mass of the measure $F_{\sharp} \nu$ can be found inside the cones, whereas if $\sigma$ and $s$ are small, then the cones are more concentrated around orthogonal vectors. The bottom line is that the smaller ${\mathcal{S}}$ and ${\mathcal{C}}/ {\Theta}$ are, the more concentrated around orthogonal vectors the measure $F_{\sharp} \nu$ is. Note also that keeping all other parameters fixed, smaller choices of $\sigma$ will force a bigger choice of $\delta$. This reflects a trade-off between the total mass of the measure $F_{\sharp} \nu$ inside the cones, and the opening angle of the cones.

To see that the measure $F_{\sharp} \nu$ has an orthogonal cone structure under the assumption of a well-separated mixture model, we first study the geometry of a related measure $F^Q_{\sharp} \nu$ where $F^Q$ is the map $$F^Q: x \in {\mathcal{M}}\longmapsto  \left( \begin{matrix} \frac{q_1(x)}{\sqrt{w_1}} \\ \vdots \\ \frac{q_N(x)}{\sqrt{w_N}} \end{matrix} \right) \in {\mathbb{R}}^N,
  \label{FQ}$$ where we recall the functions $q_k$ are as in . Note that the functions $q_k$ are normalized in the sense that $\sum_{k=1}^N |q_k(x)|^2 = 1$ for all $x\in{\mathcal{M}}$, whereas the entries of $F^Q$ are normalized in $L^2(\rho)$ since $$\|F^Q_k\|_\rho^2=\int_{\mathcal{M}}\frac{\rho_k(x)}{\rho(x)}\,\rho(x)dx= 1 \qquad \forall \, k\in\{1,...,N\}\,.$$ We will show that $F^Q_{\sharp} \nu $ has an orthogonal cone 