## The amazing SO(2,4)

Continuing the quantionic discussions, let's see how the Lie algebra so(2,4) arises naturally in this framework. First in quantum mechanics the anti-Hermitean operators form a Lie algebra and the classification of Lie algebras was obtained by Elie Cartan. But in quantum mechanics there are additional relationships obeyed by the anti-Hermitean operators because if we multiply them with the imaginary unit we get a Jordan algebra and between them there is a compatibility relationship. This compatibility relationship restricts the possible Lie algebras and as expected one gets the unitary algebras su(n). However there is also an exceptional non-unitary solution: so(2,4).

It is too complicated to follow this line of argument, and instead I want to present a more elementary argument (also due to Emile Grgin) which arrives at so(2,4). With a bit of hindsight we start from a general SO(p,q) space which as a linear space is spanned by p-positive basis vectors: $$e_1 , e_2 , \dots , e_p$$ and by q-negative basis vectors: $$e_{p+1} , e_{p+2} , \dots , e_{p+q}$$ . Then we want investigate arbitrary reflections. Why? Because we are after obtaining non-standard ways to represent sqrt(-1) using the elements of SO(p,q) which is the key to the hermitian-anti-hermitan duality in quantum mechanics. In complex numbers if we consider the complex conjugation we can represent that as a reflection on the real axis, and this is a good hint.

We are interested in continuous transformations only to the extent that they can undo a reflection. If we can find a unique reflection this can form a realization of the observables-generators duality in quantum mechanics.

Now consider a reversal of $$r$$ arbitrary basis vectors in $$e_1 , e_2 , \dots , e_p$$. If $$r$$ is even the transformation can be undone by rotations because the determinant of the transformation is positive. Similarly all reversals for $$r$$ odd are equivalent.

In general we can have $$r$$ inversions in the positive basis vectors and $$s$$ inversions in the negative basis vectors: $$J=s+r$$. Therefore in general there are $$K=n-J$$ invariant basis vectors. Let us now rename the basis vectors as: $$R_1, R_2, \dots , R_J , S_1, S_2, \dots , S_K$$ (R for reverse, and S for same). Then there are 3 kinds of bivectors:

$$R_i \wedge R_j$$
$$S_i \wedge S_j$$
$$R_i \wedge S_j$$

The first two kinds do not change sign, but the last kind does.

Let us introduce two more numbers:

N = number of bivectors of kind $$R_i \wedge S_j$$
P = number of bivectors of kind $$R_i \wedge R_j$$ + number of bivectors of kind $$S_i \wedge S_j$$ + 1 for the identity transformation

N-for negative, P-for positive

Then the following relationships hold:

N=JK
P=1/2 J(J-1) + 1/2 K(K-1)+1 = 1/2 n(n-1) - N + 1

Now r and s must be odd numbers (all even number reflections can be undone by a rotation):
r=2k+1 < p
s=2l+1 < q

and introducing m=k+l as an auxiliary notation we get:

N(m) = JK=2(m+1)(n-2m-2)

(2(m+1) = J and K = n-J=n-2m-2)

Now we need to require that the complex conjugation is uniquely defined. This means that N(m) must have the same value for all the allowed values of m:

$$N(0) =N(1)=...=N(m_{max})$$

Because N(m) is quadratic there are only two solutions for m: $$m_{max} = 1$$ and from N(0)=N(1) we get:

2*(n-2)=2*2*(n-4)

and so n=6

Therefore we can have the solutions: SO(1,5), SO(2,4), SO(3,3), SO(4,2), SO(5,1)

In the 1,5 and 3,3 case $$m_{max} = 2$$ and we cannot have a unique way to define complex conjugation!!! The only remaining case is SO(2,4) (which is isomorphic with SO(4,2)).

If we want to generalize the number system for quantum mechanics in a way that respects the tensor product and obtain a non-unitary solution, the only possibility is SO(2,4).

Why is this remarkable? Because SO(2,4) is the conformal group of the compactification of the Minkowski space. This is the first hint that ultimately we will get a relativistic quantum mechanics.