Covector and Vectors 2 Basic Questions.

[binbagsss]

I am just being introduced to the notion of covectors.

I see that the gradient of a scalar function is thought of as a covector.

Q1) On another source I have read to think of contravariant vector field as a differential operator: V^a \partial _a and to think of a covariant vector field as differential: w_adx^a.

From the fact that the gradient function is a covector, I thought this would be the other way around?

Q2) I'm trying to understand why df=\partial _a(f )dx^a is a covector?

My thoughts:

\partial _af is a covector.
dx^a is a vector.

So, isn't this just the definition of a contraction - multiplication of a covector and a vector to yield a scalar.

Thanks very much in advance.

 

[Dick]

$\partial_a$ is a contravariant vector $\partial_a(f)$ is a scalar. It's just a number. It's the coefficient of the $dx^a$. Think about it.

 

[binbagsss]

$\partial_a$ is a contravariant vector $\partial_a(f)$ is a scalar. It's just a number. It's the coefficient of the $dx^a$. Think about it.

I thought $\partial_a$=$\frac{\partial}{\partial x_a}$.
That these are components of the gradient function, which when operating on a scalar, should be regarded as a covector.

 

[Dick]

I thought $\partial_a$=$\frac{\partial}{\partial x_a}$.
That these are components of the gradient function, which when operating on a scalar, should be regarded as a covector.

It is a contravariant vector as an operator. Once it operates on a function then it just yields another function.

 

[Fredrik]

I see that the gradient of a scalar function is thought of as a covector.

It's not a cotangent vector, but it can be thought of as associating an n-tuple with each coordinate system, and those n-tuples transform covariantly. That makes "it" (actually the association of n-tuples with coordinate systems, not the original gradient) a covector according to the old-fashioned definitions. I'll quote myself:

The gradient of a 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 $x\in\mathbb R^n$. For each $i\in\{1,\dots,n\}$, $f_{,i}$ denotes the ith partial derivative of f. In differential geometry, partial derivatives are defined using both a coordinate system and the conventional type of partial derivatives. For example, if $x:U\to\mathbb R^n$ is a coordinate system on $U\subseteq\mathbb R^n$, and $p\in U$, then for all $i\in\{1,\dots,n\}$, we have
$$\frac{\partial}{\partial x^i}\bigg|_p f= (f\circ x^{-1})_{,i}(x(p)).$$ This statement defines the notation on the left.

The conventional partial derivatives in a gradient can be interpreted as partial derivatives in the sense of differential geometry, if we use the fact that the identity map $I$, defined by $I(x)=x$ for all $x\in\mathbb R^n$, is a coordinate system. We have
$$\frac{\partial}{\partial I^i}\bigg|_p f = (f\circ I^{-1})_{,i}(I(p)) = f_{,i}(p).$$

It doesn't make much sense to say that the gradient (the n-tuple with components $f_{,i}(p)$) "transforms" when we change the coordinates, because it's just one n-tuple. To discuss "transformation", we have to associate an n-tuple with each coordinate system. The formula at the end of the quote gives us an obvious way to do that. For each coordinate system x that has p in its domain, we define the "gradient of f, at p, in the coordinate system x" as the n-tuple with components $\frac{\partial}{\partial x^i}\!\big|_p f$, and we also define the ordered basis of $T_pM$ associated with x as the n-tuple $\big(\frac{\partial}{\partial x^i}\!\big|_p\big)_{i=1}^n$.

It follows immediately from our definitions that "the gradient of f at p in the coordinate system x" transforms covariantly. To transform covariantly is to transform the same way as the ordered basis, and "the gradient of f at p in the coordinate system" is just an n-tuple whose components are the elements of the ordered basis acting on f.

To be a little more explicit, the transformation of the ordered basis is given by $\frac{\partial}{\partial x^i}\!\big|_p\to \frac{\partial}{\partial y^i}\!\big|_p$. The $\frac{\partial}{\partial y^i}\!\big|_p$ are elements of $T_pM$, which is spanned by the $\frac{\partial}{\partial x^i}\!\big|_p$, so there must exist numbers $M^j{}_i$ such that $\frac{\partial}{\partial y^i}\!\big|_p=M^j{}_i \frac{\partial}{\partial x^j}\!\big|_p$. So the transformation of the ordered basis can be written as
$$\frac{\partial}{\partial x^i}\!\big|_p\to M^j{}_i \frac{\partial}{\partial x^j}\!\big|_p.$$ This implies that the transformation of the gradient at p is given by
$$\frac{\partial}{\partial x^i}\!\big|_p f\to M^j{}_i \frac{\partial}{\partial x^j}\!\big|_p f.$$ Since the numbers that appear on the right are $M^j{}_i$ (the same numbers that showed up when we transformed the ordered basis), we say that the gradient of f at p in the coordinate system x "transforms covariantly", i.e. in the same way as the ordered basis.

Q1) On another source I have read to think of contravariant vector field as a differential operator: V^a \partial _a and to think of a covariant vector field as differential: w_adx^a.

I prefer the terms "vector field" and "cotangent vector field" respectively. Cotangent vector fields are also called 1-forms. A vector field is a function that takes each point in some subset of the manifold to a tangent vector at that point. A cotangent vector field is a function that takes each point in some subset of the manifold to a cotangent vector at that point. $\big(\frac{\partial}{\partial x^i}\!\big|_p\big)_{i=1}^n$ is an ordered basis for the tangent space at p. $\big(\mathrm dx^i\big|_p\big)_{i=1}^n$ is an ordered basis for the cotangent space at p.

From the fact that the gradient function is a covector, I thought this would be the other way around?

An ordered basis of the tangent space transforms covariantly (obviously, since that means "in the same way as the ordered basis"). This implies that the components of any tangent space transforms contravariantly. This is easy to see when you expand an arbitrary tangent vector in two different bases: $v=v^i e_i =v'^ie'_i$. For this to hold, the transformation of the components has to be the "opposite" of the transformation of the ordered basis. See this post for a lot more details about this sort of thing.

Q2) I'm trying to understand why df=\partial _a(f )dx^a is a covector?

A cotangent vector is an element of the cotangent space. df is by definition a linear map from the tangent space into the real numbers (defined by $df(v)=v(f)$ for all smooth functions f), so it's by definition a cotangent vector.