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 $X,Y,Z$ .

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 $\nabla _{A}(fB)=(Af)(B)+f\nabla _{A}(B)$

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: $f:M\to \mathbb {R}$ is a $C^{\infty }$ -function.

To prove: $\!R(fX,Y)=fR(X,Y)$ , 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	Failed to parse (syntax error): {\displaystyle \nabla_{(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: $f:M\to \mathbb {R}$ is a $C^{\infty }$ -function.

To prove: $\!R(X,fY)=fR(X,Y)$ , 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 $A=X$ , $f=f$ , $B=\nabla _{Y}Z$ 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 $f:M\to \mathbb {R}$ be a scalar function. We will show that:

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

We start out with the left side:

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

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

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

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

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

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:

$\!fR(X,Y)Z$