Let's consider the points \((0,0,0), (1, 0, 0), (0, 1, 0), (0, 0, 1)\) in three dimensional space. These are the corners of a certain tetrahedron \(T\). The set of all pairwise sums we can make from this set also has a certain geometry, you can check that the vectors:
$$(0, 0, 0) + (0, 0, 0) = (0, 0, 0) \ \ \ \ (1, 0, 0) + (0, 0, 0) = (1, 0, 0)$$
$$(0, 0, 0) + (1, 0, 0) = (1, 0, 0) \ \ \ \ (1, 0, 0) + (1, 0, 0) = (2, 0, 0)$$
$$(0, 0, 0) + (0, 1, 0) = (0, 1, 0) \ \ \ \ (1, 0, 0) + (0, 1, 0) = (1, 1, 0)$$
$$(0, 0, 0) + (0, 0, 1) = (0, 0, 1) \ \ \ \ (1, 0, 0) + (0, 0, 1) = (1, 0, 1)$$
$$(0, 1, 0) + (0, 0, 0) = (0, 1, 0) \ \ \ \ (1, 0, 0) + (0, 0, 0) = (1, 0, 0)$$
$$(0, 1, 0) + (1, 0, 0) = (1, 1, 0) \ \ \ \ (0, 0, 1) + (1, 0, 0) = (1, 0, 1)$$
$$(0, 1, 0) + (0, 1, 0) = (0, 2, 0) \ \ \ \ (0, 0, 1) + (0, 1, 0) = (0, 1, 1)$$
$$(0, 1, 0) + (0, 0, 1) = (0, 1, 1) \ \ \ \ (0, 0, 1) + (0, 0, 1) = (0, 0, 2)$$
are exactly the integer points inside of the tetrahedron \(2T\) that we get from scaling the edges of \(T\) by \(2\). This actually always happens, if we let \(P_n\) be the set of points obtained as ``\(n\)-sums" of \((0,0,0), (1, 0, 0), (0, 1, 0)\), and \((0, 0, 1)\), then in fact \(P_n\) is exactly the set of integer points inside the tetrahedron \(nT\), and \(|P_n| = \binom{n+ 3}{n} = \frac{1}{3!}(n^3 + 6n^2 + 11n + 6)\). Here are some pictures of what's going on when \(n = 8\):
Generally, if we take a set of points \(S = \{p_1, \ldots, p_k\}\), the set of \(n\)-sums \(P_n(S)\) that we can make from \(S\) will always sit inside the \(n\)-th dilation of a geometric shape \(conv(S)\) called the convex hull of \(S\). This is an intriguing link between the algebra of vector addition and convex geometry.
Instead of combining points \((x, y, z)$\) and \((x', y', z')\) using addition what would happen if we used a different operation? Multiplication in the Heisenberg group looks a lot like addition, except with a slight difference in the third coordinate:
$$(x, y, z)\star(x',y',z') = (x + x', y + y', z + z' + xy').$$
If we consider the sets \(Q_n\) of all ``\(n\)-combinations" of the points \((0, 0, 0), (1, 0, 0), (0, 1, 0)\), and \((0, 0, 1)\) that we can get using the Heisenberg operation, we also see some interesting geometry. Below you can see a picture of the case \(n = 10\):
Last semester we showed that \(Q_n\) is precisely the set of points above the triangle with sidelengths \(n\) bounded by the remarkably simple quadratic function \(n - x - y + xy\). We also showed that, just as with \(|P_n|\), the number of points \(|Q_n|\) is given by a polynomial: \(\frac{1}{4!}(n^4 + 6n^3 + 23n^2 +42n + 24)\). These are the ``Heisenberg tetrahedra."
Now we'd like to try to prove things like this with other input sets, like the corners of square, or some other simple shape. In particular, we'd like to know if we always get quadratic boundaries, and what the counting functions look like.
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