# Difference between revisions of "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 $X,Y,Z$.

## 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 $f:M \to \R$ be a scalar function. We will show that: $\! R(fX,Y) = f R(X,Y)$

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 $f:M \to \R$ be a scalar function. We will show that: $\! R(X,fY) = f R(X,Y)$

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 $f R(X,Y)$

### Tensoriality in the third variable

Let $f: M \to \R$ be a scalar function. We will show that: $\! R(X,Y) (fZ) = f R(X,Y) Z$

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: $\nabla_X( (Yf)(Z) + f \nabla_YZ) - \nabla_Y ((Xf)Z + f \nabla_XZ) - f \nabla_{[X,Y]}Z - ([X,Y]f) Z$

We again apply the Leibniz rule to the first two term groups: $(XYf)(Z) + (Yf) \nabla_XZ + (Xf) \nabla_YZ + f \nabla_X\nabla_YZ - (YXf)Z - (Xf) \nabla_YZ - (Yf) \nabla_XZ -f \nabla_Y\nabla_XZ - f \nabla_{[X,Y]}Z - ([X,Y]f) Z$

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: $\! f R(X,Y)Z$