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Mar 4, 2014 at 23:57 history edited Turbo
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Mar 3, 2014 at 15:31 comment added Alex Degtyarev Of course, it's $(\delta_i^2-r)$. I don't know how to handle the integral problem. Over $\mathbb{R}$, you need the points in $X$ where the two gradients span a plane containing $U([1,\ldots,1])$: $$\lambda[v_1,\ldots,v_n]+\mu[(\delta_1^2-r)v_1,\ldots,(\delta_n^2-r)v_n]=U([1,\dots,1])$$ for some $\lambda,\mu\in\mathbb{R}$, but I do not see an easy way to solve this system of a lot of quadratic equations (in $v_1,\ldots,v_n,\lambda,\mu$).
Mar 3, 2014 at 15:25 history edited Turbo CC BY-SA 3.0
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Mar 3, 2014 at 11:55 history edited Turbo CC BY-SA 3.0
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Mar 3, 2014 at 11:37 history edited Turbo CC BY-SA 3.0
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Mar 3, 2014 at 11:33 comment added Turbo Could you develop it as an answer? I believe the terms should be $(\delta_i^2-r)v_i^2$.
Mar 3, 2014 at 10:35 comment added Alex Degtyarev First, disregard $U$; you get the quadric cone $\sum_i(\delta_i-r)v_i^2=0$, where $\delta_i$ are the eigenvalues of $D$. Intersect it with the sphere to get a codimension $2$ (typically) variety, say $X$. You want $U^{-1}(X)$. Then, use Lagrange multipliers?
Mar 3, 2014 at 10:22 history asked Turbo CC BY-SA 3.0