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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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 Full statement
1 Leibniz rule for derivations For a vector field X and functions f,g, we have: \! X(fg) = (Xf)(g) + f(Xg)
2 Corollary of Leibniz rule for Lie bracket For a function f and vector fields X,Y:

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

3 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)


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]} By definition of a connection, \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 (3), the Leibniz-like axiom for connection. \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 + [fX,Y]} \nabla is C^\infty-linear in its subscript argument <nath>(Yf)\nabla_X \to \nabla_{(Yf)X}</math>
4 f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}) Fact (2) [fX,Y] \to f[X,Y] - (Yf)X.

Tensoriality in the second variable

Let f:M \to \R be a scalar function. We will show that:

\! R(X,fY) = f R(X,Y)

We start out with the left side:

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

Applying the Leibniz rule and the property of a connection being C^\infty in its subscript variable yields:

(Xf)\nabla_Y + f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{[X,fY]}

which simplifies to:

f(\nabla_X\nabla_y - \nabla_Y\nabla_X) - \nabla_{[X,fY] - (Xf)Y}

We now use the corollary stated above:

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

substituting this gives:

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

which is f R(X,Y)

Tensoriality in the third variable

Let f:  M \to \R be a scalar function. We will show that:

\! R(X,Y) (fZ) = f R(X,Y) Z

We start out with the left side:

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

Now we apply the Leibniz rule for connnections on each term:

\nabla_X( (Yf)(Z) + f \nabla_YZ) - \nabla_Y ((Xf)Z + f \nabla_XZ) - f \nabla_{[X,Y]}Z - ([X,Y]f) Z

We again apply the Leibniz rule to the first two term groups:

(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

After cancellations we are left with the following six terms:

f (\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})Z + ((XY - YX - [X,Y])f)Z

But since [X,Y] = XY - YX, the last three terms vanish, and we are left with:

\! f R(X,Y)Z