Interpretation of the Gradient Vector?

[Mandelbroth]

I've always thought of the gradient of a scalar function (id est, $\nabla\varphi$) as being a vector field. However, I started thinking about it just now in terms of transformation with respect to coordinate changes, and I noticed that the gradient transforms covariantly. Thus, shouldn't the gradient be represented with a row vector?

I don't know why this is confusing for me. After looking through a couple websites, I saw that there were some who said "yes" and some who said "no," so I don't know what to think.

 

[WannabeNewton]

It's technically not a vector field but rather a 1-form. The components $\partial_{i}f$ transform as those of a 1-form so that's a simple reason for why i.e. $\partial_{i'}f = \frac{\partial x^{i}}{\partial x^{i'}}\partial_{i}f$. A slightly more general explanation is as follows: https://www.physicsforums.com/showpost.php?p=4374094&postcount=18

 

[Mandelbroth]

It's technically not a vector field but rather a 1-form.

I thought that $\nabla\varphi=(d\varphi)^\sharp$. Wouldn't that imply that the gradient is a vector field? Or, am I just confused as to the effects of raising the index? Or, is the equation not valid?

 

[WannabeNewton]

The musical isomorphism is trivial for $\mathbb{R}^{n}$ so it really doesn't matter but if you look at the gradient component wise as $\partial_{i}f$ then it's a 1-form.

 

[micromass]

The gradient of a scalar field is actually a section of the cotangent bundle of the manifold. However, if we are given a metric tensor, then we have a natural morphism between the cotangent bundle and the tangent bundle. As such, in Riemannian manifolds, the gradient can be seen as a section of the tangent bundle. When no natural metric is given, it can not be seen as such.

In algebraic geometry, we can see the gradient as an $A$-derivation map $d:B\rightarrow \Omega_{B\A}$, where $A$, $B$ is an $A$-algebra and where $\Omega_{B\setminus A}$ is the module of relative differential forms. We can further generalize this by defining $\Omega_{X\setminus Y}$ as a quasicoherent module of sheafs, when $X$ and $Y$ are schemes.

 

[WannabeNewton]

I thought that $\nabla\varphi=(d\varphi)^\sharp$. Wouldn't that imply that the gradient is a vector field? Or, am I just confused as to the effects of raising the index? Or, is the equation not valid?

Let me just add on a bit to what I said before. We can think of $\partial_{a}$ as a derivative operator on $\mathbb{R}^{n}$ that sends (n,m) tensor fields to (n,m+1) tensor fields (I'm sure you've seen this before in a more general setting!) so in this sense it sends a scalar field (i.e. a (0,0) tensor field) to a 1-form (i.e. a (0,1) tensor field); now let's say we have a metric tensor $g_{ab}$ then we can define the associated vector field $\partial^{a}\varphi = g^{ab}\partial_{b}\varphi$ which is just the musical isomorphism. In $\mathbb{R}^{n}$ we usually take $g_{ab} = \delta_{ab}$ so if $\partial_{i}\varphi$ are the components then $\partial^{i}\varphi = \delta^{ij}\partial_{j}\varphi = \delta^{i1}\partial_{1}\varphi + \delta^{i2}\partial_{2}\varphi + \delta^{i3}\partial_{3}\varphi$ so $\partial^{1}\varphi = \partial_{1}\varphi$ etc.