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Holonomy Lie algebras and symmetries

The method function holonomy constructs the holonomy Lie algebra of a hyperplane arrangement or a matroid given by the set of 2-flats. The input may be any set of subsets of a finite set $X$, such that all subsets have at most one element in common and are of length at least 3 (the 2-flats of size 2 are determined by those). Indeed, for any such set $A$ of subsets there is a unique simple matroid of rank at most 3 with the given set as the set of 2-flats of size at least 3, and holonomy(A) represents the holonomy Lie algebra of this matroid.

i1 : L = holonomy({{a1,a2,a3},{a1,a4,a5},{a2,a4,a6}})

o1 = L

o1 : LieAlgebra
i2 : ideal L

o2 = {(a2 a1) - (a3 a2), (a3 a1) + (a3 a2), (a4 a1) - (a5 a4), (a5 a1) + (a5
     ------------------------------------------------------------------------
     a4), (a4 a2) - (a6 a4), (a6 a2) + (a6 a4), (a4 a3), (a5 a2), (a5 a3),
     ------------------------------------------------------------------------
     (a6 a1), (a6 a3), (a6 a5)}

o2 : List

The sum of the generators is a central element. Hence, by dividing out by this element and using minimalPresentation(ZZ,LieAlgebra) one obtains a presentation of a Lie algebra with one generator less, which is isomorphic to the holonomy Lie algebra in degrees $\ge\ 2$.

i3 : L0 = L/{a1+a2+a3+a4+a5+a6}

o3 = L0

o3 : LieAlgebra
i4 : L0 = minimalPresentation(3,L0)

o4 = L0

o4 : LieAlgebra
i5 : describe L0

o5 = generators => {a2, a3, a4, a5, a6}
     Weights => {{1, 0}, {1, 0}, {1, 0}, {1, 0}, {1, 0}}
     Signs => {0, 0, 0, 0, 0}
     ideal => {(a6 a5), (a6 a3), (a6 a2) + (a6 a4), (a5 a3), (a5 a2), (a4 a3), (a4 a2) - (a6 a4)}
     ambient => LieAlgebra{...10...}
     diff => {}
     Field => QQ
     computedDegree => 0

It is possible to get this Lie algebra directly by choosing one of the variables, picking all 2-flats containing that variable, deleting the variable, putting $A$ equal to the set of deleted 2-flats and $B$ equal to the remaining 2-flats, and finally applying holonomy(A,B).

i6 : L1 = holonomy({{a2,a3},{a4,a5}},{{a2,a4,a6}})

o6 = L1

o6 : LieAlgebra
i7 : L0==L1

o7 = true

Choosing another generator to delete gives another presentation (which is still isomorphic to the holonomy Lie algebra in degrees $\ge\ 2$).

i8 : L6 = holonomy({{a2,a4}},{{a1,a2,a3},{a1,a4,a5}})

o8 = L6

o8 : LieAlgebra
i9 : describe L6

o9 = generators => {a2, a4, a1, a3, a5}
     Weights => {{1, 0}, {1, 0}, {1, 0}, {1, 0}, {1, 0}}
     Signs => {0, 0, 0, 0, 0}
     ideal => { - (a1 a2) - (a3 a2), (a3 a2) + (a3 a1),  - (a1 a4) - (a5 a4), (a5 a4) + (a5 a1), (a3 a4), (a5 a2), (a5 a3)}
     ambient => LieAlgebra{...10...}
     diff => {}
     Field => QQ
     computedDegree => 0
i10 : dims(1,6,L6)

o10 = {5, 3, 6, 9, 18, 27}

o10 : List
i11 : dims(1,6,L1)

o11 = {5, 3, 6, 9, 18, 27}

o11 : List
i12 : dims(1,6,L)

o12 = {6, 3, 6, 9, 18, 27}

o12 : List

The procedure above corresponds to the deconing process of a central hyperplane arrangement, yielding an affine hyperplane arrangement. The first input set in holonomy should be all maximal sets of parallel hyperplanes of size at least 2, and the second input set should be all maximal sets of hyperplanes of size at least 3 that intersect in an affine space of codimension 2.

A local Lie algebra of a holonomy Lie algebra, see holonomyLocal, is the Lie subalgebra generated by the generators in one of the subsets defined in the input. If this set is of size $k$, then the local Lie algebra is free on $k$ generators if the set belongs to the first input set, and it is free on $k-1$ generators in degrees $\ge\ 2$, if it belongs to the second input set (observe that the numbering of the sets begins with 0).

i13 : 
      describe holonomyLocal(1,L1)

o13 = generators => {a4, a5}
      Weights => {{1, 0}, {1, 0}}
      Signs => {0, 0}
      ideal => {}
      ambient => LieAlgebra{...10...}
      diff => {}
      Field => QQ
      computedDegree => 0
i14 : describe holonomyLocal(2,L1)

o14 = generators => {a2, a4, a6}
      Weights => {{1, 0}, {1, 0}, {1, 0}}
      Signs => {0, 0, 0}
      ideal => {(a4 a2) - (a6 a4), (a6 a2) + (a6 a4)}
      ambient => LieAlgebra{...10...}
      diff => {}
      Field => QQ
      computedDegree => 0

The kernel of the natural map, in degrees $\ge\ 2$, from $L$ to the direct sum of the local Lie algebras, see holonomyLocal, is obtained by decompose(LieAlgebra). This ideal is generated by the basis elements in degree 3 of the form (a b c), where not all of a,b,c belong to the same local Lie algebra.

