TITLE: Cup products in cohomologies of connected sum of copies of a specific manifold
QUESTION [3 upvotes]: Cup products in cohomologies of connected sum of copies of $S^3\times S^4$
I want to calculate $H^*\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)$ (that is, connected sum of 5 copies of the same space).
Since
$$H^*\left(S^3,\mathbb{Z}\right)=\mathbb{Z}<1>\oplus0\oplus0\oplus\mathbb{Z}<x>
=\mathbb{Z}[x]/_{(x^2)}$$
and
$$H^*\left(S^4,\mathbb{Z}\right)=\mathbb{Z}<1>\oplus0\oplus0\oplus0\oplus\mathbb{Z}<y>
=\mathbb{Z}[y]/_{(y^2)},$$
Künneth formula gives us $$H^*\left(S^3\times S^4,\mathbb{Z}\right)=H^*\left(S^3,\mathbb{Z}\right)\otimes H^*\left(S^4,\mathbb{Z}\right)=\mathbb{Z}[x,y]/_{(x^2y^2)},$$
where for $n=0,3,4,7~$ $H^n\left(S^3\times S^4,\mathbb{Z}\right)=\mathbb{Z}$ and for $n=1,2,5,6~$ $H^n\left(S^3\times S^4,\mathbb{Z}\right)=0.$
Cohomology group of connected sum is sum of cohomology groups of connected summands for every dimension except for $n=0,7.$ Thus,
\begin{align*}
&H^0\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}\\
&H^1\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^2\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^3\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}=\mathbb{Z}^5\\
&H^4\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}=\mathbb{Z}^5\\
&H^5\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^6\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=0\\
&H^7\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right)=\mathbb{Z}
\end{align*}
If all I did above is OK, it remains to determine cup products, which I am horrible at. Since the groups are like that, we only need to describe the map $$H^3\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right) \times H^4\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right) \xrightarrow{\smile} H^7\left( (S^3\times S^4)^{\#5},\mathbb{Z}\right).$$
How can I determine this map? I found something here https://mathoverflow.net/questions/98376/cup-products-of-connected-sum, but I understood nearly nothing.
Thank you!

REPLY [3 votes]: It would help to know what was confusing about the MO page. If we have a connected manifold $M \# N$, then the cohomologies are
$H^0(M \# N) = \Bbb Z$
$H^k(M \# N) = H^k(M) \oplus H^k(N) \;\;\;\; 0 < k < n$
$H^n(M \# N) = \Bbb Z$.
I would prove this by using the relative long exact sequence $$H^*(M \vee N) = H^*(M \# N, S^{n-1}) \to H^*(M \# N) \to H^*(S^{n-1});$$ you can find the above description by chasing through the sequence.
The cup product with $1 \in H^0(M \# N)$ is the identity. The cup product between $(m, n)$ and $(m', n')$ whose degrees sum to less than $n$ is $(m \smile m', n \smile n')$. If the degrees sum to $n$, then $$(m,n) \smile (m',n') = m \smile m' + n \smile n'.$$ This will follow again from the above long exact sequence.
So for your space one has $H^3 = \langle e_1, \cdots, e_5\rangle$ while $H^4 = \langle f_1, \cdots f_5\rangle$, with the property that $e_i \smile f_j = \delta_{ij}$, ie, it is zero if $i \neq j$ and 1 if $i = j$.
That is it's just the direct sum of 5 copies of the cup product pairing of $S^3 \times S^4$.


TITLE: Let $A$ be a square matrix, such that $A^3 = I$. If 1 is not an eigenvalue of $A$, show: $A^{2} + A + I = 0$
QUESTION [1 upvotes]: So 1 is neither a eigenvalue of $A^2$ nor of $A^3$. But I don't know what to make of that.

REPLY [0 votes]: $t^3-1$ is anihilating polynomial for $A$. I assume that $A$ is complex matrix and $1$ is not an eigenvalue, so possible minimal polynomials will be any of the $t-\omega, t-\omega^2, (t-\omega)(t-\omega^2)$.
This gives $A=\omega.I, A=\omega^2. I$ or $(A-\omega)(A-\omega^2)$ as the possibility for $A$, each of which satisfy $A^2+A+I=0$.


TITLE: Continuous function on complete bounded metric space need not be bounded
QUESTION [2 upvotes]: I came across the following old qual problem:

Suppose $(X,d)$ is a complete metric space with finite diameter.  Is every continuous function on $X$ bounded?

It seems like the function $1/x$ on the interval $(0,1)$ with the discrete metric is a counterexample.  Is it this simple, or am I missing something?

REPLY [5 votes]: If all that’s wanted is a counterexample, yours is sufficient. Using the same basic idea, you could take $\Bbb N$ with the discrete metric and use the function $f(n)=n$. A different approach is to define a metric $d$ on $\Bbb R$ by $d(x,y)=\min\{|x-y|,1\}$ and use $f(x)=x$.
A more complete answer is that if the space is not compact, such a function always exists. If $X$ is complete but not compact, then $X$ is not totally bounded. This means that there is an $\epsilon>0$ such that no finite set of $\epsilon$-balls covers $X$. If follows that $X$ has an infinite closed, discrete subset $D$. Let $D_0=\{x_n:n\in\Bbb N\}$ be a countably infinite subset of $D$. Let $f:D_0\to\Bbb R:x_n\mapsto n$; $f$ is a continuous function on $D_0$, and $D_0$ is a closed subset of $X$, so by the Tietze extension theorem $f$ has a continuous extension $F:X\to\Bbb R$. Clearly $F$ is not bounded.


TITLE: Disprove the following statement: For all real numbers $x$ and $y$, if $x + \lfloor x \rfloor = y + \lfloor y \rfloor$ then $x = y$.
QUESTION [0 upvotes]: Disprove the following statement:
For all real numbers $x$ and $y$, if $x + \lfloor x \rfloor = y + \lfloor y \rfloor$ then $x = y$.
Aka: Prove the negation:
There are real numbers $x$ and $y$, that $x + \lfloor x \rfloor = y + \lfloor y \rfloor$ and $x \neq y$.
I have tried plugging in many real number combinations to disprove it, and have tried a few properties to try and prove this but I am completely stuck!
Any hints or guidance on how to approach this question would be helpful in the least.

REPLY [1 votes]: Let $x = n + p$ and $y = m + q$, where $n,m \in \mathbb{Z}$ and $0 \leq p,q < 1$ (Note that $n = \lfloor x \rfloor$ and, similarly, $m = \lfloor y \rfloor$). Then we have:
$$
x + \lfloor x \rfloor = y + \lfloor y \rfloor \;\Rightarrow\;
2n + p = 2m +q \;\Rightarrow\;
2(n - m) = q-p
$$
The LHS is an even integer. However $(q - p) \in (-1, 1)$, since $p,q \in [0,1)$ and the only even integer in the interval $(-1,1)$ is $0$.


TITLE: Two orthonormal sets in a Hilbert space. One is complete, the other must be complete.
QUESTION [2 upvotes]: Given two orthonormal sets $\{e_k\}_{k=1,2\ldots}$, $\{e'_k\}_{k=1,2\ldots}$ in a Hilbert space $H$, which satisfy
$$
\sum_{k=1}^\infty \|e_k-e'_k\|^2 < 1. \tag{*}
$$
Prove that if $\{e_k\}_{k=1,2\ldots}$ is complete, $\{e'_k\}_{k=1,2\ldots}$ must be complete. Here by complete, we mean $\overline{span\{e_k\}}=H$.
I tried to prove that it if there is a non-zero $x \perp \overline{span\{e'_k\}}$, i.e.
$$\forall k,\quad
\langle x,e'_k \rangle = \sum_{j=1}^\infty \langle x,e_j \rangle\langle e_j, e'_k \rangle = 0,
$$
then $(*)$ would not hold. But I have not reached any useful results in that direction yet.

