Worked solution: Dehn's invariant: the tetrahedron and cube are not scissors-congruent (1900)

For a polyhedron $P$, let $\ell(e)$ and $\theta(e)$ denote the length and dihedral angle of each edge $e$. Dehn defines $\operatorname{Dehn}(P) = \sum_e \ell(e) \otimes [\theta(e)]$, an element of the tensor product $\mathbb{R} \otimes_{\mathbb{Q}} (\mathbb{R}/\pi\mathbb{Q})$ (lengths paired $\mathbb{Q}$-bilinearly with angles taken modulo rational multiples of $\pi$). Working modulo $\pi\mathbb{Q}$ is essential: it is what lets a rational multiple of $\pi$ contribute nothing, matching the fact that angles summing to a straight or full angle should not obstruct a dissection.

The tensor product $\otimes_{\mathbb{Q}}$ is what makes this a genuine, checkable invariant rather than an unwieldy pair of numbers: it declares $\ell \otimes [\theta_1 + \theta_2] = \ell \otimes [\theta_1] + \ell \otimes [\theta_2]$ and $(q\ell) \otimes [\theta] = \ell \otimes [q\theta]$ for rational $q$, so that lengths interact with angles exactly like a bilinear pairing, and two elements of $\mathbb{R} \otimes_{\mathbb{Q}} (\mathbb{R}/\pi\mathbb{Q})$ can be compared for equality using ordinary linear algebra over $\mathbb{Q}$.

The next step shows why this particular combination of length and angle, added over all edges, deserves to be called an invariant at all: it must not change when a polyhedron is cut into pieces and reassembled.