Definition of $\partial/\partial x^i$ and $dx^i$ vs. $e_i$ and $e^i$

Question

I am quite new to tensor calculus and I am facing a quite basic problem I think. While reading about tensors I came along the difference between vectors and covectors which I now understand quite well.

Vectors: Specify a direction in n dimensional space. They are specified as follows: $V = v^ie_i$

Covectors: Something that takes a vector as argument and returns a scalar. They are specified as follows: $\alpha = \alpha_i e^i$

It is also clear to me that the following condition is satisfied: $e^i(e_j) = \delta^i_j$

Now while turning to calculus I came along the following definitions: $\partial/\partial x^i = e_i$ and $dx^i =e^i$. And this is what I do not really understand.

Question 1: This would mean that

dx is a convector and

$\partial/\partial x$ is a vector as well as

$dx(\partial/\partial x)=1$ ans

$dx(\partial/\partial y)=0$

Why can we assume this? How can I imagine this?

Question 2:

Assume that f is a scalar function f(x,y) What is then the difference between $f * \partial/\partial x$ and $ \partial f/\partial x$ or $\partial/\partial x(f)$?

Thanks in advance.

Regards, Harry

Answer 1

Having to do with tensor fields, vector and covector are understood to be vector field and covector field. In such a case $$v=v^ie_i$$ and $$\alpha=\alpha_ie^i$$ are relative to a given point.

Moreover, a vector is not simply a way to represent a direction otherwise two parallel vectors at different points would be equivalent. A vector in a point instead represents a class of equivalent parametric curves passing through that point and having that vector as a derivative.

As to a covector, it is not only "something that gets a vector as an argument and gives a scalar", but it does it linearly, so they are linear functional of vectors.

Now, back to your question 1.

Consider the $x^i$-coordinate function $x^i=x^i(P)$ in a neighborhood of a point (This equation has an abuse of notation because $x^i$ is used for the function and its value, but it is useful as it will shown in a while). It gives the $x^i$-coordinate of a point $P$. Its differential at a point $dx^i$ is the $x^i$-component functional of vectors at that point, that is, it gives the $x^i$-component of a vector at a point. (Note that thanks to the previous abuse of notation such a functional can be expressively writen as $dx^i$ instead of something not immediately understandable as $x^i$-component functional as would be tha case if the $x^i$-coordinate function had been writen without notational abuse like $x^i=\varphi^i(P)$: correspondigly the $x^i$-component functional would have been written as $d\varphi^i$). Since it linearly gives a scalar on vectors, it is a covector.

$$dx^i(v)=v^i$$

It's soon understood that any linear functional (covector) $\alpha$ on vectors in a point can be represented by components that are just the values taken on the basis vectors $e_i$'s.

Indeed given a vector $v$ at a point, a covector at that same point takes on $v$ the values: $$\alpha(v)=\alpha(v^ie_i)=v^i\alpha(e_i)=\alpha_iv^i=\alpha_idx^i(v)$$

But then from here you can see that: $$\alpha=\alpha_idx^i\tag{1}$$

In particular you see that: $$dx^i(e_j)=\delta^i_j\tag{2}$$ that is similar to the relation you wrote: $$e^i(e_j)=\delta^i_j$$

For that, you can write: $$dx^i=e^i$$

In the form $(1)$, covectors are said differential $1$-forms.

Analogously for vectors: A vector $v$ at a point can be interpreted as a linear functional of covectors at that same point. Given a covector $\alpha$, the value that $v$ (as a linear functional) takes on it is given by: $$v(\alpha)=\alpha(v)$$ In particular $e_i$ gives the $x_i$-component of covectors: $$e_i(\alpha)=\alpha_i$$ But with vectors we can go even further. A vector $v$ in a point can be associated to a differential operator $\partial_v$ on functions, such that given a function $f$: $$\partial_vf=v^i\frac{\partial f}{\partial x^i}$$ and in particular $$\partial_{e_i}f=\frac{\partial f}{\partial x^i}$$

Then we can write: $$\partial_v=v^i\partial_{e_i}$$ Being $v\mapsto\partial_v$ bijective, we can identify $v$ with $\partial_v$ reinterpreting vectors as differential operators on functions and no more as linear functionals on covectors or classes of equivalent parametric curves, and then we can rewrite the last equation as $$v=v^i\frac{\partial}{\partial x^i}$$

where $$\frac{\partial}{\partial x^i}=e_i$$

Moreover you see that now components of a covector are written as: $$\alpha_i=\alpha(e_i)=\alpha\left(\frac{\partial}{\partial x^i}\right)$$ and the relation $(2)$ is rewritten as: $$dx^i\left(\frac{\partial}{\partial x^j}\right)=\delta^i_j$$

Additionally you may also see that differentials of function are covectors, that is, they are differential $1$-forms:

$$df=\frac{\partial f}{\partial x^i}dx^i$$

Be warned anyway that the converse is not true. That is, given a covector or differential $1$-form it may not exist a scalar function $f$ such that the covector is the differential of a function. When this happens the covector is said to be an exact differential $1$-form.

Look at here for further information.

By now you should be able to answer by yourself at your question 2: $$\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}f$$

Answer 2

Question 1

On a differentiable manifold $M$, a vector field $X$ is defined as an operator on $C^\infty(M)$ (i.e. it takes a smooth function and returns a smooth function) satisfying the product rule $X(fg) = f \, X(g) + g \, X(f)$ for all $f,g \in C^\infty(M).$

Given coordinates $(x^\mu)$ on $M$ it is easy to see that $\partial_\mu$ is a vector field. The set $\{ \partial_\mu \}$ is in fact also a basis in the meaning that every $X$ can be written as $f^\mu \partial_\mu$ for some functions $f^\mu.$

Covectors are defined as linear functionals taking a vector field and returning a (real or complex) number. The basis $\{ dx^\mu \}$ is defined so that $dx^\mu(\partial_\nu) = \delta^\mu_\nu.$

Question 2

There is no difference between $\frac{\partial f}{\partial x},$ $\frac{\partial}{\partial x}f$ and $\partial_x f.$ They are just different ways to write the same thing.

With $f * \partial/\partial x$ I suppose that you mean $f$ multiplied with $\frac{\partial}{\partial x}.$ This is an operator, the result of which is $f \frac{\partial g}{\partial x}$ when applied on $g$.