TL;DR: I'm confused about some very basic definitions in Differential geometry, namely how tangent vectors and vector fields are related to each other.

Let $M$ be smooth manifold. My professor defined the tangent vector$^1$ $X\in T_p M$ as a linear map $C^\infty (M) \to \mathbb{R}$ which satisfies the derivation property $$X(fg) = (Xf)g(p)+f(p)(Xg).$$

I already struggle here. Where exactly does the $p\in M$ on the rhs of the equation come from? On the lhs, isn't there some information missing, or how exactly do we know at which point of $M$ we need to evaluate the functions?

He then further says that in "in any chart (representing $p\in U_p$) we have $$X f=X^{i} f_{, i}(x): \quad X^{i}=X\left(x^{i}\right),$$ where $x^i\in C^\infty(U_p)$ denotes the coordinate function $p \mapsto x^i$. $^2$'' And here is where my real confusion starts. He then goes on to define what a vector field is, namely a linear map $X: C^\infty(M) \to C^\infty(M)$ with the derivation property $$X(fg) = (Xf)g+f(Xg).$$

According to him it should now be clear what the "connection" between tangent vectors and vector fields is, but I find it really hard to piece anything together from what I described here.

It almost seems like all a vector field does is assign to every point on the Manifold a tangent vector in the tangent space, but I doubt that this is true after seeing his description...

$^1$ My professor only called it vector, but I think it is usually called tangent vector...

$^2$ Does he mean here that $x^i(p)\in \mathbb{R}$?

  • $\begingroup$ " a vector field does is assign to every point on the Manifold a tangent vector in the tangent space" Yes, that's basically it, but normally one want to do it in a smooth, or at least a continuous, way. $\endgroup$ – Angina Seng Jul 23 '20 at 13:47
  • $\begingroup$ To be honest I cannot find a more confusing way to teach what is $T_pM$. Note also that he defined elements in $T_pM$ to be linear map $C^\infty(M) \to \mathbb R$, which is inconsistent to the next paragraph, where they localize in a chart (which is not possible, at least he has to point out there's some cutoff argument). $\endgroup$ – Arctic Char Jul 23 '20 at 13:51
  • $\begingroup$ The definition that I am aware of use "germs of smooth function at $p$" instead of the whole $C^\infty(M)$. $\endgroup$ – Arctic Char Jul 23 '20 at 13:52
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    $\begingroup$ About your first question, the $p$ appearing in the equation is the same point $p$ in $T_p M$, which is the tangent space in $p$. $\endgroup$ – FormulaWriter Jul 23 '20 at 14:07
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    $\begingroup$ The way you describe it, your professor seems to be leaving out a lot of details. I suggest you to pick up an introductory DG text and go through the definition. $\endgroup$ – Arctic Char Jul 23 '20 at 15:38

There are various approaches to introduce tangent vectors and tangent spaces. I think the approach in your question is the most abstract one and perhaps not optimal for a beginner in the motivational sense. The general idea is that a tangent vector at a point $p \in M$ is something like a direction in which you can go on $M$ starting at $p$. There are various ways to make this precise, one of them is to describe a direction as a directional derivative applied to real valued smooth functions on $M$, i.e. by a certain operator $X$ assigning to any function $f \in C^\infty(M)$ a value $X(f) \in \mathbb R$. I give you a sketch, you have to fill missing details by consulting good textbooks.

For any manifold $M$, $C^\infty(M)$ is an algebra over $\mathbb R$, that is a real vector space with an additional commutative multiplication satisfying suitable compatibility requirements. Thus we get the vector space $L(C^\infty(M),\mathbb R)$ of linear maps $X: C^\infty(M) \to \mathbb R$. For each $p \in M$ define the subset $$T_pM = \{ X \in L(C^\infty(M),\mathbb R) \mid \forall f,g :X(fg) = (Xf)g(p)+f(p)(Xg) \}.$$ It is easy to see that $T_pM$ is a linear subspace of $L(C^\infty(M),\mathbb R)$ which is called the tangent space of $M$ at $p$. Each $X \in T_pM$ is a tangent vector of $M$ at $p$. In your question you write the tangent vector which is misleading because it suggests that there is only one.

