2

As mentioned in the relevant Wikipedia page, given a differential function $f:M\to N$ between differential manifolds of dimensions $m$ and $n$, respectively, its differential is a mapping $df:TM\to TN$ such that, for all $p\in M$, we have $df(p):T_p M\to T_{f(p)}N$.

I understand the notion of differential when $f\in C^\infty(M,\mathbb R)$, that is, $f:M\to\mathbb R$. In this case, given $p\in M$ and $v\in T_p M$ we can define $df(p)(v)=(f\circ\gamma)'\rvert_0$ for any curve $\gamma:(-1,1)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v$. Here, I understand $\gamma'(0)$ as the differential of $\phi\circ\gamma:(-1,1)\to\mathbb R^m$ for any coordinate chart $\phi:U\to\mathbb R^m$, $U\subset M$.

So now, how exactly should I understand the differential $df(p)(v)$ when $f:M\to N$? Parsing e.g. this other answer I would guess something like the "standard" differential of $\phi_N\circ f$ for a coordinate chart $\phi_N:U_N\to\mathbb R^n$, $U_N\subset N$, that is, $$df(p)(v) \equiv (\phi_N\circ f\circ \gamma)'\big\rvert_0$$ for $\gamma:(-1,1)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=v\in T_p M$ defined as the derivative of $\phi_M\circ\gamma$ for some coordinate chart $\phi_M:U_M\to\mathbb R^m$, $U_M\subset M$.

Is this correct? The expression in the linked answer is a bit different, but I suppose this is just due to different notation in which we write the differential as a map between the tangent bundles $TM$ and $TN$. Is this the case?

glS
  • 7,963
  • "Here, I understand $\gamma^{\prime}(0)$ as the differential of $\phi\circ\gamma$...." This is not independent of your choice of $\phi$, furthermore $(\phi\circ\gamma)^{\prime}(0)$ is an element of $\mathbb{R}^m$, whereas $\gamma^{\prime}(0)$ ought to be an element of $T_pM$. The same goes for your suggested approach. This can be made to work by passing to an appropriate equivalence relation and keeping track of charts, but is rather tedious compared to the standard approach. – Thorgott Nov 04 '20 at 11:42
  • I thought the idea was to define $T_p M$ as the set of equivalence classes of curves $\gamma_i$ having the same $(\phi\circ\gamma_i)'\big\rvert_0$, and that this was independent from the choice of coordinate chart. At least that's one way it is described in the wiki page. Which one are you referring to as the "standard approach"? – glS Nov 04 '20 at 11:49
  • That's one possible way of defining $T_pM$, but $(\phi\circ\gamma)^{\prime}(0)$ is independent of the choice of curve $\gamma$ in that approach, not independent of the choice of chart, and the tangent vector $v$ is defined to be that equivalence class of curves, not $(\phi\circ\gamma)^{\prime}(0)$ (which isn't well-defined). There are multiple ways of defining tangent spaces, which all turn out to be equivalent, but defining the differential works differently in all of them. Which do you want to use? The standard one I mean is the approach by derivations. – Thorgott Nov 04 '20 at 12:01
  • @Thorgott I'm confused. So you are saying that if I define $v$ as the equivalence class of curves with equal $(\phi\circ\gamma)'(0)$, this equivalence class depends on the coordinate chart $\phi$? But isn't the definition that curves are equivalent when the derivatives coincide on all coordinate charts? – glS Nov 04 '20 at 12:07
  • You need to be precise. Two curves $\gamma_1,\gamma_2$ are equivalent if $(\phi\circ\gamma_1)^{\prime}=(\phi\circ\gamma_2)^{\prime}(0)$ for all coordinate charts $\phi$. But you don't require that $(\phi\circ\gamma_1)^{\prime}(0)=(\psi\circ\gamma_1)^{\prime}(0)$ for different coordinate charts $\phi,\psi$ (there is no reason for that equality to hold in general, consider $M=\mathbb{R}^n$ and a tangent vector $v$ (which we think of as element of $\mathbb{R}^n$), then any invertible linear map $f\colon\mathbb{R}^n\rightarrow\mathbb{R}^n$ is a chart and $(f\circ\gamma)^{\prime}(0)=f(v)$). – Thorgott Nov 04 '20 at 12:15
  • @Thorgott I see. So in other words, different charts will give different values for $(\phi\circ\gamma_i)'(0)$, but that doesn't affect the definition of the equivalence class of curves which share the same such derivative (i.e. using different charts still gives the same equivalence classes). Still, $(f\circ \gamma)'(0)$, remains independent of the choice of coordinate chart, right? This aside, does the rest of the construction work in the $f:M\to N$ case? – glS Nov 04 '20 at 12:26
  • Yes to the first two. No to the last, because $(\phi_N\circ f\circ\gamma)^{\prime}(0)$ still depends on the chart $\phi_N$. Even if it were well-defined, it would define an element of $\mathbb{R}^{\dim N}$, not of $T_{f(p)}N$. An element of $T_{f(p)}N$ is an equivalence class of curves in $N$ passing through $f(p)$. Can you think of a way to obtain that from an equivalence of curves in $M$ passing through $p$ (i.e. an element of $T_pM$) via $f$? – Thorgott Nov 04 '20 at 12:39
  • @Thorgott ah, I suppose then the equivalence class in $T_{f(p)}N$ should be ${f\circ\tilde\gamma ,|, (\phi_N\circ f\circ\tilde\gamma)'(0)=(\phi_N\circ f\circ\gamma)'(0)}\equiv (f\circ\gamma)'(0)\in T_{f(p)}N$ for some $\gamma\in\gamma'(0)\in T_pM$? Or maybe, equivalently, ${f\circ\gamma,|, \gamma\in\gamma'(0)}$. – glS Nov 04 '20 at 12:58
  • Indeed, if $\gamma$ determines the element $\gamma^{\prime}(0)\in T_pM$, then $f\circ\gamma$ determines the element $df_p(\gamma^{\prime}(0))$ in $T_{f(p)}N$. – Thorgott Nov 04 '20 at 13:12

0 Answers0