = ((NB) \ N) B' = ((NF) \ N) F' which is the set of locations containing the front-right edge and the front-right-bottom corner.

So P_A are front-right-top corner and top-right edge.

H_A = ((NA) \ N) A' = ((NR') \ N) R

So P_{N are front-right edge and front-right-bottom corner.}

See also

• Commutators and Conjugates
• Conjugate