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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 ?

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  • $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Commented Jun 7, 2022 at 9:18

2 Answers 2

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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.

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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 $t=0$) $$\frac{dv_k}{dt}=\sum_{j=1}^n\frac{\partial y_k}{\partial x^j}(p)\frac{du_j}{dt}.$$

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  • $\begingroup$ 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 $\psi \circ \phi^{-1}(x)=(y_1 (x), \cdots, y_n (x))$. $\endgroup$
    – daㅤ
    Commented Jun 2, 2022 at 0:22
  • $\begingroup$ 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. $\endgroup$
    – Balloon
    Commented Jun 2, 2022 at 8:54

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