Timeline for Self-linkage of the orthogonal group $O_n({\mathbb R})$.
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Oct 15, 2010 at 12:14 | comment | added | Denis Serre | David, if $n$ is not among the integers $1+\binom{m+1}{2}$, we may at least choose $m$ such that the dimension of the level set $s_1(M)=\cdots=s_m(M)=1$ is smaller than $1+2\binom{n}{2}$. Then it makes sense to speak of linking. $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$ are linked if every sub-manifold $X$ of dimension $1+\binom{n}{2}$, with boundary $SO_n({\mathbb R})$, intersects $O_n^-({\mathbb R})$. Is there any problem with this definition ? | |
| Oct 14, 2010 at 14:23 | comment | added | David E Speyer | That also works. | |
| Oct 14, 2010 at 13:57 | comment | added | Denis Serre | At first glance, the question makes sense if $n=1+\binom{m+1}{2}$ (that is $n=2,4,7,11,16,...$) and $V_n$ is defined by $s_1(M)=\cdots=s_m(M)=1$. | |
| Oct 14, 2010 at 13:48 | comment | added | David E Speyer | But when $r=0$, we have $1-kr/(n-1)=1$. Doesn't that fix the question? | |
| Oct 14, 2010 at 13:43 | comment | added | Denis Serre | @David. Actually, the choice of smaller $s_2,\ldots$ does not work, because the corresponding set does not contain $O_n({\mathbb R})$. Orthogonal matrices have $s_1=\cdots=s_n=1$, and are characterized by this property. I'll think to how fix the question. | |
| Oct 14, 2010 at 13:36 | comment | added | Denis Serre | Thank you ! I agree with your calculation and your suggestion. After all, it was important to impose $s_1=1$, but for the choice of $s_2,\ldots,s_{n-1}$, I was lazy. I change my question accordingly. | |
| Oct 14, 2010 at 13:34 | history | edited | Denis Serre | CC BY-SA 2.5 |
$V_n$ instead of $M_n(R)$ in the first sentence
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| Oct 14, 2010 at 13:25 | history | edited | David E Speyer | CC BY-SA 2.5 |
deleted 82 characters in body
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| Oct 14, 2010 at 13:08 | history | edited | David E Speyer | CC BY-SA 2.5 |
edited body
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| Oct 14, 2010 at 12:48 | history | answered | David E Speyer | CC BY-SA 2.5 |