Consider the Lorenz System
$${\displaystyle {\begin{aligned}{\frac {\mathrm {d} x}{\mathrm {d} t}}&=\sigma (y-x),\\[6pt]{\frac {\mathrm {d} y}{\mathrm {d} t}}&=x(\rho -z)-y,\\[6pt]{\frac {\mathrm {d} z}{\mathrm {d} t}}&=xy-\beta z.\end{aligned}}} {\displaystyle }$$
and its jacobi matrix
$$\boldsymbol{J} = \begin{bmatrix} -\sigma & \sigma & 0\\ \rho & -1 & 0\\ 0 & 0 & -b \end{bmatrix}$$
In Paper, Eq. 6 they claim:
Since the trace of the Jacobian is constant it follows that:
$$Tr(\boldsymbol{J})= - \sigma -1 - b = \sum_{i=1}^{3}\lambda_i$$
So the sum of the lyapunov exponents is equal to the trace of the jacobian.
Does someone know how this relation can be derived?
$\newcommand{\tr}{\operatorname{tr}}$ This is true for any matrix, as $$ \tr(A)=\tr(Q^{-1}AQ) $$ and if you take the transformation to Jordan normal form, you get the claimed identity. Also $$ \tr(A^k)=\tr((Q^{-1}AQ)^k)=\sum_{j=1}^n(λ_j)^k. $$
For your specific problem, you have a differential equation $\dot u=f(u)$ and the partial derivatives $U(t)=\frac{\partial x(t)}{\partial x_0}$ wrt. the initial point. These satisfy the differential equation $\dot U(t)=J(x(t))U(t)$. Now $$ t\sum_{j=1}^n λ_j = \ln(|\det(U(t))|) $$ by definition of the process. We know that the derivative of the determinant is $$ \frac{d}{dt}\det(U(t)) =\det(U(t))\tr(U(t)^{-1}\dot U(t)) =\det(U(t))\tr(\dot U(t)U(t)^{-1}) \\ =\det(U(t))\tr(J(x(t)) $$ Now if $\tr(J(x(t))$ is a constant, then $$ \det(U(t))=\det(U(0))e^{t\tr(J(x_0))} \implies \tr(J(x_0))=\sum_{j=1}^n λ_j, $$ using $U(0)=I\implies \ln|\det(U(0))|=0$, which proves the claim.