For an affine connection $\nabla$, prove the curvature R
$R(X,Y,Z,\alpha)=\alpha(\nabla_X \nabla_Y Z – \nabla_Y \nabla_X Z -\nabla_{[X,Y]}Z)$
with $X,Y,Z$ vector fields and $\alpha$ a co-vector, is a tensor.
So I realise the aim is probably to show that each of the three terms in the expression for $R$ are tensors themselves as then the result will obviously follow. However, I'm not too sure how to show each of these terms are tensors. Do I need to expand out each covariant derivate to get a bunch of connection coefficients and then see some cancellations? Still though, I'm not too sure what I'll want to see after all of this so that I can say "…therefore $R$ is a tensor".
Also, very much related to the question, I'm a bit confused with the difference between
$\nabla_Y Z$ $\space \space $ and $\space \space $ $\nabla_{\mu}\omega$.
The second one I know as the covariant derivative of a 1-form, but what is meant by the covariant derivate of a vector $\textit{field}$? As in, what's the difference between the $\nabla_Y$ and $\nabla_{\mu}$?
Edit: In answering this last question, I have realised that $\nabla_YZ=Y^{\mu}\nabla_{\mu}Z^{\nu}$ and $\nabla_{\mu}\omega=\nabla_{\mu}\omega_{\nu}$.
Best Answer
A $(k,l)$-tensor field $A$ on smooth manifold $M$ is a smooth section $$ A : M \longrightarrow T^{(k,l)}TM, $$ where $T^{(k,l)}TM = \coprod_{p \in M} \underbrace{T_pM \otimes \cdots \otimes T_pM}_{k \text{ times}} \otimes \underbrace{T^*_pM \otimes \cdots \otimes T^*_pM}_{l \text{ times}} $.
Any tensor field has property that for any smooth vector fields $X_1,\dots,X_l$ and smooth covector fields $\omega^1,\dots,\omega^k$ we have a smooth function $$ A(\omega^1,\dots,\omega^k,X_1,\dots,X_l) : M \longrightarrow \mathbb{R} $$ defined as $A(\omega^1,\dots,\omega^k,X_1,\dots,X_l)(p) = A_p(\omega^1|_p,\dots,\omega^k|_p,X_1|_p,\dots,X_l|_p) \in \mathbb{R}$. So $(k,l)$-tensor field $A$ induces a map $$ \tilde{A} : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M). $$ An important fact is that this map is multilinear over $C^{\infty}(M)$. It turns out that this in fact characterize tensor fields.
In the proof of the above lemma, the tensor induce from the multilinear map $\tau : \mathfrak{X}^*(M) \times \cdots \mathfrak{X}^*(M) \times \mathfrak{X}(M)\times \cdots \mathfrak{X}(M) \longrightarrow C^{\infty}(M)$ is a tensor field $A : M \to T^{(k,l)}TM$ defined as $$ A_p(w^1,\dots,w^k,v_1,\dots,v_l):= \tau(\omega^1,\dots,\omega^k,X_1,\dots,X_l)(p) $$ where $\omega^i\in\mathfrak{X}^*(M)$ is any smooth extension of $w^i\in T^*_pM$ and $X_i \in \mathfrak{X}(M)$ is any smooth extension of $v_i \in T_pM$, for each $i$ (the map $A$ independent of the extensions).
Because of this lemma, we often identify $A$ with its induced map $\tilde{A}$.
This lemma is oftenly used in Riemannian geometry text implicitly. You can see the proof in Lee's book here p.318.
Beside the type of maps in $\color{blue}{(\star)}$ above, another but similar way to spot a disguised tensor field is through these kind of maps $$ \tau_0 : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow \color{red}{\mathfrak{X}(M)}. $$ If this map is multilinear over $C^{\infty}(M)$ then we can define a map $$ \tau : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{\color{red}{k+1 \text{ times}}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M) $$ as $\tau(\omega^{1},\dots,\omega^{k}, \omega^{k+1},X_1,\dots,X_l) = \tau_0(\omega^{1},\dots,\omega^{k},X_1,\dots,X_l)(\omega^{k+1})$, which is also multilinear over $C^{\infty}(M)$. Therefore $\tau_0$ induced by a smooth $(k+1,l)$-tensor field (as its induced map $\tau$ induce them in the Tensor Characterization Lemma above) iff $\tau_0$ is multilinear over $C^{\infty}(M)$. Other respective variants of $\tau_0$ can be deduced the same way.