Some of my classmates found it challenging to grasp the concept that the torsion tensor, defined as $T(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$ is a $(1,2)$-tensor field. Therefore, the aim of this post is to elucidate this concept. I apologize in advance for the notational overload.
Let’s quickly recall few facts about tensors. For the remaining of this post $M$ will denote a smooth manifold of dimension $n$ and $V$ a finite-dimensional vector space of dimension $n$ as well.
A covariant $k$-tensor on $V$ is a multilinear map
\[F : \underbrace{V \times \dots \times V}_{k \text{ times }} \to \Bbb R.\]
Similarly a contravariant $k$-tensor on $V$ is a multilinear map
\[F : \underbrace{V^\ast \times \dots \times V^\ast}_{k \text{ times }} \to \Bbb R.\]
Often times we need to consider tensors of mixed valence so a mixed tensor of type $(k,l)$ i.e. $k$-contravariant and $l$-covariant is a multilinear map
\[F : \underbrace{V^\ast \times \dots \times V^\ast}_{k \text{ times }} \times \underbrace{V \times \dots \times V}_{l \text{ times }} \to \Bbb R.\]
If $(e_i)$ is a basis for $V$ and $(\varepsilon^j)$ the dual basis, a basis for the space tensors of type $(k,l)$ is given by the set of tensors of the form
\[e_{i_1}\otimes\dots\otimes e_{i_k} \otimes \varepsilon^{j_1} \otimes \dots \otimes \varepsilon^{j_l}.\]
It follows that any $(k,l)$-tensor $F$ can be written as
\[F = F^{i_1,\dots,i_k}_{j_1,\dots,j_l}e_{i_1}\otimes\dots\otimes e_{i_k} \otimes \varepsilon^{j_1} \otimes \dots \otimes \varepsilon^{j_l}\]
where
\[F^{i_1,\dots,i_k}_{j_1,\dots,j_l} = F(\varepsilon^{i_1},\dots,\varepsilon^{i_k}, e_{j_1},\dots,e_{j_l}).\]
Notice the switching of the indices in the basis vectors. The following calculation may or may not make this a bit more clear. Let $\omega^1,\dots,\omega^k \in V^\ast$ and $v_1,\dots,v_l \in V$. Then
\[\begin{align*}
F(\omega^1,\dots,\omega^k, v_1,\dots,v_l) &= F\left(\sum_{i_1} \omega^1_{i_1}\varepsilon^{i_1}, \dots, \sum_{i_k}\omega^k_{i_k}\varepsilon^{i_k}, \sum_{j_1} v_1^{j_1}e_{j_1},\dots,\sum_{j_l} v_1^{j_l}e_{j_l} \right) \\
&= \sum_{i_1,\dots,i_k}\sum_{j_1,\dots,j_l} \omega^1_{i_1}\dots\omega^k_{i_k} v_1^{j_1}\dots\ v_l^{j_l}F(\varepsilon^{i_1}, \dots, \varepsilon^{i_k}, e_{j_1},\dots,e_{j_l}) \\
&= \sum_{i_1,\dots,i_k}\sum_{j_1,\dots,j_l} F(\varepsilon^{i_1}, \dots, \varepsilon^{i_k}, e_{j_1},\dots,e_{j_l})e_{i_1}\otimes\dots\otimes e_{i_k} \otimes \varepsilon^{j_1} \otimes \dots \otimes \varepsilon^{j_l}(\omega^1,\dots,\omega^k, v_1,\dots,v_l).
\end{align*}\]
Note that here
\[\begin{align*}
e_{i_1}\otimes\dots\otimes e_{i_k} \otimes \varepsilon^{j_1} \otimes \dots \otimes \varepsilon^{j_l}(\omega^1,\dots,\omega^k, v_1,\dots,v_l) &= \omega^1(e_{i_1})\dots\omega^k(e_{i_k})\varepsilon^{j_1}(v_1)\dots\varepsilon^{j_l}(v_l) \\
&= \omega^1_{i_1}\dots\omega^k_{i_k}v_1^{j_1}\dots v_l^{j_l}.
