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What are strategies for generating the following types of pictures: enter image description here

Here's what's going on here. Take a toda flow in 3 variables. The equations of motion are:

$$\frac{d}{dt}a_k=2(b_k^2-b_{k-1}^2),$$ $$\frac{d}{dt}b_k=b_k(a_{k+1}-a_k),$$

where $a_i,b_i>0$, which can be represented as a commutator involving the matrix $L(a,b)$, which is a tridiagonal matrix with $a_i$ on the diagonal and $b_i$ on the superdiagonals. The hexagon above incorporates the following phenomenon of the Toda flow of $L(a,b)$:

enter image description here

where the edges and vertices are labeled by the structure of the $L$ matrix there. My question is, what are the possible ways of giving this hexagon a sense of distance (thus being able to draw Toda flow trajectories). Essentially this should somehow depend on the relative size of $a_i,b_i$'s, how close $a_i$ is to a particular eigenvalue, etc. For example, the $b_i$ are all exponentially decreasing, so it seems like one needs to look at ratios of $b_i$ to conclude which side of the hexagon we are closer to. This still seems very arbitrary and results in singularities at the boundary (which in higher dimensions become an issue since Toda flows perfectly well cross over such boundaries).

I know that the $n$ dimensional Toda flow is diffeomorphic to the $n-1$ sphere via looking at the vector $(u_{11},u_{12},\ldots,u_{1n})$ of normalized first-components of the corresponding eigenvectors of $L$. Is it as simple as projecting these onto the embedded permutahedron? Here's the picture for $n=4$,

enter image description here

What is a good coordinate system to draw these trajectories on the hexagon/permutahedron?

Reference: ODE's and the Symmetric Eigenvalue Problem

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  • $\begingroup$ The sets $b_j=0$ are of course invariant. This seems to contradict some of your remarks, but probably I am misinterpreting them. $\endgroup$
    – Christian Remling
    Commented Aug 22, 2014 at 20:13
  • $\begingroup$ @ChristianRemling: you're right but for higher dimensions you inevitably have to exit each face, and this is where the difficulty arises. See for example figure 7 of the above reference. $\endgroup$
    – Alex R.
    Commented Aug 22, 2014 at 20:52
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    $\begingroup$ It seems like you might be asking the following question: Let $X$ be the set of tridiagonal symmetric matrices with spectrum $(\lambda_1, \lambda_2, \lambda_3)$ and off diagonal elements nonnegative. Let $\delta: X \to \mathbb{R}^3$ take a tridiagonal matrix to its diagonal. Is this map a bijection onto the hexagon? The answer appears to be no! Suppose the spectrum is $(3, -1, -2)$. Then $(0,0,0)$ is in the hexagon. But a tridiagonal matrix with diagonal $(0,0,0)$ always has determinant $0$, so can't have spectrum $(3,-1,-2)$. $\endgroup$
    – David E Speyer
    Commented Aug 26, 2014 at 3:03
  • $\begingroup$ I am very puzzled trying to visualize what $\delta(X)$ does look like. Since $X$ is compact, $\delta(X)$ is closed, so there must be some open neighborhood of $(0,0,0)$ missing. And the paper you link says that $X$ is a toplogical disc. I don't have a contradiction, but this is confusing. $\endgroup$
    – David E Speyer
    Commented Aug 26, 2014 at 3:05
  • $\begingroup$ Working on it a bit more: If we ask for $\begin{pmatrix} a & x & 0 \\ x & -a-b & y \\ 0 & y & b \end{pmatrix}$ to have spectrum $(-3,1,2)$, we get $x = \sqrt{(2-a)(1-a)(3+a)/(b-a)}$ and $y=\sqrt{(2-b)(1-b)(3+b)/(a-b)}$. I find that the two quantities under the square root are positive only on two rectangles: $[-3,1] \times [1,2]$ and $[1,2] \times [-3,1]$. So $\delta(X)$ forms a "bow tie" in the hexagon. Moreover, there is a hole line segment of $X$ collapsed to the point $(1,1)$. So $\delta$ is not at all a homeomorphism. $\endgroup$
    – David E Speyer
    Commented Aug 26, 2014 at 3:44

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