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Thursday, 4 July 2013

Real representations of the Lie algebra so(3)

Recently I wondered about the real (irreducible) representations of the Lie group SO(3) and its Lie algebra so(3). It is well known, that the representations of SO(3) are odd-dimensional, there are no (irreducible) representations of SO(3) in even dimensions.

Usually, in textbooks the authors quickly move from SO(3) to the covering group SU(2) and derive the (complex) representations of the corresponding Lie algebra su(2) using quantum mechanical angular momentum algebra and ladder operators.

There appears to be surprisingly little in the published literature on real-valued representations of so(3). How do the real-valued representations look like? A google search lead me to
V. M. Gordienko
Matrix entries of real representations of the group O(3) and SO(3)
Siberian Mathematical Journal (2002), 43(1):36-46
DOI: 10.1023/A:1013816403253

who describes their construction; but see also
G. Itzkowitz, S. Rothman and H. Strassberg
A note on the real representation of SU(2,C)
Journal of Pure and Applied Algebra (1990), 69:285-294
DOI: 10.1016/0022-4049(91)90023-U
If I understand correctly, there is no ladder operator approach for real representations. Rather, Gordienko constructs real representations using homogeneous polynomials P(N)(x,y) in x and y with x,yC. I.e.
P(1)(x,y)=a1x2+a2xy+a3y2P(2)(x,y)=a1x4+a2x3y+a3x2y2+a4xy3+a5y4P(N)(x,y)=a1x2N+a2x2N1y+a3x2N2y2++aN+1xNyN++a2Nxy2N1+a2N+1y2N Gordienko chooses the following basis in the vector space of polynomials P(N)(x,y)
e(N)n(x,y)=iN+1(2N+1)!2(Nn)!(N+n)!×(xNnyN+n(1)nxN+nyNn)ifn1e(N)n=0(x,y)=iN(2N+1)!N!xNyNifn=0e(N)n(x,y)=iN(2N+1)!2(N+n)!(Nn)!×(xN+nyNn+(1)nxNnyN+n)ifn1 with n=N,,N. Now let's take a generic rotation matrix from the group SU(2), parametrized by the three real numbers α, β and γ and we write with δ1α2β2γ2
M=(δiγβiαβiαδ+iγ)(ABBA) and apply it on the 2-dimensional complex vector (x,y), i.e.
(x,y)(x,y)(ABBA)=(AxBy,Bx+Ay) For the "rotated" basis e(N)n(x,y) we obtain
e(N)n(x,y)=e(N)n(AxBy,Bx+Ay)=+Nm=NT(N)n,m(A,B)e(N)m(x,y) The (2N+1)×(2N+1) matrix T(N)m,n(A,B) is the (2N+1)-dimensional representation matrix of SO(3) we're looking for.

Why is T(N)m,n(A,B) real-valued? Turns out, the basis vectors e(N)n(x,y) are constructed in such a way that
e(N)m(x,y)=(e(N)m(y,x)) holds, where x denotes the complex conjugate of x. Gordienko shows that for basis vectors with this property the entries of the representation matrices T(N)n,m(A,B) are indeed real. Specifically,
e(N)m(x,y)=e(N)m(AxBy,Bx+Ay)=(e(N)m(BxAy,AxBy))=(e(N)m(B(x)+A(y),AxBy))=(e(N)m(y,x))=(Nn=NT(N)m,n(A,B)e(N)n(y,x))=Nn=N(T(N)m,n(A,B))(e(N)n(y,x))=Nn=N(T(N)m,n(A,B))e(N)n(x,y)Thus, T(N)m,n(A,B)=(T(N)m,n(A,B)) since the e(N)n(x,y) are linear independent; in other words, the matrix T(N)m,n(A,B) is real.

representationofso3.m a is a rough-and-ready MATLAB programme (included in LieTools), which outputs numerical approximations of the n-dimensional representations of so(3) (for n below about 90) using the method described above. For dimensions 3, 5 and 7 the output of representationofso3.m yields the following matrices. The 3-dimensional representation is familiar
t1=(010100000)t2=(001000100)t3=(000001010) the 5-dimensional representation is t1=(0001000301030001000001000)t2=(0000200010000000100020000)t3=(0100010000000300030100010) and the 7-dimensional representation turns out to be
t1=(000003/2000005/203/2000605/20006000005/2000003/205/2000003/200000)t2=(0000003000002000001000000000001000002000003000000)t3=(03/2000003/205/2000005/2000000000600000605/2000005/203/2000003/20) The zero entries in the off-diagonals look somewhat strange, but the matrices are definitely real, skew-symmetric and obey the so(3) commutation relations
[t1,t2]=t3[t2,t3]=t1[t3,t1]=t2

Saturday, 22 September 2012

Picturing structure constants

As mentioned before, in the Chevalley basis the structure constants Kij obey
[eαi,eαj]=Kijeαi+αj or vanish. I.e., the commutator of the generators eαi and eαj either is zero or collinear to one specific generator (and not a linear combination of two or more generators) with constant of proportionality Kij.

