Things to do with 120 dodecahedra

2011-12-06

Well it has been more than a year since I posted to this blog, which makes it not much of a blog. But I have big plans to get back to it, and to start the ball rolling: ...

I gave a math club talk a few weeks ago about the 120-cell, a 4-dimensional polytope with 120 dodecahedral facets. Here is the video taken by Eddie Beck of the UGA math club. Enjoy:  

Trivial homotopies between Morse functions

2010-12-01

This entry relates to a paper Katrin Wehreim, Chris Woodward and I are writing, and is also relevant to my forthcoming paper with Rob Kirby. I want to prove the following statement, which one would naturally assume is true but which is actually subtle, so I'm going to be very pedantic in the proof:

Theorem If $g_t \co X \to [0,1]$ is a $1$-parameter family of Morse functions on a cobordism $X$, with $g_t$ honestly Morse for all $t \in [0,1]$, with all critical values distinct, then there exist ambient isotopies $\phi_t \co I \to I$ and $\psi_t \co X \to X$ (fixed on boundaries and equal to the identity when $t=0$) such that $g_t = \phi_t \circ f_0 \circ \psi_t$.

Basically, the idea is that if you have a homotopy between Morse functions which "doesn't do anything", in the sense that there are no births or deaths of cancelling pairs of critical points and no crossings of critical values, then it really doesn't do anything in the sense that it can be realized by pre- and post-composing with ambient isotopies.

Proof First, because the critical values never cross and are all distinct, we can post-compose with an ambient isotopy $\alpha_t \co [0,1] \to [0,1]$ to arrange that the critical values are constant (independent of $t$). In other words, we replace $g_t$ with $f_t = \alpha_t \circ g_t$ so that, for each $t$, the critical values of $g_t$ are constants $c_1, \ldots, c_k$, independent of $t$. Now consider the function $F \co [0,1] \times X \to [0,1] \times [0,1]$ defined by $F(t,p) = (t,f_t(p))$. Note that the critical points of $F$ are pairs $(t,p)$ where $p$ is critical for $f_t$, that the set of critical points of $F$ is a collection of arcs in $[0,1] \times X$, and that along these arcs $DF$ has rank $1$. Furthermore, the image of each of these arcs is a horizontal line $[0,1] \times \{c_i\} \subset [0,1] \times [0,1]$, because the critical values of $f_t$ are independent of $t$. Thus the image of $DF$ at each critical point $(t,p)$ is spanned by $\partial_t \in T_{(t,f_t(p))} [0,1] \times [0,1]$.

I like to use coordinates $(t,z)$ on $[0,1] \times [0,1]$, since $t$ is "time" and $z$ is "height". This is a bit sloppy because I will also use the variable $t$ in the domain $[0,1] \times X$.

Now I claim that there exists a vector field $V$ on $[0,1] \times X$ such that $DF(V) = \partial_t$. To see this, we only need to argue that, around every point $(t_0,p_0) \in [0,1] \times X$, there is a ball $B$ in which such a vector field exists. After showing that such a vector field exists in each such ball, we can patch such vector fields together using a partition of unity, since, if $DF(V_1) = \ldots = DF(V_n) = \partial_t$ and $a_1 + \ldots + a_n =1$ then $DF(a_1 V_1 + \ldots + a_n V_n) = \partial_t$. When $(t_0,p_0)$ is a regular point for $F$, such a ball with such a vector field clearly exists since $DF$ is surjective. When $(t_0,p_0)$ is a critical point, with critical value $(t_0,c_i=f_{t_0}(p_0))$, note that we can find coordinates $(x_1, \ldots, x_n)$ in a neighborhood of $p_0 \in X$, depending smoothly on $t$, with respect to which $f_t(x_1, \ldots, x_n) = c_i + \sum \pm x_j^2$ (Here we are using the fact that the critical values of $f_t$ are independent of $t$). This translates into coordinates $(t',x_1,\ldots,x_n)$ on a neighborhood of $(t_0,p_0) \in [0,1] \times X$ with respect to which $F(t',x_1,\ldots,x_n) = (t',c_i+\sum \pm x_j^2)$, and thus we can take $V = \partial_{t'}$ in this coordinate system.

(Note: I am appealing to a parametrized version of the Morse lemma, that nondenerate critical points are standard. Perhaps this should be written down also, to be very precise, but since the Morse lemma boils down to Taylor's theorem and Gram-Schmidt orthonormalization, both of which behave smoothly with respect to a parameter, this should be obvious, right?)

At this point the reader can probably see the end of the proof, but just to be totally pedantic, here it is (the trick to actually writing it down is getting the isotopies going in the right directions):

Now that we have this vector field $V$, note that $DF(V) = \partial_t$ means that flow along $V$ in $[0,1] \times X$ preserves the height function $z \circ F$, where $z \circ F(t,p) = f_t(p)$. We also know that $V = \partial_t + V_X$, where $V_X$ is tangent to $\{t\} \times X$. Thus $V_X$ is actually a time-dependent vector field $W(t)$ on $X$, and thus flow along $V$ turns into flow along $W(t)$ which gives an isotopy $\beta_t \co X \to X$. The fact that flow along $V$ preserves $z \circ F(t,p) = f_t(p)$ means that when we pull $f_t$ back via the flow $\beta_t$ along $W(t)$ we get $f_0$. I.e. $f_0 = f_t \circ \beta_t$.