REPLY [2 votes]: Assuming $\{ e_n \}$ and $\{ e_n'\}$ are orthonormal, define
$$
              Lx = \sum_{n=1}^{\infty}(x,e_n)e_n'
$$
The operator $L$ is linear and satisfies
$$
             \|Lx\|^2=\sum_{n=1}^{\infty}|(x,e_n)|^2 \le \|x\|^2.
$$
Now assume that $\{e_n\}$ is a complete orthonormal basis, and assume that $d=\left(\sum_{n=1}^{\infty}\|e_n-e_n'\|^2\right)^{1/2} < 1$. Then $x=\sum_{n=1}^{\infty}(x,e_n)e_n$ and
\begin{align}
       \|(I-L)x\|^2&=\|\sum_{n=1}^{\infty}(x,e_n)(e_n-e_n')\|^2 \\
  &\le \left(\sum_{n=1}^{\infty}|(x,e_n)|\|e_n-e_n'\|\right)^{2} \\
   & \le\sum_{n=1}^{\infty}|(x,e_n)|^2\sum_{n=1}^{\infty}\|e_n-e_n'\|^2 \\
   & = \|x\|^2\sum_{n=1}^{\infty}\|e_n-e_n'\|^2 = d^2\|x\|^2.
\end{align}
Because $d < 1$ by assumption, then $\|I-L\| \le d < 1$. It follows that
$$
        L = I-(I-L)
$$
is invertible. Therefore $L$ is surjective, which implies that the closure of the linear space spanned by $\{ e_n'\}_{n=1}^{\infty}$ is $H$. Hence, $\{ e_n'\}_{n=1}^{\infty}$ is a complete orthonormal basis.
Note: I do not know if this is the standard proof for your problem, but I like the proof, and I found on this site. Unfortunately, I cannot find it again. If this is not the standard proof, it should be. :)


TITLE: Question about finite sequences
QUESTION [0 upvotes]: Suppose $(a_1,\dots,a_n)$ is a sequence of real numbers such that $$a_1\leq a_2\leq \dots \leq a_n.$$
If $(b_1,\dots b_n)$ is a rearrangement of the sequence $(a_1,\dots,a_n)$ such that $$b_1\leq b_2\leq \dots \leq b_n,$$
then does it follow $a_1=b_1,\dots,a_n=b_n$?
I know that if the sequences were strictly monotonic, then the conclusion would have been obvious. How do I prove the question here, though... please help me with a proof maybe. Thank you in advance.

REPLY [0 votes]: Apart from the claim being obvious, Suppose $b_n\ne a_n$.
If $b_n<a_n$, then $a_n=b_i$ for some $i<n$, but then $a_n=b_i\le \ldots \le b_n<a_n$, contradiction.
Similarly, $b_n>a_n$ leads to a contradiction. Hence $a_n=b_n$. Now use induction on $n$.


TITLE: there exists an orthogonal transformation such that
QUESTION [0 upvotes]: Prove or Disprove: There exists an orthogonal transformation T: $\mathbb{R}^3 \to \mathbb{R}^3 $ such that
$$
        T\left(\begin{bmatrix}
        2  \\
        3  \\
        0  \\
        \end{bmatrix}\right)=
        \begin{bmatrix}
        3 \\
        0 \\
        2 \\
        \end{bmatrix} and
        T\left(\begin{bmatrix}
        -3  \\
        2  \\
        0  \\
        \end{bmatrix}\right)=
        \begin{bmatrix}
        2 \\
        -3 \\
        0 \\
        \end{bmatrix}
$$
I am not quite sure how to begin this problem. All I know is that a transformation T: $\mathbb{R}^n \to \mathbb{R}^n $ is orthogonal if T preserves length. Not quite sure how to proceed from there.
Thanks!

REPLY [1 votes]: If only you knew something about the angle between the vectors $(2,3,0)$ and $(-3,2,0)$ and then also about the angle between their images...


TITLE: I open the fridge - do I let the heat in or the cold out?
QUESTION [0 upvotes]: Conventionally, heat flows from hot to cold. How legitimate is it to redefine things in terms of "cold" rather than "heat" eg Cold flows from cold to hot?

REPLY [0 votes]: The refrigerator works by removing heat.  When you open the door heat goes in.


TITLE: How to derive this series?
QUESTION [1 upvotes]: I am wondering how to derive this series?

I know that

But not sure how to solve the first one. Please give me some hints. Thanks a lot.
A.

REPLY [0 votes]: EDIT:
$$\sum_{r=1}^\infty  r x^{r-1}=\sum_{r=1}^\infty \frac{d (x^r)}{dx}=\frac{d\left(\sum_{r=1}^\infty x^r\right)}{dx}$$
Now, $$\sum_{r=1}^\infty x^r=\frac x{1-x}=-1+\frac1{1-x}$$ for $|x|<1$


TITLE: Unique critical point and this point is global minimum of $f$.
QUESTION [1 upvotes]: I've been working on the following exercise:
Consider constants $a, b \in \mathbb{R}$, with $a>0$. Show that the function $f:\mathbb R^2 \rightarrow \mathbb R$ given by $f(x_1, x_2)=ax_1^4+bx_2+e^{x_1^2+x^2_2}$ has a unique critical point and this is global minimum of $f$.
I tried to solve it in a practical way, taking $c= (c_1, c_2)$ a critical point and trying to show that it is unique using operations. In the end I found that $c_1=0$ and $c_2= -\displaystyle{\sqrt{\frac{W\left(\frac{b^2}{2}\right)}{2}}}$ if $b>0$, $c_2= \displaystyle{\sqrt{\frac{W\left(\frac{b^2}{2}\right)}{2}}}$ if $b<0$ and $c_2 =0$ if $b=0$, where $W$ is the product log function.
Based on this, I have two questions:

I know that $c_2$ is unique depending on the choice of $b$, but for this to be true $W$ must be injective, am I right?

It seems to me that the $f$ function is a convex function, mainly because of the graph:

I could prove the result by showing that each of the components of the function $f$, ie $4ax_1^4$, $bx_2$ and $e^{x_1^2+x_2^2}$ are convex, and using that the sum of convex functions is convex? (and then use this result For a strictly convex function $f:\mathbb{R}^n\to \mathbb{R}$ critical point is unique.)

REPLY [1 votes]: It's not too hard to directly compute the Hessian matrix of $f$ and use the result you linked to.
$$
    \mathscr{H}f(x_1,x_2) = \left[\begin{matrix}2 \left(6 a x_{1}^{2} + 2 x_{1}^{2} e^{x_{1}^{2} + x_{2}^{2}} + e^{x_{1}^{2} + x_{2}^{2}}\right) & 4 x_{1} x_{2} e^{x_{1}^{2} + x_{2}^{2}}\\4 x_{1} x_{2} e^{x_{1}^{2} + x_{2}^{2}} & 2 \left(2 x_{2}^{2} + 1\right) e^{x_{1}^{2} + x_{2}^{2}}\end{matrix}\right]
$$
Its determinant is:
$$
\det \mathscr{H}f(x_1,x_2) = 48 a x_{1}^{2} x_{2}^{2} e^{x_{1}^{2} + x_{2}^{2}} + 24 a x_{1}^{2} e^{x_{1}^{2} + x_{2}^{2}} + 8 x_{1}^{2} e^{2 x_{1}^{2} + 2 x_{2}^{2}} + 8 x_{2}^{2} e^{2 x_{1}^{2} + 2 x_{2}^{2}} + 4 e^{2 x_{1}^{2} + 2 x_{2}^{2}}
$$
This is clearly positive for all $x_1$ and $x_2$, as is the trace of $\mathscr{H}f(x_1,x_2)$  So both eigenvalues of $\mathscr{H}f(x_1,x_2)$ are positive, which means that $\mathscr{H}f(x_1,x_2)$ is positive-definite.


TITLE: Relation of bijective functions and even functions?
QUESTION [0 upvotes]: Every point of the domain of the bijective function generates only one point of the range, one-to-one, and every one-to-one function has an inverse function.
The even function is symmetric on the y-axis, such that f(-x) = f(x).
Therefore, the bijective function can't be even function. Is that conclusion right?
and Is there any other relationship between the inverse and the symmetry property?

REPLY [1 votes]: No even function from $\mathbb R$ into $\mathbb R$ is bijective because, for instance, $f(1)=f(-1)$. Almost the same argument shows that, if $a>0$, no even function from $(-a,a)$ into $\mathbb R$ is bijective (or, indeed, injective).
Note that some odd functions from $\mathbb R$ into $\mathbb R$ are not bijective too. Take, say, $f(x)=x^3-x$.


TITLE: Inverse Function theorem on closed sets
QUESTION [1 upvotes]: Let $U \subseteq \mathbb{R}^n$ open subset of $\mathbb{R}^n$ and $f:U\to\mathbb{R}^n$ be a $C^\infty$- function. Suppose for every $x\in U$ the derivative of at $x$, $df_x$ is non singular. Which of the following are true ?
$1)$ If $V \subset U$ is open then $f(V)$ is open in $\mathbb{R}^n$
$2)$ $f:U\to f(U)$ is a homeomorphism
$3)$ $f$ is one-one
$4)$ If $V \subset U$ is closed then $f(V)$ is closed in $\mathbb{R}^n$

As a consequence of inverse function theorem, $1)$ is true infact $f(U)$ is open in $\mathbb{R}^n$ and $F(x,y)=\begin{bmatrix}e^x\cos y \\ e^x \sin y\end{bmatrix}$
for $F:\mathbb{R}^2\to\mathbb{R}^2$ is a counter example to $2)$ and $3)$ since $F$ is not injective infact periodic hence not a homeomorphism.
For $4)$ if we consider $V=U$ then it is false. But I guess we should not assume $V=U$ since then $V\subsetneq U$. I am not sure if $\subsetneq$ and $\subset$ mean the same here.
I am unable to show $4)$ is not true for a proper closed subset of $U$. Any help will be appreciated.

