Curvature is tensorial

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

Proof

To prove tensoriality in a variable, it suffices to show $C^{\mathrm{\infty }}$-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.

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:

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

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

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

Now applying the Leibniz rule for connections, we get:

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

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

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

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:

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

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:

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

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:

$fR(X,Y)Z$