\end{align*}\]
The key thing to keep in mind is that we have a one-to-one correspondence between multilinear maps
\[\underbrace{V^\ast \times \dots \times V^\ast}_{k \text{ times }} \times \underbrace{V \times \dots \times V}_{l \text{ times }} \to \Bbb R\]
and elements in
\[\underbrace{V \otimes \dots \otimes V}_{k \text{ times }} \otimes \underbrace{V^\ast \otimes \dots \otimes V^\ast}_{l \text{ times }}.\]
For example if $F = v \otimes \omega \in V \otimes V^\ast$, then $F$ “eats” a pair $(\alpha, u)$ in $ V^\ast \times V$ not in $ V \times V^\ast$ in order for $F(\alpha,u) = v\otimes \omega(\alpha,u) = \alpha(v)\omega(u)$ to make sense.
Now, let’s discuss something related to the actual aim of this post.
There is a natural basis independent isomorphism between the space $T^{(k+1,l)}(V)$ of $(k+1,l)$-tensors over $V$ and the space of multilinear maps
$$
\underbrace{V^\ast \times \dots \times V^\ast}_{k \text{ times }} \times \underbrace{V \times \dots \times V}_{l \text{ times }} \to V.
$$
Note the target here is $V$ instead of the underlying field of $V$.
Consider a multilinear map
$$
F : \underbrace{V^\ast \times \dots \times V^\ast}_{k \text{ times }} \times \underbrace{V \times \dots \times V}_{l \text{ times }} \to V.
$$
By the universal property of tensor products this induces a unique linear map
$$
\widetilde{F} : \underbrace{V^\ast \otimes \dots \otimes V^\ast}_{k \text{ times }} \otimes \underbrace{V \otimes \dots \otimes V}_{l \text{ times }} \to V.
$$
such that $\widetilde{F}(\omega^1\otimes\dots\otimes \omega^k \otimes v_1\otimes\dots\otimes v_l) = F(\omega^1,\dots,\omega^k, v_1,\dots,v_l)$ for any $\omega^1,\dots,\omega^k, v_1,\dots,v_l \in V^\ast \times \dots \times V^\ast \times V \times \dots \times V$.
It follows that $\widetilde{F} \in \operatorname{Hom}(V^\ast \otimes \dots \otimes V^\ast \otimes V \otimes \dots \otimes V, V)$, but
$$
\begin{align*}
\operatorname{Hom}(\underbrace{V^\ast \otimes \dots \otimes V^\ast}_{k \text{ times }} \otimes \underbrace{V \otimes \dots \otimes V}_{l \text{ times }}, V) &\cong \left(V^\ast \otimes \dots \otimes V^\ast \otimes V \otimes \dots \otimes V\right)^\ast \otimes V \\
&= \underbrace{V \otimes \dots \otimes V}_{k+1 \text{ times }} \otimes \underbrace{V^\ast \otimes \dots \otimes V^\ast}_{l \text{ times }}
\end{align*}
$$
which concludes the result.
Before we march onto tensor fields let’s go over a lemma we are going to need soon.
Let $E, F$ and $G$ be vector bundles over $M$. If
$$
\Phi : \Gamma(E) \times \Gamma(G) \to \Gamma(F)
$$
is a $C^\infty(M)$-bilinear map, then for any $p \in M$ there exists a unique $\Bbb R$-linear map on the fibers
$$
\Phi_p : E_p \times G_p \to F_p
$$
such that for every $s \in \Gamma(E)$ and $t \in \Gamma(G)$ we have
$$
\Phi_p(s(p),t(p)) = \Phi(s,t)(p).
$$
Let $(v,u) \in E_p \times G_p$ and choose sections $s \in \Gamma(E)$ and $t \in \Gamma(G)$ with $s(p) = v$ and $t(p) = u$. Define
$$
\Phi_p(v,u) = \Phi(s,t)(p).
$$
We will first argue that this choice is independent of the sections $s$ and $t$. Suppose that $s'$ and $t'$ are two sections with $s'(p) = v$ and $t'(p) = u$, then $(s-s')(p) = 0$ and $(t-t')(p) = 0$ and since we know by our previous results
here that $\Phi$ is a point operator it follows that $\Phi(s-s',t-t')(p) = 0$ which yields that
$$
\Phi(s,t)(p) = \Phi(s',t')(p).
$$
I will show that it's linear in the first arguement and the argument carries over similarly to the second argument mutatis mutandis. Let $v_1,v_2 \in E_p, u \in G_p$ and $a_1,a_2 \in \Bbb R$ as well as $s_i \in \Gamma(E)$ be global sections such that $s_i(p) = v_i$ and lastly $t \in \Gamma(G)$ with $t(p) = u$. We have that
$$
\begin{align*}
\Phi_p(a_1v_1+a_2v_2, u) &= \Phi(a_1s_1 + a_2s_2, t)(p) \\
&= a_1\Phi(s_1,t)(p) + a_2\Phi(s_2,t)(p) \\
&= a_1\Phi_p(v_1,u) + a_2\Phi(v_2,u)
\end{align*}
$$
which concludes the proof.
