# Transversality condition

From my textbook:

"If we examine linearizations of $f_\mu (x)$, (where $\dot{x}=f_\mu(x) \ x\in \mathbb{R}^n, \mu\in \mathbb{R}^2$) at the equilibria $f_\mu(x)=0$, then we can formulate a transversality condition which guarantees, for example, that no linearization of $f$, has a zero eigenvalue of multiplicity greater than two and that any equilibrium which does have a zero eigenvalue of multiplicity two has a Jordan normal form with the block

$$\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$$"

What is this transversality condition? I understand that this condition should include the Jacobian of $f$, even though I do not know how to proceed. I have already asked a question kept on transversality: Transversal intersection.

Thank you very much

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The condition may have something to do with the parameter $\mu$. I cannot understand the role of this parameter from the given excerpt. Can you give a reference to the textbook? – user31373 Sep 22 '12 at 22:47
Yes. "Nonlinear oscillations dynamical systems and bifurcations of vector fields" Guckenheimer and Holmes. – Mark Sep 23 '12 at 6:36

First, the authors introduce the set $\mathcal M=\{(x,\mu)\in\mathbb R^{n+2}: f_\mu(x)=0\}$. "Using transversality, we expect that $\mathcal M$ is a smooth two-dimensional surface" near $(\mu_0,x_0)$, they say. I think that the appropriate transversality condition for this part is $$\mathrm{rank}\, \frac{\partial f_\mu}{\partial (x,\mu)} (\mu_0,x_0)=n\tag{1}$$ In words, the total derivative of $f_\mu$ with respect to both $x$ and $\mu$, represented by an $n\times (n+2)$ matrix, has maximal rank $n$, and therefore the Implicit Function theorem applies.
Then we consider the set $$\mathcal J=\left\{ \frac{\partial f_\mu}{\partial x} : (x,\mu)\in\mathcal M\right\}\tag{2}$$ of all Jacobians (derivatives with respect to $x$ only) evaluated at the points of $\mathcal M$. This is a parametric two-dimensional surface in the $n^2$-dimensional space of square matrices of size $n$.
Next, introduce a set $\mathcal B$ of bad square matrices. Specifically, $\mathcal B$ comprises of all matrices which have eigenvalue zero with multiplicity at least three, and also all matrices that have the block $\begin{pmatrix}0&0\\0&0\end{pmatrix}$ in their Jordan normal form. The authors leave it to the reader to verify that $\mathcal B$ has codimension greater than two. Informally, this is true because in order to qualify to be a member of $\mathcal B$, a matrix has to satisfy at least three different equations.
The idea of transversality now takes the following form: being two-dimensional, $\mathcal J$ will in general be disjoint from the set $\mathcal B$, because the latter set has codimension greater than two. I guess our new "transversality condition" is now $$\mathcal J\cap \mathcal B=\varnothing\tag{3}$$