Ralph Cohen (professor at Stanford) is teaching a class on algebraic topology and moduli spaces this quarter, beginning by reviewing his perspective of Morse theory. He defined "nice" metrics, proved that they are dense in the $L^2$ space of metrics on $\mathbb R^n$, proved one result using that, and doesn't need them anymore. But given the previous assumptions, I and some classmates want to know: are not all metrics "nice"?

The setup: Assume a real-valued $C^\infty$ function $f$ on a closed smooth manifold $M$ of dimension $n$ is Morse: critical points are nondegenerate in the sense of having full-rank Hessians. Given some critical point $p$ of index $k$, find a small neighborhood $U$ and a diffeomorphism $U\cong\mathbb R^n$ such that $f$ becomes a function $\sum_{i=k+1}^nx_i^2-\sum_{i=1}^kx_i^2$.

A "nice" (smooth Riemannian) metric on $M$ is defined to be one that, when restricted to $U\cong\mathbb R^n$, gives $f$ a gradient field that after some diffeomorphism of $\mathbb R^n$ can be written as $\sum c_ix_i\partial_i$ for nonzero constants $c_i$. (Edited to add the missing $x_i$s.)

Can anyone make an illuminating example of a non-nice metric on $\mathbb R^n$? In fact (because $U$ is bounded) I might prefer one in the setting where $U$ is mapped an open ball rather than all of $\mathbb R^n$.

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    $\begingroup$ Dear Elizabeth S. Q. Goodman, just a silly question, probably I have not correctly undestood, but if $df(0)=0$ then shouldn't be $\nabla f(0)=0$, independently of the metric $g$, and so unequal to $\Sum c_i\partial_i|_0$ independently by the diffeomorphism? Excuse me if I am not correct. $\endgroup$
    – agt
    Oct 2, 2011 at 8:02
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    $\begingroup$ Probably it should be $\Sum c_i x_i\partial_i$ instead of $\Sum c_i\partial_i$. $\endgroup$
    – agt
    Oct 2, 2011 at 8:15

1 Answer 1


Assuming Giuseppe's suggested correction is right, here's what you have to worry about: Consider the function $f(x,y) = \tfrac12 x^2 + y^2 + x^2 y$ and the metric $g = (1+2y)\ dx^2 + dy^2$ on the half-plane $y > -\tfrac12$. You can compute that $$ \nabla f = x\ \frac{\partial\ }{\partial x} + (2y+x^2)\ \frac{\partial\ }{\partial y} $$ and this is one of those vector fields that cannot be linearized smoothly near $(x,y) = (0,0)$, i.e., it is not equivalent, under any smooth change of variables near this point, to the vector field $$ x\ \frac{\partial\ }{\partial x} + 2y\ \frac{\partial\ }{\partial y} $$ (There is a large literature about conditions that obstruct or allow a vector field to be linearized near a singular point.)

One way you can see this is that the flow lines of the first vector field are $x=0$ and those of the form $y = (c + \ln |x|)\ x^2$ where $c$ is a constant, while the flow lines of the second vector field are $x=0$ and those of the form $y = c\ x^2$. (Of course, $(x,y) = (0,0)$ is fixed by both vector fields.) Since the flow lines of $\nabla f$ aren't smooth curves, no smooth change of variables will convert them to smooth curves.

  • $\begingroup$ Nice answer! I believe I've now checked it. $\endgroup$ Oct 2, 2011 at 19:47
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    $\begingroup$ Thanks. I probably should have mentioned that it can be linearized by a $C^1$ change of coordinates: If $({\bar x}, {\bar y}) = (x, y - x^2\ \ln|x|)$, then the given vector field~$\nabla f$ is linear in the ${\bar x}{\bar y}$-coordinates. Also, I'm surprised that Prof. Cohen needed to make the argument you mention. Isn't it easier to choose flat metrics in 'Morse coordinates' near each critical point of $f$ and then use a partition of unity to find a metric on $M$ that agrees with these metrics near the critical points? $\endgroup$ Oct 2, 2011 at 20:52
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    $\begingroup$ I'm not sure, since he did something quite similar anyway in showing they were L^2 dense. But I have a guess: the point is to redo Morse theory with non-standard arguments that generalize better to infinite dimensions, so that one can do Floer theory for example. In that case I think it makes sense to keep out partitions of unity as much as possible, though I am unused to thinking about how to set up elliptic PDEs. $\endgroup$ Oct 3, 2011 at 3:46

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