Vectors as differential operators act on functions.

Within my studies of differential geometry the concept of tangent space has just arisen. Now, a tangent vector $$v_p$$ at point $$p \in M$$ on a manifold $$M$$ is defined to be the differential operator $$v_p^i \frac{\partial}{\partial x^i}$$. Given an arbitrary smooth function $$f$$, I understands how this vector corresponds to a directional derivative of $$f$$.

However, what I don't understand -- even after consulting numerous google results and similar questions posted on this forum -- is why I need this function for my vector to be defined? To be more specific, what has this function (a specific function? a general function? a class of functions?) to do with my vector.

I suppose, from an naive point of view, when dealing with vectors I want to e.g. describe mechanical situations, maybe a collision. Perhaps I want to describe the electrical field of an electron. But what do I have to think of when talking about an action of my vector field (electric field) on some function? What physical relevance has this function. Why would I want my definition of a vector to depend on an additional function? Is there a correspondence to a more intuitive but less abstract object (e.g. in 3-d euclidean space) or maybe a low-level but didactically promising example? This is all so confusing.

$${}$$

EDIT: To formalise my concerns:

The "usual" interpretation of a vector field as I am used to it is a function $$V: \mathbb{R}^3 \longrightarrow \mathbb{R}^3$$. The "new" interpretation of a vector field as a differential operator is a function $$V: C^{\infty}(M) \longrightarrow C^{\infty}(\mathbb{R})$$.

But this is not the object I want to have, is it? In fact this is a scalar-function, where I would want to have a vector.

• I'd consider adding a "soft question" tag to this. This question is quite filosophical, as it asks not about the validity of a definition, but its deeper meaning. At least, as far as I can tell. To answer the easy (and less important) part - it's not "this function", but the class of all smooth functions. Which may make things even more confusing, I guess. Commented Feb 26, 2021 at 23:12
• I am afraid you are right with the latter assumption. :) However, I have updated the tags as suggested. Commented Feb 26, 2021 at 23:14
• Sorry to say, but your last question nobody but you can answer since it is about what you would want. Commented Feb 28, 2021 at 1:25
• Hm, replace "I" with "one" in the question. The question was intended to be read as: "One would want to have a vector instead of a scalar-function when dealing with vector fields, wouldn't he". $${}$$ The "I" was thereby meant as the lyrical subject, the reader of the mathematical definitions, etc. - maybe this is some grammatical maneuver only used in my native language and was just badly translated. Commented Mar 1, 2021 at 12:53

Suppose you want to explain the definition of a quadratic function to somebody and you give an example, $$f(x)=x^2$$. Then they ask "Why do you need this $$x$$ for the function to be defined? What does this $$x$$ have to do with my function?" What would be your answer? Mine would be that without this $$x$$ there is no function since the whole point of functions is to act on elements of the domain and the letter $$x$$ is a placeholder for a generic element of the domain of $$f$$.

The same about vectors as derivations. Yes, this definition is unintuitive, but it is technically convenient. If you prefer a more intuitive definition, think about manifolds embedded in some $$R^N$$, then vectors can be identified with directed segments, geometrically tangent to the submanifold. This is useful. But this intuitive picture has various drawback such as dependence on an embedding. Also, the picture of a tangent bundle becomes a mess since you have to worry about intersections between different tangent spaces.

Another intuitive picture is to draw vectors geometrically in domains of charts on your manifold. The problem then is that you always have to keep track of transition maps and how things depend on the choice of a chart.

How do these intuitive definitions correspond to derivations? By associating with each "geometric" vector $$v$$ the directional derivative $$D_v$$, as it is done in vector calculus classes.

If the abstract definition bothers you, when reading the textbook and working through proofs and problems, keep both formal and geometric definitions at your disposal, like training wheels on a bicycle when learning how to ride.

