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This is a follow up to my earlier question "Understanding Higher Dimensional Solution Set". The answer to that question determined that the solution set of \begin{align} x_1^2 + x_2^2 + x_3^2 + x_4^2 = 1\\ x_1 x_2 + x_3 x_4 = 0 \end{align} is given by the product of two circles (in a transformed coordinate system), and is therefore path connected. This was determined by transforming into a coordinate system that eliminated the cross-terms.

However, I am also interested in understanding the solution set of the following generalized set of polynomial equations: \begin{align} x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 1\\ x_1 x_2 + x_2 x_3 + x_4 x_5 + x_5 x_6 = 0 \\ x_1 x_3 + x_4 x_6 = 0 \end{align} Note that this reduces to the first set of equations above when $x_5=x_2 = 0$. Unfortunately, the transformation used to eliminate the cross terms in the first set of equations doesn't seem to nicely generalize to the second set of equations.

How we can characterize the solution set of these equations? In particular, I'm interested as to whether the set is path connected.

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  • $\begingroup$ the System given is non-linear! $\endgroup$ – Dr. Sonnhard Graubner Nov 9 '17 at 20:17
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We have the equations \begin{eqnarray*} x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 = 1\\ x_1 x_2 + x_2 x_3 + x_4 x_5 + x_5 x_6 = 0 \\ x_1 x_3 + x_4 x_6 = 0 \end{eqnarray*} multipy the second and third equations by $2$ and add them to the first equation gives $(x_1+x_2+x_3)^2+(x_4+x_5+x_6)^2=1$ which can be parameterised to \begin{eqnarray*} x_1+x_2+x_3=\cos \theta \\ x_4+x_5+x_6 = \sin \theta . \\ \end{eqnarray*} Now substitute these into the second equation and we have \begin{eqnarray*} x_2 (\cos \theta-x_2)+ x_5( \sin \theta-x_5)=0 , \\ \end{eqnarray*} multiply this equation by $-4$ and complete the squares, \begin{eqnarray*} (2x_2 -\cos \theta)^2+ (2x_5- \sin \theta)^2=1 , \\ \end{eqnarray*} which can be parameterised to give \begin{eqnarray*} 2x_2=\cos \theta +\cos \phi \\ 2(x_1+x_3)=\cos \theta -\cos \phi \\ 2x_5 = \sin \theta+\sin \phi \\ 2(x_4+x_6) = \sin \theta-\sin \phi . \\ \end{eqnarray*} Now substitute these into the third equation and we have \begin{eqnarray*} x_1 \left(\frac{\cos \theta-\cos \phi}{2}-x_1\right)+ x_4\left(\frac{ \sin \theta-\sin \phi}{2}-x_4\right)=0 , \\ \end{eqnarray*} multiply this equation by $-16$ and complete the squares, \begin{eqnarray*} (4x_1 -(\cos \theta-\cos \phi))^2+ (4 x_4-( \sin \theta-\sin \phi)^2=(\cos \theta-\cos \phi))^2+ ( \sin \theta-\sin \phi)^2 =2(1- \cos(\theta- \phi)). \\ \end{eqnarray*} Thus the manifold can be fully parameterised by \begin{eqnarray*} 4x_1 &=& \cos \theta -\cos \phi+\sqrt{2(1- \cos(\theta- \phi)} \cos \psi \\ 4x_3 &=& \cos \theta -\cos \phi+\sqrt{2(1- \cos(\theta- \phi)} \cos \psi \\ 4x_4 &=& \sin \theta -\sin \phi+\sqrt{2(1- \cos(\theta- \phi)} \sin \psi \\ 4x_6 &=& \sin \theta -\sin \phi+\sqrt{2(1- \cos(\theta- \phi)} \sin \psi \\ 2x_2 &=& \cos \theta +\cos \phi \\ 2x_5 &=& \sin \theta+\sin \phi. \\ \end{eqnarray*} None of these equations have any discontiunities and so to obtain a path from $(\theta, \phi, \psi)$ to $(\theta', \phi', \psi')$ the simple parameterisation will do $(t\theta +(1-t)\theta', t\phi+(1-t)\phi', t\psi +(1-t) \psi')$ for $t \in [0,1]$. So the space is path connected, As far characterisation is concerned it is $3-$dimensional apart from when $ \theta= \phi$ when the space is $1-$dimensional(forming a unit circle in the $x_2-x_5$ plane.)

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  • $\begingroup$ This parametrization suggests that $x_1 = x_3$ and $x_4 = x_6$ when the equations are satisfied. I've found solutions, however, with $x_1 \neq x_3$. Could you clarify how you derived your parametrizations for $x_3$ and $x_6$? In particular, I think the sign in front of the roots should be different. $\endgroup$ – David Egolf Nov 10 '17 at 17:19
  • $\begingroup$ With this correction, the parametrization provided seems to work, and so I've accepted the answer. $\endgroup$ – David Egolf Nov 10 '17 at 17:48

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