# Differential of a function between manifolds

The book we are using in class is Frank Warner Foundation of Differential Manifold and Lie Group. Let $M,N$ be two smooth $d$-dim manifold, the differential of a $C^\infty$ function $\phi:M\rightarrow N$ is defined by $$d\phi: M_m \rightarrow N_{\phi(m)}$$ For $v\in M_n$, and $g:N \rightarrow \mathbb{R}$ a smooth function, we define $$d\phi(v)(g) = v(g\circ \phi).$$

Now for a smooth function $f:M \rightarrow \mathbb{R}$, I don't quite see why the book states $$df(v)(g) = v(f) \frac{\partial}{\partial r}\bigg|_{r_0} g$$ from the definition above.

Edit: Reading the answer here seems that if we plug $g(r) = r$, we will get the identity $df(v) = v(f)$, but how do we know it will hold for all $g$?

• What you've written makes absolutely no sense: if $f : M \to \Bbb R$ and $v \in M_m$, then $v(f) = (\Bbb d _m f) (v)$ is a number, therefore it makes no sense to write $(\Bbb d _m f) (v) (g)$. Also, who are $r$ and $r_0$? – Alex M. Sep 14 '16 at 22:10
• Many books use shorthand notation and don't distinguish very well between a tangent vector $v \in T_m M$ (ie your $M_m$) and a vector field $v \in \Gamma(TM)$. Vectors on a manifold eat functions and spit out numbers. Vector fields on the other hand eat points on the manifold and spit out tangent vectors. Look above at the definition to see what is going on. One way you can derive the result u r after involves using the identity function on the manifold and tangent spaces (which are seperate functions!) and using $\text{id}=\psi \circ \psi^{-1}=\psi^{-1} \circ \psi$ where $\psi$ is a chart – MKF Sep 14 '16 at 22:20
• The $v$ is in $M_m$, the tangent space of at $m$, we have not learned about vector field yet. $r$ is the real variable in $\mathbb{R}$ and $r_0 = f(m)$. The first definition we have is $df(v) = v(f) \frac{d}{dr}\bigg|_{r_0}$, then the book says we usually choose to identify $v(f) \frac{d}{dr}\bigg|_{r_0}$ as $v(f)$. – Xiao Sep 14 '16 at 22:30
• Did you define tangent vectors as derivations? Then you only need to check that the second display formula yields a map in functions satisfying the derivations axioms. – Moishe Kohan Sep 14 '16 at 23:38
• @studiosus Yes, we define tangent vectors as derivations, do you mean we need to show that $v(f\circ g)$ is a derivation of $g$? Then from this, since $\frac{\partial }{\partial r}\bigg|_{r_0}$ is a basis, then $df(v) = v(f) \frac{\partial }{\partial r }\bigg|_{r_0}$ would be the tangent vector. – Xiao Sep 14 '16 at 23:53

Just to elaborate on Xiao's answer; considering the fact that we refer to tangent vectors derivations then the differential or push forward map;

$$[f_{*,p}(X_p)] g = X_p( g \circ f); \ X_p \in T_pM, g \in C_{f(p)}^{\infty}(N), f: M \to N$$

here $X_p( g \circ f) \in T_{f(p)}N$ (why?). Well if you take $g,h \in C_{f(p)}^{\infty}(N)$ then;

$$[f_{*,p}(X_p)] (gh) = X_p(gh \circ f) = X_p(g \circ f \cdot h \circ f)$$

and now since $X_p$ is a derivation;

$$X_p(g \circ f \cdot h \circ f) = X_p(g \circ f) \cdot h(f(p)) + g(f(p)) \cdot X_p(h \circ f)$$

$$\hspace{1.2in}= [f_{*,p}(X_p)]g \cdot h(f(p)) + g(f(p)) [f_{*,p}(X_p)] h$$

The linearity piece is also clear since $X_p$ is linear. Therefore; if you take $f: M \to \mathbb{R}$ and $(U,x^1,...,x^d)$ to be a chart about $p$ then;

$$\left\{\frac{\partial}{\partial x^1}\Bigr|_p,...,\frac{\partial}{\partial x^d}\Bigr|_p\right\}$$

is a basis for $T_pM$. Similarly, we can use the coordinate $t$ to parametrize a neighborhood of $f(p) \in \mathbb{R}$ and so $T_{f(p)}\mathbb{R}$ has basis vector;

$$\frac{\partial}{\partial t}\Bigr|_{f(p)} := \frac{d}{dt}\Bigr|_{f(p)}$$

Since $f_{*,p}$ is linear, it maps tangent vectors to tangent vectors i.e;

$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) = \alpha \frac{d}{dt}\Bigr|_{f(p)}$$

If we evaluate both sides at $t$ and use the definition of the differential we have;

$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) t = \alpha \frac{d}{dt}\Bigr|_{f(p)} t \Rightarrow \frac{\partial}{\partial x^i}\Bigr|_p (t \circ f) = \frac{\partial}{\partial x^i}\Bigr|_p f = \alpha$$

The above follows from the fact that the coordinate function $t$ picks out the first coordinate of the map $f$, which is real-valued, so that if just $f$. It now follows that;

$$f_{*,p}\left(\frac{\partial}{\partial x^i}\right) =\frac{\partial}{\partial x^i}\Bigr|_p f \frac{d}{dt}\Bigr|_{f(p)}$$

By definition $df(v)$ is a tangent vector in $\mathbb{R}_{r_0}$, since $\bigg\{\frac{\partial}{\partial r}\bigg|_{r_0}\bigg\}$ is a basis of the tangent space $\mathbb{R}_{r_0}$, then it can be written as $$df(v) (\cdot) = K\frac{\partial}{\partial r}\bigg|_{r_0}(\cdot)\quad \text{ for some } K\in \mathbb{R}.$$ To solve for this $K$, we can plug the identiy function into the tangent vector, that will tell us $K = v(f)$.