In the context of Bell nonlocality, is the space of behaviours a hypersphere? I am currently reading Bell Nonlocality review paper by Brunner et al. (2014). On page 424,

A behaviour can be viewed as a point $\mathbf{p} \in \mathbb{R}^{\Delta^2 m^2}$ belonging to the probability space $\mathcal{P} \subset \mathbb{R}^{\Delta^2 m^2}$ defined by the positivity constraints $p(ab|xy) \geq 0$ and the normalization constraints $\sum_{a,b=1}^{\Delta} p(ab|xy) = 1$. Due to the normalization constraints $\mathcal{P}$ is a subspace of $\mathbb{R}^{\Delta^2 m^2}$ of dimension $\dim \mathcal{P} = (\Delta^2 - 1)m^2$.

Does that actually mean the probability space is a $(\Delta^2-1)m^2$-sphere?
 A: To clarify a little bit:

*

*In general, a (finite-dimensional) "probability space" is a set of positive reals $p_i\ge0$ such that $\sum_i p_i=1$. This is generally called a simplex, and you can imagine it as a section of a hyperplane passing through the basis vectors $\mathbf e_i$.


*It's worth noting that technically speaking the space of behaviour is not the same thing as the space of probability distributions, because some of the indices, here $x,y$, are not bound by normalisation condition like the others. The defining constraints are precisely as you wrote. Geometrically, each linear constraint $\sum_{a,b} p(ab|xy)=1$ defines a hyperplane. Imagine that, for every choice of $x,y$, you get a probability space on the remaining two variables $a,b$. You have one such constraint for each $x,y$, thus the space of behaviours is bound to be in the intersection of all the associated hyperplanes. There are $m^2$ such linear constraints, and without the constraint the space would be $\mathbb{R}^{\Delta^2 m^2}$, hence the dimension once the constraints are taken into account is $\Delta^2 m^2-m^2$.


*This is not a sphere. A (hyper)sphere is characterised by a quadratic constraint of the form $\sum_i x_i^2=1$.
