# Let $p=(5,0,-4)$ and $v \in T_{(5,0,-4)}M$. Compute $(F^{*}\omega)_p(v)$.

Let me show my work before presenting the problem itself.

Let $M=\{(x,y,z) \in \mathbb{R}^3 : x+y=5, x+z=cos^2y\}$.

We can easily see that $M$ is a submanifold of $\mathbb{R}^3$ of dimension $1$.

We can also see that we can construct a global atlas for $M$, say $\{(M,\varphi)\}$, where $\varphi:M\rightarrow\mathbb{R}$ is given by $(x,y,z) \mapsto y$.

If $P=(p_1,p_2,p_3) \in M$, then $T_PM=\langle (-1,1,1-2sin(p_2)cos(p_2)) \rangle$. In particular, for $P=(5,0,-4) \in M$, $T_{(5,0,-4)}M = \langle (-1,1,1) \rangle$.

Consider the morphism (smooth map) $F:M \rightarrow S^1$ given by $(x,y,z) \mapsto (\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}})$.

Remember that $S^1=\{(x,y) \in \mathbb{R}^2 : x^2+y^2=1\}$, so the given morphism is well defined.

Expressing $F$ in the atlas of $M$, we get $$F(x,y,z)=F(\varphi^{-1}(y))=(\frac{5-y}{\sqrt{2y^2-10y+25}},\frac{y}{\sqrt{2y^2-10y+25}})$$

We can see that $F(5,0,-4)=(1,0)$.

Now we choose a coordinate chart $\psi$ of $S^1$ in a neighborhood of $(1,0)$. For example, $\psi:U=\{(x,y) \in S^1 : x > 0\} \longrightarrow ]-1,1[$ given by $(x,y) \mapsto y$.

We have also that $\omega=-\frac{1}{x}dy$ is one $1$-form which is an orientation of $S^1$ in a neighborhood of the point $(1,0)$.

Let $P=(5,0,-4)$ and $v \in T_{(5,0,-4)}M$.

I want to compute $(F^{*}\omega)_P(v)$.

I'll show my attempt:

We put $x:=\frac{u}{\sqrt{u^2+v^2}}$ and $y:=\frac{v}{\sqrt{u^2+v^2}}$.

So $-\frac{1}{x}dy=\frac{v}{u^2+v^2}du-\frac{u}{u^2+v^2}dv$ (I don't present the calculations because they're somewhat boring).

So $F^*\omega = \frac{y}{x^2+y^2}dx -\frac{x}{x^2+y^2}dy$ and $(F^*\omega)_p=-\frac{1}{5}dy$, so $(F^*\omega)_p(v)=-\frac{1}{5}C$, where $v=(-C,C,C) \in T_{(5,0,-4)}M = \langle (-1,1,1) \rangle$.

Is this right?

Some help would be appreciated. Thanks in advance.

Your notation is a bit strange. Let's write $F(x,y,z) = (u(x,y,z),v(x,y,z))$. Then you want to calculate $F^*ω$ with $ω = -1/u dv$. Now, $F^*(1/u) = 1/u(x,y) = \sqrt{x^2 + y^2}/x$ and $F^*dv = \frac{∂v}{∂x} dx + \frac{∂v}{∂y} dy + \frac{∂v}{∂z} dz$. So it seems that you missed a term.
Then you plug you values for $(x,y,z)$ in. Concerning the tangent vector: If you write $v = a∂_x + b∂_y + c∂_z$ then by definition of the dual basis $dx(v) = a, dy(v) = b, dz(v) = c$.