# Perturbation of Morse function

The following question is probably classic in Morse theory, so a reference to an existing result should be sufficient. I don't know much about Morse theory and I am dealing with the following situation. I have a compact manifold $X$ and I have a Morse function $f$ on it with a saddle point at $x_0$. I don't like the properties of $f$(for some "mysterious" reason) so I take a parametrization around $x_0$, taking $x_0$ to the origin and containing no additional critical points, and add to $f$ a function of the type $\epsilon \rho Q$, where $Q$ is a quadratic polynomial, $\epsilon$ is a small number and $\rho$ is a bump function living in the coordinate patch that is identically equal to 1 in a neighborhood of the origin.

Clearly this new function still has a critical point at $x_0$. My question is: for small enough $\epsilon$ is this new function still Morse, with the same critical points as the original function $f$?

-

## 4 Answers

Sure. For any sequence $\epsilon_n\to 0$ a sequence of critical points of $f_n:=f+\epsilon_n \rho Q$ would have a subsequence which converges to a critical point of $f$, so in particular would eventually lie in a neighborhood on which the function equals either the original $f$ (if the limit isn't $x_0$) or $f+\epsilon_n Q$ (if the limit is $x_0$). But in such neighborhoods the only critical points are the critical points of the original function $f$, at least if $\epsilon_n$ is small enough.

If the answer to your question was "no" then you could find $\epsilon_n\to 0$ contradicting the above.

-
Many thanks for your time. – The Common Crane Jul 1 '12 at 16:52
@Mike: The critical point might shift when you perturbed the function. Just think about quadratic functions of one variable. – Misha Jul 1 '12 at 16:55
@Misha: My interpretation of the question is that we are using a coordinate system centered at $x_0$, and the unique critical point of $Q$ is the center of the coordinate system. So, as noted in the original question, $x_0$ will continue to be a critical point of the perturbation. (It goes without saying that the critical points would shift under generic perturbations, but the question asked about a perturbation of a rather particular kind.) – Mike Usher Jul 1 '12 at 17:11
"But in such neighborhoods the only critical points are the critical points of the original function f, at least if $\epsilon_n$ is small enough". Why is this obvious? – The Common Crane Jul 1 '12 at 17:25
Working in a (suitably small) coordinate system with $x_0$ identified with the origin and using the standard metric on this coordinate system, the nondegeneracy of the Hessian of $f$ at $x_0$ gives you an estimate $|(\nabla f)(v)|\geq C_1|v|$ for $v$ in some fixed neighborhood of the origin. Meanwhile there's also an estimate $|(\nabla Q)(v)|\leq C_2|v|$. So $|\nabla(f+\epsilon Q)(v)|\geq (C_1-\epsilon C_2)|v|$, showing that, in this fixed neighborhood and for $\epsilon$ sufficiently small, the only critical point of $f+\epsilon Q$ is the origin. – Mike Usher Jul 1 '12 at 17:55

There is classic result (I think going back to H. Whitney) that states that if $f\to \mathbb{R}$ is a Morse funtion on a compact smooth manifold $M$, thent there exist a neighborhood $\mathscr{N}$ of $f$ in $C^\infty(M)$ sucht that if $g\in \mathscr{N}$, then $g$ is equivalent to $f$, i.e., one can obtain $g$ from $f$ by conjugating $f$ with a diffeomorphism $\Phi$ of $M$ and a diffeomorphim $\varphi$ of $\mathbb{R}$. More precisely, this means that

$$g = \varphi \circ f \circ \Phi.$$

For a proof see the book of Golubitski and Guillemin Stability of smooth mappings. Loosely speaking this result says that if $g$ is not too far from $f$, then $g$ is obtained from $f$ via a change of "coordinates" on $M$ and a change in the target space $\mathbb{R}$.

-
intersting. Thanks! – The Common Crane Jul 7 '12 at 5:48

Lemma B on page 12 of Milnor's Lectures on h-cobordism theorem. I am sure it is also in his book on Morse functions, I just do not have it with me now.

-
thanks. A reference is just what I wanted. – The Common Crane Jul 1 '12 at 16:52
I think I declared victory too soon :(. How does it follow that for small $\epsilon$ the new function will have the same critical points as the original $f$. – The Common Crane Jul 1 '12 at 17:23

Are you sure you cannot achieve what you want by simply changing coordinates? Answers are about adding general small function but you want to add a quadratic polynomial only on a neighborhood of $x_0$.

-
No. I am dealing with quantities that are coordinate independent! – The Common Crane Jul 1 '12 at 17:12
You add a "quadratic polynomial Q". This only make sense in a choosen local coordinates. – Maciej Starostka Jul 1 '12 at 21:34