Given a cylinder $M=S^1\times [0,1]$ with the counterclockwise orientation viewed from exterior as shown below:
I want to find orientation at $C_0$ and at $C_1$.
Note that at any $p=(x,y,0)\in C_0$, orientation in $T_p M$ is given by the ordered basis $B=\{(-y,x,0), e_3, (x,y,0)\}$. An outward pointing vector at $p$ is given by $-e_3=(0,0,-1)$. Let $\omega$ be a non vanishing form on $M$ that represents the basis $B$, i.e., $\omega (B)>0$.
Now, the boundary orientation is given by contraction of $\omega$ with outward pointing vector $-e_3$, i.e., $i_X\omega$:
$(i_X\omega)_p ((-y,x,0),(x,y,0))=\omega_p(-e_3, (-y,x,0),(x,y,0))=\omega_p(B)>0$
So the basis $(-y,x,0),(x,y,0)$ is positively oriented. But this means that $C_0$ is clockwise oriented when viewed from the top.
But an answer key shows that $C_0$ is anticlockwise oriented when viewed from the top. Please explain what went wrong in the above. Thanks.
I'm a little comfused with your notation 'orientation in $T_pM$ is given by the ordered basis $B=\{(-y,x,0),e_3,(x,y,0)\}$'. The tangent space $T_pM$ is of dimension 2 and with orientation $\{(-y,x,0),e_3=(0,0,1)\}$.