A paradox based on Simons cones

Question

Let $\mathbf{C}_S \subset \mathbf{R}^{2n}$ be a Simons cone, where the dimension is large enough that it is area-minimizing: $n \geq 4$. Let $T$ be a leaf of the Hardt–Simon foliation with $\operatorname{dist}(T,0) = 1$.

For all $R > 1$, $T$ intersects $\partial B_R$ transversely. Thus $T \cap \partial B_R$ is a smoothly embedded submanifold of $\partial B_R$; rescale this homothetically to $\Sigma_R \subset \partial B_1$.

• When $R$ is close to $1$, $\Sigma_R$ is diffeomorphic to the boundary of an $2n-1$-dimensional disk, so $\Sigma_R \simeq \mathbf{S}^{2n-2}$.
• As $R \to \infty$, $\frac{1}{R} T \to \mathbf{C}_S$, so when $R$ is large enough then $\Sigma_R \simeq \mathbf{S}^{n-1} \times \mathbf{S}^{n-1}$.

Question. The paradox is that the family $(\Sigma_R)$ gives an isotopy from $\mathbf{S}^{2n-2}$ to $\mathbf{S}^{n-1} \times \mathbf{S}^{n-1}$, where we let $R$ vary across $[1+\frac{1}{N},N]$ for example. What am I missing here?

Edit. As pointed out in the answers below, the first point is incorrect. I mixed up two arguments I thought up: the first working with balls centered around a point of the surface—this guarantees the validity of both bullet points—, and the second working with balls centered at the origin—guaranteeing the transversality. Presumably in the former approach the transversality would be false for some radii, which would allow the topology to change.

Answer 1

The issue here is that (I suspect) your intuition is failing you as your first bullet point is incorrect.

A good first visualization is the Clifford torus in $\mathbf{S}^3$ all the parallel surfaces are tori, until a critical distance is reached and one gets two circles on each side.

Another, way to think about the Simons' cone is to reduce by the symmetry group. After take the appropriate quotient one is left with the (closed) upper quadrant. Distance to the origin is the same in either picture. Any point away from the axes corresponds to a torus in the original picture, but the points on the axis are collapsed and are either collapsed version so the tori (i.e. like the circles above). Obviously at the origin everything is collapsed.

In the simple picture, the Simons' cones are just the line $y=x$. However the curves in the foliation lie on one side of this curve and look (qualitatively) like $y=x^2/(x+1)$ so meet the axis perpendicularly. The point is that the intersection with the sphere of radius 1 is just a point (in the reduced space) i.e. a "circle" in the original picture.