The differential $df:T(M)\rightarrow\mathbb{R}$ and the differential map $df:T(M)\rightarrow T(\mathbb{R^1})$ differ by a canonical isomorphism.

Question

This is a problem from Semi-Riemannian Geometry by Barrett O'Neill.

For a smooth manifold $M$, we denote by $\mathcal{F}(M)$ the set of all smooth functions $f:M\rightarrow \mathbb{R}$, and by $T(M)$ the tangent bundle of $M$. I'm asked to prove the following:

Let $f \in \mathcal{F}(M)$. Then, the differential $df:T(M)\rightarrow\mathbb{R}$ and the differential map $df:T(M)\rightarrow T(\mathbb{R^1})$ differ only by a canonical isomorphism.

I find this a bit confusing since, initially I thought the differential and the differential map were obviously the same thing but I guess this is the whole point of the problem. Also I'm not familiar with the definition of a canonical isomorphism, the book gives two examples which I don't find to be all that illustrative. Any help/tips are appreciated.

Answer 1

Given a smooth map $f : M \to N$, its differential map $df$ is a map from the tangent bundle $T(M)$ to the tangent bundle $T(N)$. It is given fiberwise by the differential maps $df_p : T_p(M) \to T_{f(p)}(N)$ at the points $p \in M$. In the sequel let us write $T_p(f)$ instead of $df_p$ for distinction.

The differential of $f \in \mathfrak{F}(M)$ is the one-form $df$ which assigns to each point $p \in M$ the element $df_p$ of the cotangent space $T_p(M)^*$ given by $df_p(v) = v(f)$ for all $v \in T_p(M)$. Hence $df_p$ is a linear map $T_p(M) \to \mathbb{R}$ , and we may regard the collection of these $df_p$ as a map $df : T(M) = \bigcup_{p\in M} T_p(M) \to \mathbb{R}$. In my opinion this is not explained very well in your textbook.

Exercise 13 therefore does not really make sense. In fact each $f \in \mathfrak{F}(M)$ is nothing else than a smooth map $f : M \to \mathbb{R}$ so that it has a differential map $T(f) : T(M) \to T(\mathbb{R})$. However, $\mathbb{R}$ cannot be identified with the tangent bundle $T(\mathbb{R})$, hence the maps in question do not differ only by a canonical isomorphism.

What is meant is this: For each $p \in M$ we can compare the maps $T_p(f) : T_p(M) \to T_{f(p)}(\mathbb{R})$ and $df_p : T_p(M) \to \mathbb{R}$. But for each $y \in \mathbb{R}$ there is a canonical isomorphism $h_y :\mathbb{R} \to T_y(\mathbb{R})$ (see the section "Vector Spaces as Manifolds"). Then we get $h_{f(p)} df_p = T_p(f)$. This is now really a nice exercise in putting together all definitions in your textbook.

Answer 2

I don't have this text, but often the differential $df$ of a smooth function $f : M \to \Bbb R$, which can be viewed as a map $TM \to \Bbb R$ is introduced before the differential map of a smooth map $g : M \to N$ between smooth manifolds, which can be viewed as a map $TM \to TN$.

The purpose of this exercise is to show that these concepts coincide in the case $N = \Bbb R$: In order for this to make sense we must be able to identify $T_x \Bbb R$ with $\Bbb R$ for all $x \in \Bbb R$, so that for any point $p \in M$, we have $$df_p : T_p M \to T_{f(p)} \Bbb R \cong \Bbb R .$$ This identification is exactly furnished by the canonical isomorphism: For any real (or complex) vector space $\Bbb V$ and any $v \in \Bbb V$, we get a (canonical) identification $$T_v \Bbb V \cong \Bbb V .$$

The coincidence of these two concepts justifies the reuse of the term differential (map) for the map $TM \to TN$ as well as the notation $dg$ sometimes used for that map.

Answer 3

I would show that the following diagramm is commutative and the lower map is an isomorphism of vector spaces:

$$\require{AMScd} \begin{CD} T_pM @>{df}>> \Bbb R\\ @V{df}VV @VV{\text{id}}V \\ T_{f(p)}\Bbb R @>{d\,{\text{id}}}>> \Bbb R \end{CD}$$

By definition $d\,{\text{id}}(v)=v(\text{id})$ and if $v$ is given by a curve $\gamma$ then $d\,{\text{id}}(v)=\gamma'(0)$.