0

Let $f:\mathbf R^n\to\mathbf R^m$ be differentiable. The total derivative at a point $p$ is a linear map $D_pf:T_p\mathbf R^n\to T_{f(p)}\mathbf R^m$, where $T_p\mathbf R^n$ denotes the tangent space of $\mathbf R^n$ at $p$.

We can take the disjoint union of these tangent spaces over all points $p\in\mathbf R^n$ to form the tangent bundles $T\mathbf R^n$ and $T\mathbf R^m$. Then the derivative of $f$ (not just at a point) can be defined as one of the following two things:

  • a map $Df:\mathbf R^n\to\bigsqcup_{p\in\mathbf R^n}\text{Hom}(T_p\mathbf R^n, T_{f(p)}\mathbf R^m)$ that maps each $p$ to $D_pf$

  • a map $Df:T\mathbf R^n\to T\mathbf R^m$, that when restricted to $T_p\mathbf R^n$ recovers the original derivative $D_pf$

Is one of these wrong? Are they equivalent definitions? They seem like two very different objects to me.

node196884
  • 193
  • 1
    Typo: it should be $T_{f(p)}\Bbb{R}^m$, not $T_p$. Next, both these maps contain the same amount of information. – peek-a-boo Feb 09 '24 at 18:23
  • 1
    They are equivalent if you know $f$. The second map is essentially given by $Df(x,v) = (f(x),df_x(v))$ for $v\in T_x\Bbb R^n$. So it turns your first (pointwise) construction into a bundle map. – Ted Shifrin Feb 09 '24 at 18:23
  • Sorry, I'm a little confused as to what $df_x(v)$ is, could you please elaborate? – node196884 Feb 09 '24 at 18:33
  • 1
    Call your first map $F_1$ and the second one $F_2$. The relationship between these two is that $F_2(x,v)=(f(x),F_1(x))$. Anyway, this is a general set-theoretic fact, not much differential geometry/calculus going on. I’ll update with an answer soon. – peek-a-boo Feb 09 '24 at 18:45
  • this may need to be another question, but are there also two essentially equivalent maps for the second derivative? as in, a pointwise map whose image takes two vectors as an argument, and then a map from $TT\mathbf R^n$ to $TT\mathbf R^m$? – node196884 Feb 09 '24 at 19:01

1 Answers1

0

Start with two sets $X,Y$. And suppose we have two families of sets $\{V_x\}_{x\in X}$ and $\{W_y\}_{y\in Y}$. Suppose also that for each $x\in X$ and $y\in Y$, we are given a certain family of functions $\mathscr{F}_{x,y}\subset\text{set of all functions $$}V_x\to W_y$. Suppose you now are given a map $f:X\to Y$.

Then, there is a bijective correspondence between the two types of maps you mentioned:

  • maps of the type $\Phi:X\to \bigsqcup\limits_{x\in X}\mathscr{F}_{x,f(x)}$, such that for each $x\in X$, $\Phi(x)$ lies in the function space $\mathscr{F}_{x,f(x)}$.
  • maps of the type $\Psi:\bigsqcup\limits_{x\in X}V_x\to\bigsqcup\limits_{y\in Y}W_y$ such that for each $x\in X$, $\Psi$ restricts to a mapping $V_x\to W_{f(x)}$ which lies in the given function space $\mathscr{F}_{x,f(x)}$

To go from $\Phi$ to $\Psi$, you simply set $\Psi(x,v_x):=\left(f(x),[\Phi(x)](v_x)\right)$. Conversely, given $\Psi$, you define $\Phi(x):=\Psi(x,\cdot)|_{V_x}:V_x\to W_{f(x)}$.

In your example of course $X=\Bbb{R}^n,Y=\Bbb{R}^m$ and $V_x=T_x\Bbb{R}^n$, $W_y=T_y\Bbb{R}^n$ and $\mathscr{F}_{x,y}=\text{Hom}(V_x,W_y)$ is the set of all linear maps $V_x\to W_y$ (or more generally you can let $X,Y$ be manifolds and consider their tangent bundles $TX,TY$, or even more generally just consider two vector bundles over bases $X,Y$ respectively, or even more generally, consider fiber bundles etc).

In the language of (vector/fiber) bundles, the first type of map is a section of a (certain Hom-type) bundle. While the second type of map is a bundle morphism. This reasoning shows that these two pieces of data are essentially equivalent.

peek-a-boo
  • 55,725
  • 2
  • 45
  • 89