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I learnt that a contravariant 1-tensor is equivalent to a vector (am I right?). I am confused about this. Now a contravariant 1-tensor is a function having a covector as its argument. But a vector, if seen as a derivation, is a function having another function as its argument. Is there any relation between these two roles of functions (i.e. a function having a covector as its argument on the one hand, and a function having another function as its argument on the other hand)?

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  • $\begingroup$ Covectors are functions. $\endgroup$
    – anon
    Jul 16, 2022 at 5:06
  • $\begingroup$ It is true that covectors area functions. What I mean is that a contravariant 1-tensor is a function having a covector as its argument., while a vector, if seen as derivation, is a function having a zero-form as its argument. But a covector is different from a zero-form. $\endgroup$ Jul 16, 2022 at 6:16
  • $\begingroup$ Yeah, you can evaluate linear forms at vectors or you can use vectors to define directional derivatives of scalar functions. I don't see a relation. $\endgroup$
    – anon
    Jul 16, 2022 at 8:54

2 Answers 2

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That vectors are linear functions on the space of covectors boils down to the fact that any finite dimensional vector space is reflexive. Hence in particular it also applies to derivations living on tangent spaces of a finite dimensional manifold. You can define a derivation at point $p$ to act on the basis one-forms via:

$$\dfrac{\partial}{\partial x_i}|_p (dx_j) = \dfrac{\partial x_j}{\partial x_i}|_p = \delta_{ji}$$

In coordinate-free language this means:

$$v_p(w_p) = w_p(v_p) $$ where $w_p$ is a one form at point $p$.

And you recover the notion of vectors being linear functions on covectors.

So the point I'm trying to make is that any vector (living on a finite dimensional vector space) can be seen as a linear function acting on covectors/one-forms. This is also true for derivations.

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After pondering about the problem, I seem to have discovered the following relationship between the two roles. For simplicity, here I use a 2-dimensional example to illustrate my idea. Let $p$ be a point and $v_p = v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p$ be a vector. On the one hand, if we consider $v_p$ as a contravariant 1-tensor and apply it to a covector $\alpha_p = \alpha_1(dx_1)_p + \alpha_2(dx_2)_p$. Now by the canonical pairing between a covector and a vector, we have $v_p(\alpha_p) = \alpha_p(v_p)$, and so we have \begin{eqnarray*} & & \left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right)(\alpha_1(dx_1)_p + \alpha_2(dx_2)_p) \\ & = & (\alpha_1(dx_1)_p + \alpha_2(dx_2)_p)\left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right) \\ & = & \alpha_1v_1 + \alpha_2v_2 \\ & = & \alpha_1\left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right)[x_1] + \alpha_2\left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right)[x_2] \\ & = & \left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right)[\alpha_1x_1 + \alpha_2x_2] \end{eqnarray*} In the above, the $x_1$ are $x_2$ in square brackets $[\,\,\,]$ are coordinate functions. Therefore, given $v_p(\alpha_p)$, i.e. a contravariant 1-tensor acting on a covector, we can transform it to a derivation acting on a function $\alpha_1x_1 + \alpha_2x_2$.

On the other hand, if we consider $v$ as a derivation and apply it to a function $f$, then we have

\begin{eqnarray*} & & \left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right)[f] \\ & = & df\left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right) \\ & = & \left(\left.\frac{\partial f}{\partial x_1}\right|_p(dx_1)_p + \left.\frac{\partial f}{\partial x_2}\right|_p(dx_2)_p\right)\left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right) \\ & = & \left(v_1\left.\frac{\partial}{\partial x_1}\right|_p + v_2\left.\frac{\partial}{\partial x_2}\right|_p\right)\left(\left.\frac{\partial f}{\partial x_1}\right|_p(dx_1)_p + \left.\frac{\partial f}{\partial x_2}\right|_p(dx_2)_p\right) \end{eqnarray*} In the last step above, we have used the canonical pairing between vectors and covectors again. Therefore, given $v_p[f]$, i.e. a derivation acting on a function, we can transform it to a contravariant 1-tensor acting on a covector $\left.\frac{\partial f}{\partial x_1}\right|_p(dx_1)_p + \left.\frac{\partial f}{\partial x_2}\right|_p(dx_2)_p$.

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