In a sense, calculus is all about the study of critical points of functions on flat space $\mathbb{R}^N$ (e.g. here). Let's try a different venue, the unit tangent bundle of the sphere. $$ T^1(S^2) = \big\{ (x,\vec{v}): x \in S^2,\, \vec{v} \in T_x(S^2),\, \big|\big|\,\vec{v}\,\big|\big|=1 \big\} $$

And there are other example of circle bundles over the sphere, such as tensor products of this bundle with itself. I was able to obtain on Math.SE a rather general result: cohomology of circle bundles can be found via the Gysin exact sequence. Can obtain this result via Morse theory?

  • $H^0\big( T^1(S^2) \big) \simeq \mathbb{Z} $
  • $H^1\big( T^1(S^2) \big) \simeq 0 $
  • $H^2\big( T^1(S^2) \big) \simeq \mathbb{Z}/2\mathbb{Z} $
  • $H^3\big( T^1(S^2) \big) \simeq \mathbb{Z} $

I post this question here because I know explicit Morse functions always exist but can be difficult to find. And I'm trying to work out the critical points and visualize the Morse flow.

And perhaps I should clarify what I mean by "explicit". I believe the unit tangent bundle could be given the structure of a variety. Looking at our construction, we could try to embed

$$ T^1(S^2) \subseteq \mathbb{R}^3 \times \mathbb{R}^3$$

I don't think his is quite the "universal bundle" construction, but it's something. And we could write down the constraints:

  • $x_1^2 + x_2^2 + x_3^2 = 1$
  • $v_1^2 + v_2^2 + v_3^2 = 1$
  • $\vec{v} \in T_x(S^2) \subseteq \mathbb{R}^3$ which could be a hyperplane in 3-space.
  • $\langle x, v \rangle = x_1 v_1 + x_2 v_2 + x_3 v_3 = 0$ this shows that $\vec{x} \perp \vec{v} $ and that $\vec{v}$ is tangent to $S^2$.

And if we pin down all the relations we obtain the structure of an algebraic variety. Therefore, could it be possible to write down polynomial morse functions of this kind of space? And work out the critical points?

Even a derivation of the Gysian homomorphism via Morse theory could be interesting. Certainly I've never seen it.

We could try to build Morse functions of out of polynomials in $x$ and $v$. Does the set Morse functions of given degree form a vector space?

  • $\begingroup$ Is not this just $S^3$? $\endgroup$ – მამუკა ჯიბლაძე Oct 18 '17 at 18:36
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    $\begingroup$ @მამუკაჯიბლაძე $\mathbb{R}P^3$ ? $\endgroup$ – john mangual Oct 18 '17 at 18:45
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    $\begingroup$ You are right, it is $SO(3)$ $\endgroup$ – მამუკა ჯიბლაძე Oct 18 '17 at 19:11
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    $\begingroup$ Isn't $x_1+2v_2$ a Morse function? I haven't done the computation, but I'd say it has exactly $4$ critical points, namely $(\pm1,0,0;0,\pm1,0)$, of index $0,1,2,3$ $\endgroup$ – Pietro Majer Oct 18 '17 at 19:55
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    $\begingroup$ Most linear functions will be Morse. Given that the unit tangent bundle is quadratic, this is a very reasonable Morse-theory setup. $\endgroup$ – Ryan Budney Oct 18 '17 at 21:35

To expand the comment: the function $f(x,v):=x_1+2v_2$ is a Morse function on $M:=T^1 S^2$, in your coordinates $(x,v)$. Denoting $(e_j)_{1\le j\le3}$ the standard basis of $\mathbb{R}^3$, it is easy to see that the only critical points $ p:=( x,v)\in M$ of $f$ are $p_0:=(-e_1,-e_2)$, $p_1:=(e_1,-e_2)$, $p_2:=(-e_1,e_2)$, and $p_3:=(e_1,e_2)$, and that $\text{ind}(p_k)=k$. This is more or less evident from geometrical considerations. To make a formal computation, define for $q:=(q_1,q_2,q_3)$ in a nbd of $0\in\mathbb{R}^3$ the skew simmetric matrix $$Q=Q_q:=\left[ \begin {array}{ccc} 0&-q_{{3}}&q_{{2}}\\ q_{ {3}}&0&-q_{{1}}\\ -q_{{2}}&q_{{1}}&0\end {array} \right], $$ and consider a local chart at $p$ of the form $(q_1,q_2,q_3)\mapsto (e^Qx,e^Qv)$. In this chart the function $f$ reads $\tilde f(q)=[e^Qx]_1+2[e^Qv]_2$; since $e^Q=I+Q+Q^2/2+o(Q^2)$ at $Q=0$, by easy computations this gives the second order expansion at $q=(0,0,0)$; precisely $$\nabla \tilde f(0)= (-2v_3 ,\ x_3 ,\ 2v_1-x_2) $$ $$\text{Hess }\tilde f(0):=\ \left[ \begin {array}{ccc} -4\,v_{{2}}&2\,v_{{1}}+x_{{2}}&x_{{3}} \\ 2\,v_{{1}}+x_{{2}}&-2\,x_{{1}}&2\,v_{{3}} \\ x_{{3}}&2\,v_{{3}}&-4\,v_{{2}}-2\,x_{{1}} \end {array} \right] \ .$$ So if $\nabla\tilde f(0)=0$ then $x_3=v_3=0$ and $x_2=2v_1$; since $x\cdot v=0$ also $|v_1|=|x_2|$, so that $v_1=x_2=0$ and since $\|x\|=\|v\|=1$ we also get $x_1=\pm1$, and $v_2=\pm1$, that is $p$ is one of $p_0,\dots, p_3$. For each of these values $\text{Hess }\tilde f(0)$ is a diagonal matrix with respectively $0,\dots,3$ negative elements, ending the computation.

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  • $\begingroup$ This looks nice. I didn't pick a complicated space either $T^1 S^2 \simeq SO(3)$ is very natural. There are many natural situations where $f: M \to \mathbb{R}$ won't be quite Morse. Or we may not get to choose $f$ at all. $\endgroup$ – john mangual Oct 20 '17 at 19:02

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