Timeline for Toda Flow Embeddings
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| S Sep 10, 2014 at 23:46 | history | bounty ended | CommunityBot | ||
| S Sep 10, 2014 at 23:46 | history | notice removed | CommunityBot | ||
| Sep 7, 2014 at 21:36 | comment | added | Lucas Seco | I think that, geometrically, this hexahedron is the image of the moment map from the flag manifold $\mathcal{F}$ of complete flags of 3-space to the traceless 3x3 matrices, where the Toda flow comes from the action of the diagonal matrix A in $\mathcal{F}$ (actually, action of exp(tA) or looking at A as induced vector field). The tridiagonal matrices form an invariant manifold and it seems that the moment map restricted to them is a diffeomorphism: maybe one can get more concrete coordinates from this. I can try to find some references for this if you think it is a direction worth pursuing. | |
| S Sep 2, 2014 at 21:38 | history | bounty started | Alex R. | ||
| S Sep 2, 2014 at 21:38 | history | notice added | Alex R. | Canonical answer required | |
| Sep 2, 2014 at 21:36 | comment | added | Alex R. | @DavidSpeyer: I am indeed interested in the coordinate system on the hexagon. I've added this as emphasis to the question. | |
| Sep 2, 2014 at 21:36 | history | edited | Alex R. | CC BY-SA 3.0 |
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| Aug 26, 2014 at 15:10 | comment | added | Christian Remling | @DavidSpeyer: Actually, what I wrote is not completely correct. The parametrization by the $w_j$'s works if $b_k>0$, and then $w_j>0$. The degenerate cases (when $e_1$ is not cyclic) need to be looked at separately (I don't think it can be difficult). | |
| Aug 26, 2014 at 11:46 | comment | added | David E Speyer | @ChristianRemling Well, the obvious reason to draw it as a hexagon is that there are six special points on the boundary (the diagonal matrices). Indeed, I'm curious: Where are these six points in the $w_i$ coordinates? I think the original poster's question (which I have not answered) is what coordinate system on the hexagon is being used to draw flow diagrams like the first figure. | |
| Aug 26, 2014 at 4:14 | comment | added | Christian Remling | Elaborating on this, it is pretty obvious (given some background in this kind of inverse spectral theory, that is) that this map identifies $X$ with a "hexagon" (why not call it a disk, though??). A Jacobi matrix with prescribed spectrum is determined by also choosing three weights in the spectral measure that satisfy $w_1+w_2+w_3=1$, $w_j\ge 0$. | |
| Aug 26, 2014 at 4:14 | comment | added | David E Speyer | @ChristianRemling If I understood the paper correctly, that makes a map to a spherical triangle (and three of the edges of the hexagon get collapsed to points). I think Alex R's question was how to naturally get hexagonal coordinates on $X$. The map $\delta$ seemed the obvious approach, so it was interesting to me that it didn't work. I might have misunderstood the question, though. | |
| Aug 26, 2014 at 4:11 | comment | added | Christian Remling | @DavidSpeyer: Your $\delta$ is not the map studied in the linked paper. Rather, $J\in X$ gets mapped to the first components of the eigenvectors (it's really better though to view this as the vector $\rho(\{\lambda_j\})$, where $\rho$ is the "natural" spectral measure). | |
| Aug 26, 2014 at 4:01 | comment | added | David E Speyer | Assuming that the $a_i$ are positive doesn't help; you can always add a large multiple of the identity. So tridiagonal matrices with spectrum $(7,11,12)$ would have diagonal missing a neighborhood of $(10,10,10)$, just as tridiagonal matrices with spectrum $(-3,1,2)$ missed a neighborhood of $(0,0,0)$. | |
| Aug 26, 2014 at 3:53 | comment | added | Alex R. | @DavidSpeyer: I believe one assumes that $a_i,b_i>0$ and so the Toda flow then preserves positivity. I've added this detail, thank you! I've also added the 4-d toda flow picture. | |
| Aug 26, 2014 at 3:48 | history | edited | Alex R. | CC BY-SA 3.0 |
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| Aug 26, 2014 at 3:44 | comment | added | David E Speyer | 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. | |
| Aug 26, 2014 at 3:05 | comment | added | David E Speyer | 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. | |
| Aug 26, 2014 at 3:03 | comment | added | David E Speyer | 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 $(3,-1,-2)$. | |
| Aug 22, 2014 at 20:52 | comment | added | Alex R. | @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. | |
| Aug 22, 2014 at 20:13 | comment | added | Christian Remling | The sets $b_j=0$ are of course invariant. This seems to contradict some of your remarks, but probably I am misinterpreting them. | |
| Aug 22, 2014 at 17:27 | history | edited | Alex R. | CC BY-SA 3.0 |
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| Aug 22, 2014 at 17:22 | history | asked | Alex R. | CC BY-SA 3.0 |