# Why would we expect the pushforward to encode the total derivative of a smooth map?

According to Lee, the pushforward was invented to give a coordinate independent definition of the total derivative of a smooth function between two smooth manifolds. To each smooth map $F:M \to N$ and each point $p \in M$ we associate a linear map $F_*:T_pM\to T_{F(p)}N$ defined by $F_*X(f) = X(f \circ F)$ where $X$ is any derivation in $T_pM$ and $f:N \to \mathbb{R}$ is any smooth function. Given a smooth chart $\phi:M \to \mathbb{R}^m$ near $p$ we have a basis for $T_pM$ given by $\frac{\partial}{\partial x^i}|_{p} = ({\phi ^{-1}}_*)\frac{\partial}{\partial x^i}|_{\phi(p)}$. Similarly given a smooth chart $\psi$ near $F(p)$ we have a basis for $T_{F(p)}N$ given by $\frac{\partial}{\partial y^i}|_{F(p)} = ({\psi ^{-1}}_*)\frac{\partial}{\partial y^i}|_{\psi(p)}$. A calculation in Lee shows that the matrix representation of $F_*$ with respect to these bases is the total derivative of the coordinate representation $\hat{F} = \psi \circ F \circ \phi ^{-1}$ evaluated at $\phi(p)$.

My question is, is there some intuitive reason why we would expect this to be true? This all seems very abstract to me. I can't tell if it is supposed to be obvious that this definition should be a coordinate independent way of encoding the total derivative of $F$ and I am just missing something, or if it is just difficult to understand. How should I think about the pushforward?

• Do you know how to relate derivations to directional derivatives? That's the intuition here. – Qiaochu Yuan Oct 16 '12 at 23:08
• Well I think we can think of the basis element $\frac{\partial}{\partial x^i}$ above as "taking the derivative in the direction of $\frac{d\phi^{-1}}{dt}(p^1,...,p^{i-1},p^i + t,p^{i+1}, ... p^m)$" is that what you mean? – Seth Oct 16 '12 at 23:16

As I understand your question, you want to know why the definition $F_*X(f) := X(f \circ F)$ is an appropriate generalization of the total derivative. In other words, knowing only the definition of the total derivative, how would one come to this definition of pushforward?
Let $F\colon \mathbb{R}^m \to \mathbb{R}^n$ be smooth, and let $D_pF\colon \mathbb{R}^m \to \mathbb{R}^n$ denote the total derivative at $p \in \mathbb{R}^n$. To each vector $w \in \mathbb{R}^n$ (at $F(p)$), we associate the derivation at $F(p) \in \mathbb{R}^n$ via: $$w \in \mathbb{R}^n \mapsto w^j \left.\frac{\partial}{\partial x^j}\right|_{F(p)}.$$ In particular, for $v \in \mathbb{R}^m$ (at $p$), $$D_pF(v) \in \mathbb{R}^n \mapsto D_pF(v)^j \left.\frac{\partial}{\partial x^j}\right|_{F(p)}.$$ And in fact, this derivation on the right-hand side is none other than $$\left.v^i\frac{\partial}{\partial x^i}\right|_p(-\circ F).$$
To see this, we just use the chain rule: \begin{align*} v^i \left.\frac{\partial}{\partial x^i}\right|_p(-\circ F) & = v^i \left.\frac{\partial F^j}{\partial x^i}\right|_p \left.\frac{\partial}{\partial x^j}\right|_{F(p)} \\ & = v^i D_pF(e_i)^j \left.\frac{\partial}{\partial x^j}\right|_{F(p)} \\ & = D_pF(v)^j \left.\frac{\partial}{\partial x^j}\right|_{F(p)} \end{align*}
Alternatively, I believe it also suffices to note that both derivations give the same value when applied to the coordinate function $x^k$: $$D_pF(v)^j\frac{\partial x^k}{\partial x^j} = D_pF(v)^k = v^i D_pF(e_i)^k = \left.v^i\frac{\partial F^k}{\partial x^i}\right|_p = v^i\left.\frac{\partial}{\partial x^i}\right|_p(x^k \circ F).$$
Point: The derivation at $F(p) \in \mathbb{R}^n$ given by $$v^i\partial_i|_{p}(-\circ F)$$ is exactly $$D_pF(v)^j \left.\frac{\partial}{\partial x^j}\right|_{F(p)}$$