Before asking my question, I put the necessary definitions and some context. If you are used with Morse Theory, you can skip the text within [[[...]]].

[[[Let me first define what I mean by gradient flow. Let $X$ be a smooth vector field on some orientable closed manifold $M$. For $\xi \in M$, let's consider the initial value problem for smooth curves $\eta:\mathbb{R}\to M$ given by

$$\eta'(t)=X(\eta(t)), \ \ \ \ \eta(0)=\xi.$$

Clearly, since $M$ is closed, the solution $\eta=\eta_\xi$ exists for all $t\in \mathbb{R}$. My diff geo professor called it the trajectory of flow line through $\xi$. The flow generated by $X$ is defined as the smooth map $((\phi: \mathbb{R}\times M\to M):(t,\xi)\mapsto \eta_\xi(t))$. Thus, for every $t\in \mathbb{R}$ this gives rise to a diffeomorphism $((\phi_t:M\to M):\xi \mapsto\phi_t(\xi))$, which my diff geo prof called time-$t$ map.

Let me, furthermore, put a bit more of context to my question (Morse Theory notation). Let Crit$_k(f)$ be the set of critical points of index $k$ of a Morse function $f$. Let then, $x\in$Crit$_{k+1}(f)$ and let $y\in$Crit$_k(f)$. Let

$$W^u(x):=\{a\in M | \lim_{t\to +\infty}\phi_t(a)=x\} \ \ \ and \ \ \ W^u(y):=\{b\in M | \lim_{t\to -\infty}\phi_t(b)=y\},$$

denote respectively the unstable manifold of $x$ and the stable manifold of $y$. Then, one might define the connecting manifold from x to y as

$$_xM_y:=W^u(x)\cap W^s(y).$$

If $\theta$ is a regular value of $f$ lying in $(f(x),f(b))$, we can construct the space of connecting orbits from $x$ to $y$ as

$$_x\tilde{M}_y=_x\tilde{M}_y|_\theta:=W^u(x)\cap W^s(y)\cap f^{-1}(\theta).$$

This set represents precisely the orbits of the negative gradient flow ($X=-\nabla f$) running from x to y, because every orbit intersects the level hypersurface exactly once. Note that for the critical points we choose, $\dim(_x\tilde{M}_y)=0$ so that $_x\tilde{M}_y$ corresponds to a set of points.]]]

Now, let $u\in \ _x\tilde{M}_y$. Let's identify $_x M_y^u$ with the connected of component of $_x M_y$ containing $u$. Note that $\dim(_x M_y^u)=1$.

Finally, here's my question. The author of the paper claims that there's is an orientation for $_x M_y^u$ provided by the flow -- i.e. there's something in the relation


that provides a way to get an orientation on that space.

I just do not see why this is true. Is there some theorem on gradient inducing orientation on the corresponding flow I am missing? Please help.

Thank you!

  • $\begingroup$ I think I got it with a dimensionality argument. $\endgroup$ Jul 19, 2018 at 20:48

1 Answer 1


You will need to make some choices. Let us choose orientations for each unstable manifold $W^u(x)$. This is possible, because the $W^u(x)$ are contractible (hence orientable). This choice of orientation of the unstable manifolds provides an coorientation of all the stable manifolds, i.e. provides an orientation of the normal bundle of the stable manifold.

Now ${}_xM_y:=W^u(x)\cap W^s(y)$ is a transverse intersection of an oriented and a co-oriented manifold, hence is oriented. It is a nice exercise to check that this is the case, and that the intersection of oriented submanifolds need not be oriented!

Now if the index of $x$ and $y$ differs by one, ${}_xM_y$ is a one dimensional oriented manifold. On this space there is a non-vanishing vector field: the gradient flow. A non-vanishing vector field on a one dimensional manifold also provides an orientation. Thus we can compare both orientations and get $+$ or $-$ signs.

  • $\begingroup$ That’s good and was pretty much the same argument I developed yesterday. I’m just not sure about the « a non vanishing vector field on 1 dim manifold also provides an orientation » part. Do you have any hints or links to the proof? Also, is a non vanishing vf on two dimensional manifold providing an orientation? Say that the index of x and y differs by 2 instead. I’m just curious! :-) $\endgroup$ Jul 20, 2018 at 15:47
  • $\begingroup$ This might be because the space of one forms is isomorphic to the vector space of smooth vector fields on the same subspace of $\mathbb{R}^3$. $\endgroup$ Jul 20, 2018 at 15:54
  • $\begingroup$ @HolomorphicGuy: What is your notion of orientation? Maybe a non-vanishing volume form? On a one-dimensional manifold a non-vanishing vector field give rise to such: Just take a form that maps the vector field to 1 pointwise. This will be a non-vanishing volume form $\endgroup$
    – Thomas Rot
    Jul 23, 2018 at 8:20

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