Timeline for The geometry of the solution set of a symmetric equation in four symmetric matrices
Current License: CC BY-SA 3.0
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| Sep 25, 2017 at 18:06 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 24, 2017 at 15:34 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 24, 2017 at 12:20 | comment | added | David E Speyer | The above suggestion does not quite work as written: When $n=1$, we have the solution $w=x=y=z=0$, $a+b+c+d=0$ which has dimension $3>1^2+1$. And we can direct sum this solution with other solutions. Still, I am hopeful that there is some insight to be had by looking at this deformation. | |
| Aug 23, 2017 at 21:35 | comment | added | David E Speyer | Surprisingly, the equations $AW=0$ with $A$ and $W$ symmetric still have a $\binom{n+1}{2}$ dimensional space of solutions -- it breaks up into $n+1$ components according to the ranks of $A$ and $W$. It would be enough to show that, for each of the $(n+1)^4$ components of $AW=BX=CY=DZ=0$, the intersection with the linear space $A+B+C+D=W+X+Y+Z=0$ has dimension $n^2+n$. | |
| Aug 23, 2017 at 21:35 | comment | added | David E Speyer | @WillSawin Here is a strategy for proving this is a complete intersection: Rewrite the equations using $8$ symmetric matrices: $W+X+Y+Z=A+B+C+D=0$, $AW=BX=CY=DZ=1$; this gives us a closed subvariety of $4n^2+4n$ dimensional space. Taking the limit as we "zoom out" turns these equations into $W+X+Y+Z=A+B+C+D=0$, $AW=BX=CY=DZ=0$. (continued) | |
| Aug 23, 2017 at 20:38 | comment | added | Will Sawin | Interesting! I think I can show this gives all the two by two quadruples with fixed determinant. I also agree with your suspicion that this extends to higher dimension. | |
| Aug 23, 2017 at 20:05 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 23, 2017 at 13:48 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 23, 2017 at 2:03 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 23, 2017 at 2:03 | comment | added | David E Speyer | Oh, grr. I'm wrong. $XY^{-1}ZX^{-1}$ does not equal $ZY^{-1}$. I'll delete that part shortly. | |
| Aug 23, 2017 at 2:01 | comment | added | MTyson | Where does that last equality come from? | |
| Aug 23, 2017 at 1:30 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 23, 2017 at 0:42 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Aug 23, 2017 at 0:22 | history | answered | David E Speyer | CC BY-SA 3.0 |