Lorenz Equations, embedding and Takens’ theorem

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.

Solution 1:

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 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.