Determinant and Trace

So far we have discovered 2 scalars that can characterize a particular element of the algebra. The first is given by

AA = a

If we were to represent our algebra by square matrices, then this quantity would be the determinant of the matrix which represents A.

The second characterizing scalar is called the "scalar part" of A, and is given by

<A>_S = 1/2(A + A)

To see how these two scalars relate to each other, we need to introduce the exponential function. Consider that we have an element that is expressed as the exponent of A

e^A

Taking the exponent of an algebraic element that we don't know much about seems kind of ambiguous. However, if you must, consider that the exponent converges from the infinite sum

e^A = ∑ (1/n!) Aⁿ

Now let us determine what happens if we take the algebraic conjugate of e^A.

First, we know that the conjugate commutes with respect to addition, so the conjugate gets applied to each term of the infinite sum seperately.

Next, we know that the conjugate only affects Aⁿ, since 1/n! is a scalar.

Finally, we know that the conjugate of Aⁿ is just the n^th power of the conjugate of A

Aⁿ = Aⁿ

The end result of all of this is

e^A = e^A

Multiplying e^A by its conjugate produces a scalar, which we have identified with the determinant of e^A.

e^A e^A = e^Ae^A = det(e^A)

Now, if we multiply the exponent of two real numbers together, this is the same as exponentiating the sum of the two numbers. For instance

e^xe^y = e^x+y

This rule does not apply generally since the algebraic elements do not commute with respect to multiplication. However, the elements A and A DO commute, and so the exponential rule CAN be applied in this instance.

e^A e^A = e^A+A

The factor in the exponent is a multiple of the scalar part of A as we have previously defined.

Thus we can relate the determinant of A with the scalar portion of A through the exponential function

det(e^A) = e^a e^A = e^{A + A} = e^2<A>s

This equation can be compared to a parallel equation from the determinant theory of square matrices

det(e^A) = e^Tr(A)

Thus we see that if we were to use square matrices to represent the elements of our algebra, then the "scalar part" is merely a multiple of the trace of the matrix.

In the theory of Lie Algebra, a matrix that has a trace of zero is used to represent the generator of a specific transformation. In our algebra, such a generator would be an element whose scalar portion is zero.