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Let $A(x)$ be a symmetric negative semi-definite matrix which depends continuously on the parameter $x\in\mathbb{R}^{d}$. We consider the differential equation $$\dot{x} = (I-xx^*)A(x)x$$ on the unit sphere. Is there anything known about the convergence behaviour of the trajectories for $t\to\infty$?

If the matrix is constant it converges to a stable equilibrium.

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  • $\begingroup$ This question is very interesting for me. Can I ask you how to prove that, if $A$ is constant, then $x(t)$ converges to a stable equilibrium? I guess that such equilibrium is related to the dominant eigenvector of $A$. $\endgroup$ – Giuseppe Negro Jul 9 '20 at 13:07
  • $\begingroup$ I asked a related question on Math.SE. $\endgroup$ – Giuseppe Negro Jul 10 '20 at 11:14
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I think you can get any flow on the unit sphere in this way. Take $A(x)$ to be the rank-one matrix $$A(x) = -(x-b)(x^*-b^*)$$ with $b = b(x)$ tangent to the sphere (that is, orthogonal to $x$). Then $x^*b=b^*x=0$ and $x^*x = 1$, so that $$\begin{aligned}(I - xx^*)A(x)x & = xx^*(x-b)(x^*-b^*)x-(x-b)(x^*-b^*)x \\ & = x - (x-b) = b .\end{aligned}$$ Therefore, the differential equation under discussion takes form $$\dot{x} = b(x).$$

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  • $\begingroup$ This is some how similar to the following fact(in the case of 2-sphere): If we identify $S^2$ with the space of projectiions of $M_2(\mathbb{C})$ then every vector field on $S^2$ can be written in the form $\dot X= [A,X]=AX-XA$ for some antisymmetric 2x2 matrix A. I learned this fact in page 13 of the book Non commutative Geometry by Alain Connes(When he was interpreting the consequence of the Gauss Bonnet theorem in terms of cyclic cocycles) $\endgroup$ – Ali Taghavi Oct 4 '17 at 12:08

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