Page 351 - DMTH502_LINEAR_ALGEBRA

Unit 32: Groups Preserving Bilinear Forms

Notes
n
Example 4: Let f be the symmetric bilinear form on R with quadratic form
2
q(x, y, z, t) = t – x – y – z .
2
2
2
4
A linear operator T on R which preserves this particular bilinear (or quadratic) form is called a
Lorentz transformation, and the group preserving f is called the Lorentz group. We should like
to give one method of describing some Lorentz transformations.
Let H be the real vector space of all 2  2 complex matrices A which are Hermitian, A = A*. It is
easy to verify that
t x y iz
(x, y, z, t) =
y iz t x
4
defines an isomorphism of R onto the space H. Under this isomorphism, the quadratic form q
is carried onto the determinant function, that is
t x y iz
q(x, y, z, t) = det
y iz t x
or
q( ) = det ( ).
4
This suggests that we might study Lorentz transformations on R by studying linear operators
on H which preserve determinants.
Let M be any complex 2  2 matrix and for a Hermitian matrix A define
U (A) = MAM*.
M
Now MAM* is also Hermitian. From this it is easy to see that U is a (real) linear operator on H.
M
Let us ask when it is true that U 'preserves' determinants, i.e., det [U (A)] = det A for each A
M M
in H. Since the determinant of M* is the complex conjugate of the determinant of M, we see that
2
det [U (A)] = [det M| det A.
M
Thus U preserves determinants exactly when del M has absolute value 1.
M
So now let us select any 2 2 complex matrix M for which [det M| = 1. Then U is a linear
M
operator on H which preserves determinants. Define
T = –1 U .
M M
4
Since is an isomorphism, T is a linear operator on R . Also, T is a Lorentz transformation; for
M M
q(T ) = q( –1 U )
M M
= det ( –1 U )
M
= det (U )
M
= det ( )
= q( )
and so T preserves the quadratic form q.
M
By using specific 2 2 matrices M, one can use the method above to compute specific Lorentz
transformations.

Self Assessment

1. Suppose M belongs O(n, C). Let
n
y = M x
ik k
i
k 1

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