I have read that the degree induced by a linear map on a torus $T^{n}$ is its determinant. How could one prove that result? Does this rule extend to any linear map on a space X?
Thanks
I have read that the degree induced by a linear map on a torus $T^{n}$ is its determinant. How could one prove that result? Does this rule extend to any linear map on a space X?
Thanks
In general, if $X$ is any smooth, oriented manifold with a volume form with finite total volume, then the degree of any smooth map $f\colon X\to X$ is given by the formula $ \deg f\;=\; \frac{1}{\text{vol}(X)}\int_X J(f) $ where $\text{vol}(X)$ is the total volume of $X$, and $J(f)$ denotes the Jacobian determinant of $f$, i.e. the unique real-valued function on $X$ satisfying $ f^*(\omega) = J(f)\, \omega $ where $\omega$ is the volume form. For a linear map, the Jacobian determinant $J(f)$ is the same as the determinant of the map. Note that this formula doesn't work if $X$ has infinite volume, in which case the degree might be different than the average value of the Jacobian.
More generally, if $X$ any $Y$ are oriented manifolds with finite volume, then the degree of a smooth map $f\colon X\to Y$ is given by the formula $ \deg f \;=\; \frac{1}{\text{vol}(Y)}\int_X J(f) $ where $\text{vol}(Y)$ denotes the total volume of $Y$.