Why is $T$ in $\mathcal T_2^1(M)$?

Let $M$ be smooth manifold and $\nabla$ an affine connection on $M$. Then the torsion tensor of $\nabla$ is the map $T:\mathcal T(M)\times\mathcal T(M)\to\mathcal T(M)$ and $$T(X,Y)=\nabla_XY-\nabla_YX-[X,Y]$$ The author of book claims that $T\in\mathcal T_2^1(M)$ but this is very strange for me, since by definition $\mathcal T_2^1(M)$ is the set of smooth sections from $M\to T_2^1M=\coprod_{p\in M}T_2^1(T_pM)$ where $T_2^1(T_pM)$ is the space of $(2+1)$-linear maps from $T_pM\times T_pM\times T_p^*M\to\mathbb R$, but $T$ is is a map $\mathcal T(M)\times\mathcal T(M)\to\mathcal T(M)$.

Can anyone explain for me, that how this is possible, please?

Also my note in order to prove that $T\in\mathcal T_2^1(M)$, shows the following relations $$(f_1w_1+f_2w_2)(V)=f_1w_1(V)+f_2w_2(V)\\ T(X+Y,V)=T(X,V)+T(Y,V)\\ T(fX,gY)=fgT(X,Y)\\$$ if $T\in\mathcal T_2^1(M)$ then for all $q\in M$ $$T(q)\in T_2^1M$$ so we need to show that $T(q)(X_p,Y_p,w_p)$ is linear with respect to every variable.

• What is $\mathcal{T}_2^1(M)$ for you? Is it the collection of tensors of the form $\mathcal{T}(M)\times\mathcal{T}(M)\times\mathcal{T}^*(M) \to \mathcal{C}^{\infty}(M)$? – Michael Albanese Jun 12 '16 at 16:13
• Check your definition. There should be no fixed $p$ ... – Ted Shifrin Jun 12 '16 at 16:18
• @MichaelAlbanese By definition $\mathcal T_2^1(M)$ is the set of smooth maps from $M\to T_2^1M=\coprod_{p\in M}T_2^1(T_pM)$ where $T_2^1(T_pM)$ is the space of $(2+1)$-linear maps from $T_pM\times T_pM\times T_p^*M\to\mathbb R$. – user302007 Jun 12 '16 at 16:30
• @user302007: It should be defined as the set of smooth sections, not maps. Which book are you using? – Michael Albanese Jun 12 '16 at 16:41
• @MichaelAlbanese Riemannian Manifolds by J. Lee. – user302007 Jun 12 '16 at 16:45

Associated to the map $$T : \mathcal{T}(M)\times\mathcal{T}(M) \to \mathcal{T}(M),$$ we have the map $$\hat{T} : \mathcal{T}(M)\times\mathcal{T}(M)\times\mathcal{T}^*(M) \to \mathcal{C}^{\infty}(M)$$ given by $\hat{T}(X, Y, \omega) = \omega(T(X, Y))$. If $T$ is $\mathcal{C}^{\infty}(M)$-linear in both arguments, then $\hat{T}$ will be $\mathcal{C}^{\infty}(M)$-linear in all of its arguments (it's automatically $\mathcal{C}^{\infty}(M)$-linear in the third argument). If this is the case, then $\hat{T}$ is a $2$-covariant, $1$-contravariant tensor field on $M$.
What they mean is that $\hat{T} \in \mathcal{T}_2^1(M)$.