896 28. DIFFERENTIAL
GEOMETRY
whichis
precisely
(the
negative
of) the
exterior
product
n A
fl,
asthe
reader
maywishto
verify.
Thus,e =dn+
nAn
=0, andthe spaceis indeed
fiat!
III
In
all the foregoing examples, the curvatnre was calculated intrinsically. We
neverhadtoleavethespaceandgo to a higherdimension to
"see"
the
curvature.
For example, in the case of the sphere, the only information we had was the line
element in terms
of
the coordinates on the sphere. We never had to resort to any
three-dimensional analysis to discover a globe embedded in the Euclidean
IR
3.
As
mentioned
earlier,
if
aspacehaslineelements with
constant
gi
j,
then
the
Riemann
curvature vanishes trivially. Wehave also seen examples in which the components
of a metric tensor were by no means trivial, but
e was smart enough to detect the
flatness in disguise. Under what conditions can we choose coordinate systems in
terms of which the line elements have
gij = ±8ij? To answer this question we
need the following lemma (proved in [Flan 89, pp.
135-136]):
28.2.8.
Lemma.
If
0 is a matrix
of
i-formssuch that
dO
+0 /\ (GO) = 0, then
there exists an orthogonal matrix
Asuch that dA = AGO.
The questionraised above is intimatelyrelatedto the connectionbetweencoor-
dinate and orthonormal frames. We have seen the usefulness of both. Coordinate
frames, due to the existence of the related coordinate
functions, are useful for
many analytical calculations, for example in Hamiltoniandynamics. Orthonormal
frames are useful because of the simplicity of expressions inherentin all orthonor-
mal vectors. Furthermore, we saw how curvatnre was easily calcnlated once we
constructed orthonormal dual frames. Naturally, we would like to have both. Is
it possible to construct frames that are both coordinate and orthonormal? The
following theorem answers this question:
28.2.9.
Theorem.
Let M be a Riemannian manifold. Then M is flat, i.e., e = 0
if
and only
if
there exists a local coordinate system {xi} for which
{ail
is an
orthonormal basis.
Proof
The existence of orthonormal coordinate frames implies that {dxi} are
orthonormal. Thus, we can use them to find the curvature. But since
d(dx
i)
= 0
for all
i, it follows from Equation(28.12) that w
i
j
=
OandO
=
O.
So the curvature
must vanish. Conversely, suppose that e =
O.
Then by Lemma28.2.8, there exists
an orthogonalmatrix Asuch that
dA = AGO.Now we define the one-formcolunm
matrix
T by T = Ae, where e is the one-form colunm matrix of Equation (28.14).
. Then, using (28.19), we have
dT
= d(Ae) = dA A e +Ado = (AGO) /\ 0 - A(GO /\
0)
=
o.
Thus,
dT
i
= 0 for all i,
By
Theorem 26.5.14 there must exist zero-forms (func-
tions)
xi such that T
i
=
dx',
These xi are the coordinates we are after. The basis
{ail
is obtained using the inverse
of
A (see the discussion following Proposition
25.1.1). Since Ais orthogonal, both
{dxi} and {ail are orthononnal bases. D