Burning G2 At Both Ends

Attached are the images most associated with G2. By treating the problem on multiple levels we will have an complete and intuitive understanding of where these images came from, and how we can produce them ourselves.

CM:=(({{2, -3}, {-1, 2}}))

This is all fine and well but where do we look to find this Algebra? Since the classification of the classical Lie Algebras is so straight forward we can look there for inspiration. Notice the Dynkin Diagram for B3 is oddly similar to that of G2. From here we can begin mining B3 for a basis for G2. Below is the Dynkin diagram and basis for B3.

A1:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, -1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}})); A2:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, -1, 0, 0}})) ; A3:= (({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, -1, 0}})); A4:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, -1, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}})) ;  A5:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, -1}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  A6:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, -1}, {0, 0, 0, 0, 0, 0, 0}}));  h1:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, -1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}})); h2:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, -1, 0}, {0, 0, 0, 0, 0, 0, 0}}));  h3:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, -1}}));  b1:=(({{0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  b2:=(({{0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  b3:=(({{0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-1, 0, 0, 0, 0, 0, 0}}));  b4:=(({{0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  b5:=(({{0, 0, 1, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-1, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  b6:=(({{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-1, 0, 0, 0, 0, 0, 0}}));  c1:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, -1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  c2:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, -1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  c3:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, -1, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  c4:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0, 0}, {0, -1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  c5:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, -1, 0, 0, 0, 0, 0}}));  c6:=(({{0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0}, {0, 0, -1, 0, 0, 0, 0}}));

As we can guess (but won' t prove!) this is a basis for so(7) and all the good things that come along for the ride. For instance we can see c5 has an eigenvalue of - 1 when bracketed with h1. It should be noted that h1, h2 and h3 are the Cartan Subalgebra for so(7).

MatrixForm[h1.c5-c5.h1] 
 MatrixForm[-1*c5]

If we can recognize that the 2 Dynkin diagrams are similar we should be able to find a copy of G2 inside of B3.   [Sage Demo]
What we need to do is rotate B3 and project the b’s and c’s to collapse onto each other. This same rotation will force one dimension from the CSA to disappear. We’re basically geometrically rotating the root space image for so(7) and projecting the roots into a two dimensional plane. This collapse the b’s and c’s into a new root space shown below along with the CSA for G2. For instance: B1 = (Sqrt[2])b4 - c3

B1:=(({{0, (Sqrt[2]), 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, -1}, {0, 0, 0, 0, 0, 1, 0}, {-(Sqrt[2]), 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}})); B2:=(({{0, 0, (Sqrt[2]), 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, -1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-(Sqrt[2]), 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}}));  B3:=(({{0, 0, 0, (Sqrt[2]), 0, 0, 0}, {0, 0, 0, 0, 0, -1, 0}, {0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {-(Sqrt[2]), 0, 0, 0, 0, 0, 0}}));

C1:=-Transpose[B1]; C2:=-Transpose[B2];  C3:=-Transpose[B3];

H1:=h1-2*h2+h3;
H2:=h2-h1;

MatrixForm[H1]
MatrixForm[H2]

So now we have a basis for G2 consisting of {H1, H2, A1, A2, A3, A4, A5, A6, B1, B2, B3, C1, C2, C3}. Let’s take a look at G2’s multiplication table.

({{[. ,.], H1, H2, A1, A2, A3, A4, A5, A6, B1, B2, B3, C1, C2, C3}, {H1, 0, 0, 3A1, 0, -3A3, -3A4, 0, 3A6, -B1, 2B2, -B3, C1, -2C2, C3}, {H2, 0, 0, -2A3, -A2, A3, 2A4, A5, -A6, B1, -B2, 0, -C1, C2, 0}, {A1, -3A1, 2A3, 0, 0, A2, -H2, -A6, 0, -B2, 0, 0, 0, C1, 0}, {A2, 0, A2, 0, 0, 0, -A3, -H2, A1, -B3, 0, 0, 0, 0, C1}, {A3, 3A3, -A3, -A2, 0, 0, 0, A4, -H1-H2, 0, -B3, 0, 0, 0, C2}, {A4, 3A4, -2A4, H2, A3, 0, 0, 0, -A5, 0, -B1, 0, C2, 0, 0}, {A5, 0, -A5, A6, H2, -A4, 0, 0, 0, 0, 0, -B1, C3, 0, 0}, {A6, -3A6, A6, 0, -A1, H1+H2, A5, 0, 0, 0, 0, -B2, 0, C3, 0}, {B1, B1, -B1, B2, B3, 0, 0, 0, 0, 0, 2C3, 0, -H1-3H2, 3A4, 3A5}, {B2, -2B2, B2, 0, 0, B3, B1, 0, 0, -2C3, 0, 2C1, 3A1, -H1, 3A6}, {B3, B3, 0, 0, 0, 0, 0, B1, B2, 0, -2C1, 0, 3A2, 3A3, 2H1+3H2}, {C1, -C1, C1, 0, 0, 0, -C2, -C3, 0, H1+3H2, -3A1, -3A2, 0, 2B3, -2B2}, {C2, 2C2, -C2, -C1, 0, 0, 0, 0, -C3, -3A4, H1, -3A3, -2B3, 0, 2B1}, {C3, -C3, 0, 0, -C1, -C2, 0, 0, 0, -3A5, -3A6, -2H1-3H2, 2B2, -2B1, 0}})

Below is a simulation for the commutator bracket as well as a matrix check so we can see what these elements bracket to. Have at it!

MatrixForm[A3.A6-A6.A3]
MatrixForm[-H1-H2]

So we can see that this thing, whatever it is, is closed under the brackets and moreover each element when bracketed with H1 and H2 have integer eigenvalues. We will show that A4 and B2 correspond to simple roots.

Let α = B2 and β = A4. A rather unfortunate naming coincidence. Note that for x in the α root space, and y in the β root space we have [α,β] is in the α+β root space. [B2,A4]=B1. We can keep whacking successive outputs with α’s and β’s until we fall off the map to find the highest weight root vector. 
[B2,B1] = -2C3 ϵ 2α+β root space.
[B2,C3] = 3A6 ϵ 3α+β root space.
[B2,A6] = 0 because there is no 4α+β root space. However, there is a 3α+2β root space. Let’s figure out what’s in it.
[A4,A6] = -A5 ϵ 3α+2β root space. Applying either A4 or B2 to this will nullify it since it is the highest weight vector.

Let’s develop the Cartan Matrix from these simple roots. Simply taking the eigenvalues from the brackets below. Note the re-ordering is merely an optimization of the basis. It was chosen in an ad hoc fashion prior.

(({{[H2,A4], [H1,A4]}, {[H2,B2], [H1,B2]}}))=(({{2, -3}, {-1, 2}}))

From here we can see the ‘short’ root is α and the ratio of the lengths between the roots is (Sqrt[3]). Referencing the table given to us in class we now know that the angle between the simple roots is 5π/6. Let’s draw a picture!

Let α =<1,0> and β =<((-3)/2), ((Sqrt[3])/2)> and watch the magic unfold.

There are several other ways to construct G2 including but not limited to
	Derivations on the Octonions;
	Outer automorphisms on D4;
	Permutation games on vector operators;
	Probably more - but what we’ve done is a very classical construction.