REPLY [1 votes]: Answer for 4): Let $V=\mathbb N$ in $\mathbb R$ and $f(x)=\arctan x$. Then $\frac {\pi} 2$ is  a limit point of $f(V)$ which does not belong to $f(V)$.


TITLE: Let f(z) be analytic on the unit disk D.f(r) = max{| f(z)| : |z| = r}. Does f'(r) > 0, if f is not a constant?
QUESTION [1 upvotes]: Let $f(z)$ be analytic on the unit disk $D.$
$f(r) = \max\{| f(z)| : |z| = r\}.$
Why $f'(r) > 0,$ if $f$ is not a constant.
And why if $f(0) = 0,$ then $r f ′
(r) \geq f(r)$ and the equality holds if and only if $f(z) = cz$ for
some nonnegative constant $c.$
It looks familiar but I just cannot recall any theorem dealing with this.

REPLY [0 votes]: For harmonic functions:

because the maximum is always on the boundary
is a monotonicity formula: you obtain it by comparing the Dirichlet energy of your function with the Dirichlet energy of the 1-homogeneous function with the same value on a circle.

For analytic functions you know that real and imaginary part are harmonic... don't know if this is the best way to obtain these informations.


TITLE: Find the eigenvalues of a projection operator
QUESTION [26 upvotes]: A projection operator $P$ is defined as $P^2$=$P$. Use this definition to find the eigenvalues of this operator.

In this question is it necessary to define what the projection operator is? And won't the eigenvalue just be zero?

REPLY [0 votes]: Geometrically, eigenvalues are the scaling factors by which particular vectors are scaled when multiplied by the respective matrix. Since the projection matrix projects a vector to its column space (or you can think of some nonzero subspace $W$), one such type of vector is the vector that is already living in that subspace. (Projecting a vector to its plane/subspace does not change it right?)
However, when doing so, the length doesn't change, so one eigenvalue must be $1$.
What about the vector that is perpendicular to $W$? Where will it go after the projection? Yes, it goes to the origin and hence the second eigenvalue must be $0$.
Although this approach does not prove anything, it gives a nice neat geometric intuition.


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\title{An exposition on finite dimensionality of Chow groups}
\author{Kalyan Banerjee}

\email{kalyan@math.tifr.res.in, kalyan.b@srmap.edu.in}
\begin{abstract}
In this exposition  we understand when the natural map from the Chow variety parametrizing codimension $p$ cycles on a smooth projective variety $X$ to the Chow group $\CH^p(X)$ is surjective. We derive some consequences when the map is surjective.
\end{abstract}
\maketitle

\section{Introduction}
The representability question in the theory of Chow groups is an important question. Precisely it means the following: let $X$ be a smooth projective variety and let $\CH^p(X)$ denote the Chow group of codimension $p$ algebraic cycles on $X$ modulo rational equivalence. Let $\CH^p(X)_{\hom}$ denote the subgroup of $\CH^p(X)$ consisting of homologically trivial cycles. Suppose that there exists a smooth projective curve $C$ and a correspondence $\Gamma$ on $C\times X$ such that $\Gamma_*$ from $J(C)\cong \CH^1(C)_{\hom}$ to $\CH^p(X)_{\hom}$ is surjective. Then we say that the group $\CH^p(X)_{\hom}$ is representable. The most interesting and intriguing is the case of highest codimensional cycle on a smooth projective variety $X$, i.e. the zero cycles on $X$. The first breakthrough result in this direction is the result by Mumford: the group $\CH^2(S)_{\hom}$ of a smooth projective complex algebraic surface with geometric genus greater than zero is not representable. It was further generalised by Roitman \cite{R1} for higher dimensional varieties proving that: if the variety $X$ has a globally holomorphic $i$-form on it, then the group $\CH^n(X)_{\hom}$ is not representable. Here $n$ is the dimension of $X$ and we have $i\leq n$. Then there is the famous converse question due to Spencer Bloch saying that: for a smooth projective complex algebraic surface $S$ with geometric genus equal to zero, the group $\CH^2(S)_{\hom}$ is representable. This question has been answered for the surfaces not of general type with geometric genus equal to zero by Bloch-Kas-Liebarman, in \cite{BKL}. The conjecture is still open in general for surfaces of general type but it has been solved in many examples in \cite{B},\cite{IM},\cite{V},\cite{VC}.

In  the case of highest codimensional cycles on a smooth projective variety $X$, the notion of representability can also be defined in another way. That is we consider the natural map from the symmetric power $\Sym^d X$ to $\CH^n(X)_{\hom}$, for $d$ positive and $n$ being the dimension of $X$. Suppose that this map is surjective for some $d$ then we say that the group $\CH^n(X)_{\hom}$ is representable. It can be proved as in \cite{Vo}, that this notion of representability implies the first notion of representability and vice-versa.

So following the approach of Voisin as in \cite{Vo}, it is natural to ask is there a second notion of representability for lower co-dimensional cycles. Precisely it means the following: Let us consider the two-fold product of the Chow variety $C^p_{d}(X)\times C^p_{d}(X)$. Consider the natural map from this product two the group $\CH^p(X)_{\hom}$ and we ask the question: does the surjectivity of this map implies the representability in the first sense of $\CH^p(X)_{\hom}$. First we prove the following in this direction:

\smallskip

\textit{ The Chow group of codimension $p$-cycles are generated by linear subspaces. That is there exists a surjective map from $\CH_0(F(X))$ to $\CH^p(X)$. Here $F(X)$ is the Fano variety of linear subspaces of codimension $p$. Suppose that $\CH^p(X)_{\hom}$ is representable in the sense that the map from the two-fold product of the Chow variety to $\CH^p(X)_{\hom}$ is surjective. Then there exists a smooth projective curve $C$ and a correspondence $\Gamma$ on $C\times X$, such that
$\Gamma_*:\CH^1(C)_{\hom}\to \CH^p(X)_{\hom}$
is surjective.
}

\smallskip

As an application we show that the natural map from $C^3_{d}(X)\times C^3_d(X)$ to $\CH^3(X)_{\hom}$ is not surjective for any $d$, where $X$ is a cubic fourfold embedded in $\PR^5$.

Our argument in this direction is minor modification of the argument present in the approach of Voisin in \cite{Vo}, where she deals with the case of zero cycles. First we recall various notions representability in the second sense, denoted as "finite dimensionality" of Chow groups of codimension $p$-cycles and show their equivalence. The key point is to use the Roitman's result on the map from the two-fold product of the Chow variety to $\CH^p(X)_{\hom}$ saying that the fibers of this map is a countable union od Zariski closed subsets in the product of Chow varieties. Then after having this equivalent notions of "finite dimensionality" in hand we proceed to the main theorem.

\smallskip

{\small \textbf{Acknowledgements:} The author would like to thank the hospitality of Tata Institute Mumbai, for hosting this project. }

\section{Finite dimensionality of the Chow group of codimension $i$ cycles}
Let $X$ be a smooth projective variety defined over the ground field $k$. Let $C^p_d(X)$ denote the Chow variety of $X$ parametrizing all codimension $p$ cycles on $X$ of a certain degree $d$. To consider the degree we fix an embedding of $X$ into some projective space $\PR^N$. Consider the $k$-points of the variety $C^p_{d}(X)$. Then consider the map
$$\theta_p^d:C^p_d(X)\times C^p_d(X)\to \CH^P(X)$$
given by
$$(Z_1,Z_2)\mapsto [Z_1-Z_2]$$
where $[Z_1-Z_2]$ is the class of the cycle $Z_1-Z_2$ in the Chow group. By abusing notation we will denote the class $[Z_1-Z_2]$ as $Z_1-Z_2$.
\begin{definition}
We say that the group $\CH^p(X)$ is representable if there exists $d$ such that $\theta^p_d$ is surjective.
\end{definition}
Now the natural question is that, what are the fibers of $\theta^p_d$, for a fixed $p,d$. Here is a theorem about that:
\begin{theorem}
\label{theorem1}
The fibers of  the map $\theta^p_d$ are countable union of Zariski closed subsets of $C^p_d(X)\times C^p_d(X)$.
\end{theorem}
\begin{proof}
To prove this we consider the following reformulation of the definition of rational equivalence. Let $Z_1,Z_2$ be two codimension $p$-cycles. They are rationally equivalent if there exists a positive cycle $Z'$, such that $Z_1+Z',Z_2+Z'$ belong to $C^p_d(X)$ for some fixed $d$, and there exists a regular morphism $f$ from $\PR^1$ to $C^p_d(X)$, such that
$$f(0)=Z_1+Z',\quad f(\infty)=Z_2+Z'\;.$$

