Tensor saturation for even spin groups

The explanation

Start with the simplest symmetry group in this story, SL(2). Its basic representations are labelled by whole numbers 0, 1, 2, and so on. Combining labels a and b gives the labels from |a − b| up to a + b, in steps of 2. Three representations a, b, c combine to leave something unchanged when c appears in a combined with b.

Try 1, 1, 1. Combining 1 and 1 gives 0 and 2, but not 1. So there is no invariant. Now stretch everything by 2, to get 2, 2, 2. Combining 2 and 2 gives 0, 2 and 4, which includes 2. Now there is an invariant. Stretching created one where none existed. The reason is parity: 1 + 1 + 1 is odd, while 2 + 2 + 2 is even.

So any honest version of the question needs a parity test. In general this test asks that the three labels add up to an element of the root latticeSource 1. The root lattice is the set of labels that the group's basic directions, its roots, can reach. Without it, the group's center acts and kills any invariantSource 1. For SL(2), the test is that a + b + c is even.

The saturation problem asks this. If stretched labels give a tensor invariant, do the original labels give one tooSource 2? A tensor invariant here is a vector in the combined representation that the whole group leaves fixed. The question links to the geometry of eigenvalues, through work of KlyachkoSource 3. Knutson and Tao answered it for type A, the groups SL(n), with their honeycomb theorem in 1999Source 4. Kapovich and Millson then conjectured the same, with the parity test, for all simply laced groupsSource 5.

Before this paper, type D was known only in small cases. Kapovich, Kumar and Millson did Spin(8)Source 6. Kiers did Spin(10) and Spin(12). Other work showed that a stretch by four always suffices for these groups. Those results do not give the factor-one statement under the parity testSource 7.

This paper claims the type D caseSource 8. For Spin(2n) with n at least 2, three labels that pass the parity test give an invariant whenever some stretch of them doesSource 9. The claim includes the half-integer spin labelsSource 10. For n = 2, Spin(4) is two copies of SL(2), so the opening example is a piece of it.

The proof turns the question into one about paths through the space of labels. Real-number paths exist by a geometric argument. The one obstacle is making them integral. The directions used may span the space without reaching every lattice pointSource 11. In type D the paper shows only one obstruction remains, a parity mismatch between two piecesSource 12. It moves the path to fix that mismatch, while keeping the path admissible throughoutSource 13. A separate construction then turns integral paths into actual invariantsSource 14.

How sure is this? No Lean formalization is listed for this paper. The claim rests on the paper's own long argument. No specialist review of the paper has been published.

What it doesn't show. This does not cover the exceptional simply laced groups E6, E7 and E8.