# Difference between revisions of "Curvature is tensorial"

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 $X,Y,Z$.

## 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 $\nabla_A(fB) = (Af)(B) + f\nabla_A(B)$

## 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 $f\nabla_X\nabla_Y - \nabla_Y (f \nabla_X) - \nabla_{[fX,Y]}$ 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 $f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{(Yf)X} - \nabla_{[fX,Y]}$ $\nabla$ is $C^\infty$-linear in its subscript argument $(Yf)\nabla_X \to \nabla_{(Yf)X}$
4 $f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{(Yf)X + [fX,Y]}$ $\nabla$ is additive in its subscript argument $(Yf)}X} + \nabla_{[fX,Y]} = \nabla_{(Yf)X + [fX,Y]$
5 $f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]})$ Fact (2) $[fX,Y] + (Yf)X \to f[X,Y]$.

### 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 $\nabla_X(f\nabla_Y) - f\nabla_Y\nabla_X - \nabla_{[X,fY]}$ By definition of a connection, $\nabla$ is $C^\infty$-linear in its subscript argument $\nabla_{fY} \to f\nabla_Y$.
2 $(Xf)\nabla_Y + f(\nabla_X\nabla_Y) - f\nabla_Y\nabla_X - \nabla_{[X,fY]}$ Fact (3), the Leibniz-like axiom for connection $\nabla_X(f\nabla_Y) \to (Xf)\nabla_Y + f(\nabla_X\nabla_Y)$. 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 $f(\nabla_X\nabla_Y - \nabla_Y\nabla_X) - \nabla_{[X,fY]} + \nabla_{(Xf)Y}$ $\nabla$ is $C^\infty$-linear in its subscript argument. $(Xf)\nabla_Y \to \nabla_{(Xf)Y}$
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 $f(\nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X,Y]}$ Fact (2) $[X,fY] - (Xf)Y \to f[X,Y]$

### 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:

$\! f R(X,Y)Z$