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Daniele Tampieri
Daniele Tampieri
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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 $S^3$$\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.

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 $S^3$ 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.

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.

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RBega2
RBega2
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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 $S^3$ 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.