Prove that the Riemann curvature tensor is a tensor

Question

How would I prove that the Riemann curvature tensor $R: \scr X(M)^3 → \scr X(M)$, $R(X, Y )Z := \nabla_X\nabla_Y Z − \nabla_Y \nabla_XZ − \nabla_{[X,Y ]}Z$, is indeed a tensor ?

I thought I could use this:

(pag 41 in https://radbouduniversitypress.nl/site/books/m/10.54195/EFVF4478/)

and simply prove that

$R(fX_1+gX_2, Y )Z=fR(X_1, Y )Z+gR(X_2, Y )Z$ ...(1)

$R(X, fY_1+gY_2 )Z=fR(X,Y_1)Z+gR(X, Y_2 )Z$ ...(2)

and

$R(X, Y )(fZ_1+gZ_2)=fR(X,Y)Z_1+gR(X, Y )Z_2$ ...(3)

Am I on the right track?Isn't doing this indeed using proposition 2.7? My T.A said I cannot use 2.7 it as it is stated, but I don't see why And then how do I use proposition 2.7 then?

Answer 1

In proposition 2.7 the map $\tau$ has values in $C^\infty (M)$.

But the curvature tensor is a map $\mathfrak{X}(M)^3 \to \mathfrak{X}(M)$ as you have defined it. Therefore you can not apply proposition 2.7 in the way you want to.

To apply proposition 2.7 we first need to transform the Riemann "Tensor" (note that what is given in your definition is not at all a tensor field!!!) to a $C^\infty$-multilinear map $\mathfrak{X}(M)^k \times \Omega(M)^l \to C^\infty (M)$ in some "natural" way and then transform it to an actual tensor field using proposition 2.7. What are the right $k$ and $l$ and how can we transform it?

When speaking of the Riemann "tensor" (or when saying that it is a tensor(-field)) as defined in the post it is implicitly assumed that the multilinear map is transformed to a tensor field using this "natural way" together with proposition 2.7.

Hint:

Given a $C^\infty$-multilinear map $R :\mathfrak{X}(M)^3 \to \mathfrak{X}(M)$ we can define a $C^\infty$-multilinear map $\tilde{R} : \mathfrak{X}(M)^3 \times \Omega (M)^1 \to C^\infty (M)$ by $$\tilde{R} (X,Y,Z,\omega) = \operatorname{trace}( R(X,Y,Z)\otimes \omega).$$

Answer 2

In the Riemannian context you always have a metric to identify vector fields with 1-forms. Proposition 2.7 says you that a tensor of type $(k,l)$ is nothing but a $\mathcal C^\infty(M)$-multilinear map $$ \mathfrak X^k(M) \times \Omega(M)^l \to \mathcal C^\infty(M) $$ By the universal property of tensor product, any map with this property factors to a $\mathcal C^\infty(M)$-linear map $$ \mathfrak X^{\otimes k}(M) \otimes_{\mathcal C^\infty} \Omega^1(M)^{\otimes l} \to \mathcal C^\infty(M) $$

Now, if $g : \mathfrak X^{\otimes 2}(M) \to \mathcal C^\infty(M)$ is a Riemannian metric, for any vector field $X \in \mathfrak X(M)$ you can obtain a 1-form by considering the linear functional $X^\flat: Y \to g(X,Y)$. This transformation (and its inverse) is called musical isomorphisms.

To get back to your case, you have the curvature $$ R : \mathfrak X(M)^3 \to \mathfrak X(M) $$ defined as you stated using the Levi-Civita connection. You may check manually that this is $\mathcal C^\infty(M)$-multilinear in all of three entries (while the connection is not $\mathcal C^\infty(M)$-bilinear, so why?), and so it defines a tensor of type (3,1). However, if you want to use proposition 2.7, you just have to use the musical isomorphism to get a totally covariant tensor. Explicitly you have to consider $$ \tilde R(X,Y,Z,W) = g(W,R(X,Y)Z) $$ this defines a map $\mathfrak X(M)^4 \to \mathcal C^\infty(M)$ which can be checked to be $\mathcal C^\infty(M)$-multilinear, and so a tensor of type (4,0).