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"A tensor (k,l) is a multilinear map from k dual vectors and l vectors to R (...) The gradient, ..., is an honest (0,1) tensor."

These citations are retired from Sean Carrol Spacetime and Geometry. But i can't understand one thing: Why is the gradient a (0,1) tensor? Is not the gradient a differential 1-form (co vetors)? So, shouldn't it be (1,0), a map from the space dual vector to real?

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    $\begingroup$ Can you choose a meaningful question title? $\endgroup$ Feb 21, 2021 at 23:18

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As you have correctly seen, the differential $\mathrm{d}f$ of some function $f$ is a one-form. As such, it linearly maps one vector to the base field (here: $\mathbb{R}$), $$ \mathrm{d}f:\ TM \rightarrow \mathbb{R}. $$ Thus, $l=1$ and $\mathrm{d}f$ is a $(0,1)$-tensor.

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Terminology can always be confusing: given a smooth function $f:M\to\Bbb{R}$, we can always consider the object $df$. You can refer to this by many names: exterior derivative of $f$/differential of $f$ or even gradient covector field of $f$, though I particularly despise this last term. This is a true $(0,1)$ tensor field on $M$ (also known as a (exterior differential) $1$-form on $M$ or as a covector field on $M$). This means for each $p\in M$, we have the object $df_p\in T_p^*M$, i.e $df_p:T_pM\to \Bbb{R}$ is a linear map.

Now, if you're on a pseudo-Riemannian manifold $(M,g)$ (for relativity you're going to have $M$ be a 4-dimensional smooth manifold and $g$ having Lorentzian signature), then you can use the musical isomorphism $g^{\flat}:TM\to T^*M$ and its inverse $g^{\sharp}:T^*M\to TM$ to "convert" back and forth between vector fields and covector fields. In particular, given a smooth function $f:M\to\Bbb{R}$, we define \begin{align} \text{grad}_g(f):= g^{\sharp}\circ df \end{align} Notice what kind of object this is: $df$ is a mapping $M\to T^*M$, and $g^{\sharp}$ is a mapping $T^*M\to TM$, so $\text{grad}_g(f)$ is a mapping $M\to TM$; and it is easily verified that for each $p\in M$, $\text{grad}_gf(p)\in T_pM$, so this truly is a vector field on $M$ (equivalently a $(1,0)$ tensor field on $M$).

The appropriate terminology for this would be the gradient vector field of $f$, with respect to $g$. No one likes to say so much, hence we simply refer to this as "the gradient of $f$". But it is important to realize that this is only meaningful once we have a particular choice for a pseudo metric tensor $g$ in place.


Now that you know what the two different objects are, you're free to call them by whatever name you wish; just don't get confused by which object is which, and never ever confuse the two objects as being the same thing.

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