Is $\frac{\partial}{\partial x_1}=\frac{\partial}{\partial y_1}$ if $x_1=y_1$?

Question

Let $M$ be a manifold, $p\in M$ and $(U,\varphi=(x_1,\dots,x_n))$, $(V,\psi=(y_1,\dots,y_n))$ two local charts such that $p\in U\cap V$ and $x_1=y_1$. Is it true that $(\frac{\partial}{\partial x_1})_p=(\frac{\partial}{\partial y_1})_p$?

My professor says that the answer is yes, and so I build a proof. But then I found the following counterexample:

Let $M=\mathbb{R}^2$, $p=(0,0)$, $U=V=\mathbb{R}^2$, $\varphi(u,v)=(u,v)$ and $\psi(u,v)=(u,u+v)$. Then $x_1(y_1,y_2)=y_1$ and $x_2(y_1,y_2)=y_2-y_1$. Using the formula for the change of basis I've got

$$(\frac{\partial}{\partial y_1})_p=\sum_{i=1}^2(\frac{\partial x_i}{\partial y_1})_p(\frac{\partial}{\partial x_i})_p=(\frac{\partial}{\partial x_2})_p-(\frac{\partial}{\partial x_1})_p\neq (\frac{\partial}{\partial x_1})_p$$

What's wrong with this counterexample?

Answer 1

Think the answer is no. There Is nothing wrong with your example. There are plenty of examples where this could happen, in fact this is $\textbf{warned}$ in standard text in differential geometry e.g $\textbf{Lee's Smooth Manifold}$ page 65, by consider the coordinates $(x,y)$ and $(\tilde{x},\tilde{y})$ in $\mathbb{R}^2$ related by $$ \tilde{x} = x, \qquad \tilde{y} = y+x^3 $$ Evaluate at $p=(x,y)=(1,0) \in \mathbb{R}^2$ will gives $$ \frac{\partial}{\partial x}\Big|_p \neq \frac{\partial}{\partial \tilde{x}}\Big|_p. $$ Which is tell us that $\frac{\partial}{\partial x^i}\Big|_p$ depend on the entire coordinate system, not just $x^i$.

Answer 2

You have re-discovered a confusion with the Leibniz notation $$\frac{\partial u}{\partial x}$$ which has caught students for years. The answer depends not only on the variable $x$, but also on what the other variables are. So if you change the other variables, the value of $\frac{\partial u}{\partial x}$ could change, even if $x$ stays the same.

Hopefully, mathematicians are aware of this and avoid the problem. But students, as well as users of math (such as physicists and engineers) do sometimes make such glaring errors.

Added: The standard changes of coordinates are fixed so that this problem does not come up. For example, the change between cartesian coordinates $(x,y,z)$ and cylindrical coordinates $(r,\theta, z)$. Here, the coordinates $(r,\theta,z)$ are orthogonal in the $(x,y,z)$ coordinate system (and vice versa), so $$ \frac{\partial u}{\partial z} $$ is the same for both coordinate systems. Challenge: Try proving this, as an exercise to see that you understand.

Answer 3

The both variables can be reduced to the same function germ on the intersection of the chart, so the Members of the dual-space in the tangentialmanifold do coincide. The terminology function germ according to p in M , means to functions are in the same function germ equivalence class if they do coincide in an Environment of this point. So [x1]=[y1] as function germs and according to the definition the duals in Tangenspace do coincide. There is no need of a chart transformation anymore due to the fact that the functions are in the different Environment equivalent in the intersection of two open sets U and V.