Why is a gradient not always a vector

[Alain De Vos]

I learned gradient in 3D space. And gradients where always vectors, pointing in the direction of steepest ... and normal to the surface where the functions is constant.
But reading one-forms , a gradient of a function is not always a vector and it has something to do with metric... Can you proof this mathematically? Or an example which disproves that a gradient is a vector? Or visualise it?
If it is not a vector a change of coordinate base is does not change the coordinates in the "correct way"?

 

[lavinia]

I learned gradient in 3D space. And gradients where always vectors, pointing in the direction of steepest ... and normal to the surface where the functions is constant.
But reading one-forms , a gradient of a function is not always a vector and it has something to do with metric... Can you proof this mathematically? Or an example which disproves that a gradient is a vector? Or visualise it?
If it is not a vector a change of coordinate base is does not change the coordinates in the "correct way"?

When there is an inner product on the tangent spaces one can define the gradient by the relation, <grad(f),v> = df(v) where df is the differential of f and <,> is the inner product. With a different inner product, there is a different grad(f).

- df(v) is defined as the derivative of f along a curve fitting the vector,v.

_ The important conceptual point is that one can define the derivative without using an inner product.

 

[Ssnow]

Yes the crucial point is the metric (or scalar product) $g$, if $\frac{\partial}{\partial x^{i}}|_{p}$ is a basis of the tangent space in $p$ and $dx^{i}|_{p}$ is a basis of the cotangent space in $p$ we have using the metric that

g\left(\frac{\partial}{\partial x^{i}}|_{p},dx^{j}|_{p}\right)=\delta^{i}_{j}

where $\delta^{i}_{j}=\left\{\begin{array}{rl} 1 \ \ if \ \ i=j \\ 0 \ \ if \ \ i\not=j \end{array}\right.$. Given a differentiable function $f$ (on a manifold or on the Euclidean space of dimension $n$) you can in general consider the differential of $f$ that is

df=\sum_{i=1}^{n}\frac{\partial f}{\partial x^{i}}dx^{i}

that is a 1 form (that is an element of the cotangent space) and has as components the partial derivatives (elements of the tangent space) associated to the gradient vector. All depends if you think to the gradient as vector $\nabla f|_{p}$ or as linear application on vectors $g\left(\nabla f|_{p}, \cdot \right)$ ...

 

[Fredrik]

Yes the crucial point is the metric (or scalar product) $g$, if $\frac{\partial}{\partial x^{i}}|_{p}$ is a basis of the tangent space in $p$ and $dx^{i}|_{p}$ is a basis of the cotangent space in $p$ we have using the metric that

g\left(\frac{\partial}{\partial x^{i}}|_{p},dx^{j}|_{p}\right)=\delta^{i}_{j}

The metric takes two tangent vectors as input, not a tangent vector and a cotangent vector. You can however do this (without involving a metric):
$$(\mathrm dx^i)_p\left(\frac{\partial}{\partial x^j}\right)_p =\left(\frac{\partial}{\partial x^j}\right)_p(x^i) = (x^i\circ x^{-1})_{,j}(x(p)) =(x\circ x^{-1})^i{}_{,j}(x(p)) =\delta^i_j.$$

 

[Fredrik]

I learned gradient in 3D space. And gradients where always vectors, pointing in the direction of steepest ... and normal to the surface where the functions is constant.
But reading one-forms , a gradient of a function is not always a vector and it has something to do with metric... Can you proof this mathematically? Or an example which disproves that a gradient is a vector? Or visualise it?
If it is not a vector a change of coordinate base is does not change the coordinates in the "correct way"?

I will use the notation $f{}_{,i}$ for the $i$th partial derivative of a function $f:\mathbb R^n\to\mathbb R$. In differential geometry, partial derivatives are defined using coordinate systems (charts). If $M$ is a smooth manifold, $p\in U\subseteq M$, and $x:U\to\mathbb R^n$ is a coordinate system, then we define $\frac{\partial}{\partial x^i}\big|_p$ by
$$\frac{\partial}{\partial x^i}\bigg|_p f =(f\circ x^{-1})_{,i}(x(p))$$ for all smooth functions $f:M\to\mathbb R$. The notation $\big(\frac{\partial}{\partial x^i}\big)_p$ can be used instead of $\frac{\partial}{\partial x^i}\big|_p$, and the notation $\frac{\partial f(p)}{\partial x^i}$ can be used instead of $\frac{\partial}{\partial x^i}\big|_p f$.

The gradient of a differentiable function $f:\mathbb R^n\to\mathbb R$ is the function $\nabla f:\mathbb R^n\to\mathbb R^n$ defined by $\nabla f (x)=(f_{,1}(x),\dots,f_{,n}(x))$ for all $x\in\mathbb R^n$. The right-hand side can be rewritten using the differential geometry definition of partial derivative, if we use the fact that the identity map on $\mathbb R^n$ is a coordinate system. I will denote the identity map by $I$. We have
$$\nabla f(x)=(f_{,1}(x),\dots,f_{,n}(x)) =\left((f\circ I^{-1})_{,1}(I(x)),\dots,(f\circ I^{-1})_{,n}(I(x))\right) =\left(\frac{\partial}{\partial I^1}\bigg|_x f,\dots,\frac{\partial}{\partial I^n}\bigg|_x f \right).$$ This is the n-tuple of components of the cotangent vector $\mathrm (df)_x$ in the coordinate system $I$, since
$$(\mathrm df)_x =\left((df)_x \frac{\partial}{\partial I^i}\bigg|_x\right) \mathrm (dI^i)_x =\left(\frac{\partial}{\partial I^i}\bigg|_x f\right) \mathrm (dI^i)_x.$$ This is one reason to think of the gradient of f as a cotangent vector, specifically as the cotangent vector $(\mathrm df)_x$.

Another reason is that we can interpret the formula for $\nabla f(x)$ above as associating an n-tuple with each coordinate system. The right-hand side is the n-tuple associated with the coordinate system $I$. To get the n-tuple associated with an arbitrary coordinate system $J$, just make the substitution $I\to J$. Now that we have an n-tuple associated with each coordinate system, we can investigate how they're related to each other, i.e. we can investigate how an n-tuple "transforms" under a change of coordinates. If you know that $\big(\frac{\partial}{\partial J^1}\big|_x,\dots,\frac{\partial}{\partial J^1}\big|_x\big)$ is an ordered basis for the tangent space at $x$, and that "transforms covariantly" means "transforms in the same way as the ordered basis", then you should see that it follows almost immediately that our n-tuple $\big(\frac{\partial}{\partial J^1}\big|_x f,\dots,\frac{\partial}{\partial J^1}\big|_x f\big)$ "transforms covariantly" (unlike n-tuples of components of tangent vectors, which "transform contravariantly").