How to generalize transversality condition? Consider the system
$$ \dot{x} = f(x,y)$$
$$\dot{y} = g(x,y)$$
where $x \in \mathbb{R}, y \in \mathbb{R}^2$, with the following properties:


*

*The y-hyperplane consists of equilibria, $f(0,y) = g(0,y) \equiv 0$. 

*The y-plane loses normal hyperbolicity at a point $y^*$, $\frac{\partial f}{\partial x} (0,y^*)=0$. 

*This loss of normal hyperbolicity is caused by the transverse eigenvalue crossing zero transversally, $\nabla_y \frac{\partial f}{\partial x} (0,y^*) \neq 0$. 


Now, I want to generalize these conditions to the case $x,y \in \mathbb{R}^N$. Conditions 1 and 2 are straightforward, but I'm having trouble figuring out condition 3 because now $\frac{\partial f}{\partial x} (0,y^*)$ is not a scalar but a matrix.
How could I generalize condition 3 for the $N$-dimensional case?  
 A: I came up with way to generalize the transversality condition by borrowing ideas from Theorem 3.4.1 in Guckenheimer and Holmes. Basically, the transversality condition is the same transversality condition required for a transcritical bifurcation ($\partial^2 f /\partial \mu \partial x (v)$ where $v$ is the right eigenvector of the zero eigenvalue of the Jacobian at $(0,y^*)$) with $ \mu = y$ as the bifurcation parameter. The problem here is that we have $x \in \mathbb{R}^N$ so the dimensions don't work. However, we can extend the condition to higher dimensions by projecting the 3-tensor $\partial^2 f /\partial \mu \partial x$ onto both the right $v$ and the left $w^T$ eigenvectors of the zero eigenvalue (see SN3 in Theorem 3.4.1):
$$w^T \frac{\partial^2 f}{\partial y \partial x} (v,v)$$
to make it easier to compute, we can rewrite the transversality condition as $w^T u$ where $u \in \mathbb{R}^N$ $N$ has entries
$$u_i = v^T \frac{\partial r_i^T}{\partial y} v,$$
where $\partial r_i^T/\partial x$ is an $N \times N$ matrix and $r_i$ is the $i$-th row of $\partial f /\partial x$ at the bifurcation point. 