Let us consider two cycles $Z_1,Z_2$ belonging to ${\theta^p_d}^{-1}(z)$ for some rational equivalence class $z$. Then $Z_1,Z_2$ belong to ${\theta^p_d}^{-1}(z)$ means that $Z_1,Z_2$ are rationally equivalent. So there exists $Z',f$ as above such that
$$f(0)=Z_1+Z',\quad f(\infty)=Z_2+Z'\;.$$
So it is natural to consider the following subvarieties of $C^p_d(X)\times C^p_d(X)$ denoted by $W^{u,v}_d$, given as the collection of all $(Z_1,Z_2)$ so that there exist $Z'\in C^p_{u}(X)$ $f$ in $\Hom^v(\PR^1,C^p_{d+u}(X))$, for some positive integer $u$ satisfying
$$f(0)=Z_1+Z',\quad f(\infty)=Z_2+Z'\;.$$
Here $\Hom^v(\PR^1,C^p_{d+u}(X))$ is the Hom scheme of degree $v$ morphisms from $\PR^1$ to $C^p_{d+u}(X)$.
For working purpose denote $\prod_{i=1}^n C^p_{d_i}(X)$ as $C^p_{d_1,\cdots,d_n}(X)$.

  Let
  $$
  e:\Hom ^v(\PR ^1,C^p_{d+u,d+u}(X))\to
  C^p_{d+u,d+u}(X)
  $$
be the evaluation morphism sending $f:\PR ^1\to C^p_{d+u,d+u}(X)$ to the ordered pair $(f(0),f(\infty ))$, and let us consider the diagonal in $C^p_{u,u}(X)$ and multiply with $C^p_{d,d}(X)$, call it $F$ and consider:
  $$
  s:F\to C^p_{d+u,d+u}(X)
  $$
be the regular morphism sending $(Z_1,Z_2)$ to $(Z_1+Z',Z_2+Z')$. The two morphisms $e$ and $s$ allow to consider the fibred product
  $$
  V=\Hom ^v(\PR ^1,C^p_{d+u,d+u}(X))\times _{C^p_{d+u,d+u}(X)}F\; .
  $$
This $V$ is a closed subvariety in the product
  $$
  \Hom ^v(\PR ^1,C^p_{d+u,d+u}(X))\times F
  $$
over $\Spec (k)$ consisting of quintuples $(f,Z_1,Z_2,Z')$ such that
  $$
  e(f)=s(Z_,Z_2,Z')\; ,
  $$
i.e.
  $$
  (f(0),f(\infty ))=(Z_1+Z',Z_2+Z')\; .
  $$
The latter equality gives
  $$
  V= W_d^{u,v}\; .
  $$

Vice versa, if $(Z_1,Z_2)$ is a closed point of $W_d^{u,v}$, there exists a regular morphism
  $$
  f\in \Hom ^v(\PR ^1,C^p_{d+u,d+u}(X))
  $$
with $ f(0)=Z_1+Z'$ and $ f(\infty )=Z_2+Z'$. Then $(f,Z_1,Z_2,Z')$ belongs to $V$.

So the set $W_d^{u,v}$ is itself a quasi-projective variety.

Suppose that $(Z_1,Z_2)$ is in $W_d^{u,v}$. Then there exists $f$ in $\Hom^v(\PR^1,C^p_{d+u}(X))$, $Z'$ in $C^p_u(X)$ such that

$$f(0)=Z_1+Z',\quad f(\infty)=Z_2+Z'\;.$$
Then this immediately imply that $(Z_1+Z',Z_2+Z')$ is in $W_{d+u}^{0,v}$. On the other consider the map
$$\tilde{s}:C^p_{d,d}(X)\times \Delta_{C^p_{u,u}(X)}\to C^p_{d+u,d+u}(X)$$
given by
$$(Z_1,Z_2,Z')\mapsto (Z_1+Z',Z_2+Z')\;.$$
By the above we have that
$$W_d^{u,v}\subset \pr_{1,2}(\tilde{s}^{-1}(W_{d+u}^{0,v}))\;.$$
Conversely suppose that $(Z'_1,Z'_2)$ belongs to $\pr_{1,2}(\tilde{s}^{-1}(W_{d+u}^{0,v}))$. Then $(Z'_1,Z'_2)$ is of the form $(Z_1+Z',Z_2+Z')$, such that there exists $f$ in $\Hom^v(\PR^1,C^p_{d+u}(X))$ satisfying
$$f(0)=Z_1+Z',\quad f(\infty)=Z_2+Z'\;.$$
This tell us that $(Z_1,Z_2)$ belongs to $W_d^{u,v}$.
Hence we have that
  $$
  W_d^{u,v}=\pr _{1,2}(\tilde s^{-1}(W_{d+u}^{0,v}))\; .
  $$

Since $\tilde s$ is continuous and $\pr _{1,2}$ is proper,
  $$
  \bar W_d^{u,v}=
  \pr _{1,2}(\tilde s^{-1}(\bar W_{d+u}^{0,v})\; .
  $$
This gives that to prove the second assertion of the proposition it is enough to show that $\bar W_d^{0,v}$ is contained in $W_d$.

Let $(Z_1,Z_2)$ be a closed point of $\bar W_d^{0,v}$. If $(Z_1,Z_2)$ is in $W_d^{0,v}$, then it is also in $W_d$. Suppose
  $$
  (Z_1,Z_2)\in \bar W_d^{0,v}\smallsetminus W_d^{0,v}\; .
  $$
Let $W$ be an irreducible component of the quasi-projective variety $W_d^{0,v}$ whose Zariski closure $\bar W$ contains the point $(Z_1,Z_2)$. Let $U$ be an affine neighbourhood of $(Z_1,Z_2)$ in $\bar W$. Since $(Z_1,Z_2)$ is in the closure of $W$, the set $U\cap W$ is non-empty.

Let us show that we can always take an irreducible curve $C$ passing through $(Z_1,Z_2)$ in $U$. Indeed, write $U$ as $\Spec (A)$. It is enough to show that there exists a prime ideal in $\Spec (A)$ of height $n-1$, where $n$ is the dimension of $\Spec (A)$, where $A$ is Noetherian. Since $A$ is of dimension $n$ there exists a chain of prime ideals
  $$
  \gop _0 \subset \gop _1 \subset \cdots \subset \gop_n=\gop
  $$
such that this chain can not be extended further. Now consider the subchain
   $$
   \gop _0 \subset \gop _1 \subset \cdots \subset \gop _{n-1}\; .
   $$
This is a chain of prime ideals and $\gop _{n-1}$ is a prime ideal of height $n-1$, so we get an irreducible curve.

Let $\bar C$ be the Zariski closure of $C$ in $\bar W$. Two evaluation regular morphisms $e_0$ and $e_{\infty }$ from $\Hom ^v(\PR ^1,C_d^p(X))$ to $C_d^p(X)$ give the regular morphism
  $$
  e_{0,\infty }:\Hom ^v(\PR ^1,C_d^p(X))\to C_{d,d}^p(X)\; .
  $$
Then $W_d^{0,v}$ is exactly the image of the regular morphism $e_{0,\infty }$, and we can choose a quasi-projective curve $T$ in $\Hom ^v(\PR ^1,C_d^p(X))$, such that the closure of the image $e_{0,\infty }(T)$ is $\bar C$.

For that consider the curve $C$ in $W$ so it is contained in $W_d^{0,v}$. We know that the image of $e_{0,\infty }$ is $W_d^{0,v}$. Consider the inverse image of $\bar C$ under the morphism $e_{0,\infty }$. Since $\bar C$ is a curve, the dimension of $e_{0,\infty }^{-1}(C)$ is greater than or equal than $1$. So it contains a curve. Consider two points on $\bar C$, consider their inverse images under $e_{0,\infty }$. Since $\Hom ^v(\PR ^1,C^p_d(X))$ is a quasi  projective variety, $e_{0,\infty }^{-1}(\bar C)$ is also projective, we can embed it into some $\PR ^m$ and consider a smooth hyperplane section through the two points fixed above. Continuing this procedure we get a curve containing these two points and contained in $e_{0,\infty }^{-1}(C)$. Therefore we get a curve $T$ mapping onto $\bar C$. So the closure of the image of $T$ is $\bar C$.

