If $f:S^1 \to \mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2})$

Question

If $f:S^1 \to \mathbb{R}$, defined by $f(x,y)=x-y$ and $X_p$ be a tangent vector at $p=(\frac{1}{\sqrt 2}, \frac{1}{\sqrt 2})$ given by $X_p=4 (\frac{\partial}{\partial t})_p$ where $t$ is the local coordinate on the chart $(U, \phi)$, where $U=S^1-\{(0,0)\}$, and $\phi(x,y)=\frac{x}{1-y}$, then find $X_p(f)$.

Answer:

$X_p(f)=4(\frac{\partial}{\partial t})_pf=4(\frac{\partial}{\partial x})_p (x-y)+4(\frac{\partial}{\partial y})_p(x-y)=4-4=0$

But where is the need of $ \phi(x,y)=\frac{x}{1-y}$ ?

Answer 1

When you wrote $ f(x,y)=x-y $, you have already put $S^1$ into $R^2$ and used a chart of $ R^2 $. $(x,y)$ here is a point in $R^2$ chart. If you want to redefine a chart $ (U, \phi) $, I think you have to rewrite $ f$ by $\phi$.

Answer 2

The easiest way to proceed here is to solve for $x(t)$ and $y(t)$ [Hint: Use $x^2+y^2=1$.] and then use $$\frac{\partial}{\partial t} = \frac{\partial x}{\partial t}\frac{\partial}{\partial x} + \frac{\partial y}{\partial t}\frac{\partial}{\partial y}.$$ In this case, at the point $(1/\sqrt2,1/\sqrt2)$, you should find that $$\frac{\partial}{\partial t} = \frac{1-\sqrt2}2\,\frac{\partial}{\partial x} + \frac{\sqrt2-1}2\,\frac{\partial}{\partial y}.$$ Note that this vector is a multiple of $-\frac{\partial}{\partial x}+\frac{\partial}{\partial y}$, as is appropriate at a point on the circle with $x=y$.

Even easier, once you know $x(t)$ and $y(t)$, then rewrite $f$ as $f(x(t),y(t))$ and take $\partial/\partial t$ !!