Curvature is tensorial

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This article gives the statement, and possibly proof, that a map constructed in a certain way is tensorial
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Statement

Let \nabla be a connection on a vector bundle E over a differential manifold M. The Riemann curvature tensor of \nabla is given as a map \Gamma(TM) \otimes \Gamma(TM) \otimes \Gamma(E) \to \Gamma(E) defined by:

R(X,Y)Z = \nabla_X\nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[X,Y]} Z

We claim that R is a tensorial map in each of the variables X,Y,Z.

Related facts

Facts used

Fact no. Name Statement with symbols
1 Any connection is C^\infty-linear in its subscript argument \nabla_{fA} = f\nabla_A for any C^\infty-function f and vector field A.
2 The Leibniz-like axiom that is part of the definition of a connection For a function f and vector fields A,B, and a connection \nabla, we have \nabla_A(fB) = (Af)(B) + f\nabla_A(B)
3 Corollary of Leibniz rule for Lie bracket (in turn follows from Leibniz rule for derivations For a function f and vector fields X,Y:


\! f[X,Y] = [fX,Y] + (Yf)X
\! f[X,Y] = [X,fY] - (Xf)Y

Proof

To prove tensoriality in a variable, it suffices to show C^\infty-linearity in that variable. This is because linearity in C^\infty-functions guarantees linearity in a function that is 1 at exactly one point, and zero at others.

The proofs for X and Y are analogous, and rely on manipulation of the Lie bracket [fX,Y] and the property of a connection being C^\infty in the subscript vector. These proofs do not involve any explicit use of Z. The proof for Z relies simply on repeated application of the product rule, and the fact that XY - YX = [X,Y].

Tensoriality in the first variable

Given: f:M \to \R is a C^\infty-function.

To prove: \! R(fX,Y) = f R(X,Y), or more explicitly, \! \nabla_{fX}\nabla_Y - \nabla_Y \nabla_{fX} - \nabla_{[fX,Y]} = f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}

We start out with the left side:

\nabla_{fX}\nabla_Y - \nabla_Y \nabla_{fX} - \nabla_{[fX,Y]}

Each step below is obtained from the previous one via some manipulation explained along side.

Step no. Current status of left side Facts/properties used Specific rewrites
1 f\nabla_X\nabla_Y - \nabla_Y (f \nabla_X) - \nabla_{[fX,Y]} Fact (1): \nabla is C^\infty-linear in its subscript argument. \nabla_{fX} \to f\nabla_X
2 f\nabla_X\nabla_Y - (Yf)\nabla_X - f \nabla_Y\nabla_X - \nabla_{[fX,Y]} Fact (2) \nabla_Y(f \nabla_X) \to (Yf)\nabla_X + f\nabla_Y\nabla_X. To understand this more clearly imagine an input Z to the whole expression, so that the rewrite becomes \nabla_Y(f \nabla_X(Z)) \to (Yf)\nabla_X(Z) + f\nabla_Y\nabla_X(Z). In the notation of fact (3), A = Y, f = f, and B = \nabla_X(Z).
3 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{(Yf)X} - \nabla_{[fX,Y]} Fact (1) (Yf)\nabla_X \to \nabla_{(Yf)X}
4 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{(Yf)X + [fX,Y]} \nabla is additive in its subscript argument \nabla_{(Yf)X} + \nabla_{[fX,Y]} = \nabla_{(Yf)X + [fX,Y]}
5 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{f[X,Y]} Fact (3) [fX,Y] + (Yf)X \to f[X,Y]
6 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}) Fact (1) \nabla_{f[X,Y]} \to f\nabla_{[X,Y]}

Tensoriality in the second variable

Given: f:M \to \R is a C^\infty-function.

To prove: \! R(X,fY) = f R(X,Y), or more explicitly, \nabla_X\nabla_{fY} - \nabla_{fY}\nabla_X - \nabla_{[X,fY]} = f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}.

We start out with the left side:

\nabla_X\nabla_{fY} - \nabla_{fY}\nabla_X - \nabla_{[X,fY]}

Each step below is obtained from the previous one via some manipulation explained along side.

Step no. Current status of left side Facts/properties used Specific rewrites
1 \nabla_X(f\nabla_Y) - f\nabla_Y\nabla_X - \nabla_{[X,fY]} Fact (1) \nabla_{fY} \to f\nabla_Y.
2 (Xf)\nabla_Y + f(\nabla_X\nabla_Y) - f\nabla_Y\nabla_X - \nabla_{[X,fY]} Fact (2) \nabla_X(f\nabla_Y) \to (Xf)\nabla_Y + f(\nabla_X\nabla_Y). To make this more concrete, imagine an input Z. Then, the rewrite becomes \nabla_X(f\nabla_Y(Z)) \to (Xf)\nabla_Y(X) + f(\nabla_X\nabla_Y(Z)). This comes setting A = X, f = f, B = \nabla_YZ in Fact (3).
3 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{[X,fY]} + \nabla_{(Xf)Y} Fact (1) (Xf)\nabla_Y \to \nabla_{(Xf)Y}
4 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{[X,fY] - (Xf)Y} \nabla is additive in its subscript argument. \nabla_{[X,fY]} - \nabla_{(Xf)Y} \to \nabla_{[X,fY] - (Xf)Y}.
5 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{f[X,Y]} Fact (3) [X,fY] - (Xf)Y \to f[X,Y]
6 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}) Fact (1) \nabla_{f[X,Y]} \to f\nabla_{[X,Y]}

Tensoriality in the third variable

Given: A C^\infty-function f:M \to \R.

To prove: \! R(X,Y) (fZ) = f R(X,Y) Z. More explicitly, \! \nabla_X\nabla_Y(fZ) - \nabla_Y\nabla_X(fZ) - \nabla_{[X,Y]}(fZ)  = f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z + ((XY - YX - [X,Y])f)Z.

We start out with the left side:

\nabla_X\nabla_Y(fZ) - \nabla_Y\nabla_X(fZ) - \nabla_{[X,Y]}(fZ)

Each step below is obtained from the previous one via some manipulation explained along side.

Step no. Current status of left side Facts/properties used Specific rewrites
1 \! \nabla_X( (Yf)(Z) + f \nabla_YZ) - \nabla_Y ((Xf)Z + f \nabla_XZ) - f \nabla_{[X,Y]}Z - ([X,Y]f) Z Fact (2) \nabla_Y(fZ) \to (Yf)(Z) + f\nabla_YZ and \nabla_X(fZ) \to (Xf)Z + f\nabla_XZ
2 \! (XYf)(Z) + (Yf) \nabla_XZ + (Xf) \nabla_YZ + f \nabla_X\nabla_YZ - (YXf)Z - (Xf) \nabla_YZ - (Yf) \nabla_XZ -f \nabla_Y\nabla_XZ - f \nabla_{[X,Y]}Z - ([X,Y]f) Z Fact (2) \nabla_X((Yf)Z) \to X((Yf)Z) + (Yf)\nabla_XZ, etc.
3 f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z + ((XY - YX - [X,Y])f)Z -- cancellations
4 f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z + ((XY - YX - [X,Y])f)Z use [X,Y] = XY - YX, definition cancellation