Timeline for Higher order generalization of Cauchy-Schwarz?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2020 at 23:20 | comment | added | Robert Bryant | I guess another way to express what Terry is saying is that the identity $$ (v,v)(w,w) = (v,w)^2 + |v\wedge w|^2 = (v,w)^2+(v{\wedge}w,v{\wedge}w)$$ (with the natural inner product on $\Lambda^2(V)$) already shows that $(v,v)(w,w) - (v,w)^2$ is a sum of squares anyway, so the natural quartic polynomial would be $$P(v_1,\ldots,v_n) = \sum_{1\le i < j\le n} |v_i\wedge v_j|^2.$$ In particular, it is expressed as a sum of squares. | |
| Aug 1, 2020 at 3:24 | comment | added | Terry Tao | If one removes the square, then this (now quartic) polynomial also has a geometric interpretation: it is the trace of the exterior square $\bigwedge^2(T^* T)$ of the square $T^* T$ of the linear transformation $T: {\bf R}^n \to V$ that maps the standard basis to $v_1,\dots,v_n$. (It is also the square of the Frobenius norm of $\bigwedge^2 T$.) | |
| Jul 31, 2020 at 21:28 | comment | added | Robert Bryant | @Malkoun: Yes, you are absolutely correct. I didn't think of that, but, yes, simply the sum of the terms would be enough. Of course, this implies that this construction would work for any ordered field. | |
| Jul 31, 2020 at 20:28 | comment | added | Malkoun | Come to think about it, are the squares necessary? We already know that the expressions inside parentheses are nonnegative. | |
| Jul 31, 2020 at 20:09 | vote | accept | Malkoun | ||
| Jul 31, 2020 at 20:08 | history | answered | Robert Bryant | CC BY-SA 4.0 |