# How is $\Delta: T(M)\times T^*(M) \rightarrow C^{\infty}(M)$ a section

I have a map $\Delta: T(M)\times T^*(M) \rightarrow C^{\infty}(M)$, where T(M) is the set of all smooth vector fields on a smooth manifold, M, and $T^*(M)$ is the set of all smooth covector fields on M, such that $\Delta(X, \omega) = \omega(X)$.

I am to show that this map is a smooth section of $T^{(1,1)} TM$, the (1,1) tensor bundle on M. First, as I understand, a section, $\sigma$, is a continuous map s.t. if we have a projection map $\pi : E \rightarrow B$, where E is some total space and B the base space, the map $\sigma : B \rightarrow E$ satisfies $\pi(\sigma(b)) = b, \forall b\in B$. For instance, a vector field on a smooth manifold can be seen as a section of the tangent bundle by $\pi (\sigma(p))=p$ where we think of $\sigma$ as sending a point $p$ to some vector in $T_pM$ and $\pi$ just projecting back to the point.

I think I want to combine this fact with the characteristic property of tensor product, namely: (no idea how to draw a commutative diagram here)

Let V and W be finite-dimensional real vector spaces. If $A: V \times W \rightarrow X$, is a bilinear map into any vector space X, there is a unique linear map $\tilde{A}: V \otimes W \rightarrow X$ such that $\tilde{A} = P\circ A$ where $P$ is the map taking $V\times W \rightarrow V\otimes W$. Then maybe identify X with $C^\infty (M)$ and $V,W$ with $T(M)$ and $T^*(M)$.

As I've understood from Lee's introduction to smooth manifolds, a section of $T^{(1,1)}TM$ is typically of the form: $\sigma = T^j_{i} dx^i \otimes \frac{\partial}{\partial x^j}$ for some basis {$\frac{\partial}{\partial x^j}$} of $T_pM$ with dual basis {$dx^i$} and $T^i_j$ are the component functions.

The problem I have with both these constructions is to identify the spaces E and B in the definition of $\Delta$ as section of $T^{(1,1)}TM$, In my head, just setting E as $C^\infty (M)$ doesn't make sense, because I don't see how to construct a projection map from $C^\infty (M)$ to $T(M) \times T^*(M)$, any direction or piece of intuition to the matter would be appreciated.

## 1 Answer

Let ${\rm Bil}_{\mathbb R}(TM\times T^*M)$ be the vector bundle of bilinear (over $\mathbb R$) maps over $M$ whose fiber over $p\in M$ is ${\rm Bil}_{\mathbb R}T_pM\times T_p^*M$. You have a map $\Delta:M\to {\rm Bil}_{\mathbb R}(TM\times T^*M)$ whose action is the following: For each $p\in M$, $\Delta(p)$ is a map on $T_pM\times T_p^*M\to\mathbb R$ given by $\Delta(p)(X_p, \omega_p) = \omega_p(X_p)$. This gives $\pi(\Delta(p)) = p$, where $\pi:{\rm Bil}_{\mathbb R}(TM\times T^*M)\to M$ is the projection.

To show that $\Delta$ (still viewed as a map from $M$ into the bilinear maps on the product bundle) is tensorial you can show that $\Delta$ is bilinear over $C^\infty(M)$ in the sense that both $\Delta(fX + Y, \omega) = f\Delta(X,\omega) + \Delta(Y, \omega)$ and $\Delta(X, f\omega + \eta)= f\Delta(X, \omega) + \Delta(X, \eta)$ for all $f\in C^\infty(M)$, all smooth vector fields $X,Y$ and all smooth covector fields $\omega, \eta$. The equalities are interpreted as equality in $C^\infty(M)$. The proofs of these equalities follows routinely from the definition of the action of a covector on a vector.

Now we know that $\Delta$ is induced by a tensor field. That is $\Delta:M\to \Gamma(T^*M\otimes TM)$, where $\Gamma(E)$ denotes the sections of vector bundle $E$ over $M$. To show smoothness it is enough to pass to local coordinates $\{x^i\}$ and show that all coefficient functions $\Delta_i^j$ relative to the coordinate representation $\Delta = \Delta_i^j \; d x^i\otimes \partial_j$ are smooth $M\to \mathbb R$.