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I have two terminology questions regarding the method of steepest descent:

  1. This section of the Wikipedia page defines

A non-degenerate saddle point, $z^0 ∈ \mathbb{C}^n$, of a holomorphic function $S(z)$ is a point where the function reaches an extremum (i.e., $∇S(z^0) = 0$) and has a non-vanishing determinant of the Hessian (i.e., $\det S_{zz}″(z^0) ≠ 0$).

I understand the "non-degenerate" part regarding the invertibility of the Hessian matrix. But why is a point satisfying $∇S(z^0) = 0$ called a "saddle point" instead of a "critical point", as is standard in other contexts? This terminology conflicts with the usual definition of a "saddle point", which is basically a special case of a critical point which is not a local extremum (although the notion of an "extremum" is somewhat ambiguous for complex-valued functions).

  1. Why is it called the method of "steepest descent"? What is descending steeply? It seems to me that the whole point of the method is to deform the contour to a stationary point, where the integrand is not descending (or ascending) at all.
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  1. $∇S(z^0) = 0$ automatically implies a saddle point for holomorphic functions because of the maximum modulus principle if $z^0$ is not a zero of $S$. Consider any open set $U$ containing $z^0$, then $S(z^0)$ cannot be a minimum or a maximum of $Z$ in $U$ according to that theorem. If $z^0$ is indeed a zero of $S$, then thanks to the fact the hessian has a non vanishing determinant at this point implies that $z^0$ is still a saddle point.

  2. That point is more complex and would require a long answer. You can find some information with this post .

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  • $\begingroup$ Your point 1 is a bit misleading as currently phrased. The maximum modulus principle only rules out local maxima for the modulus $|f(z)|$ itself - you can certainly have local minima, around isolated zeros of $f$. However, similar reasoning can be used to rule out both local minima and maxima for the separate function $|e^{f(z)}|$. Only the latter quantity is restricted to saddle points. $\endgroup$
    – tparker
    Jul 4 '18 at 20:18
  • $\begingroup$ I went a little fast about my explanations indeed. Edited. $\endgroup$
    – nicomezi
    Jul 5 '18 at 6:27
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I realized that my confusion with both pieces of terminology stemmed from the same root. For simplicity, consider the single-variable case of a contour integral of the form $$ I(\lambda) := \int_\gamma e^{\lambda f(z)}\ \mathrm{d}z$$ with $f(z)$ analytic and $\lambda$ real and large.

I was thinking of the "primary" quantity as the exponent $f(z)$, an analytic function $f: \mathbb{C} \to \mathbb{C}$. From this perspective the terminology is very confusing, because the critical point $z_0$ satisfying $f'(z_0) = 0$ is not a saddle point of $f(z)$ in any sense, nor is $f(z)$ descending steeply near it - as I said in the question, it is actually stationary there, which is in some sense the opposite of "steeply descending".

But whoever came up with the terminology was thinking of the modulus of the integrand $$M({\bf x}) = \left| e^{\lambda f(z({\bf x}))} \right| = e^{\lambda\, \mathrm{Re}[f(z({\bf x}))]},$$ as the "primary" quantity, and considering $M$ as a function $M: \mathbb{R}^2 \to \mathbb{R}^{\geq 0}$, using the natural isomorphism $\mathbb{R}^2 \leftrightarrow \mathbb{C}$ given by $z({\bf x}) = x + i y$. This composed function has saddle points at the the critical points $z_0$ of $f(z)$, so $z_0$ is not a saddle point of $f$ but rather of $M$. And the quantity that is "descending steeply" along the optimal stationary-phase contour is not $f$ (which, again, is actually stationary there), but rather $M$. Indeed, I believe that for a non-degenerate saddle point the optimal contour actually points along the "direction" given by $f''(z_0)$.

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