# Tag Info

## Hot answers tagged tensors

4

There is an identification taking place. 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, ...

2

The comment as it is in your post is taken out of context: in your post, it seems to be a part of some considerations about vectors - in which case one naturally asks how could scalars be particular cases of vectors (exactly your question). In reality, this comment in its original context is a part of some considerations about tensors, and in this case ...

2

Disclaimer: this is just a (graduate) educated guess based on this. I've never really worked with such objects. One way to think of finite dimensional tensors is as multilinear operators $$T:V^*\times\cdots\times V^*\times V\times\cdots\times V\rightarrow W$$ where $V,W$ are finite dimensional vector spaces and $V^*$ is the dual space of $V$. So, why ...

1

A section of a bundle $E$ (with various properties) is a fancier way of referring to a "smoothly varying" choice of $s_p\in E_p$ (with the same properties) as $p$ varies over $M$. So (1) and (2) are identical. With regard to (2) and (3), we're just using the isomorphism (truly a definition) $\text{Hom}(E,\Bbb R) = E^*$ (where here $E=TM\otimes TM$). Notice ...

1


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