Differential with Chain rule. (Relating to manifold.) Where did I make a mistake?

Let $$M$$ be $$n$$-dimensional manifold, and $$I\subset \mathbb R$$ open interval.

For $$p\in M$$, define $$L_p=\{ l : (-\epsilon, \epsilon)\to M \mid \epsilon >0, l(0)=p \}.$$

Fix $$p\in M$$ and $$l\in L_p.$$

There exists $$\epsilon>0$$ s.t. $$l:(-\epsilon, \epsilon)\to M$$.

Let $$(U, \phi)$$ and $$(V, \psi)$$ be charts around $$p$$.

($$\phi$$ is a homeomorphism from $$U$$ to $$\mathbb R^n$$, and the same is true for $$\psi.$$)

Then, the coordinate transformation $$\psi \circ \phi^{-1} : \phi (U\cap V) (\subset \mathbb R^n) \to \psi(U\cap V) (\subset \mathbb R^n)$$, by $$(x_1, \cdots , x_n)\mapsto (\psi \circ \phi^{-1})(x_1, \cdots, x_n)=:(y_1, \cdots, y_n)$$ is homeomorphism.

Choose $$\epsilon_0\in (0,\epsilon)$$ s.t. $$l((-\epsilon_0, \epsilon_0))\subset U$$ and $$l((-\epsilon_0, \epsilon_0))\subset V$$ and assume the domain of $$l$$ is $$(-\epsilon_0,\epsilon_0)$$.

And let us write

$$(\phi \circ l )(t)=(u_1(t), \cdots, u_n(t))$$, $$(\psi \circ l) (t)=(v_1(t), \cdots, v_n(t))$$.

Then, show that $$\dfrac{dv_k}{dt}=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} \dfrac{du_j}{dt}.$$

Here is my calculation and maybe I mistake somewhere.

Let $$\psi \circ \phi^{-1}=:F, \phi \circ l=G.$$

First, $$\dfrac{d}{dt}(\psi \circ l)(t)=(\frac{d}{dt}v_1(t),\cdots, \frac{d}{dt}v_n(t))$$.

On the other hand, \begin{align} &\quad \dfrac{d}{dt}(\psi \circ l)(t)\\ &=\dfrac{d}{dt}(F\circ G)(t)\\ &\underset{\mathrm{chain\ rule}}=(DF)(G(t))\cdot \dfrac{d}{dt}G(t)\ (DF:\mathrm{Jacobi\ matrix \ of\ } F)\\ &=\begin{pmatrix} \frac{\partial y_1}{\partial x_1}(G(t)) & \cdots & \frac{\partial y_1}{\partial x_n}(G(t))\\ \vdots & & \vdots \\ \frac{\partial y_n}{\partial x_1}(G(t)) & \cdots & \frac{\partial y_n}{\partial x_n}(G(t)) \end{pmatrix} \begin{pmatrix} \frac{d}{dt}u_1(t)\\ \vdots \\ \frac{d}{dt}u_n(t)\end{pmatrix}\\ &=\begin{pmatrix} \displaystyle\sum_{j=1}^n \dfrac{\partial y_1}{\partial x_j}(G(t)) \dfrac{du_j}{dt}(t)\\ \vdots\\ \displaystyle\sum_{j=1}^n \dfrac{\partial y_n}{\partial x_j} (G(t))\dfrac{du_j}{dt}(t) \end{pmatrix}. \end{align}

Thus I get $$\dfrac{dv_k}{dt}(t)=\displaystyle\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} (G(t))\dfrac{du_j}{dt}(t) \tag{i}$$

I think this is a bit different from the goal $$\displaystyle\dfrac{dv_k}{dt}=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} \dfrac{du_j}{dt} \tag{ii}$$

Of course, the LHS of (i) :$$\displaystyle\dfrac{dv_k}{dt}(t)$$ is the same as the LHS of (ii) :$$\dfrac{dv_k}{dt}$$.

Similarly, $$\dfrac{du_j}{dt}(t)=\dfrac{du_j}{dt}$$.

But I don't think $$\dfrac{\partial y_k}{\partial x_j} (G(t))=\dfrac{\partial y_k}{\partial x_j}$$.

$$\dfrac{\partial y_k}{\partial x_j}$$ is written as $$\dfrac{\partial y_k}{\partial x_j}(x_1, \cdots, x_n)$$ but $$\dfrac{\partial y_k}{\partial x_j} (G(t)) =\dfrac{\partial y_k}{\partial x_j} (u_1(t),\cdots, u_n(t)).$$

So I think I mistook somewhere. Do you find where I mistook ?

