Flips and blowups induce semiorthogonal decompositions (conjecturally in the case of flips), not derived equivalences. The true birational invariant should be the "Griffiths component" of the derived category, see e.g. Kuznetsov's notes arxiv.org/abs/1509.09115v1, which is to my knowledge still not satisfactorily defined.

@DavidBen-Zvi To my knowledge, the only place where this is written is in Raskin's "W-algebras and Whittaker categories", arxiv.org/abs/1611.04937v1, where it is briefly mentioned in Example 1.23.1. Probably why it wasn't stated earlier is because the results of that paper are needed to make sure the statement is actually reasonable?

@user94041 That's a good question - that reminds me that 1. I forgot to mention the multiplication on R is by convolution 2. I should probably be considering the Weil group in the definition of $S$. I think with those two fixes the $\mathbb{G}_m$ case should just be a Fourier transform?

Call it $X$. By the Lefschetz hyperplane theorem $X$ will have Picard rank $1$ and $4$th Betti number equal to $2$ (assuming large degree in a large enough Grassmannian.) Under that assumption, X will be its own canonical model. Therefore $X$ can't be birational to a product of two varieties (or, taking canonical models of both, $X$ would need to have Picard rank at least $2$.) And it can't be birational to one of your specified varieties - the only general type (of large dimension) possibility is a hypersurface in projective space, also its own canonical model, but with $H^4$ rank $1$.

Does the following argument work? By Larsen-Lunts, $K_0(Var)/[\mathbb{A}^1]$ is the free abelian group on stable birational classes. It thus suffices to show that there is some variety which is not stably birational to a variety in class $P$. In fact, as the MRC base of any variety in $P$ is in $P$ (up to birational equivalence), it suffices to show that there is a general type variety not birational to a variety in P. To do this, consider a high degree hypersurface in a Grassmannian. (cont.)

Here is my understanding of the story: if you are doing 4-dimensional GR, then indeed the relevant manifold is not a Calabi-Yau but an Einstein Lorentzian manifold (I might not have the terminology quite right.) However if you are doing some form of 10d supergravity/string theory the 6d manifold you are compactifying on gets forced to be a Calabi-Yau manifold for some reason. I really have no understanding of the physics so take this with a large grain of salt.

What is the rational bounded derived category (as opposed to the usual one?)

There is an argument that such a threefold cannot exist in section 3 of "Uniformization of Fake Projective Four Spaces", by Sai-Kee Yeung.

I think this actually should be much easier than the other problem, just because $\frac{\pi^2}{6}-1<\frac{3}{4},$ which gives you a chance of reducing this to a finite computation. In particular if there is an $n$ so that you can pack squares of side lengths $1/2\cdots, 1/(n-1)$ and $n$ right triangles of side length $2/n$, then you can also pack the squares in the question. This is because you can pack squares of side length $1,1/2,\cdots$ in a right triangle of side length $2$.

If I understand your question correctly, the answer is yes (apply a translation functor to reduce the case of general integral antidominant $\lambda$ to any individual such weight.)

@js21: Not so! Every irreducible component of $S$ is a copy of $\operatorname{Spec}{\mathbb{Z}}$, but they may be joined at some finite primes. As I see it, this is the essential source of difficulty for the elementary solutions to this problem.

Out of curiosity, would you place Hrushovski's proof of function-field Mordell-Lang in this category? Personally, I find these transfer principle results (I'm thinking of e.g., Ax-Kochen here) to be "reasonable", in the sense that while the central idea is certainly surprising, once one thinks about it one sees why the idea should work. On the other hand, though I can't claim to fully understand the proof, I really find it hard to explain why Hrushovski's line of argument should work.