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).

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