## Kaib Kba0233

Conversely, if K is a vector field which satisfies Killing's equation, then Xgg = 0, so

Thus K is a Killing vector field if and only if it satisfies Killing's equation. Then one can locally choose coordinates = (3?,t)(v = lton- 1)

such that Ka = 8xa/8t = San\ in these coordinates Killing's equation takes the form „ ,, , „.

A general space will not have any symmetries, and so will not admit any Killing vector fields. However a special space may admit r linearly independent Killing vector fields Ka (a = 1 ,...,r). It can be shown that the set of all Killing vector fields on such a space forms a Lie algebra of dimension r over E, with the algebra product given by the Lie bracket [ , ] (see (2.16)), where 0 < r < £n(n+ 1). (The upper limit may be lessened if the metric is degenerate.) The local group of diffeomorphisms generated by these vector fields is an r-dimensional Lie group of isometries of the manifold The full group of isometries of may include some discrete isometries (such as reflections in a plane) which are not generated by Killing vector fields; the symmetry properties of the space are completely characterized by this full group of isometries.

### 2.7 Hypersurfaces

If Sf is an (w — 1 )-dimensional manifold and 6: is an imbedding, the image 6(£f) of is said to be a hypersurface in If pe SP, the image of Tp in under the map will be a (n — 1 )-dimensional plane through the origin. Thus there will be some non-zero form neT*^p) such that for any vector X 6 Tp, <n, 0* X) = 0. The form n is unique up to a sign and a normalizing factor, and if 6(£f) is given locally by the equation / = 0 where df # 0 then n may be taken locally as df. If 6(Sf) is two-sided in Jt, one can choose n to be a nowhere zero one-form field on 6(£f). This will be the situation if and are both orientable manifolds. In this case, the choice of a direction of n will relate the orientations of 6(5?) and of if {a;1} are local coordinates from the oriented atlas of such that locally 6(£?) has the equation a;1 = 0 and n = a da;1 where a > 0, then (a;2,...,xn) are oriented local coordinates for 6{5?).

If g is a metric on the imbedding will induce a metric 6*g on where if X,Ye Tp, 6*g(X, Y)^ = g(6* X, tf* Y)^. This metric is sometimes called the first fundamental form of SIf g is positive definite the metric 6*g will be positive definite, while if g is Lorentz, 0*g will be

(a) Lorentz if gabnanb > 0 (in this case, 6(£?) will be said to be a timelike hypersurface),

(6) degenerate if = 0 (in this case, 6{£f) will be said to be a null hyper surf ace),

(c) positive definite if < 0 (in this case, 6(£?) will be said to be a spacelike hyper surf ace).

To see this, consider the vector Nb = nagab. This will be orthogonal to all the vectors tangent to 6(£f), i.e. to all vectors in the subspace H = 6*(Tp) in Tg(p). Suppose first that N does not itself lie in this subspace. Then if (E2,..., En) are a basis for Tp, (N, E2),..., 0*(En)) will be linearly independent and so will be a basis for T#p). The components of g with respect to this basis will be

As the metric g is assumed to be non-degenerate, this shows that gr(N, N) # 0. If g is positive definite, g(N, N) must be positive and so the induced metric 6*g must also be positive definite. If g is Lorentz and <7(N, N) = gabnanb < 0, then 6*g must be positive definite since the matrix of the components of g has only one negative eigenvalue. Similarly if g(N, N) = gabnanb > 0, then 6*g will be a Lorentz metric. Now suppose that N is tangent to 6{£f). Then there is some non-zero vector XeTp such that 6*(X) = N. But g(N,6*Y) = 0 for all YeTp, which implies 6*g(X, Y) = 0. Thus 6*g is degenerate. Also, taking Y to be X, 0(N, N) = gabnanb = 0.

If gabnanb + 0, one can normalize the normal form n to have unit magnitude, i.e. gabnanb = ± 1. In this case the map 6*:T*^p)-^T*p will be one-one on the (w— 1)-dimensional subspace H*#p) of T*#p) consisting of all forms u> at 6(p) such that gabna(ub = 0, because 6*n = 0 and n does not lie in H*. Therefore the inverse (0*)-1 will be a map of T*p onto H*#p), and so into T*#p).

This map can be extended in the usual way to a map of covariant tensors on to covariant tensors on 6(£?) in as there already is a map 6m of contravariant tensors on to 6{£f), one can extend to a map of arbitrary tensors on to 6{£f). This map has the property that 8,„T has zero contraction with n on all indices, i.e.

