Lie Derivatives of Tensor Fields • Any Lie derivative on vector fields automatically induces Lie derivatives on all tensor bundles over M, and thus gives us a way to compute Lie derivatives of all tensor fields. • We want L obeys the following product rule w.r.t. the natural pairing between a covector field ω and a vector field Y :

LX (ω(Y )) = LX ω(Y )+ω(LX Y ).

Thus, it seems reasonable to define LX ω as

LX ω(Y )=LX (ω(Y )) − ω(LX Y ).

Lemma 1. Let L be a Lie derivative on Γ(TM). There is a unique Lie derivative k in each tensor bundle T` M, also denoted by L, such that the following conditions are satisfied. (a) LX f = Xf, for all smooth real-valued functions f; (b) LX (fσ)=(LX f)σ + fLX σ, for all smooth tensor filds σ, τ; (c) LX (σ ⊗ τ)=(Lσ) ⊗ τ + σ ⊗LX τ, for smooth tensor fields σ, τ; (d) If Y1, ··· ,Yk are smooth vector fields and σ is a smooth k-tensor field, then

LX (σ(Y1, ··· ,Yk)) =(LX σ)(Y1, ··· ,Yk)

+ ···+ σ(Y1, ··· , LX Yk).

Corollary 2. If X is a smooth vector field and σ is a smooth covariant k-tensor field, then for any smooth vector fields Y1, ··· ,Yk,

(LX σ)(Y1, ··· ,Yk)) =X(σ(Y1, ··· ,Yk)) − σ([X, Y1],Y2, ··· ,Yk) −···

− σ(Y1, ··· ,Yk−1, [X, Yk]).

For any p ∈ M, it is sufficiently close to zero, θt is a diffeomorphism from a nbhd ∗ of p to a nbhd of θt(p), so θt pulls back tensors at θt(p) to ones at p. Lemma 2. Let X be a smooth vector field on smooth manifold M, and let θ be its flow. Given a smooth covariant tensor field τ on M, the Lie derivative of τ with respect to X d θ∗(τ ) − τ L ∗ t θt(p) p ( X τ)p = (θt τ)p = lim . dt t=0 t→0 t Proof. We have ∗ θ τ (Yp) − τp(Yp) τ ((θt(p))∗Y ) − τp(Yp) t θt(p) = θt(p) t t τ ((θt(p))∗Yp − Y )+(τ Y − τp(Yp)) = θt(p) θt(p) θt(p) θt(p) t (p)0 →−(τ(Lθ(p)0(0)Y ))p +(θ (0))(τ(Y ))p, as t → 0,

= − (τ(LX Y ))p + X(τ(Y ))p. 

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1 2

Covariant Deivatives of Tensor Fields • By definition, a connection on M is a way to compute covariant derivatives of vector fields. • In fact, any connection automatically induces connections on all tensor bundles over M, and thus gives us a way to compute covariant derivatives of all tensor fields. • We want D obeys the following product rule w.r.t. the natural pairing between a covector field ω and a vector field Y :

DX (ω(Y )) = DX ω(Y )+ω(DX Y ).

Thus, it seems reasonable to define DX ω as

DX ω(Y )=DX (ω(Y )) − ω(DX Y ).

Lemma 1. Let D be a connection on M. There is a unique connection in each k tensor bundle T` M, also denoted by D, such that the following conditions are satisfied. (a) On TM, D agrees with given connection. 0 ∞ (b) On T M = C (M), D is the ordinary differentiation of functions: DX f = Xf. (c) D obeys the following product rule w.r.t. tensor products:

DX (F ⊗ G)=(DX F ) ⊗ G + F ⊗ (DX G).