So onto tensor fields then.
A $(k,l)$-tensor field on a manifold $M$ is a section of the vector bundle
$$
\left(\bigotimes^k TM\right) \otimes \left(\bigotimes^l T^\ast M \right)
$$
An example you might have already encountered is the metric tensor $g$ on a Riemannian manifold $M$. Infact $g$ is a smooth $(0,2)$-tensor field.
Similarly any differential $k$-form is a smooth section of the exterior bundle $\bigotimes^k(T^\ast M)$ that corresponds to an alternating $k$-linear and is thus a smooth $(0,k)$-tensor field. To be more specific a differential $k$-form is a section of the exterior bundle $\Lambda^k T^\ast M$.
An important property that tensor fields have is that they are multilinear over $C^\infty(M)$. If $F$ is a $(k,l)$-tensor field, then for $1$-forms $\omega^1,\dots,\omega^k$ and vector fields $X_1,\dots,X_l$ the function $F(\omega^1,\dots,\omega^k, X_1,\dots,X_l)$ from $M$ to $\Bbb R$ is smooth.
All in all we see that $F$ induces a map multilinear map
\[\widetilde{F} : \underbrace{\Gamma(T^\ast M) \times \dots \times \Gamma(T^\ast M)}_{k \text{ times }} \times \underbrace{\Gamma(TM) \times \dots \times \Gamma(TM)}_{l \text{ times }} \to C^\infty(M)\]
over $C^\infty(M)$. The gist of this post is that the converse is true as well.
(Tensor Characterization Lemma) A map
$$
F : \underbrace{\Gamma(T^\ast M) \times \dots \times \Gamma(T^\ast M)}_{k \text{ times }} \times \underbrace{\Gamma(TM) \times \dots \times \Gamma(TM)}_{l \text{ times }} \to C^\infty(M)
$$
is induced by a smooth $(k,l)$-tensor field if and only if it is multilinear over $C^\infty(M)$. Likewise a map
$$
F : \underbrace{\Gamma(T^\ast M) \times \dots \times \Gamma(T^\ast M)}_{k \text{ times }} \times \underbrace{\Gamma(TM) \times \dots \times \Gamma(TM)}_{l \text{ times }} \to \Gamma(TM)
$$
is induced by a smooth $(k+1,l)$-tensor field if and only if it is multilinear over $C^\infty(M)$.
Suppose that
$$
F : \Gamma(T^\ast M) \times \dots \times \Gamma(T^\ast M) \times \Gamma(TM) \times \dots \times \Gamma(TM) \to C^\infty(M)
$$
is multilinear over $C^\infty(M)$. For each $p \in M$ there is a unique $\Bbb R$-multilinear map
$$
F_p : T^\ast_p M \times \dots \times T^\ast_p M \times T_pM \times \dots \times T_pM \to \Bbb R
$$
such that for all $\omega^1,\dots,\omega^k \in \Gamma(T^\ast M) \times \dots \times \Gamma(T^\ast M)$ and $X_1,\dots,X_l \in \Gamma(TM) \times \dots \times \Gamma(TM)$
$$
F_p(\omega^1_p,\dots,\omega^k_p, X_{1,p},\dots,X_{l,p}) = F(\omega^1,\dots,\omega^k, X_{1},\dots,X_{l})(p).
$$
Now the universal property of tensor products yields that the multilinear map $F_p$ corresponds to a unique multilinear map
$$
\widetilde{F}_p : \underbrace{T^\ast_p M \otimes \dots \otimes T^\ast_p M}_{k \text{ times }} \otimes \underbrace{T_pM \otimes \dots \otimes T_pM}_{l \text{ times }} \to \Bbb R
$$
such that $\widetilde{F}_p(\omega^1_p \otimes \dots \otimes \omega^k_p, X_{1,p} \otimes \dots \otimes X_{l,p}) = F(\omega^1,\dots,\omega^k, X_{1},\dots,X_{l})(p)$.