Now, as we have mentioned above, $\Hom ^v(\PR ^1,\bcC _d^p(X))$ is a quasi-projective variety. This is why we can embed it into some projective space $\PR ^m$. Let $\bar T$ be the closure of $T$ in $\PR ^m$, let $\tilde T$ be the normalization of $\bar T$ and let $\tilde T_0$ be the pre-image of $T$ in $\tilde T$. Consider the composition
  $$
  f_0:\tilde T_0\times \PR ^1\to T\times \PR ^1\subset
  \Hom ^v(\PR ^1,C_d^p(X))\times \PR ^1
  \stackrel{e}{\to }C_d^p(X)\; ,
  $$
where $e$ is the evaluation morphism $e_{\PR ^1,C_d^p(X)}$. The regular morphism $f_0$ defines a rational map
  $$
  f:\tilde T\times \PR ^1\dasharrow C_d^p(X)
  $$
Then by resolution of singularities we get that $f$ could be extended to a regular map from $(\tilde T\times \PR^1)'$ to $C_d^p(X)$, where $(\tilde T\times \PR^1)'$ denote the blow up of $\tilde T\times \PR^1$ along the indeterminacy locus which is a finite set of points. Continue to call the strict transform of $\tilde T $ in the blow up as $\tilde T$, and the pre-image of $T$ as $\tilde {T_0}$

The regular morphism $\tilde T_0\to T\to \bar C$ extends to the regular morphism $\tilde T\to \bar C$. Let $P$ be a point in the fibre of this morphism at $(Z_1,Z_2)$. For any closed point $Q$ on $\PR ^1$ the restriction $f|_{\tilde T\times \{ Q\} }$ of the rational map $f$ onto $\tilde T\times \{ Q\} \simeq \tilde T$ is regular on the whole curve $\tilde T$, because $\tilde T$ is non-singular. Then
  $$
  (f|_{\tilde T\times \{ 0\} })(P)=Z_1
  \qquad \hbox{and}\qquad
  (f|_{\tilde T\times \{ \infty \} })(P)=Z_2\; .
  $$
$$f:\{P\}\times \PR^1\to C^p_d(X)$$
has the property that
$$f(0)=Z_1,\quad f(\infty)=Z_2\;.$$

Hence we have that $W_d^{0,v}$ is Zariski closed. So $W_d^{u,v}$ is Zariski closed, and hence we have that ${\theta^p_d}^{-1}(z)$ is a countablel union of Zariski closed subsets in the product $C^p_{d,d}(X)$.
\end{proof}
By the previous Theorem \ref{theorem1} we can define the dimension of the fiber of $\theta^p_d$ to be the maximum of the dimensions of the Zariski closed subsets occuring in ${\theta^p_d}^{-1}(z)$, for $z$ in $\CH^p(X)$. Now consider the  subset of $C^p_{d,d}(X)$ consisting of  points  such that  the dimension of ${\theta^p_d}^{-1}(\theta^p_d)(Z)$ is not  constant as $Z$ varies. By the existence of Hilbert schemes this is a countable union of Zariski closed subsets of $C^p_{d,d}(X)$. Call this subset $B$. Then for $Z$ in the complement of $B$, we have that the dimension of ${\theta^p_d}^{-1}(\theta^p_d)(Z)$ is constant and say $r$.

\begin{definition}
The dimension of the image of $\theta^p_d$ is defined to be equal to $2\dim(C^p_{d}(X))-r$.
\end{definition}
Suppose that there exists a codimension $p$ prime cycle on $X$ of degree $e$. Then this prime cycle gives rise to an embedding of $C^p_{d}(X)$ into $C^p_{d+e}(X)$. Hence we have
$$\dim(C^p_d(X))\leq \dim(C^p_{d+e}(X))\;.$$
Hence we can define the limit superior of the
$$\dim(\im(\theta^p_d))\;.$$
We say that $\CH^p(X)$ is infinite dimensional if
$$\limsup_d \dim(\im(\theta^p_d))=\infty$$
and finite dimensional otherwise.

\begin{theorem}
The group $\CH^p(X)$ is representable if and only if it is finite dimensional.
\end{theorem}
\begin{proof}
Suppose that $\CH^p(D)$ is representable. Then there exists $d$ such that $\theta^p_d$ is surjective. For every integer $n$ consider the subset
$$R\subset C^p_n(X)\times C^p_n(X)\times C^p_d(X)\times C^p_d(X)$$
consisting of quadruples
$$(Z_1,Z_2,Z_1',Z_2')$$
such that
$$\theta^p_n(Z_1,Z_2)=\theta^p_d(Z'_1,Z'_2)\;.$$
As we have that $\theta^p_d$ is surjective, it follows that the projection
$$\pr_1:R\to C^p_n(X)\times C^p_n(X)$$
is surjective. Now by Theorem \ref{theorem1}, $R$ is a countable union of Zariski closed subsets in the ambient variety $C^p_n(X)\times C^p_n(X)\times C^p_d(X)\times C^p_d(X)$. We prove it as a separate lemma:

\begin{lemma}
The set $R$ is a countable union of Zariski closed subsets in $C^p_n(X)\times C^p_n(X)\times C^p_d(X)\times C^p_d(X)$.
\end{lemma}

\begin{proof}
To prove this we consider the following reformulation of the definition of rational equivalence. Let $Z_1,Z_2$ be two codimension $p$-cycles. They are rationally equivalent if there exists a positive cycle $Z'$, such that $Z_1+Z',Z_2+Z'$ belong to $C^p_d(X)$ for some fixed $d$, and there exists a regular morphism $f$ from $\PR^1$ to $C^p_d(X)$, such that
$$f(0)=Z_1+Z',\quad f(\infty)=Z_2+Z'\;.$$

Let us consider two cycles $(Z_1,Z_2,Z_1',Z_2')$ belonging to $R$. That would mean that the cycle class of $Z_1-Z_2$ is rationally equivalent to that of $Z_1'-Z_2'$.. So there exists a positive cycle $Z$ and a regular map $f$ from $\PR^1$ to $C^p_{d+n+u}(X)$  such that
$$f(0)=Z_1+Z_2'+Z,\quad f(\infty)=Z_1'+Z_2+Z'\;.$$
So it is natural to consider the following subvarieties of $C^p_{d+n+u}(X)\times C^p_{d+n+u}(X)$ denoted by $W^{u,v}_{d+n}$. It is  given by the collection of all elements  in the image of $C^p_d(X)\times C^p_d(X)\times C^p_n(X)\times C^p_n(X)$ in $C^p_{d+n+u}(X)\times C^p_{d+n+u}(X)$, under the natural map, given by
$$(Z_1,Z_2,Z_1',Z_2')\mapsto (Z_1+Z_2',Z_1'+Z_2)$$
Call this image as $F$.

so that there exist $Z\in C^p_{u}(X)$ and  $f$ in $\Hom^v(\PR^1,C^p_{d+n+u}(X))$, for some positive integer $u$ satisfying
$$f(0)=Z_1+Z_2'+Z,\quad f(\infty)=Z_2+Z_1'+Z'\;.$$
Here $\Hom^v(\PR^1,C^p_{d+n+u}(X))$ is the Hom scheme of degree $v$ morphisms from $\PR^1$ to $C^p_{d+n+u}(X)$.
For working purpose denote $\prod_{i=1}^n C^p_{d_i}(X)$ as $C^p_{d_1,\cdots,d_n}(X)$.

  Let
  $$
  e:\Hom ^v(\PR ^1,C^p_{d+n+u,d+n+u}(X))\to
  C^p_{d+n+u,d+n+u}(X)
  $$
be the evaluation morphism sending $f:\PR ^1\to C^p_{d+n+u,d+n+u}(X)$ to the ordered pair $(f(0),f(\infty ))$, and let us consider the diagonal in $C^p_{u,u}(X)$ and multiply with  $F$ and consider:
  $$
  s:F\times \Delta_{C^p_{u,u}(X)}\to C^p_{d+n+u,d+n+u}(X)
  $$
be the regular morphism sending $(Z_1+Z_2',Z_2+Z_1')$ to $(Z_1+Z_2'+Z,Z_2+Z_1'+Z')$. The two morphisms $e$ and $s$ allow to consider the fibred product
  $$
  V=\Hom ^v(\PR ^1,C^p_{d+n+u,d+n+u}(X))\times _{C^p_{d+n+u,d+n+u}(X)}(F\times \Delta_{C^p_{u,u}(X)}\; .
  $$
This $V$ is a closed subvariety in the product
  $$
  \Hom ^v(\PR ^1,C^p_{d+n+u,d+n+u}(X))\times F \times \Delta_{C^p_{u,u}(X)}
  $$
over $\Spec (k)$ consisting of tuples $(f,Z_1+Z_2',Z_2+Z_1',Z)$ such that
  $$
  e(f)=s(Z_1+Z_2',Z_2+Z_1',Z)\; ,
  $$
i.e.
  $$
  (f(0),f(\infty ))=(Z_1+Z_2'+Z,Z_2+Z_1'+Z)\; .
  $$
The latter equality gives
  $$
  V= W_{d+n}^{u,v}\; .
  $$

Vice versa, if $(Z_1+Z',Z_2+Z_1')$ is a closed point of $W_{d+n}^{u,v}$, there exists a regular morphism
  $$
  f\in \Hom ^v(\PR ^1,C^p_{d+n+u,d+n+u}(X))
  $$
with $ f(0)=Z_1+Z_2'+Z$ and $ f(\infty )=Z_2+Z_1'+Z$. Then $(f,Z_1+Z_2',Z_2+Z_1',Z)$ belongs to $V$.

