Does there exist a nonconstant smooth map $f:\mathbb{R}^4\setminus\{0\}\to L\left(\mathbb{R}^4,\mathbb{R}^5\right)$ such that each $f(x):\mathbb{R}^4\to\mathbb{R}^5$ is an isometric linear map and $x\in\bigcap_{i=1}^4\ker\frac{\partial f}{\partial x_i}(x)$ for all $x\in\mathbb{R}^4\setminus\{0\}$?

  • $\begingroup$ I'm so used to people writing $f(x)$ to refer to a function rather than its values that it took me a few re-readings to parse that the question is meaningful exactly as written. $\endgroup$ – LSpice 2 days ago
  • $\begingroup$ Can you give an example that works if 5 is replaced by 8? $\endgroup$ – Matt F. 2 days ago
  • $\begingroup$ @MattF. Yes. Take $f:\mathbb{R}^4-\{0\}\to L\left(\mathbb{R}^4,\mathbb{R}^8\right)$ such that the matrix $\left(F_{jk}(x)\right)$ of the linear map $f(x):\mathbb{R}^4\to\mathbb{R}^8=\mathbb{C}^4$ with respect to the canonical bases of $\mathbb{R}^4$ and $\mathbb{C}^4$ is given by $$F_{jk}(x)=\left(\frac{x_k}{\|x\|}\left(\frac{1}{2}-i\frac{x_j}{\|x\|}\right)+i\delta_{jk}\right)e^{2i\frac{x_j}{\|x\|}}.$$ $\endgroup$ – Submanifolds in Space Forms 2 days ago
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    $\begingroup$ @RobertBryant : It turns out that such a map $f$ gives rise to an isometric immersion of $\mathbb{S}^{n-1}$ into $\mathbb{S}^{n+m-1}$, which is totally geodesic if and only if $f$ is constant. Since every isometric immersion of $\mathbb{S}^3$ into $\mathbb{S}^4$ is totally geodesic, the answer for my question is no. But I was wondering if perhaps there might be an analytical proof of this that could offer some insights into the open problem of whether there exists a nontotally geodesic isometric immersion of $\mathbb{S}^n$ into $\mathbb{S}^{2n}$ for $n\geq3$. $\endgroup$ – Submanifolds in Space Forms 2 days ago
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    $\begingroup$ @SubmanifoldsinSpaceForms: I see; I suspected as much. I think that, in general, it's not good practice to pose questions on MO for which you know the answer without at least revealing that you know the answer and explaining why the argument you have is not satisfactory for your purposes. $\endgroup$ – Robert Bryant 2 days ago

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