With the help of LieTools I created plots of Kij for some classical and the five exceptional Lie algebras. The following figures show the Kij's as a function of root numbers i and j corresponding to the (non-zero) roots αi and αj.
Fig. 1: Structure constants for Lie algebra A looping thru rank 1 to 13.
In the case of An (n=1,2,3,) the structure constants Kij are 1, 0 or +1. Since Kij=Kji, the figures are anti-symmetric with respect to the main diagonal. In addition, for root numbers close to zero (and larger ranks) there appears to be an approximate symmetry with respect to the horizontal and vertical lines.

Fig. 2: Structure constants for Lie algebra B looping thru rank 2 to 13.

Fig. 3: Structure constants for Lie algebra C looping thru rank 3 to 13.
The corresponding plots for the Lie algebras Bn (n2) and Cn (n3) (Figs. 2 and 3) show structure constants ranging from2 to +2. It appears that values ±2 occur solely within a square region centered at the origin with edge length 1.5imax and a circle centered at the origin with radius imax for the algebras B and C, respectively. Here, imax denotes the maximum root number (half of the total number of non-zero roots). 
Fig. 4: Structure constants for Lie algebra D looping thru rank 4 to 13.

The color-coded structure constants for the exceptional Lie algebras E6, E7, E8, F4 and G2 are shown in the remaining three figures with fig. 5 looping thru E6, E7 and E8. Surprisingly (to me), the plots for the exceptional algebras lack the somewhat regular patterns which are found Figs. 1 - 4.
Fig. 5: Structure constants for Lie algebra E looping thru rank 6 to 8.

Fig. 6: Structure constants for Lie algebra F (rank4)

Fig. 7: Structure constants for Lie algebra G (rank 2)

The Lie algebra G2 is the only one with structure constants varying between 3 and +3.

Monday, 17 September 2012

Structure constants

Previously on Visual Lie Theory it was mentioned, that the commutator of two basis elements of a Lie algebra L is again an element of L. I.e., it can be written as a linear combination of the basis vectors tk,k=1,,d; i.e.,
[ti,tj]=dk=1fkijtk From the literature I learn, that the parameters fkij ("structure constants") (almost?) completely describe the Lie algebra L.

What's interesting about the structure constants is, that once the fkij are known, one immediately can write down a d-dimensional matrix representation of L, the adjoint representation. Concretely, the adjoint representation of ti, the i-th generator, is given by
(Ti)k,j=fkij where j=1,,d and k=1,,d denote the row and column index of the matrix Ti.

Clearly,  the structure constants depend on the chosen basis vectors tk,k=1,,d. Viewed from another basis tk˜tkdl=1alktl the structure constants will change as well;
[˜ti,˜tj]=dk=1˜fkij˜tk It turns out, that there is a particular basis, known as the Chevalley basis, in which the fkij are integers and assume only values between 3 and +3; in addition, the sum in [ti,tj]=dk=1fkijtk contains at most one non-zero element.

In general, the generators ti,i=1,,d can be divided in dr generators eα corresponding to non-zero roots and r generators hα corresponding to zero roots; d being the dimension and r being the rank of L. Here, the label i of the generator ti is replaced by the root α, which we may regard as a r-dimensional index of eα and hα. For the generators eα we find
[eαi,eαj]=Kijeαi+αj where Kij is non-zero if the sum αi+αj is a non-zero root. If αi+αj is not a root, the commutator vanishes (Kij=0). If αi+αj is a zero-root (i.e. αj=αi) the commutator is
[eαi,eαi]=hαi where hαi is an element of the Cartan subalgebra. All generators contained within the Cartan subalgebra commute, i.e.
[hαi,hαj]=0. Furthermore,
[hαi,eαj]=Nijeαj with Nij being the entries of the Cartan matrix.

Since all information on L supposedly is encoded in the fkij, naively I had expected that there existed tables of  fkij (or Kij) for all (classical and exceptional) Lie algebras. My web search, however, turned up surprisingly little - perhaps I looked at all the wrong places.

So how does one (or a computer) then actually calculate the fkij? The following two papers I found most useful for finding an answer :
  1. V. K. Agrawala and Johan G. Belinfante: Weight diagrams for lie group representations: A computer implementation of Freudenthal's algorithm in ALGOL and FORTRAN, BIT (1969) 9(4):301-314. doi: 10.1007/BF01935862
  2. R.B Howlett, L.J Rylands, D.E Taylor: Matrix Generators for Exceptional Groups of Lie Type, Journal of Symbolic Computation, volume 31, issue 4, April 2001, pages 429-445, doi: 10.1006/jsco.2000.0431
Based on algorithms and source code contained therein I wrote some MATLAB programs (the only computer language I speak somewhat fluently) for the calculation of the coefficients Kij; these tools ("LieTools") are available for download here. (At this point Octave is not (yet) supported due to Octave's missing implementation of nested functions.)

Fig. 1: Structure constants Kij for the Lie algebra F4 (52 dimensions, i.e. 52-4=48 non-zero roots)
As an example figure 1 shows a graphical representation of (what "LieTools" thinks are) the coefficients Kij for the exceptional Lie algebra F4. In the Chevalley basis the Kij assume one of five possible value, 2,1,0,+1,+2, here plotted in 5 different colours. (You've probably noticed that I used part of this figure for the blog header.)