So the set $W_{d+n}^{u,v}$ is itself a quasi-projective variety.

Suppose that $(Z_1+Z_2',Z_2+Z_1')$ is in $W_{d+n}^{u,v}$. Then there exists $f$ in $\Hom^v(\PR^1,C^p_{d+n+u}(X))$, $Z$ in $C^p_u(X)$ such that

$$f(0)=Z_1+Z_2'+Z,\quad f(\infty)=Z_2+Z_1'+Z\;.$$
Then this immediately imply that $(Z_1+Z_2'+Z,Z_2+Z_1'+Z)$ is in $W_{d+n+u}^{0,v}$. On the other consider the map
$$\tilde{s}:F\times \Delta_{C^p_{u,u}(X)}\to C^p_{d+n+u,d+n+u}(X)$$
given by
$$(Z_1+Z_2',Z_2+Z_1',Z)\mapsto (Z_1+Z_2'+Z,Z_2+Z_1'+Z)\;.$$
By the above we have that
$$W_{d+n}^{u,v}\subset \pr_{1,2}(\tilde{s}^{-1}(W_{d+n+u}^{0,v}))\;.$$
Conversely suppose that $(Z_1+Z_2',Z_1'+Z_2)$ belongs to $\pr_{1,2}(\tilde{s}^{-1}(W_{d+n+u}^{0,v}))$. Then there exists $f$ in $\Hom^v(\PR^1,C^p_{d+n+u}(X))$ satisfying
$$f(0)=Z_1+Z_2'+Z',\quad f(\infty)=Z_2+Z_1'+Z'\;.$$
This tell us that $(Z_1+Z_2',Z_2+Z_1')$ belongs to $W_{d+n}^{u,v}$.
Hence we have that
  $$
  W_{d+n}^{u,v}=\pr _{1,2}(\tilde s^{-1}(W_{d+n+u}^{0,v}))\; .
  $$

Since $\tilde s$ is continuous and $\pr _{1,2}$ is proper,
  $$
  \bar W_{d+n}^{u,v}=
  \pr _{1,2}(\tilde s^{-1}(\bar W_{d+n+u}^{0,v})\; .
  $$
This gives that to prove the second assertion of the proposition it is enough to show that $\bar W_{d+n}^{0,v}$ is contained in $W_{d+n}$.

Let $(Z_1+Z_2',Z_1'+Z_2)$ be a closed point of $\bar W_{d+n}^{0,v}$ (here the closure is taken with respect to $F$).  Suppose
  $$
  (Z_1+Z_2',Z_1'+Z_2)\in \bar W_{d+n}^{0,v}\smallsetminus W_{d+n}^{0,v}\; .
  $$
Let $W$ be an irreducible component of the quasi-projective variety $W_{d+n}^{0,v}$ whose Zariski closure $\bar W$ contains the point $(Z_1+Z_2',Z_1'+Z_2)$. Let $U$ be an affine neighbourhood of $(Z_1+Z_2',Z_2+Z_1')$ in $\bar W$. Since $(Z_1+Z_2',Z_1'+Z_2)$ is in the closure of $W$, the set $U\cap W$ is non-empty.

Let us show that we can always take an irreducible curve $C$ passing through $(Z_1+Z_2',Z_1'+Z_2)$ in $U$. Indeed, write $U$ as $\Spec (A)$. It is enough to show that there exists a prime ideal in $\Spec (A)$ of height $n-1$, where $n$ is the dimension of $\Spec (A)$, where $A$ is Noetherian. Since $A$ is of dimension $n$ there exists a chain of prime ideals
  $$
  \gop _0 \subset \gop _1 \subset \cdots \subset \gop_n=\gop
  $$
such that this chain can not be extended further. Now consider the subchain
   $$
   \gop _0 \subset \gop _1 \subset \cdots \subset \gop _{n-1}\; .
   $$
This is a chain of prime ideals and $\gop _{n-1}$ is a prime ideal of height $n-1$, so we get an irreducible curve.

Let $\bar C$ be the Zariski closure of $C$ in $\bar W$. Two evaluation regular morphisms $e_0$ and $e_{\infty }$ from $\Hom ^v(\PR ^1,C_{d+n}^p(X))$ to $C_{d+n}^p(X)$ give the regular morphism
  $$
  e_{0,\infty }:\Hom ^v(\PR ^1,C_{d+n}^p(X))\to C_{d+n,d+n}^p(X)\; .
  $$
Then $W_{d+n}^{0,v}$ is exactly the image of the regular morphism $e_{0,\infty }$, and we can choose a quasi-projective curve $T$ in $\Hom ^v(\PR ^1,C_{d+n}^p(X))$, such that the closure of the image $e_{0,\infty }(T)$ is $\bar C$.

For that consider the curve $C$ in $W$ so it is contained in $W_{d+n}^{0,v}$. We know that the image of $e_{0,\infty }$ is $W_{d+n}^{0,v}$. Consider the inverse image of $\bar C$ under the morphism $e_{0,\infty }$. Since $\bar C$ is a curve, the dimension of $e_{0,\infty }^{-1}(C)$ is greater than or equal than $1$. So it contains a curve. Consider two points on $\bar C$, consider their inverse images under $e_{0,\infty }$. Since $\Hom ^v(\PR ^1,C^p_{d+n}(X))$ is a quasi  projective variety, $e_{0,\infty }^{-1}(\bar C)$ is also quasi projective, we can embed it into some $\PR ^m$ and consider a smooth hyperplane section through the two points fixed above. Continuing this procedure we get a curve containing these two points and contained in $e_{0,\infty }^{-1}(C)$. Therefore we get a curve $T$ mapping onto $\bar C$. So the closure of the image of $T$ is $\bar C$.

Now, as we have mentioned above, $\Hom ^v(\PR ^1,C _{d+n}^p(X))$ is a quasi-projective variety. This is why we can embed it into some projective space $\PR ^m$. Let $\bar T$ be the closure of $T$ in $\PR ^m$, let $\tilde T$ be the normalization of $\bar T$ and let $\tilde T_0$ be the pre-image of $T$ in $\tilde T$. Consider the composition
  $$
  f_0:\tilde T_0\times \PR ^1\to T\times \PR ^1\subset
  \Hom ^v(\PR ^1,C_{d+n}^p(X))\times \PR ^1
  \stackrel{e}{\to }C_{d+n}^p(X)\; ,
  $$
where $e$ is the evaluation morphism $e_{\PR ^1,C_{d+n}^p(X)}$. The regular morphism $f_0$ defines a rational map
  $$
  f:\tilde T\times \PR ^1\dasharrow C_{d+n}^p(X)
  $$
Then by resolution of singularities we get that $f$ could be extended to a regular map from $(\tilde T\times \PR^1)'$ to $C_{d+n}^p(X)$, where $(\tilde T\times \PR^1)'$ denote the blow up of $\tilde T\times \PR^1$ along the indeterminacy locus which is a finite set of points. Continue to call the strict transform of $\tilde T $ in the blow up as $\tilde T$, and the pre-image of $T$ as $\tilde {T_0}$

The regular morphism $\tilde T_0\to T\to \bar C$ extends to the regular morphism $\tilde T\to \bar C$. Let $P$ be a point in the fibre of this morphism at $(Z_1+Z_2',Z_2+Z_1')$. For any closed point $Q$ on $\PR ^1$ the restriction $f|_{\tilde T\times \{ Q\} }$ of the rational map $f$ onto $\tilde T\times \{ Q\} \simeq \tilde T$ is regular on the whole curve $\tilde T$, because $\tilde T$ is non-singular. Then
  $$
  (f|_{\tilde T\times \{ 0\} })(P)=Z_1+Z_2'
  \qquad \hbox{and}\qquad
  (f|_{\tilde T\times \{ \infty \} })(P)=Z_2+Z_1'\; .
  $$
$$f:\{P\}\times \PR^1\to C^p_{d+n}(X)$$
has the property that
$$f(0)=Z_1+Z_2',\quad f(\infty)=Z_2+Z_1'\;.$$

Hence we have that $W_{d+n}^{0,v}$ is Zariski closed. So $W_{d+n}^{u,v}$ is Zariski closed. Therefore we have a countable union of Zariski closed subsets in $F$, whose pull-back to $C^p_d(X)\times C^p_d(X)\times C^p_n(X)\times C^p_n(X)$ is exactly $R$. Therefor $R$ itself is a countable union of Zariski closed subsets in   $C^p_d(X)\times C^p_d(X)\times C^p_n(X)\times C^p_n(X)$.

