0708-1300/Class notes for Tuesday, October 9

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Class Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

Scanned Notes

Scanned notes for today's class can be found here

(see discussion)

Typed Notes - First Hour

Reminder:

1) An immersion locally looks like $\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}$ given by $x\mapsto (x,0)$

2) A submersion locally looks like $\mathbb {R} ^{m}\rightarrow \mathbb {R} ^{n}$ given by $(x,y)\mapsto x$

Today's Goals

1) More about "locally things look like their differential"

2) The tricky Sard's Theorem: "Evil points are rare, good points everywhere"

Definition 1

Let $f:M^{m}\rightarrow N^{n}$ be smooth. A point $p\in M^{m}$ is critical if $df_{p}$ is not onto $\Leftrightarrow$ rank $df_{p}<n$. Otherwise, p is regular.

Definition 2 A point $y\in N^{n}$ is a critical value of f if $\exists p\in M^{m}$ such that p is critical and $f(p)=y$. Otherwise, y is a regular value

Example 1

Consider the map $f:S^{2}\rightarrow \mathbb {R} ^{2}$ given by $(x,y,z)\mapsto (x,y)$. I.e., the projection map. The regular points are all the points on $S^{2}$ except the equator. The regular values, however, are all $(x,y)$ such that $x^{2}+y^{2}\neq 1$

Example 2

Consider $f:\mathbb {R} ^{3}\rightarrow \mathbb {R}$ given by $p\mapsto ||p||^{2}$. That is, $(x,y,z)\mapsto x^{2}+y^{2}+z^{2}$. Clearly $df|_{p=(x,y,z)}=(2x,2y,2z)$ and so p is regular $\Leftrightarrow df_{p}\neq 0\Leftrightarrow p\neq 0$

So, the critical values are the image of zero, thus only zero. All other $x\in \mathbb {R}$ are regular values.

Note: In both the last two examples there were points in the target space that were NOT hit by the function and thus are vacuously regular. In the previous example these are the point x<0.

Example 3

Consider a function $\gamma$ from a segment in $\mathbb {R}$ onto a curve in $\mathbb {R} ^{2}$ such that $d\gamma$ is never zero. Thus, rank($d\gamma )=1$ and so $d\gamma$ is never onto. Hence, ALL points are critical in the segment. The points on the curve are critical values, as they are images of critical points, and all points in $\mathbb {R} ^{2}$ NOT on the curve are vacuously regular.

Theorem 1

Sard's Theorem

Almost every $y\in N^{n}$ is regular $\Leftrightarrow$ the set of critical values of f is of measure zero.

Note: The measure is not specified (indeed, for a topological space there is no canonical measure defined). However the statement will be true for any measure.

Theorem 2

If $f:M^{m}\rightarrow N^{n}$ is smooth and y is a regular value then $f^{-1}(y)$ is an embedded submanifold of $M^{m}$ of dimension m-n.

Re: Example 2

$f^{-1}(y)$ is a sphere and hence (again!) the sphere is a manifold

Re: Example 3

$f^{-1}(y)$ for regular y is empty and hence we get the trivial result that the empty set is a manifold

Proof of Theorem 2

Let $f:M^{m}\rightarrow N^{n}$ is smooth and y is a regular value. Pick a $p\in f^{-1}(y)$. p is a regular point and thus $df_{p}$ is onto. Hence, by the submersion property (Reminder 2) we can find a "good charts" thats maps a neighborhood U of p by projection to a neighborhood V about y. Indeed, on U f looks like $\mathbb {R} ^{n}\times \mathbb {R} ^{m-n}\rightarrow \mathbb {R} ^{n}$ by $(x,z)\mapsto x$.

So$f^{-1}(0)=\{(0,z)\}=\mathbb {R} ^{m-n}$. Q.E.D

Diversion

Arbitrary objects can be described in two ways:

1) With a constructive definition

2) with an implicit definition

For example, a constructive definition of lines in $\mathbb {R} ^{3}$ is given by $\{v_{1}+tv_{2}\}$ but implicitly they are the solutions to the equations $ax+by+cz=d$ and $ez+fy+gz+h$.

Hence in general, a constructive definition can be given in terms of an image and an implicit definition can be given in terms of a kernal.

Homological algebra is concerned with the difference between these philosophical approaches.

Remark

For submanifolds of smooth manifolds, there is no difference between the methods of definition.

Definition 3

Loosely we have the idea that a concave and convex curve which just touch at a tangent point is a "bad" intersection as it is unstable under small perturbation where as the intersection point in an X (thought of as being in $\mathbb {R} ^{2}$) is a "good" intersection as it IS stable under small perturbations.

Precisely,

Let $N_{1}^{n_{1}},N_{2}^{n_{2}}\subset M$ be smooth submanifolds. Let $p\in N_{1}^{n_{1}}\bigcap N_{2}^{n_{2}}$

We say $N_{1}$ is transverse to $N_{2}$ in M at p if $T_{p}N_{1}\subset T_{p}M$ and$T_{p}N_{2}\subset T_{p}M$ satisfy $T_{p}N_{1}+T_{p}N_{2}=T_{p}M$

Example 4

Our concave intersecting with convex curve example intersecting tangentially has both of their tangent spaces at the intersection point being the same line and thus does not intersect transversally as the sum of the tangent spaces is not all of $\mathbb {R} ^{2}$.

Our X example does however work.

Typed Notes - Second Hour

Assistance needed: There is a symbol for "intersects transversally" but I am not sure of the latex command. See the scanned notes for what this looks like, I will just write it.

RE: Assistance needed: Try the command \pitchfork. I don't think you need any special packages, and it's pretty close to the "intersects transversally" symbol.