i15 : I1=decompose(L1)

o15 = I1

o15 : FGLieIdeal
i16 : dim(3,I1)

o16 = 0

It follows from the output displayed above that $L1$, in degrees $\ge\ 2$, is the direct sum of its local Lie algebras: $L1$ is "decomposable". This is not true for the "quadrangel", i.e., the graphical arrangement of the complete graph on four vertices, which is also the braid arrangement of dimension 4.

i17 : Q = holonomy({{a1,a2,a3},{a1,a4,a5},{a2,a4,a6},{a3,a5,a6}})

o17 = Q

o17 : LieAlgebra
i18 : decompose Q

o18 = finitely generated ideal of Q

o18 : FGLieIdeal
i19 : basis(3,oo)

o19 = {(a6 a5 a4), (a5 a6 a4)}

o19 : List
i20 : Q1 = holonomy({{a2,a3},{a4,a5}},{{a2,a4,a6},{a3,a5,a6}})

o20 = Q1

o20 : LieAlgebra
i21 : decompose Q1

o21 = finitely generated ideal of Q1

o21 : FGLieIdeal
i22 : basis(3,oo)

o22 = {(a6 a5 a4), (a5 a6 a4)}

o22 : List

Here is a way to obtain decompose(LieAlgebra) (which is not used in the program). The direct sum of the local Lie algebras of $Q1$ may be obtained as follows

i23 : L0 = holonomyLocal(0,Q1)

o23 = L0

o23 : LieAlgebra
i24 : L1 = holonomyLocal(1,Q1)

o24 = L1

o24 : LieAlgebra
i25 : L2 = holonomyLocal(2,Q1)

o25 = L2

o25 : LieAlgebra
i26 : L3 = holonomyLocal(3,Q1)

o26 = L3

o26 : LieAlgebra
i27 : M = L0++L1++L2++L3

o27 = M

o27 : LieAlgebra
i28 : gens M

o28 = {pr_0, pr_1, pr_2, pr_3, pr_4, pr_5, pr_6, pr_7, pr_8, pr_9}

o28 : List

and the map from $Q1$ to $M$ is given as

i29 : f = map(M,Q1,{pr_0+pr_4,pr_1+pr_7,pr_2+pr_5,pr_3+pr_8,pr_6+pr_9})
warning: the map might not be well defined, 
         use isWellDefined

o29 = f

o29 : LieAlgebraMap
i30 : describe f

o30 = a2 => pr_0 + pr_4
      a3 => pr_1 + pr_7
      a4 => pr_2 + pr_5
      a5 => pr_3 + pr_8
      a6 => pr_6 + pr_9
      source => Q1
      target => M

and the ideal decompose(LieAlgebra) may be obtained as the kernel of $f$:

i31 : kernel f

o31 = ideal of Q1

o31 : LieIdeal
i32 : basis(3,oo)

o32 = {(a6 a5 a4), (a5 a6 a4)}

o32 : List

The symmetric group S_4 operates on the vertices of K_4, and this induces an action of S_4 on the six edges, which in turn induces an action of S_4 on $Q$ as automorphisms. One such permutation of the edges is (231645) but not (231564). Using isIsomorphism(LieAlgebraMap), it is possible to check if a permutation of the generators, written as a rearrangement of the generators, defines an automorphism of the Lie algebra.

i33 : use Q
i34 : f=map(Q,Q,{a2,a3,a1,a6,a4,a5})
warning: the map might not be well defined, 
         use isWellDefined

o34 = f

o34 : LieAlgebraMap
i35 : g=map(Q,Q,{a2,a3,a1,a5,a6,a4})
warning: the map might not be well defined, 
         use isWellDefined

o35 = g

o35 : LieAlgebraMap
i36 : isIsomorphism f

o36 = true
i37 : isIsomorphism g

o37 = false
i38 : describe f

o38 = a1 => a2
      a2 => a3
      a3 => a1
      a4 => a6
      a5 => a4
      a6 => a5
      source => Q
      target => Q

The ideal decompose(Q) is invariant under all automorphisms of $Q$. We may use trace(ZZ,LieSubSpace,LieAlgebraMap) and a character table for S_4 to determine its irreducible representation constituents. There are four conjugacy classes (except $id$). Representatives for them as permutations of the six generators are, in cycle presentation, (23)(45), (123)(465), (16)(2354) and (16)(25) corresponding, in S_4, to one 2-cycle, one 3-cycle, one 4-cycle and a product of two 2-cycles.

i39 : I=decompose Q

o39 = I

o39 : FGLieIdeal
i40 : use Q
i41 : f1=map(Q,Q,{a1,a3,a2,a5,a4,a6})
warning: the map might not be well defined, 
         use isWellDefined

o41 = f1

o41 : LieAlgebraMap
i42 : f2=map(Q,Q,{a2,a3,a1,a6,a4,a5})
warning: the map might not be well defined, 
         use isWellDefined

o42 = f

o42 : LieAlgebraMap
i43 : f3=map(Q,Q,{a6,a3,a5,a2,a4,a1})
warning: the map might not be well defined, 
         use isWellDefined

o43 = f3

o43 : LieAlgebraMap
i44 : f4=map(Q,Q,{a6,a5,a3,a4,a2,a1})
warning: the map might not be well defined, 
         use isWellDefined

o44 = f4

o44 : LieAlgebraMap
i45 : trace(4,I,f1)

o45 = -1

o45 : QQ
i46 : trace(4,I,f2)

o46 = 0

o46 : QQ
i47 : trace(4,I,f3)

o47 = -1

o47 : QQ
i48 : trace(4,I,f4)

o48 = 1

o48 : QQ

Making calculations with the character table for S_4, we see that $I$ in degree 4 is the sum of the nontrivial irreducible representations.

See also