# why does contour lines intersect at saddle points?

Consider functions of two variables. A saddle point of $$f(x,y)$$ is a point in the domain of $$f$$ (or on the graph of $$f$$ by an abuse of language) where the gradient is $$\bf 0$$, but which is not a local extremum of the function. A classic example is $$(0,0)$$ for $$f(x,y)=y^2-x^2$$. However, there are two types of saddle points.

Type I: the graph looks like a saddle.

Type II: the graph does not look like a saddle. e.g. $$f(x,y)=x^3$$, and $$f(x,y)=x^2+y^3$$.

I have two questions:

Question 1: How to rigorously define saddle points of the above two types?

Question 2: A folklore theorem says $$P$$ is a saddle point of type I if and only if the contour lines (aka level curves) intersect at $$P$$. I don't find a proof of such a result, and I'm not sure if the theorem holds as "if and only if" or just holds in one direction.

• One natural classification would be to declare a saddle point $p$ of $f$ to be of type I if it is a nondegenerate critical point, that is, $\det \operatorname{Jac}(f)(p) \neq 0$ and type II if it is degenerate, but that definition would make both directions of the folklore theorem false and would exclude examples like $g(x, y) = x^4 - y^4$. As C. Falcon mentions, what you might be looking for is a special case of Morse's Theorem. Mar 10 at 23:35

A reasonable definition for saddle points of type I is the following:

Let $$f\colon\mathbf{R}^2\to\mathbf{R}$$ be a smooth function, a saddle point of type I of $$f$$ is a point $$p\in\mathbf{R}^2$$ such that $$\frac{\partial f}{\partial x}(p)=0=\frac{\partial f}{\partial y}(p)$$ and $$\begin{bmatrix}\frac{\partial^2 f}{\partial x^2}(p) & \frac{\partial^2 f}{\partial x\partial y2}(p)\\\frac{\partial^2 f}{\partial y\partial x}(p) & \frac{\partial^2 f}{\partial y^2}(p)\end{bmatrix}$$ has nonpositive determinant.

Regarding the saddle points of type II, you can always say that the previous matrix has zero determinant (it cannot have nonnegative determinant, otherwise $$p$$ would be a maximum or a minimum). However, from the differential topology view point, this is not such a great definition, since the level sets around such critical points can be very different looking.

Using a standard result called Morse's lemma, around a saddle point of type I, there exist local coordinates such that $$f(x,y)=x^2-y^2$$, in this local model, it is clear that the level curves $$f^{-1}(0)$$ intersects at $$(0,0)$$. The reverse implication is false, a counter example is given by $$f(x,y)=x^3-3xy^2$$ for which $$(0,0)$$ is a saddle point of type II, even though $$f^{1}(0)$$ consists of three transversally intersecting curves.

• In the case of $0$ determinant, it needn’t be a saddle point at all. You really need higher order Taylor info. Mar 10 at 21:45
• @TedShifrin I agree, this was my point saying that it is not a great definition, but perhaps I was unclear (I wrote type I instead of type II). I found the OP definition for saddle points way too wide. Mar 10 at 22:29
• @TravisWillse You are right for both points, thank you! I forgot the $y$-axis as part of the $0$-level set of the monkey saddle, it was late here. However, it still provides a counterexamples to the OP's statement. Mar 10 at 23:01
• I'm still somewhat confused by the last paragraph: Since $f(x, y) = x^3 - 3 x y^2$ has zero Jacobian at the origin (it is a polynomial only with terms of degree $\geq 3$), so $\det \operatorname{Jac}(f)(0, 0) = 0$ and hence that saddle point has (in the sense of the definition you proposed) type I. Mar 10 at 23:28
• @TravisWillse It does not have type I, since the Hessian at the origin is degenerate, this was precisely my point. If the level sets intersect transversally, then the critical point not necessarily has type I. Nonetheless, I wrote it has "type I" instead of "type II" in my answer as I can't get my head around this terminology. Mar 11 at 1:19