Now recall that $f_t = \alpha_t \circ g_t$, and $f_0 = g_0$ because all our isotopies start with the identity. We want $g_t = \phi_t \circ g_0 \circ \psi_t$. Well, $g_t = \alpha_t^{-1} \circ f_t = \alpha_t^{-1} \circ f_0 \circ \beta_t^{-1} = \alpha_t^{-1} \circ g_0 \circ \beta_t^{-1}$, so we let $\phi_t = \alpha_t^{-1}$ and $\psi_t = \beta_t^{-1}$.

QED (I don't know an html symbol for a little square.)

Answer to the rectangle homotopy question

2010-11-15

OK, I think the answer is "yes" to the question I posed yesterday. Consider the following picture:

RectangleHomotopy2

The red curve is the original $\sigma = G \circ \gamma$, extended a little before $0$ and after $1$, hence the crossing at the point $q$, which is the middle blue dot. The two green curves are $\overline{\sigma}[0-\epsilon,1+\epsilon]$, and note that, because of the orientation, they can always be taken to be push-offs of the red curve passing through one of the other two blue dots. The point is, then, that the mapped-in disk bounded by the red curve can itself be pushed either up or down to give the a mapped-in disk bounded by one or the other green curves, and thus the green curves can be homotoped to short straight lines joining their end points via a homotopy that misses $q$ (the middle blue dot). That finishes the proof, but I'll be writing more details in the paper.

Extending homotopies of arcs to homotopies of rectangles

2010-11-14

I am going to continue using this space as a sort of stream-of-consciousness place to sort out things that are confusing me. This is coming from my work with Rob Kirby. Here is the issue: Consider a smooth map $G \co X^n \to \Sigma^2$ from an $n$--manifold $X$ to a surface $\Sigma$; make it generic, i.e. a Morse $2$--function, with $G^{-1}(\partial \Sigma) = \partial X$. (Although I'm not sure that those requirements are really important here.) Now consider an embedding $\gamma \co [0,1] \to X$ such that $G \circ \gamma(0) = G \circ \gamma(1) = q \in \Sigma$, and assume that $\gamma$ avoids all critical points of $G$, and that $q$ is a regular value. Now further suppose that $\sigma = G \circ \gamma$ is an immersion and is homotopically trivial in $\Sigma$, via a homotopy $\sigma_t$ fixed at the end points, and suppose that $\sigma_t^{-1}(q) = \{0,1\}$ for all $t$ other than $t=1$, with, of course, $\sigma_1^{-1}(q) = [0,1]$. 

In this situation I want to argue that there exists an embedding $\overline{\gamma} \co [-\epsilon,\epsilon] \times [0-\epsilon,1+\epsilon] \to X$ with $\overline{\gamma}(0,x) = \gamma(x)$ for $x \in [0,1]$, rectanglehomotopy

avoiding all critical points of $G$, with:

  • $G \circ \overline{\gamma}(y,x)$ a regular value for all $y \in [-\epsilon,\epsilon]$ and $x \in \{0,1\}$, 
  • $\overline{\sigma} = G \circ \overline{\gamma}$ a generic map onto its image, i.e. locally a Morse $2$--function, and such that
  • there is a generic homotopy $\overline{\sigma}_t \co [-\epsilon,\epsilon] \times [0-\epsilon,1+\epsilon] \to \Sigma$ from $\overline{\sigma}_0 = \overline{\sigma}$ to $\overline{\sigma}_1$, fixing the boundary pointwise, and such that 
  • for the associated map $\tilde{\sigma} \co [0,1] \times [-\epsilon,\epsilon] \times [0-\epsilon,1+\epsilon]$ defined by $\tilde{\sigma}(t,y,x) = \overline{\sigma}_t(y,x)$, the set $\tilde{\sigma}^{-1}(q)$ is a single arc with endpoints $(0,0,0)$ and $(0,0,1)$, and otherwise in the interior of $[0,1] \times [-\epsilon,\epsilon] \times [0-\epsilon,1+\epsilon]$.

One observation is that, for this to happen, I think that $\overline{\sigma}$ needs to do the opposite thing to the orientation of $[-\epsilon,\epsilon] \times [0-\epsilon,1+\epsilon]$ at $(0,0,0)$ than it does at $(0,0,1)$.

Why do I think this can be done? You need to split the normal bundle of $\gamma([0,1])$ into a $2$-dimensional part, containing the tangents to the curve itself, and a complementary part, such that the $2$-dimensional part projects positively to $T_q \Sigma$ at $\gamma(0)$ and negatively at $\gamma(1)$. Then exponentiating this $2$-dimensional part for a short distance gives you $\overline{\gamma}$. The only hard part of the claim is that you can arrange for the homotopy $\tilde{\sigma}$ to have $\tilde{\sigma}^{-1}(q)$ a single arc with endpoints $(0,0,0)$ and $(0,0,1)$, while fixing the boundary pointwise. Well, I'm not totally sure it's true, so let me go work with pen and paper, and maybe leave this entry here.