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Show that observation of $z$ component alone of the Lorenz equations \begin{align} \dot x& = \sigma (y-x)\\ \dot y &= x (\rho -z) -y\\ \dot z &= xy-\beta z \end{align} does not lead to an embedding. Why does this not violating Takens' Theorem? What modification is required to generate an embedding?

Given the Lorenz equations, how come when I only observe the $z$ component alone i.e: ($\frac{\mathrm{d}z}{\mathrm{d}t} = xy - Bz$), this does not lead to an embedding. By Whitney’s embedding theorem, I need an embedding $Q$ to map from a finite $d$-dimensional attractor $A$ to $\mathbb R^k$ with $k \ge 2d+1$. In this case is $Q$ mapping from 2 dimensions to 1 dimension? Hence $k = 1$, $d = 2$ thus not satisfying $k \ge 2d+1$? How does this not violate Takens’ theorem, which states:

If $A$ is a finite-dimensional manifold and $q : A \to \mathbb R$ is differentiable then $v : A \to \mathbb R^k$ where $k > 2d$ is either an embedding or an arbitrarily small differentiable perturbation of $q$ and/or $\tau$ makes $v$ an embedding.

Arctic Char
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    How do these theorems construct/characterize the maps to the $k$-dimensional space? – Lutz Lehmann Mar 20 '20 at 10:55
  • How did you determine that the map to $[z(t),z(t-τ),z(t-2τ)]$ for instance does not reflect the original dynamic? – Lutz Lehmann Mar 20 '20 at 11:05
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    I fail to understand your problem. Please [edit] your question to clarify: What is your $Q$? Do you use a delay embedding at any point? In which case do you consider the conditions of which theorem to be fulfilled? Where exactly do you see a contradiction? (Note that the crucial difference between Takens’ and Whitney’s theorems is that Takens considers a delay map; except for that they are pretty analogous. Therefore I fail to see any incompatibility.) – Wrzlprmft Mar 20 '20 at 12:49
  • You should add that $\phi:\Bbb R^n \to \Bbb R^n$ is a map that leaves $A$ invariant (or even has $A$ as attractor), here it is the propagator of the ODE by time step $-\tau$, and that $v(x)=[q(x),q(\phi(x)),...,q(\phi^{k-1}(x))]$. Else the mention of the perturbation of $q$ or $\tau$ makes little sense. – Lutz Lehmann Mar 20 '20 at 16:15
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    Please cease vandalizing your question. – Gerry Myerson Mar 21 '20 at 08:56

1 Answers1

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You can easily check that if $(x(t), y(t), z(t))$ is a solution of the Lorenz system, then also $(-x(t), -y(t), z(t))$ is a solution. Just observing the $z$ component of a solution does not allow you to distinguish these two solutions. Essentially, the two wings of the attractor "butterfly" get mapped into one.

stereo plots of Takens embedding

Stereo pairs for the embedding $\small [u(t-\tau),u(t),u(t+\tau)]$ displayed via differences $\small[u(t),u(t+\tau)-u(t-\tau),u(t+\tau)-2u(t)+u(t-\tau)]$ for $\small u=x,z$ and $\small\tau=0.1$

From the values $$v([x(t),y(t),z(t)])=[z(t),z(t-τ),...,z(t-4τ)]$$ it is thus impossible to determine the sign of the $x$ and $y$ component, there will always be 2 solutions, $v$ is not injective, it is impossible to get an embedding, increasing the number $k$ of components of $v$ or changing $\tau$ will not change that.

In a complete formulation of the theorem you should also find this exclusion of symmetries of the dynamic. This symmetry gets broken if $q(x,y,z)=z$ is slightly perturbed, for instance adding linear terms in the other coordinates with small coefficients.

Lutz Lehmann
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