1.1. LOCAL ISOMETRIC EMBEDDING O F ANALYTIC METRICS 5

Such a definition is independent of coordinates.

DEFINITION

1.1.1. The map u is free at the point p G Mn if dim(T^(iz)) =

sn + n, or diu(p),dijii(p), i,j — 1, ...,n, are linearly independent as vectors in R9.

Moreover, u is a free map if u is free at each point in

Mn.

It is easy to see that if u :

Mn

—

R9

is free, then q sn + n and w must be an

immersion. Finally, if 0 : M n — Nn is a C2 diffeomorphism and u : 7Vn — Rq is

free, then the composition u o / : M

n

—

R9

is also free.

The map

(1.1.2) (zi, • • • , xn) G Rn •—• (xl5 • • • , x

n

, -x 2 , xix2, • • • , -x 2 ) G R s - + n

gives the simplest example of a free map from Rn to R S n + n . Prom (1.1.2) and the

local charts of manifolds, it is easy to see that every C2 differential manifold Mn

has a local free map into R

S n + n

.

Now we proceed to deal with (1.0.4). If u satisfies (1.0.4), then {d\u, • • • , dnu}

are linearly independent.

The differential system (1.0.4) is highly degenerate. We will transform it to an

equivalent differential system which is easier to analyze. We first express the metric

in a special form, adopting the notation x —

(xf,xn)

= (xi, • • • , x

n

_i, xn).

LEMMA 1.1.2. Let (Mn,g) be a smooth n-dimensional Riemannian manifold.

Then for any p G

Mn,

there exists a local coordinate system (xi,--« ,x

n

) in a

neighborhood N(p) of p where g is of the form

n-l

(1.1.3) g= ^2

gki(xf,Xn)dxkdxi

+

dx2n,

k,l=l

with

(1.1.4) 9ki(0) = Skh dngki(0)=0 for any k,l = 1, • •• , n - 1.

PROOF .

We start with a normal coordinate system (xi, • • • , xn) centered at p,

and let M

n _ 1

= {xn = 0} and e be the unit normal field along M

n _ 1

in

Mn.

For

any q G M

n _ 1

, consider the geodesic c = c(t) in

Mn

with the initial conditions

c(0) = q and c7(0) = e(q). Then (xi, • • • ,xn_i,£) forms a local coordinate system

in a neighborhood of p.

First, we note g(dt^dt) = 1 since each t-curve is an arc-length parametrized

geodesic. Next, we have for any k = 1, • • • , n — 1

dtg(du dk) = g(Vtdt, dk) + g(du Vtdk) = g(du Vfeft) = \dkg(du dt) = 0.

Hence g(dt, dk) = 0 since it is zero at t = 0. Therefore, the metric g is of the form

(1.1.3) in the coordinates (xi, • • • ,x

n

_i,£).

To prove (1.1.4), we have for any /c, / = 1, • • • , n — 1

dtg(dk, d,) = g(Vtdk, dt) + g(dk, Vtdt)

= g(Vkdt,dl)+g(dk,Vldt)

( L 1

'

5 )

= dkg(dt, dt) + dl9(dk, %) - g(dt, vfcft) - g(Vtdk, dt)

= -g(duVkdt)-g(Vidk,dt).