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

• Curvature is antisymmetric in first two variables

Facts used

• Leibniz rule for derivations: This states that for a vector field $X$ and functions $f,g$, we have:

$X(fg)=(Xf)(g)+f(Xg)$

• Corollary of Leibniz rule for Lie bracket: This states that for a function $f$ and vector fields $X,Y$:

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

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

• The Leibniz rule axiom that's part of the definition of a connection, namely:

$\nabla _{X}(fZ)=(Xf)(Z)+f\nabla _{X}(Z)$

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.

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

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:

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

By the corollary stated above, we have:

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

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:

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

We start out with the left side:

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

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

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

which simplifies to:

$f(\nabla _{X}\nabla _{y}-\nabla _{Y}\nabla _{X})-\nabla _{[X,fY]-(Xf)Y}$

We now use the corollary stated above:

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

substituting this gives:

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

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:

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

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:

$\!fR(X,Y)Z$