Timeline for A combinatorial question about orthonormal bases
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27, 2015 at 1:09 | comment | added | Will Sawin | @DouglasZare For any such function, pullback by $O(n)$ gives another such function. So the space of such functions is a representation of $O(n)$, and because continuous functions decompose into spherical harmonics, this will decompose into different irreducible representations of $O(n)$. So you just have to test the different spherical harmonics. | |
| Apr 26, 2015 at 23:15 | comment | added | Douglas Zare | With the axiom of choice, there are lots of discontinuous automorphisms of $\mathbb{R}$. However, it might be feasible to classify the continuous functions $S^{n-1} \to \mathbb{R}$ satisfying the condition. Are they all in the closure of the linear combinations of constants and Cranch's functions? | |
| Apr 26, 2015 at 22:51 | comment | added | Francesco Polizzi | What do you mean by "generate"? The abelian group $A$ is not fixed. For instance, if $f \colon S^{n-1} \to \mathbb{R}$ is any function satisfying your property, then we can construct another function $g \colon S^{n-1} \to S^1$ by setting $g(x) = e^{if(x)}$, where $S^1 \subset \mathbb{C}$ is seen as a subgroup of the abelian multiplicative group $\mathbb{C}^*$. | |
| Apr 26, 2015 at 21:19 | comment | added | Tom Goodwillie | That's what I want to know now. | |
| Apr 26, 2015 at 21:13 | comment | added | Douglas Zare | Do these and constants generate all such functions? | |
| Apr 26, 2015 at 21:12 | comment | added | James Cranch | Oh, I did! I'll also be passing it on to others. | |
| Apr 26, 2015 at 21:09 | comment | added | Tom Goodwillie | James, I hope you enjoyed it. It's such an attractive and edifying answer, too. | |
| Apr 26, 2015 at 21:05 | vote | accept | Tom Goodwillie | ||
| Apr 26, 2015 at 21:03 | comment | added | James Cranch | I sometimes wish MathOverflow was the kind of forum where one could just say "Great question! Ask your best undergraduates: they'll enjoy telling you how to do it more than I would." | |
| Apr 26, 2015 at 20:44 | history | answered | James Cranch | CC BY-SA 3.0 |