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Let $S$ be a sphere centered at origin in $\Bbb R^{2n}$ of radius $\sqrt{2n}$. Let $D$ be a diagonal matrix. Let $U$ be an orthogonal matrix. Let $r\in\Bbb Z_+$ be a fixed integer.

Let vector $v\in\Bbb Z^{2n}$ on the sphere at origin which gives $Uv$ as another vector on the sphere at the origin and such that $\langle Uv,Uv\rangle=\langle v,v\rangle\in\Bbb Z$.

How do you characterize all the vectors $v\in\Bbb Z^{2n}$ on the sphere at origin such that $$\langle DUv,DUv\rangle=r\langle Uv,Uv\rangle=r\langle v,v\rangle\in r\Bbb Z$$ $$\langle\tilde{D}Uv,\tilde{D}Uv\rangle=\langle Uv,Uv\rangle=\langle v,v\rangle\in\Bbb Z\mbox{ with } \tilde{D}=\frac{1}{\sqrt{r}}D?$$

How do you find a vector among all such vectors such that $|v|$ is maximized ($|v|$ is sum of entries of vector $v$)?

A general procedure by relaxing $v$ to $\Bbb R^{2n}$ would be a good starting help.

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  • $\begingroup$ 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? $\endgroup$
    – Alex Degtyarev
    Commented Mar 3, 2014 at 10:35
  • $\begingroup$ Could you develop it as an answer? I believe the terms should be $(\delta_i^2-r)v_i^2$. $\endgroup$
    – Turbo
    Commented Mar 3, 2014 at 11:33
  • $\begingroup$ 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 $v_1,\ldots,v_n,\lambda,\mu$). $\endgroup$
    – Alex Degtyarev
    Commented Mar 3, 2014 at 15:31

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