Equivalent definitions of tensor field

[cianfa72]

TL;DR

About the equivalence of two different definitions of tensor field

As far as I know, there are two definitions of tensor field on a differential manifold $M$. Just to fix ideas consider the two-times covariant tensor field $T_{ab}$ on it.

First definition: take the tensor product $T_p^{*}M \otimes T_p^{*}M$ as the fiber over the point $p \in M$ and define/build the fiber bundle over $M$. Then $T_{ab}$ is a smooth section of such tensor product bundle on $M$.

Second definition: given the tangent and cotangent bundles over $M$, then $T_{ab}$ is a $C^{\infty}$-multilinear map $$T_{ab} : \Gamma(T^{*}M) \times \Gamma(T^{*}M) \to C^{\infty}(M)$$ where $\Gamma(T^{*}M)$ is the $C^{\infty}(M)$-module of the set of smooth sections of $T^{*}M$.

Are the two definitions actually equivalent ?

 

[weirdoguy]

Why wouldn't they be?

 

[cianfa72]

Why wouldn't they be?

Can you formally prove it ?

 

[jbergman]

TL;DR: About the equivalence of two different definitions of tensor field

As far as I know, there are two definitions of tensor field on a differential manifold $M$. Just to fix ideas consider the two-times covariant tensor field $T_{ab}$ on it.

First definition: take the tensor product $T_p^{*}M \otimes T_p^{*}M$ as the fiber over the point $p \in M$ and define/build the fiber bundle over $M$. Then $T_{ab}$ is a smooth section of such tensor product bundle on $M$.

Second definition: given the tangent and cotangent bundles over $M$, then $T_{ab}$ is a $C^{\infty}$-multilinear map $$T_{ab} : \Gamma(T^{*}M) \times \Gamma(T^{*}M) \to C^{\infty}(M)$$ where $\Gamma(T^{*}M)$ is the $C^{\infty}(M)$-module of the set of smooth sections of $T^{*}M$.

Are the two definitions actually equivalent ?

The second doesn't look correct. It should be a multilinear function of sections of the tangent space, not the cotangent space.

It's helpful to just look at a vector space since fields over a bundle don't really add that much.

But for a vector space $V$ over the field $\mathbb R$, $V^* \otimes V^* \cong L(V, V) \rightarrow \mathbb R$ where $L(V, V) \rightarrow \mathbb R$ is the space of bilinear functions on $V$.

 

[cianfa72]

The second doesn't look correct. It should be a multilinear function of sections of the tangent space, not the cotangent space.

Oops yes, a twice covariant tensor field eats two vector fields :-)

But for a vector space $V$ over the field $\mathbb R$, $V^* \otimes V^* \cong L(V, V) \rightarrow \mathbb R$ where $L(V, V) \rightarrow \mathbb R$ is the space of bilinear functions on $V$.

Yes, definitely.

Look now at the writing $T^*M \otimes T^*M$. What does it refer to ?
I believe it is just a notation for the fiberwise tensor product of cotangent bundles. Indeed a (vector) fiber bundle doesn't carry any vector space structure, hence the symbol of tensor product between them is meaningless. So it basically means taking the bundle over $M$ of the fiberwise tensor product $T^*_pM \otimes T^*_pM$ for any $p \in M$.

 

[jbergman]

Oops yes, a twice covariant tensor field eats two vector fields :-)


Yes, definitely.

Look now at the writing $T^*M \otimes T^*M$. What does it refer to ?
I believe it is just a notation for the fiberwise tensor product of cotangent bundles. Indeed a (vector) fiber bundle doesn't carry any vector space structure, hence the symbol of tensor product between them is meaningless. So it basically means taking the bundle over $M$ of the fiberwise tensor product $T^*_pM \otimes T^*_pM$ for any $p \in M$.

Yes. In Lee's book, he defines a smooth functor with which you can lift constructions on vector spaces to vector bundles. https://en.wikipedia.org/wiki/Smooth_functor