This is fairly abstract and does not give much motivation. Let us therefore consider the simple special case where $M$ is an open subset of $\mathbb R^n$. Then it is well-know from multivariable calculus what a directional derivative is: For each vector ("direction") $v \in \mathbb R^n$ and each differentiable $f : M \to \mathbb R$ we define the directional derivative of $f$ at $p$ by $\dfrac{\partial f}{\partial v}(p) = \lim_{h\to 0}\dfrac{f(p + hv) - f(p)}{h}$. This gives us a linear map $X_v =\dfrac{\partial }{\partial v}\mid_p : C^\infty(M) \to \mathbb R$. Clearly $X_v$ satisfies the derivation property (in that case this is nothing else than the product rule for differentiation). It is well-known that $X_v = \sum_{i=1}^n v_i \dfrac{\partial }{\partial x^i}\mid_p$, where $v = (v_1,\ldots,v_n)$. The assocation $v \mapsto X_v$ gives us a linear map $\alpha: \mathbb R^n \to T_pM$ and it is not hard to show that $\alpha$ is a linear isomorphism. That is, we have a $1$-$1$-correspondence between directions $v \in \mathbb R^n$ (in which we can go on $M \subset \mathbb R^n$ starting at $p$) and elements of $T_pM$ (which are determined by their action on the functions in $C^\infty(M)$). Thus, although the elements of $T_pM$ at first glance seem to be very abstract and perhaps even incomprehensible, they are nothing else than ordinary directional derivatives which can be interpreted as directions on $M$ at $p$.

I hope this explains the above definition of $T_pM$ for a general $M$. In fact, the elements of $T_pM$ are directional derivatives in a generalized setting. This yields a concept of directions for a general $M$.

Your remark concerning the local representation of $X$ is rather vague, but I think a chart around $p \in M$ is a homeomorphism $x : U_p \to V$ (where $U_p$ is an open neighborhood of $p$ in $M$ and $V$ is open in $\mathbb R^n$) which belongs to the smooth structure of $M$. Then the $x^i : U_p \to \mathbb R$ are the $n$ coordinate functions of $x$. Now you have the problem that $x^i$ is in general not defined on all of $M$, but let us ignore it (it can be settled). Thus $X^i = X(x^i)$ is a real number. The meaning of $f_{,i}(x)$ is not clear, but I guess it is the $i$-th partial derivative of $f \circ x^{-1} : V \to \mathbb R$ at $x(p)$.

As you say, a vector field should associate to each $p \in M$ a tangent vector in $ T_pM$. This is what a linear map $\xi: C^\infty(M) \to C^\infty(M)$ with the derivation property does (I wrote $\xi$ instead of $X$ to avoid confusion - $X$ is used already for tangent vectors). In fact, given $p \in M$, define $\xi_p \in T_pM$ by $\xi_p(f) = \xi(f)(p)$. Note that $\xi(f) \in C^\infty(M)$. It is easy to verify that $\xi_p$ is linear and satisfies the derivation property required for tangent vectors of $M$ at $p$. The assignment $p \mapsto \xi_p$ is what we intuitively understand as a vector field. Working with $\xi$ is just another point of view.

So where is the direction information in $\xi_p$? We have $\xi_p \in T_pM$, and the elements of $T_pM$ are abstract directional derivatives, i.e. directions on $M$.

  • $\begingroup$ Thanks a lot for the answer! A short question: In your first paragraph, do I understand correctly that the analogy is that if $M\subset\mathbb{R}^n$ we have $\xi_p = \left.\frac{\partial}{\partial v}\right|_p$? If so, I'm a bit confused to where exactly the "direction" information went in your definition of $\xi_p$? I mean, in $\left.\frac{\partial}{\partial v}\right|_p$ it is clear that $v$ determines the "direction", but what determines the "direction" of $\xi_p$? $\endgroup$ – Marius Jaeger Jul 23 '20 at 17:34
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    $\begingroup$ @MariusJaeger See my edit. $\endgroup$ – Paul Frost Jul 23 '20 at 22:21

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