Curvature is tensorial

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 indeed a tensor, viz it is tensorial in each of $X,Y,Z$.

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.

Tensoriality in the first variable

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

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

We start out with the left side:

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

Now by the definition of a connection, $\nabla$ is $C^{\infty }$-linear in its subscript argument. Thus, the above expression can be written as:

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

Now applying the Leibniz rule for connections, we get:

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

We can rewrite $(Yf)\nabla _{X}=\nabla _{(Yf)X}$ and we then get:

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

Now it remains to simplify $(Yf)X+[fX,Y]$. Observe that:

$[fX,Y](h)=(fX)(Yh)-Y((fX)h)=(fX)(Yh)-(Yf)(Xh)-(fY(Xh)=f([X,Y]h)-(Yf)(Xh)$

This tells us that:

$(Yf)X+[fX,Y]=f[X,Y]$

which, substituted back, gives:

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

Tensoriality in the second variable

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

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:

$fR(X,Y)Z$