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At the moment I am learning about manifolds. I am trying to better understand the differential of a smooth map between manifolds.

It is said that "we have to explore the way smooth maps affect tangent vectors".

But I am stuck on the euclidean case.

Let $U \subset \mathbb{R}^m,V \subset \mathbb{R}^n$ be open subsets. Let $f: U \rightarrow V$ be a smooth map.

Then for any point $x$ in $U$, the Jacovian of $f$ at $x$ is the matrix representation of the total derivative of $f$ at $x$, which is a lienar map

$$df_x: T_x\mathbb{R}^m \rightarrow T_{f(x)}\mathbb{R}^n$$

Since $T_x\mathbb{R}^m$ and $T_{f(x)}\mathbb{R}^n$ are isomorphic to $\mathbb{R}^m$ and $\mathbb{R}^n$, we get the familiar case of the total derivative in (multivariate) calculus.

What I do not understand is how we can view $df_x$ as a map $T_x\mathbb{R}^m \rightarrow T_{f(x)}\mathbb{R}^n$. I do not get the tangent space viewpoint in multivariate calculus.

I do get that using an isomorphism we can move between $T_x\mathbb{R}^m$ and $\mathbb{R}^m$, but I am curious how one can view the total derivative from multivariate calculus as a map between tangent spaces.

Eidt:

I should have mentioned it sooner, but I use derivations to define the tangent space.

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There are different possible definitions for the tangent space of a manifold at a point $x$. Let us take the following one: $T_xM$ is the set of equivalence classes of differentiable curves $c: (-1,1) \to M$ with $c(0)=x$. Here, two such curves $c_1$ and $c_2$ are equivalent, if for every coordinate chart $\varphi: U \to \mathbb R^m$ ($m$ being the dimension of the manifold and $U \subset M$ an open subset containing $x$) the derivatives $(\varphi \circ c_1)'(0)$ and $(\varphi \circ c_2)'(0)$ coincide. We denote such an equivalence class by $c'(0)$. A tangent vector $c'(0)$ (such an equivalence class) then acts on smooth maps $f: M \to \mathbb R$ as $$c'(0)(f) = (f \circ c)'(0).$$ Note that $c'(0)$ is NOT a derivative in the classical sense, but just a symbolic notation.

If your manifold is $\mathbb R^m$ as in the question, you can forget about $\varphi$ and just define the tangent space as all tangent vectors $c'(0)$, where again $c'(0)$ is a symbolic notation for the velocity vector of all curves $\gamma: (-1,1) \to \mathbb R^m$ with $\gamma(0)=x$, for which the derivatives at zero are the same. Then again: $c'(0)$ is a functional operating on the smooth maps $f: \mathbb R^m \to \mathbb R$.

Now back to your case: we have a smooth map $f: U \to V$. It's differential by definition acts as follows: $df_x(c'(0)) = (f \circ c)'(0).$ Here again, the right hand side is a symbolic notation and stands for the tangent vector obtained by taking the equivalence class containing the element $f \circ c: (-1,1) \to \mathbb R^n$. In that way you can see that $df_x$ operates on the tangent spaces.

How is all that related to the classical total differential? We know for any $v \in \mathbb R^m$ that $df(x)v = \frac{\partial f}{\partial v}(x)$ (directional derivative). But of course, the curve $c(t):= x+tv$ satisfies $c(0)=x$ and $c'(0)=v$, which means that $v$ is a representative of the equivalence class of $c$. In other words we have $$\frac{\partial f}{\partial v}(x)=df(x) v= df(x) c'(0)= df_x(c'(0)) = (f \circ c)'(0) = df(c(0))c'(0)$$ (the last equality being by chain rule) so the viewpoint of the differential acting on tangent spaces is just a generalization of the concept of directional derivatives.

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  • $\begingroup$ Thank you for the help. But I do use derivations to define the tangent space. $\endgroup$ Commented Nov 5 at 12:50
  • $\begingroup$ In that case, for an equivalence class of velocity vectors $c'(0)$, you can define the derivation $D_{c'(0)}$ via $D_{c'(0)}(f):= (f \circ c)'(0)$, the right hand side being the classical derivative for real-valued functions. The map $c'(0) \mapsto D_{c'(0)}$ gives rise to an isomorphism, showing that both definitions of tangent space are equivalent. $\endgroup$ Commented Nov 5 at 12:54
  • $\begingroup$ I am not entirely sure what you mean by velocity vectors. The Definition I use for derivation is: Let $M$ be a smooth manifolds and let $p \in M$. A linear map $v: C^{\infty} \rightarrow \mathbb{R}$ is called a derivation at $p$ if $v(fg)=f(p)vg+g(p)vf$ for all $f,g \in C^{\infty}(M)$. $\endgroup$ Commented Nov 5 at 12:58
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It is a standard lemma stated and proved in most modern differential geometry textbooks that if $v \in T_pM$, defined as a derivation, then there exists a curve $c: I \rightarrow M$ such that $c(0)=p$ and for any smooth scalar function $\phi$ on a neighborhood of $p$, $$ v(\phi) = \left.\frac{d}{dt}\right|_{t=0}\phi(c(t)). $$ If this holds, then we say that the velocity of $c$ at $p$ is $v$ and write this as $$ c'(0) = v. $$

Let $f: M \rightarrow N$ be a map between manifolds. Given a curve $c$ in $M$ with velocity $c'(0)=v$ at $p \in M$, the composition $f\circ c$ is a curve in $N$ with velocity $(f\circ c)'(0)$ at $f(p)$. The Jacobian of a map $f: M \rightarrow N$ at $p \in M$ is the map $$ df_p: T_pM \rightarrow T_{f(p)}N, $$ where $$ df_p(v) = (f\circ c)'(0). $$ Using local coordinates, you can verify that $df_p$ is a linear map and therefore can be written as a matrix using the coordinate bases of $T_pM$ and $T_{f(p)}N$. It is straightforward to show that each entry of this matrix is the partial derivatives of a component $f$ with respect to a component of the local coordinates on $M$.

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Here's an intuitive picture of the geometric situation. Consider a surface $M$ in Euclidean space. It should be obvious what the tangent plane $T_pM$ should be at the point $p \in M$. Now consider another surface $N$ in the same Euclidean space and suppose we have a map $f: M \rightarrow N$. Now this map obviously maps the point $p \in M$ to $f(p) \in N$. But more is true, it should be geometrically obvious that we should expect that the tangent plane at $p \in M$ is mapped to the tangent plane at $f(p) \in N$. Let us write this map as $T_pf$. Then in symbols, the preceding map is:

$T_pf: T_pM \rightarrow T_f(p)N$.

Then for manifolds $M, N$, we expect to have a similar theorem. Then as Euclidean spaces are manifolds, we get the formula you've stated.

Differential geometry turns this intuitive description into rigorous mathematics.

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