Edit. Once you have the standard definition of vector fields on $$M$$ as certain maps $$C^\infty(M)\to C^\infty(M)$$, you can define their action on $$C^\infty(M, {\mathbb R}^k)$$ for any $$k$$, simply by $$X(f_1,...,f_k)= (X(f_1),...,X(f_k)).$$ On the other hand it would not make sense to define vector fields by their action on $${\mathfrak X}(M)$$, where the latter is the space of vector fields on $$M$$, since that would be circular. However, again, once you defined $${\mathfrak X}(M)$$ you may want to define a notion of "differentiation" $${\mathfrak X}(M)\to {\mathfrak X}(M)$$. This can be done by introducing affine connections $$\nabla$$ on $$M$$, they allow you to define $$\nabla_X: {\mathfrak X}(M)\to {\mathfrak X}(M)$$ for every vector field $$X\in {\mathfrak X}(M)$$. As you learn more differential geometry, you will find out what role do these play.

• Considering the example of the quadratic function, I can formalize my confusion about vector fields as differential operators as follows: The usual interpretation of a vector field as I am used to it, is a function $V: \mathbb{R}^3 \longrightarrow \mathbb{R}^3$. The "new" interpretaion of a vector field as a differential operator now is a function $V: C^{\infty}(\mathbb{R}^3) \longrightarrow \mathbb{R}$. Abstract or not, a scalar is not the object I want to have, is it? Commented Feb 27, 2021 at 16:07
• @Octavius: Vector fields are linear maps (satisfying the Leibnitz Rule) of the space of smooth functions to the space of smooth functions, not to scalars. In contrast, elements of the tangent space $T_pM$ are (certain) maps of the space of smooth functions at $p$ to real numbers (scalars). Commented Feb 27, 2021 at 16:31
• You are right, sorry. What I had in mind was, that vector fields map smooth functions to objects that transform LIKE scalars. Sorry for this misconception. Commented Feb 27, 2021 at 16:36
• @Octavius: Yes, such objects are called functions, so let's call them by their proper name. (Scalars do not transform: The value of $\pi$ is the same regardless of your coordinate system.) Commented Feb 27, 2021 at 16:44

I hope someone here can write about the motivations and physical interpretations for this definition.

I would like to point out that there are more characteristics of a good definition than these (you can probably see I'm a fan of Terrence Tao and his essay). Off the top of my head, I can name a few:

1. It doesn't require elaborate tools in order to phrase it.
2. One can clearly see how it generalizes another, more standard definition.
3. It's directly related to our physical our geometric intuition about the object it's defining.
4. It's operatonally useful, i.e., allows us to write easy/short proofs with it.
5. It highlights the relations between the object we're defining and some other objects.

Sticking to our example of defining tangent vectors, one can consider various (commonly taught) definitions:

• In a chart, a part of $$M$$ is actually a part of $$\mathbb{R}^n$$, in which case there's a standard (almost tautological) notion of a tangent vector. Identifying properly tangent vectors obtained in different charts, we can make this definition work. I think it fulfills requirements 1 and 2.
• We define tangent vectors at $$p \in M$$ as "velocity vectors at $$t=0$$" of smooth curves $$\gamma \colon \mathbb{R} \to M$$ satisfying $$\gamma(0) = 0$$. Of course, one needs to clarify when two curves give us the same velocity (and we may use charts for this). Since this definition is directly related to trajectories on $$M$$, I'd say it satisfies 3.
• What about our definition of tangent vectors as "differentiations"? I think it works nicely for 4 and 5.

As for 4, you can probably already give a few examples of how this definition saves us time. One instance is the Lie bracket: for two tangent vector fields $$X,Y$$ on $$M$$ we define a new tangent vector field $$[X,Y]$$ as $$[X,Y] f := X(Y(f)) - Y(X(f)) \quad \text{for } f \in C^\infty(M).$$ One needs to check that it's linear and satisfies the Leibniz rule, which is easy. And that's all! Since $$X$$ and $$Y$$ act on functions (by definition), and for defining $$[X,Y]$$ we only need to describe its action on functions, we're done.

Number 5 is maybe the deepest point here. It may be confusing that the definition of one tangent vector involves all smooth functions on $$M$$, right? But at the same time, there's something deep in the fact that our definition itself relates tangent vectors to functions. This kind of definitions is quite common in more geometrically oriented parts of mathematics (e.g. definitions by universal properties), but I don't feel competent enough here to say much more.