\end{proof}

Write $R=\cup_i R_i$, where each $R_i$ is a Zariski closed subset in $C^p_n(X)\times C^p_n(X)\times C^p_d(X)\times C^p_d(X)$. Considering the projection from $\pr_1$ from $R$ to $C^p_n(X)\times C^p_n(X)$, we have $\cup_i \pr_1(R_i)=C^p_n(X)\times C^p_n(X)$. But $C^p_n(X)\times C^p_n(X)$ can be uniquely decomposed into finitely many Zariski closed irreducible subsets of maximal dimension. Using the fact that the ground field $k$ is uncountable, it will follow that there exists finitely many components $R_1,\cdots,R_m$ of $R$, such that $\cup_i R_i=R'$ surjects onto $C^p_n(X)\times C^p_n(X)$. So we have that
$$\dim R'\geq 2\dim C^p_d(X)\;.$$
Now consider $(Z_1,Z_2,Z'_1,Z'_2)$ in $\cup R_i$, then we have
$$\dim_{(Z_1,Z_2)}R'\cap C^p_n(X)\times C^p_n(X)\times (Z'_1,Z'_2)\geq 2\dim C^p_n(X)-2\dim C^p_d(X)\;,$$
this number on the right hand side is bigger than zero if we take sufficiently large $n$ such that $C^p_n(X)$ contains $C^p_d(X)$. Then the above is an algebraic set contained in
$${\theta^p_{n}}^{-1}(\theta^p_d(Z'_1,Z_2'))\;.$$
As $(Z_1,Z_2)$ is arbitrary and the projection
$$R'\to C^p_n(X)\times C^p_n(X)$$
is surjective, we have that dimension of ${\theta^p_n}^{-1}(Z_1,Z_2)$ is atleast $2(\dim (C^p_n(X))-\dim(C^p_d(X)))$. Hence the image of $\theta^p_n$ is bounded by $2\dim C^p_d(X)$. Hence $\CH^p(X)$ is finite dimensional.
Now suppose that $\CH^p(X)$ is finite dimensional. We have to prove that there exists $d$ such that $\theta^p_d$ is surjective. Let $d$ be such that
$$\dim(\im(\theta^p_d))=\dim(\im(\theta^p_{d+e}))$$
for some positive integer $e$. Let $V$ be a subvariety of degree $e$ and codimension $p$ giving an embedding of $C^p_{d}(X)\times C^p_d(X)$ into $C^p_{d+e}(X)\times C^p_{d+e}(X)$. Call  the embedding as $i_V$, then we have
$$\theta^p_{d+e}\circ i_V=\theta^p_d\;.$$
Let $F$ be the fiber of $\theta^p_d$ passing through a general point $(Z_1,Z_2)$ of $C^p_d(X)\times C^p_d(X)$, let $F'$ be the fiber of $\theta^p_{d+e}$ through a general point in $C^p_{d+e}(X)\times C^p_{d+e}(X)$. By assumption we have that
$$2\dim(C^p_d(X))-r_1=2\dim(C^p_{d+e}(X))-r_2$$
where $r_1,r_2$ are dimensions of $F,F'$. So we have
$$r_2-r_1=2n$$
where $n=\dim(C^p_{d+e}(X))-\dim(C^p_d(X))\;.$
Let $F''$ be the fiber of a $\theta^p_{d+e}$ such that it passes through $(Z_1+V,Z_2+V)=i_V(Z_1,Z_2)$. Then
Then by the definition of dimension of the fiber of $\theta^p_{d+e}$ we have that
$$\dim(F'')\geq \dim(F')\;.$$
Now consider the subset
$$R=\{(Z_1,Z_2,Z'_1,Z'_2):\theta^p_{d+e}(Z_1,Z_2)=\theta^p_d(Z'_1,Z'_2)\}\;.$$
The projection from $R$ to $C^p_d(X)\times C^p_d(X)$ is surjective, so there exists finitely many irreducible subsets containing $(Z_1+V,Z_2+V,Z_1,Z_2)$ such that there union $R'$ dominates  $C^p_d(X)\times C^p_d(X)$. Fiber of this projection from $R'$ is of dimension greater or equal than that of $F''$. So it is greater than or equal to $\dim(F)+2\dim C^p_d(X)$. The fibers of the first projection
$$p:R'\to C^p_{d+e}(X)\times C^p_{d+e}(X)$$
are of dimension atmost $\dim(F)$. So we have that
$$\dim p(R')\geq \dim R'-\dim F\geq \dim F''+2\dim (C^p_d(X))-\dim(F)$$
$$\geq \dim F'-\dim F+2\dim(C^p_d(X)=\dim (C^p_{d+e}(X)\times C^p_{d+e}(X))$$
\end{proof}
Now our aim is to detect the kernel of the Abel-Jacobi map for higher dimensional cycles. Let's recall that the Abel-Jacobi map has domain ${\CH^p_(X)}_{\hom}$ and target the Intermediate Jacobian $IJ^p(X)$ given by
$$H^{2p-1}(X,\CC)/F^p H^{2p-1}(X,\CC)\oplus H^{2p-1}(X,\ZZ)\;.$$
The first theorem in this direction is to relate the representablity of the Chow group of codimension $p$ cycles with zero cycles on Jacobian.
\begin{theorem}
Suppose that there exists a smooth projective curve $C$, and a correspondence $\Gamma$ on $C\times X$ such that
$$\Gamma_*:\CH^1(C)_{\hom}\to \CH^p(X)_{\hom}$$
is surjective. Then $\CH^p(X)_{\hom}$ is representable.
\end{theorem}
\begin{proof}
Note that the natural map from $\Sym^g C\times \Sym^g C$ to $\CH^1(C)_{\hom}$ is surjective, where $g$ is the genus of the curve $C$. Therefore  $\CH^1(C)_{\hom}$ is finite dimensional. Therefore the image of $\Gamma_*$ is finite dimensional. But $\Gamma_*$ is surjective. So $\CH^p(X)_{\hom}$ is finite dimensional hence representable.
\end{proof}

It is difficult to prove the converse, that is : suppose $\CH^p(X)_{\hom}$ is representable then does there exists a curve $C$ and a correspondence $\Gamma$ on $C\times X$, such that $\Gamma_*$ is onto. Let us consider the following situation when:

I) The Chow group of codimension $p$-cycles are generated by linear subspaces. That is there exists a surjective map from $\CH_0(F(X))$ to $\CH^p(X)$. Here $F(X)$ is the Fano variety of linear subspaces of codimension $p$.

\begin{theorem}
\label{theorem2}
Let $X$ be as above. Suppose that $\CH^p(X)_{\hom}$ is representable. Then there exists a smooth projective curve $C$ and a correspondence $\Gamma$ on $C\times X$, such that
$$\Gamma_*:\CH^1(C)_{\hom}\to \CH^p(X)_{\hom}$$
is surjective.
\end{theorem}
\begin{proof}
Consider the map $\theta^p_d$ from $C^p_{d}(X)$ to $\CH^p(X)_{\hom}$ given by
$$Z\mapsto Z-dL_0$$
where $L_0$ is a fixed linear subspace of $X$. Since $\CH^p(X)_{\hom}$ is actually generated by linear subspaces. The above map restricted to $\Sym^d F(X)$ is surjective, continue to call it $\theta^p_d$. Let us consider large $d$ such that $\dim(\im(\theta^p_d))$ is constant and equal to $K$. Then the dimension of a general fiber is equal to
$$\dim(\Sym^d F(X))-K$$.

Now we prove that an irreducible component $Z$ of maximal dimension of a general fiber of $\theta^p_d$ cannot be contained in a set of the form
$$\Sym^{d-i}F(X)+ W$$
where $W$ is in $\Sym^i F(X)$, $\dim W<i$ and the above $+$ means the image of the natural map from
$$\Sym^{d-i}F(X)\times W\to \Sym^{d}F(X)\;.$$
If possible assume that $Z$ is contained in such a set. Note that the dimension of $Z$ is $\dim(\Sym^{d}F(X))-K$. So we have
$$\dim(\Sym^{d-i}F(X))\geq \dim(\Sym^d F(X))-K-i+1\;.$$
Let the dimension of $F(X)$ be $n$. Then the above says
$$n(d-i)\geq nd-K-i+1$$
which implies
$$i<K/n-1\;.$$
Consider the subset
$$Z'=\{(z,w)|z+w\in Z\}\subset \Sym^{d-i}F(X)\times W\;.$$
By definition this set dominates $Z$, hence is of dimension greater or equal than $nd-K$. So the general fibers of the second projection $\pr_2:Z'\to W$ are of dimension atleast
$$nd-K-i+1\;.$$
Also note that
$$\theta^p_{d}(z+w)=\theta^p_{d-i}(z)+\theta^p_{i}(w)\;.$$
Since $\theta^p_d$ is constant along $Z$ we have $\theta^p_{d-i}$ is constant along $Z'_w$. So if $Z$ passes through a very general point of $\Sym^d F(X)$, then $Z'_w$ passes through a very general point of $\Sym^{d-i}F(X)$. So we have $\dim(Z'_w)$ is less than the dimension of a general fiber of $\theta^p_{d-i}$ for generic $w$.  Now the dimension of $Z'_w$ is greater than or equal to
$$nd-K-i+1$$
but the dimension of the fiber of $\theta^p_{d-i}$ is equal to $(d-i)n-K$ because $d-i>d-K/n-1$ can be chosen to be arbitrarily large.

