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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.