(@*T)a—bcdna = 0 and (8*Tr~\..dtf°ne = 0 for any tensor Te^).

The tensor h on 6{£f) is defined by h = B*{6*g). In terms of the normalized form n (remember = ± 1),

Kb = 9ab + nanb since this implies 6*h = 6*g and habgbcnc = 0.

The tensor hab = gachcb is a projection operator, i.e. habhbc = hac. It projects a vector X e T#p) into its part lying in the subspace H — 6^(Tp) of Tmv) tangent to 6{5f),

Xa = habXb±nanbXb, where the second term represents the part of X orthogonal to 6(£f). Also hab projects a form u>eT*^p) into its part lying in the subspace H*

Similarly one can project any tensor T eTrs(6(p)) into its part in Hrs(6(p)) = Hm®... ®Hgp}®H*(ll)®... ®E\ph r factors s factors i.e. its part which is orthogonal to n on all indices.

The map is one-one from Tp to H^p). Therefore one can define a map B* from T#p) to Tp by first projecting with hab into and then using the inverse As one already has a map 6* of forms on 6(5f) to forms on Sf, one can extend the definition of 6* to a map B* of tensors of any type on 6(£?) to tensors on This map has the property that = T for any tensor TeTrt(j>) and B*(B*T) = T for any tensor T eHTs(6(p)). We shall identify tensors on Sf with tensors in Hra on 6(5?) if they correspond under the maps B*. In particular, h can then be regarded as the induced metric on 6(£f).

If n is any extension of the unit normal n onto an open neighbourhood of 6{£f) then the tensor x defined on 6(£f) by

Xob = hcahdbnc.d is called the second fundamental form of It is independent of the extension, since the projections by hab restrict the covariant derivatives to directions tangent to6(£f). Locally the field n can be expressed in the form fi = ad/ where/ and a are functions on and/ = 0 on 6{£f). Therefore %ab must be symmetric, since f.ab = f.baa,ndf.ahab = 0.

The induced metric h = 6*g on defines a connection on We shall denote covariant differentiation with respect to this connection by a double stroke, ||. For any tensor TeHra,

T"-bo...dte = hb,h><c... h<dh™, where T is any extension of T to a neighbourhood of This definition is independent of the extension, as the hs restrict the covariant differentiation to directions tangential to 6(Sf). To see this is the correct formula, one has only to show that the covariant derivative of the induced metric is zero and that the torsion vanishes. This follows because

Kmc = (gef + nenf),Bheahfbh«c = 0, and ftab = heah°bf.eg = heah«bf.oe = flba.

The curvature tensor B'0^ of the induced metric h can be related to the curvature tensor Babcd on 6(£?) and the second fundamental form x as follows. If Ye H is a vector field on 6(£f), then

= Y<.,khaeh'dhkc + Ye.fneriP.tkhtdhaghkc + Ye.fnfni.hhaehidhkc and Y'.fnehfd = (Y*ne).fhfd- Y<ne.fhfd = - Y<rie.fh/d, since Yene = 0 on 6(5?), therefore

R'abci y = (frbkfhachkchfd ± XbdXac + XbcXaa) Y". Since this holds for all YeH,

This is known as Gauss' equation.

Contracting this equation on a and c and multiplying by hbd, one obtains the curvature scalar B' of the induced metric:

One can derive another relation between the second fundamental form and the curvature tensor Babcd on 6(£f) by subtracting the expressions {W,dhda),eh%

This is known as Codacci's equation.

2.8 The volume element and Gauss' theorem

If {E°} is a basis of one-forms, one can form from it the n-form e = nIE1 a E2 a ... a En.

If {Ea'}, related to {E°} by E°' = \$°'aEa, is another basis, the n-form e' defined by this basis will be related to e by

£' = det (&a'a) e, so this form is not unique. However, one can use the existence of the metric to define (in a given basis) the form where g = det (gat,). This form has components

The transformation law for g will just cancel the determinant, det (G>°'a), provided that det (<J>a'Q) > 0. Therefore if Ji is orientable the n-forms >j defined by coordinate bases of an oriented atlas will be identical, i.e. given an orientation of Ji, one can define a unique n-form field rj, the canonical n-form, on Ji. The contravariant antisymmetric tensor