Winding number of a linear transformation? I know that I am computing something incorrectly.
I am trying to compute the index of a positive determinant linear bijection.
The form I am using is $\omega = \frac{-y dx + x dy}{x^2 + y^2}$. I compute the pullback of this and get, where $T_x$, $T_y$ denote the $x$ and $y$th components of $T$, $T^* \omega = \frac{ (-T_y T_x + T_x T_y) dt}{T_x^2 + T_y^2} = 0$. For, $dT = T dt$, so $dx (dT) = T_x dt$.
This is obviously wrong, since the identity transformation winds the circle once around itself, for instance, but the integral of 0 is 0. (Since the index can be interpreted as the winding number of a $T \dot C$, for $C$ a positive parametrization of the unit circle in the domain.)
I would appreciate it if someone could point out my mistake, since I am very new to doing computations with differential forms, and I am finding it hard to find my mistake.
Thank you.
 A: Edited, since there was some misunderstanding about what you are after.
A map ${\bf f}:\ D\to D$ with ${\bf f}({\bf 0})={\bf 0}$ has a degree (called index, or winding number,  by some) which is given by the winding number of ${\bf f}(\partial D)$ around the origin:
$$n({\bf f}(\partial D),{\bf 0}):={1\over 2\pi}\int_{\partial D} \nabla{\arg}({\bf z})\cdot\>d{\bf z}\ .$$
When ${\bf f}$ is a linear transformation with positive determinant then ${\bf f}$ maps the unit circle $\partial D$ onto an ellipse that goes counterclockwise once around the origin. It follows without any tedious computation that the degree of such an ${\bf f}$ is $1$. 
For the amusement of the reader I leave my original answer here.
A linear transformation $T:\>{\mathbb R}^2\to{\mathbb R}^2$ with positive determinant induces a map $\dot T:\>S^1\to S^1$ via
$$\dot T(x):={Tx\over|x|}\qquad(x\in S^1)\ .$$
This map $\dot T$ has a rotation number $\rho_{\dot T}\in{\mathbb R}/{\mathbb Z}$ , sometimes also called winding number. This rotation number has nothing to to with the winding number of curves around the origin alluded to above.
In some cases the rotation number is easy to compute, in other cases it is the result of an intricate limiting process. Here are some examples: When $T$ has a real eigenvector, i.e., $\dot T$ has a fixed point, then $\rho_{\dot T}=0$. When $T$ is a euclidean rotation by the angle $\phi$ then $\rho_{\dot T}={\phi\over2\pi}$. When some power of $T$ has a real eigenvector then $\rho_{\dot T}$ is rational.
A good source about the rotation number is Coddington/Levinson: Theory of differential equations, pp. 406–408.