This connection satisfies the following properties: (i) D obeys the following product rule w.r.t. the natural pairing between a covector field ω and a vector field Y :

DX (ω(Y )) = DX ω(Y )+ω(DX Y ).

k j (ii) For any F ∈ Γ(T` M), vector fields Yi and 1-form ω ,

1 ` 1 ` (DX F )(ω , ··· ,ω ,Y1, ··· ,Yk)=X(F (ω , ··· ,ω ,Y1, ··· ,Yk)) ` 1 j ` − X F (ω , ··· ,DX ω , ··· ,ω ,Y1, ··· ,Yk) j=1 k 1 ` − X F (ω , ··· ,ω ,Y1, ··· ,DX Yi, ··· ,Yk). i=1 3

• r r If S is a 0
or 1
tensor field, then we can define a covariant derivative DS r+1 r+1 that we interpret as a 0
or 1
-tensor field. — The main idea is to make sure that the product rule holds. 1 Thus if S is a 1
tensor field, we want to have

DX (S(Y )) = (DX S)(Y )+S(DX Y ).

Thus, it seems reasonable to define D as

DS(X, Y )=(DX S)(Y )

=DX (S(Y )) − S(DX Y ).

More generally, we define

DS(X, Y1, ··· ,Yr)=(DX S)(Y1, ··· ,Yr) r =DX (S(Y1, ··· ,Yr)) − X S(Y1, ··· ,DX Yi, ··· ,Yr), i=1

where DX is interpreted as the directional derivative when applied to a function, while we use it as covariant differentiation on vector fields. k Lemma 2. If D is a linear connection on M, and S ∈ Γ(T` M), then the map

1 ` 1 ` DS(ω , ··· ,ω ,Y1, ··· ,Yk,X)=DX S(ω , ··· ,ω ,Y1, ··· ,Yk),

k+1 defines a `
-tensor field. Definition. The tensor field DS is called the total covariant derivative.

Lemma 3. Let D be a linear connection. The components of the total covariant k derivative of a `
-tensor field F with respect to a coordinate system are given by

` k ··· ··· ··· ··· ··· F j1 j` = ∂ F j1 j` + X F j1 p j` Γjs − X F j1 j` Γp . i1···ik ;m m i1···ik i1 ···ik mp i1···p···ik mis s=1 s=1

Definition. A tensor field S is said to be parallel if DS =0.

• On a Riemannian manifold (M,g), we always have that Dg =0, since (Dg)(X, Y1,Y2)=DX (g(Y1,Y2)) − g(DX Y1,Y2) − g(Y1,DX Y2)=0. 4

• Let f be a smooth function on M. Then Df ∈ T ∗M is just the 1-form df , because both tensors have the same action on tensors:

Df(X)=DX f = Xf = df (X).

• 2 2 The 0
-tensor D f = D(Df) is called the covariant Hessian of f. The covariant Hessian is symmetric:

2 D f(X, Y )=DX (df (Y )) − df (DX Y )

=X(Yf) − df (DX Y )

=X(Yf) − df (DY X) − df ([X, Y ])

=X(Yf) − df (DY X) − [X, Y ]f

=Y (Xf) − df (DY X) =D2f(Y,X).

r Definition. For a ·
-tensor field S we define the second covariant derivative 2 r+2 D S as the ·
-tensor field

(D2 S)(Y , ··· ,Y )=(D (DS))(X ,Y , ··· ,Y ) X1,X2 1 r X1 2 1 r

=DX1 (DS(X2,Y1, ··· ,Yr)) − (DS)(DX1 X2,Y1, ··· ,Yr)

=DX1 DX2 S(Y1, ··· ,Yr) − (DS)(DX1 X2,Y1, ··· ,Yr) =(D D S)(Y , ··· ,Y ) − (D S)(Y , ··· ,Y ). X1 X2 1 r DX1 X2 1 r

• 2 With this we obtain the 0
version of the Hessian of a function defined as

2 DX,Y f =DX DY f − DDX Y f

=DX (Df(Y )) − Df(DX Y )

=DX Df(Y ) =D2f(X, Y ),

which is symmetric in X and Y .