Definition 4

$N_{1}$ intersects $N_{2}$ transversally if $N_{1}$ is transversal to $N_{2}$ at every point

Theorem 3

If $N_{1}^{n_{1}}\cap N_{2}^{n_{2}}\subset M$ transversally then

1) $N_{1}\cap N_{2}$ is a manifold of dimension $n_{1}+n_{2}-m$

2) Locally can find charts so that $N_{1}=\mathbb {R} ^{n_{1}}\times 0^{m-n_{1}}$ and $N_{2}=0^{m-n_{2}}\times \mathbb {R} ^{n_{2}}$ with $N_{1}\cap N_{2}=0^{m-n_{2}}\times \mathbb {R} ^{n_{1}+n_{2}-m}\times 0^{m-n_{1}}$

Recall a Thm from Linear Algebra

If $W_{1}$ and $W_{2}$ subspaces of V then $dim(W_{1}+W_{2})+dim(W_{1}\cap W_{2})=dim(W_{1})+dim(W_{2})$

In particular if $W_{1}+W_{2}=V$ then $dim(W_{1}\cap W_{2})=-dim(V)+dim(W_{1})+dim(W_{2})$

Proof Scheme for Theorem 3

We can write $N_{i}=\varphi _{i}^{-1}(0)$ for some such $\varphi _{i}:U\subset M\rightarrow \mathbb {R} ^{m-n_{i}=cod(N_{i})}$

We then write $\varphi =\varphi _{1}\times \varphi _{2}:M\rightarrow \mathbb {R} ^{m-n_{1}}\times \mathbb {R} ^{m-n_{2}}=\mathbb {R} ^{2m-n_{1}-n_{2}}$

Hence, $N_{1}\cap N_{2}=\varphi ^{-1}(0)$

We want $rank(d\varphi _{p})=2m-n_{1}-n_{2}$. To prove this, we consider the aforementioned theorem from linear algebra with respect to the vector spaces obeying $ker(d\varphi )=ker(d\varphi _{1})\cap ker(d\varphi _{2})$

hence, and by rank nullity, $rank(d\varphi _{p})=m-dim(ker(d\varphi ))=m-\left(dim(TN_{1})+dim(TN_{2})-dim(TN_{1}+TN_{2})\right)=$

$m-(n_{1}+n_{2}-m)=2m-n_{1}-n_{2}$ as we wanted.

This shows that 0 is a regular value and hence by our previous theorem $N_{1}\cap N_{2}$ is a submanifold.

Now, we know we can construct the following diagram,

${\begin{matrix}N_{1}\cap N_{2}&\rightarrow ^{\varphi }&\mathbb {R} ^{2m-n_{1}-n_{2}}\\\downarrow &&\downarrow ^{\iota }\\M&\rightarrow ^{\lambda }&\mathbb {R} ^{m}\\\end{matrix}}$

where $\iota (x)=(x,0)$

We then set $\psi =$ (the function that takes the first $n_{1}+n_{2}-m$ coordinates only)$\circ \lambda$

hence, $\psi :M\rightarrow \mathbb {R} ^{n_{1}+n_{2}-m}$ and $\psi |_{N_{1}\cap N_{2}}$ is a chart for $N_{1}\cap N_{2}$

We now consider $\zeta :M\rightarrow \mathbb {R} ^{m}$ given by $(\varphi _{2},\psi ,\varphi _{1})$

i.e. operating via the following table,

${\begin{matrix}&N_{1}&N_{2}\\\varphi _{2}&\zeta &0\\\psi &\zeta &\zeta \\\varphi _{1}&0&\zeta \\\end{matrix}}$

Then,

$d\zeta ={\begin{bmatrix}I&0\\I&I\\0&I\\\end{bmatrix}}$

for blocks I of the appropriate sizes.

Thus (loosely) Q.E.D

Now on to some examples and comments about why Sard's Theorem is expected, but not obvious:

Example 5

Consider a standard "first year" smooth function $f:\mathbb {R} \rightarrow \mathbb {R}$. The "critical points" are in the first year calculus sense where the derivative is zero and the critical values are the images of these points. hence, the set of critical values we expect to be "small"

Example 6

Consider the function that folds the plane in half. The critical points are along the fold, as are the critical values and this line has 1 dimension and so of trivial measure in the plane (not that we have not given it a measure yet!)

Claims:

1) $\exists f:\mathbb {R} \rightarrow \mathbb {R}$ whose critical values are homeomorphic to a cantor set.

2) $\exists$ cantor sets with measure arbitrarily close to 1

3) $\exists g\in C^{1}:\mathbb {R} ^{2}\rightarrow \mathbb {R}$ whose critical points are a cantor set cross a cantor set and whose critical values are everything. Hence we will need our functions to be $C^{\infty }$ in the theorem.

Evil functions

Example 1

Nota benne: Here we are using the name Cantor set for any perfect set with empty interior.

There exists a function $f:\mathbb {R} \rightarrow \mathbb {R}$ smooth such that its set of critical values is homeomorphic to a Cantor set.

Remember that $g_{a,b}:[a,b]\rightarrow \mathbb {R}$

$g_{a,b}(x)=e^{-1/(x-a)^{2}}e^{-1/(x-b)^{2}}$

is a smooth function such that $g_{a,b}(a)=g_{a,b}(b)=0$

We can define the function $h$ in the complement of a Cantor set using the appropriate $g_{a,b}$ in the intervals of the complement.

Notices that $f(x)=\int _{0}^{x}h(t)dt$ holds the conditions of the example.

We can divide the new $g_{a,b}$ in each step of the construction by $1/2^{n}$ just to make the integral converge. And of course define h(t)=0 on the Cantor set.

Observe that, since $h$ is non negative, $f$ is increasing (observe it is strictly increasing)(it is continuous too!). Since increasing continuous functions have continuous inverses it is a homeomorphism.