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I have some confusion understanding the motivation in constructing tensors (or tensor fields). On a differentiable manifold $\mathcal{M}$ consider a vector field $X$. At any point $p\in \mathcal{M}$, we have the tangent space $T_p\left(\mathcal{M}\right)$ where we define the space of one-forms as linear maps $\omega_p : T_p\left(\mathcal{M}\right) \rightarrow {\rm I\!R}$. This suggests that if I want to construct higher rank tensors, then I always need to construct some vector field which is acted on by a one form, so as to finally give a number - since tensors are defined as multilinear maps with ${\rm I\!R}$ as the range. As an example, consider the definition of the curvature tensor $R$ with vector fields $X,Y,Z$ and a one form (field) $\omega$ :

$$R\left(\omega, X,Y,Z\right) = \omega\left(\nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{\left[X,Y\right]}Z\right)$$

Is this supposed to be interpreted as $\omega$ 'acting on' some vector field, such that we get a number? I am not convinced of this for two reasons:

  1. In this case, the 'some vector field' would have to be $\left(\nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{\left[X,Y\right]}Z\right)$ and I am not sure if this object is a vector field at all.
  2. I know that there exist tensors that do not even act on vectors, such as the energy momentum tensor, which is a map $T_p: T_p^*\left(\mathcal{M}\right) \otimes T_p^*\left(\mathcal{M}\right) \rightarrow \rm I\!R$. In this case I would have to choose a specific one form, if my interpretation of the tensor is as an object that constructs a vector field for a one form to act on, and ultimately take me to $\rm I\!R$.

Hence, do I just understand tensors to be operators that act on whatever objects they take as input? Or do we construct tensors based on how we define a one form - tensors as operations to construct a vector field which is then acted on by a one form, to ultimately give a number.

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Is this supposed to be interpreted as $\omega$ 'acting on' some vector field, such that we get a number?

Yes.

[...] I am not sure if this object is a vector field at all.

It is.

I'm not entirely sure what your question is. A $(p,q)$-tensor is a multilinear map which eats $p$ covectors and $q$ vectors and spits out a real number. There's no reason to make it more complicated than this. It's true that if you have a $(p,q)$-tensor and leave one of the covector slots empty while filling the rest, then the result will be a vector, so you could interpret the tensor as first making that vector and then acting on it with the final covector - but whether that is a "natural" interpretation depends on the situation at hand. And of course, it's possible to make tensors which don't eat any covectors at all, e.g. the metric tensor.

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  • $\begingroup$ My question is that do I need to think of tensors as somehow constructing a vector field which is acted on by a covector to give a number, since in general tensors give a number. Initially I was also thinking of tensors as multilinear maps like you described. But a discussion with a friend confused me. The interpretation aspect of this is worrying, because effectively what I am asking is, can I write the Riemann tensor as: $R\left(\omega, X,Y,Z\right) = \left(\nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{\left[X,Y\right]}Z\right)\omega$ $\endgroup$
    – newtothis
    Commented May 8, 2022 at 21:40
  • $\begingroup$ Does this then mean that $\langle dx^\mu,\partial_\nu \rangle = \langle \partial_\nu, dx^\mu \rangle$ or does the RHS have no meaning, since the LHS basically defines the space of one forms, in some coordinate basis. $\endgroup$
    – newtothis
    Commented May 8, 2022 at 21:42
  • $\begingroup$ @newtothis The action of a vector on a covector is, by definition, the same as the covector acting on the vector, i.e. $X(\omega):=\omega(X)$, so you could write it that way if you want to. There’s no particular benefit of doing so, of course. $\endgroup$
    – J. Murray
    Commented May 8, 2022 at 21:44
  • $\begingroup$ @newtothis Assuming that the angle brackets denote the inner product, neither side has any meaning, because the inner product is taken between two vectors or two covectors, not one of each. $\endgroup$
    – J. Murray
    Commented May 8, 2022 at 21:46
  • $\begingroup$ This is quite new to me. I was of the impression that one forms are defined as dual spaces to (tangent) vector spaces. But I do not need one forms to define vectors. Secondly, at least the way it was used in my class, the angle bracket represented the map between the basis of one forms and that of the tangent vectors. I did not mean for it to be the inner product. I meant $\langle dx^\mu,\partial_\nu \rangle = \delta^\mu_\nu$ $\endgroup$
    – newtothis
    Commented May 8, 2022 at 21:49
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If you understand tensors as they ate usually described in GR texts, then understanding differential forms is straightforward. Theyy are exactly the antisymmetric covariant tensors. They do not need to 'act' on anything (thatvis, contract with).

More, these tensors are real, being defined over a real manifold. And they contract with other real tensors, including real vector fields.

(There are such things as complex manifolds amd complex differential forms (amd tensors) as well as complex vector fields. But this isn't merely changing the term 'real' to 'complex'. It's a whole new ball game as is the change from real to complex analysis is).

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