• I think writing $(\psi \circ \phi^{-1})(x_1, \cdots, x_n)=:(y_1, \cdots, y_n)$ may be misleading because it suggests that $y_i \in \mathbb R$. The $y_i$ are the coordinate functions of $\psi \circ \phi^{-1}$, and as such they are used later. Commented Jun 7, 2022 at 9:18

What you have done is absolutely correct. The problem is that the formula $$\dfrac{dv_k}{dt}=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} \dfrac{du_j}{dt} \tag{1}$$ does not give you precise information at which points the derivatives $$\dfrac{dv_k}{dt}$$, $$\dfrac{\partial y_k}{\partial x_j}$$ and $$\dfrac{du_j}{dt}$$ have to be evaluated. It is clear that both $$\dfrac{dv_k}{dt}$$ and $$\dfrac{du_j}{dt}$$ have to be evaluted at the same $$t$$, but what about $$\dfrac{\partial y_k}{\partial x_j}$$? The partial derivatives $$\dfrac{\partial y_k}{\partial x_j}$$ are functions of the generic variable $$\mathbf x = (x_1,\ldots,x_n)$$, but it is not expedient to write $$\dfrac{dv_k}{dt}(t)=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} (\mathbf x) \dfrac{du_j}{dt}(t) \tag{2}$$ because it does not tell us which specific $$\mathbf x$$ has to be taken. In fact we must take $$\mathbf x = G(t)$$: $$\dfrac{dv_k}{dt}(t)=\sum_{j=1}^n \dfrac{\partial y_k}{\partial x_j} (G(t)) \dfrac{du_j}{dt}(t) \tag{3}$$ This is your formula $$(i)$$ and it is the only correct interpretation of $$(1)$$. It generalizes the chain rule from elementary calculus which says $$(g \circ f)'(x) = g'(f(t)) f'(x)$$ or $$(g \circ f)' = (g' \circ f) \cdot f'$$ You would never write $$(g \circ f)'(x) = g'(y) f'(x)$$. But note that sometimes people write $$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx} .$$ Here $$y = y(x)$$ and $$z = z(y)$$. This notation has the same problem as we encountered above.
It is a matter of definition: remember that when you have coordinates $$\phi=(x_1,\dots,x_n):U\to\mathbb{R}^n$$, $$U\subset M$$, and a function $$f:M\to\mathbb{R}$$, then $$\frac{\partial f}{\partial x^i}$$ is a function defined on $$U$$ by $$\frac{\partial f}{\partial x^i}(p)=\frac{\partial (f\circ\phi^{-1})}{\partial r^i}(\phi(p)),$$ where $$\frac{\partial}{\partial r^i}$$ denotes the usual $$i$$-th partial derivative for functions from $$\mathbb{R}^n$$ to $$\mathbb{R}$$. Here, the function $$F:\mathbb{R}^n\to\mathbb{R}^n$$ you are trying to differentiate checks that $$F(x)=\psi\circ\phi^{-1}(x)=(y_1\circ\phi^{-1}(x),\dots,y_n\circ \phi^{-1}(x)).$$ Thus, its Jacobi matrix at $$x\in\mathbb{R}^n$$ will be given by $$\begin{pmatrix} \frac{\partial (y_1\circ\phi^{-1})}{\partial r^1}(x)&\dots&\frac{\partial (y_1\circ\phi^{-1})}{\partial r^n}(x)\\ \vdots&&\vdots\\ \frac{\partial (y_n\circ\phi^{-1})}{\partial r^1}(x)&\dots&\frac{\partial (y_n\circ\phi^{-1})}{\partial r^n}(x) \end{pmatrix}.$$ Then, by evaluating it at $$G(t)=\phi(l(t))$$ you get (for example ) $$\frac{\partial (y_1\circ\phi^{-1})}{\partial r^n}(\phi(l(t)))=:\frac{\partial y_1}{\partial x^n}(l(t)),$$ and then proceeding as you did, you make explicit what was implicit in your given formula, which is intended to be read as: $$\frac{dv_k}{dt}(t)=\sum_{j=1}^n\frac{\partial y_k}{\partial x^j}(l(t))\frac{du_j}{dt}(t)$$ or (evaluating at $$t=0$$) $$\frac{dv_k}{dt}=\sum_{j=1}^n\frac{\partial y_k}{\partial x^j}(p)\frac{du_j}{dt}.$$
• Why $\psi \circ \phi^{-1}(x)=(y_1 \circ \phi^{-1}(x), \cdots, y_n \circ \phi^{-1}(x))$ ? $(y_1,\cdots,y_n)$ is defined as $(\psi \circ \phi^{-1})(x)=(y_1,\cdots,y_n)$ for $x\in \mathbb R^n$, so I think $\psi \circ \phi^{-1}(x)=(y_1 (x), \cdots, y_n (x))$.
• Yes, you are right! In this case, you still have the $G(t)$ at the end ou your computations, and the closest you can get to find back the given formula is by evaluating the one you get at $t=0$, which gives (since $G(0)=\phi(p)$) $\frac{dv_k}{dt}=\sum_{j=1}^n\frac{\partial y_k}{\partial x^j}(\phi(p))\frac{du_j}{dt}.$ It is no surprise that you have to add something after $\frac{\partial y_k}{\partial x^j}$, since this is a function which has to be evaluated at some point to give you a real number. Commented Jun 2, 2022 at 8:54