Now since $\widetilde{F}_p \in \operatorname{Hom}(T^\ast_p M \otimes \dots \otimes T^\ast_p M \otimes T_pM \otimes \dots \otimes T_pM, \Bbb R)$ and
$$
\begin{align*}
\operatorname{Hom}(\underbrace{T^\ast_p M \otimes \dots \otimes T^\ast_p M}_{k \text{ times }} \otimes \underbrace{T_pM \otimes \dots \otimes T_pM}_{l \text{ times }}, \Bbb R) &\cong \left(T^\ast_p M \otimes \dots \otimes T^\ast_p M \otimes T_pM \otimes \dots \otimes T_pM \right)^\ast \\
&= \underbrace{T_p M \otimes \dots \otimes T_p M}_{k \text{ times }} \otimes \underbrace{T^\ast_pM \otimes \dots \otimes T^\ast_pM}_{l \text{ times }}.
\end{align*}
$$
In other words $\widetilde{F}$ is a section of $\left(\bigotimes^k TM\right) \otimes \left(\bigotimes^l T^\ast M \right)$ i.e. a $(k,l)$-tensor field.
Similarly if
$$
F : \Gamma(T^\ast M) \times \dots \times \Gamma(T^\ast M) \times \Gamma(TM) \times \dots \times \Gamma(TM) \to \Gamma(TM)
$$
is multilinear over $C^\infty(M)$, then for each $p \in M$ we get a unique $\Bbb R$-multilinear map on the fibers
$$
F_p : T_p^\ast M \times \dots \times T^\ast_p M \times T_pM \times \dots \times T_pM \to T_pM.
$$
Again the universality of the tensor product yields a unique map
$$
\widetilde{F}_p : \underbrace{T_p^\ast M \otimes \dots \otimes T^\ast_p M}_{k \text{ times }} \otimes \underbrace{T_pM \otimes \dots \otimes T_pM}_{l \text{ times }} \to T_pM
$$
i.e. $\widetilde{F}_p \in \operatorname{Hom}(T_p^\ast M \otimes \dots \otimes T^\ast_p M \otimes T_pM \otimes \dots \otimes T_pM, T_pM)$, but
$$
\begin{align*}
\operatorname{Hom}(\underbrace{T_p^\ast M \otimes \dots \otimes T^\ast_p M}_{k \text{ times }} \otimes \underbrace{T_pM \otimes \dots \otimes T_pM}_{l \text{ times }}, T_pM) &\cong \left(T_p^\ast M \otimes \dots \otimes T^\ast_p M \otimes T_pM \otimes \dots \otimes T_pM \right)^\ast \otimes T_pM \\
&= \underbrace{T_p M \otimes \dots \otimes T_p M}_{k+1 \text{ times }} \otimes \underbrace{T^\ast_pM \otimes \dots \otimes T^\ast_pM}_{l \text{ times }}
\end{align*}
$$
and so $\widetilde{F}$ is a $(k+1,l)$-tensor field. We could have just as easily deduced this from the proposition above since we reduced this to the realm of vector spaces.
So after all this trouble it should be quite clear now why the torsion $T : \Gamma(TM) \times \Gamma(TM) \to \Gamma(TM)$ is a $(1,2)$-tensor field. We just identify it with the induced map we get from the Tensor Characterization Lemma.
Before I end this there might be a question on your mind that’s bugging you (at least it was bugging me), what if we consider for example the curvature tensor $R$ of a vector bundle $E$ over $M$? We know that, for $X,Y \in \Gamma(TM)$ and $s \in \Gamma(E)$
\[R(X,Y)s = \nabla_X\nabla_Y s - \nabla_Y\nabla_X s - \nabla_{[X,Y]}s \in \Gamma(E)\]
So $R$ is a map $\Gamma(TM) \times \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$. The issue now is that our lemma doesn’t say anything about maps with target space something else than $\Gamma(TM)$ or $C^\infty(M)$. What can we do here? Looking at this on the fibers we have a map $R_p : T_pM \times T_pM \times E_p \to E_p$ which we can curry to get an alternating bilinear map $R_p : T_pM \times T_pM \to \operatorname{Hom}(E_p,E_p)$. So it would seem that $R$ is a section of the bundle $\left(\bigotimes^2 T^\ast M \right) \otimes \operatorname{End}(E)$. We call such a thing an $\operatorname{End}(E)$-valued $(0,2)$-tensor field.
In general if $E$ is a vector bundle over $M$, then an $E$-valued tensor field of type $(k,l)$ is a section of the bundle
\[\left(\bigotimes^k TM \right) \otimes \left(\bigotimes^l T^\ast M \right) \otimes E.\]
This post ended up being quite a bit longer than I expected, but hopefully you’ll be able to gain some insights from this.