@YCor Infinite-dimensional unitary representations on a Hilbert space.

@ToddTrimble In defence of the opening sentence, in noncommutative spaces, where the algebra comes first, people did think "connected spaces" (C$^*$-algebras whose only projections are $0$ and $1$) were pathological, and Kaplansky doubted that a simple C$^*$-algebra with no nontrivial projections could exist (they do).

I think the question is more or less answered by this question: mathoverflow.net/questions/38863/re-seating-a-monad The finite distribution monad (what you call $H$) is a finitary monad, and the category of groupoids is locally finitely presentable, so we can, using the Lawvere theory of $H$, re-seat it onto the category of groupoids in the manner described in the linked question.

@darijgrinberg These spaces are repeatedly rediscovered under different names, as is the characterization of those that embed in vector spaces by a cancellation property. See the first part of this: arxiv.org/pdf/0903.5522.pdf

I saw your answer while typing essentially the same thing -- the norm of the $2 \times 2$ Hadamard matrix is 2 as a bilinear form on $\ell^\infty(2) \times \ell^\infty(2)$ (and hence in the dual space of $\ell^\infty(2) \hat{\otimes} \ell^\infty(2)$) but 4 as an element of $\ell^\infty(2 \times 2)^*$.

@TomLaGatta The tensor product $\sigma$-algebra is equal to the Baire $\sigma$-algebra. In general, on a compact Hausdorff space, Baire probability measures extend to Radon probability measures on the Borel sets (i.e. inner regular with respect to compact sets) and Radon probability measures on the Borel sets are equal iff they agree on Baire sets.

@benblumsmith You are right. The probabilities $\frac{1}{2}$ and $\frac{1}{4}$ should be the other way round, as they are in John's linked blog post.

@val72 Every topological vector space over the complex numbers is a topological vector space over the reals, and this doubles the dimension, so preserves finiteness of the dimension. For every finite-dimensional topological vector space over the reals, we can pick a finite basis, and the rational linear combinations of that basis form a countable dense subset (by continuity of addition and multiplication by reals).

The other thing to mention is that the octonion 16-loop is not just any kind of loop, but a Moufang loop (and the Ježek-Kepka loops are not Moufang). Moufang loops have a Lagrange theorem, so the argument about the maximal size of the nucleus (the associativity analogue of the centre) might work. I have no idea about the second part of the argument.

lies on the line and four that don't, so the probability of associativity given distinct elements different from 1 is $\frac{1}{5}$. All together, the answer is $\frac{169}{512} + \frac{343}{512} \cdot \left(\frac{19}{49} + \frac{30}{49} \cdot \frac{1}{5}\right) = \frac{43}{64}$.

For what it's worth, the probability is $\frac{43}{64}$. Proof: We can restrict to dealing with positive signs in a triple (a,b,c) because the signs don't affect whether it associates or not. The probability of having at least one 1 is $1 - \left(\frac{7}{8}\right)^3 = \frac{169}{512}$. The probability, given that there's no 1, that at least two elements of the triple are the same is $1 - 1 \cdot \frac{6}{7} \cdot \frac{5}{7} = \frac{19}{49}$. If all three elements are different, then they associate iff they lie on a line in the Fano plane. Given the first two points, there is one point that..

I think you are looking for Bratteli diagrams. A nice textbook is Davidson's C$^*$-algebras by example.