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

$R(X,Y)Z={\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-{\mathrm{\nabla }}_{[X,Y]}Z$

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

Related facts

• Curvature is antisymmetric in first two variables
• Curvature is antisymmetric in last two variables
• Curvature is symmetric in the pairs of first and last two variables

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 $\mathrm{\nabla }$, we have ${\mathrm{\nabla }}_A(fB)=(Af)(B)+f{\mathrm{\nabla }}_A(B)$

Proof

To prove tensoriality in a variable, it suffices to show $C^{\mathrm{\infty }}$-linearity in that variable. This is because linearity in $C^{\mathrm{\infty }}$-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 $C^{\mathrm{\infty }}$ 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: $f:M\to \mathbb{R}$ is a $C^{\mathrm{\infty }}$-function.

To prove: $R(fX,Y)=fR(X,Y)$, or more explicitly, ${\mathrm{\nabla }}_{fX}{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{fX}-{\mathrm{\nabla }}_{[fX,Y]}=f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]}$

We start out with the left side:

${\mathrm{\nabla }}_{fX}{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{fX}-{\mathrm{\nabla }}_{[fX,Y]}$

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	$f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y(f{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{[fX,Y]}$	By definition of a connection, $\mathrm{\nabla }$ is $C^{\mathrm{\infty }}$-linear in its subscript argument.	${\mathrm{\nabla }}_{fX}\to f{\mathrm{\nabla }}_X$
2	$f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-(Yf){\mathrm{\nabla }}_X-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[fX,Y]}$	Fact (3), the Leibniz-like axiom for connection.	${\mathrm{\nabla }}_Y(f{\mathrm{\nabla }}_X)\to (Yf){\mathrm{\nabla }}_X+f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X$. To understand this more clearly imagine an input $Z$ to the whole expression, so that the rewrite becomes ${\mathrm{\nabla }}_Y(f{\mathrm{\nabla }}_X(Z))\to (Yf){\mathrm{\nabla }}_X(Z)+f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X(Z)$. In the notation of fact (3), $A=Y$, $f=f$, and $B={\mathrm{\nabla }}_X(Z)$.
3	$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{(Yf)X+[fX,Y]}$	$\mathrm{\nabla }$ is $C^{\mathrm{\infty }}$-linear in its subscript argument	<nath>(Yf)\nabla_X \to \nabla_{(Yf)X}</math>
4	$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]})$	Fact (2)	$[fX,Y]\to f[X,Y]-(Yf)X$.

Tensoriality in the second variable

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

$R(X,fY)=fR(X,Y)$

We start out with the left side:

${\mathrm{\nabla }}_X{\mathrm{\nabla }}_{fY}-{\mathrm{\nabla }}_{fY}{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,fY]}$

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

$(Xf){\mathrm{\nabla }}_Y+f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{[X,fY]}$

which simplifies to:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X)-{\mathrm{\nabla }}_{[X,fY]-(Xf)Y}$

We now use the corollary stated above:

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

substituting this gives:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]}$

which is $fR(X,Y)$

Tensoriality in the third variable

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

$R(X,Y)(fZ)=fR(X,Y)Z$

We start out with the left side:

${\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y(fZ)-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X(fZ)-{\mathrm{\nabla }}_{[X,Y]}(fZ)$

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

${\mathrm{\nabla }}_X((Yf)(Z)+f{\mathrm{\nabla }}_{Y}Z)-{\mathrm{\nabla }}_Y((Xf)Z+f{\mathrm{\nabla }}_{X}Z)-f{\mathrm{\nabla }}_{[X,Y]}Z-([X,Y]f)Z$

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

$(XYf)(Z)+(Yf){\mathrm{\nabla }}_{X}Z+(Xf){\mathrm{\nabla }}_{Y}Z+f{\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}Z-(YXf)Z-(Xf){\mathrm{\nabla }}_{Y}Z-(Yf){\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}Z-f{\mathrm{\nabla }}_{[X,Y]}Z-([X,Y]f)Z$

After cancellations we are left with the following six terms:

$f({\mathrm{\nabla }}_X{\mathrm{\nabla }}_Y-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_X-{\mathrm{\nabla }}_{[X,Y]})Z+((XY-YX-[X,Y])f)Z$

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

$fR(X,Y)Z$