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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

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    Intuitively: the determinant of a linear map $F : X \to X$ tells you the ratio between the volume of a set $U \subset X$ and its image $F(U)$. The idea is to lift the map $T^n \to T^n$ to the universal cover and look at what happens to a fundamental polyhedron. And no, the rule doesn't extend to all linear maps: a surjective linear map $\mathbb{R}^n \to \mathbb{R}^n$ can never have degree anything other than $1$.2011-05-30

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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 $Y$.