@TedShifrin sure, but once I’ve got the rotation symmetry along one axis (which you were right to call me on) then the permutation symmetry generates rotations around the other axes
one thing i notice that i think might have a chance at possibly having a possible use if im correct in this case is that the minimum of two functions $f$ and $g$ is greater than or equal to the sum of their minima (provided they exist)
If it is approximately its tangent cone at $(1,1,1)$, that cone has two nappes, so the surface will as well. But you're throwing one of them away by staying inside the cube?
I am puzzled by what is going on at the other three points when we look at the tangent cone at $(1,1,1)$. There must be congruent cones at the other points, too.
I'm trying to show that $PSL(2,\Bbb F_3) \cong A_4$ geometrically. My idea is: showing that $A_4$ is the tetrahedral group is not difficult. $(1,1,1), (1,−1,−1), (−1,1,−1), (−1,−1,1)$ is a tetrahedron centered at the origin, thus we may represent $A_4$ by matrices with coefficients in $SL_3(\Bbb Z) \cap O(3)$, thus we may reduce the coefficients modulo $3$ and the reduced matrices still preserve the quadratic form $x^2+y^2+z^2$
Don't know why I've been looking at the quasi-projective variety: is the set of all $[x:y:z] \in \Bbb F_3 \Bbb P^2$ such that $x^2+y^2+z^2=0$ a projective curve?
Consider the solid pinched torus $S: (\sqrt{x^2+y^2}-1)^2 + z^2 \le 1^2$. Its boundary is a pinched torus which can be foliated by horizontal circles. Now, I glue each circle to one point, so that the boundary becomes homeomorphic to $\Bbb S^1$. Do I get $\Bbb S^3$?
@Meow: The circle is tangent to the angle at $P$ and the other circle at $Q$. But the line that is tangent to both circles meets the angle at $A$ and $B$, respectively.
There are hessians that are indefinite, and it corresponds to undulation points which are not extrema. For example this $x^4+x$ math.stackexchange.com/questions/1093323/…
If I parametrize $S(\alpha) : (\sqrt{x^2+y^2}-1)^2+z^2=\alpha^2$ with $x=(1+\cos\omega)\cos\theta$ and $y=(1+\cos\omega)\sin\theta$ and $z=\sin\omega$, then $\omega$ is the poloidal direction, $\alpha$ is the radial direction, and $\theta$ is the toroidal direction. am i right?
If I parametrize $S(\alpha) : (\sqrt{x^2+y^2}-1)^2+z^2=\alpha^2$ with $x=(1+\alpha\cos\omega)\cos\theta$ and $y=(1+\alpha\cos\omega)\sin\theta$ and $z=\alpha\sin\omega$, then $\omega$ is the poloidal direction, $\alpha$ is the radial direction, and $\theta$ is the toroidal direction.
