# Gradient, one-form and Sean Carroll

"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?

• Can you choose a meaningful question title? Feb 21, 2021 at 23:18

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.
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.