Timeline for On the determination of a quadratic form from its isotropy group
Current License: CC BY-SA 3.0
5 events
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Feb 8, 2012 at 5:21 | comment | added | zapkm | @Ralph, You are right,... I cnt see reason for $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ to be independent of $v$. | |
Feb 8, 2012 at 5:19 | history | edited | zapkm | CC BY-SA 3.0 |
added 98 characters in body; added 23 characters in body
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Feb 7, 2012 at 13:42 | comment | added | Ralph | Why is $\frac{F(g^{-1}(v))}{G(g^{-1}(v))}$ independent from $v$, i.e. constant ? | |
Feb 7, 2012 at 12:29 | history | edited | KConrad | CC BY-SA 3.0 |
added 2 characters in body
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Feb 7, 2012 at 4:20 | history | answered | zapkm | CC BY-SA 3.0 |