Let us assume that $d\geq 2$, and we have $nd-K\geq d$. Consider the following lemma.

\begin{lemma}
\label{lemma1}
Let $Y$ be an ample hypersurface of $F(X)$ and let $Z$ be an irreducible subset of $\Sym^d F(X)$ not contained in any subset of the form
$$\Sym^{d-i}F(X)+W$$
with $W\subset \Sym^i F(X)$ and dimension of $W$ is less than $i$. Then $Z$ intersects $\Sym^d Y$, provided that $\dim(Z)\geq m$.
\end{lemma}
Therefore by applying the lemma we see that a general fiber of $\theta^p_d$ intersects $\Sym^d Y$, for sufficiently large $d$ and provided that $n\geq 2$. Therefore $\theta^p_d$ and $\theta^p_d|_{\Sym^d Y}$ has same image and the later has image of bounded dimension. So we can apply the lemma again and finally get that $\theta^p_d$ and $\theta^p_d|_{\Sym^d C}$ has same image, where $C$ is a smooth projective curve obtained by intersecting $n-1$ many ample hypersurfaces. This proves theorem.
\end{proof}

\smallskip

\textit{proof of Lemma \ref{lemma1}}: Consider the quotient map $r:F(X)^d\to \Sym^d F(X)$. Let $r^{-1}(Z)=\wt{Z}$, let $\wt{Z_0}$ be a component of $\wt{Z}$ dominating $Z$. By the hypothesis we have the following:

for every $i\geq 1$ and every subset $I$ of cardinality $i$, we have $\dim p_I(\wt{Z_0})\geq i$, where $p_I$ is the projection from $F(X)^d$ to $F(X)^i$ corresponding to the set of indices. Since $\wt{Z_0}$ dominates $Z$, it is sufficient to prove that $\wt{Z_0}$ intersect $Y^d$, for an ample hypersurface $Y $  in $F(X)$. Consider a complete intersection $V$ in $\wt{Z_0}$, which is obtained by intersecting $\wt{Z_0}$ with finitely many ample hypersurfaces, so that dimension of $V$ is $d$. Then the hypotheses on $\wt{Z_0}$ implies that same would be true for  $V$. So without loss of generality we can assume that dimension of $\wt{Z_0}=d$. Consider a desingularization $Z'$ of $\wt{Z_0}$. Consider the divisors
$$D_i:=(\pr_i\circ \tau)^{-1}(Y)$$
where $\tau$ is the natural map from $Z'$ to $\wt{Z_0}$. Now $Y$ is ample, so we have
$$(\pr_i\circ \tau)_*((\pr_i\circ\tau)^*(Y).C)=Y.(\pr_i\tau)_*C\geq 0$$
which means that $D_i$'s are numerically effective. Our claim will follow from the fact that $D_1\cap\cdots\cap D_d$ is non-empty. So we prove that $D_1\cap\cdots\cap D_d$ is non-empty. First suppose that $d=2$. We have $D_1,D_2$ two divisors numerically effective. Hence we have
$$D_1^2\geq 0;, \quad D_2^2\geq 0$$
Suppose that $D_1.D_2=0$. Then the intersection matrix of $(D_1,D_2)$ is semipositive. So by the Hodge index theorem we have $D_1=rD_2$ for some integer $r$. Hence $(D_1+D_2)^2=0$.  But $D_1+D_2$ is the pull-back of an ample divisor on $F(X)\times F(X)$, under a generically finite map. So it is ample. Therefore $(D_1+D_2)^2>0$, which is a contradiction. The general case follows from this.

So when, $\CH^p(X)$ is representable and $X$ satisfies the assumption of Theorem \ref{theorem2}, then the above Theorem \ref{theorem2} and arguments present in \cite{Vo} give that there exists an abelian variety $A$ and a correspondence $\Gamma$ supported on $A\times F(X)$, such that $L_*\Gamma_*:A\to \CH_0(F(X))_{\hom}\to\CH^p(X)_{\hom}$ is surjective, where $L_*$ is the universal incidence correspondence given by
$$\{(x,L):x\in L\}\;.$$

This leads us to the following result.

\begin{theorem}
\label{theorem3}
Let $X$ be smooth projective and it satisfies the hypotheses of Theorem \ref{theorem2}. Suppose that $\CH^2(X)_{\hom}$ is representable. Then the kernel of the Abel-Jacobi map is torsion.
\end{theorem}

\begin{proof}
By Theorem \ref{theorem2} there exists a smooth projective curve $C$ in $F(X)$ and a correspondence $\Gamma$ on $C\times F(X)$ such that $L_*\Gamma_*:J(C)\to \CH^2(X)_{\hom}$ is surjective. This yields further, a correspondence on $J(C)\times F(X)$, such that
$$L_*\Gamma_*:J(C)\to \CH^2(X)_{\hom}$$
is surjective. By Theorem \ref{theorem1} we have that kernel of $L_*\Gamma_*$ is a countable union of Zariski closed subsets of $J(C)$. Since kernel of $L_*\Gamma_*$ is a subgroup of $J(C)$ and we work over an uncountable ground field, the kernel is a countable union of translates of an abelian variety $A$ sitting in $\ker(L_*\Gamma_*)$. Now consider the supplementary abelian variety $B$, such that $A\times B\to J(C)$ is an isogeny.  Now replacing $J(C)$ by $B$, and $L\circ\Gamma$ by $(L\circ\Gamma)_{B\times X}$, we get that the kernel of $L_*\Gamma_*$ is countable.

Fix $x_0$ in $F(X)$, consider the subset $R$ of $F(X)\times B$ given by
$$\{(x,a):L_*\Gamma_*(a)=L_*(x)-L_*(x_0)\}$$
By Theorem \ref{theorem1} we have $R$ is a countable union of Zariski closed subsets in $F(X)\times B$. Since $\Gamma_*L_*$ is surjective, the projection from $R$ onto $F(X)$ is onto. Hence there exists a component $R_0$ of $R$ surjecting onto $F(X)$. Since $\ker(L_*\circ \Gamma_*) $ is countable, the map is actually finite of say degree $r$. Thus $R_0$ gives rise to a correspondence of dimension equal to $\dim(F(X))$ between $F(X)$ and $B$, this provides a morphism
$$\alpha:F(X)\to B$$
given by
$$\alpha(x)=\alb_B(R_{0*}(x-x_0))\;.$$
By definition of $R_0$ we have that
$$L_*\Gamma_*\alpha(x)=r(L_*(x)-L_*(x_0))\;.$$
Now this $\alpha$ gives rise to a regular homomorphism from $\CH^2(X)_{\hom}$ to $B$. Hence by the universality the intermediate Jacobian $IJ^2(X)$, there exists a unique regular map $\beta:IJ^2(X)\to B$, such that
$$\alpha=\beta\circ \Phi_2$$
where $\Phi_2$ is the Abel-Jacobi map.
So we have
$$L_*\Gamma_*\beta\Phi_2(z)=rz\;.$$
This proves that kernel of $\Phi_2$ is torsion.
\end{proof}

\subsection{Application of the above result}

Consider $X$ to be a smooth cubic fourfold embedded in $\PR^5$. Then we know that the group of homologically trivial one cycles $\CH^3(X)_{\hom}$ is generated by lines on $X$. So  the criterion for Theorem \ref{theorem2} is satisfied. We know, by \cite{Sc}, that there does not exists a smooth projective curve $C$ and a correspondence $\Gamma$ on $C\times X$, such that $\Gamma_*$ is surjective. Hence it follows that the natural map from the Chow varieties parametrising one cycles on $X$ does not surject onto $\CH^3(X)_{\hom}$.

\begin{theorem}
The group $\CH^3(X)_{\hom}$ is not representable. That is, the natural map from the product $C^3_d(X)\times C^3_d(X)$ to $\CH^3(X)_{\hom}$ is not surjective for any $d$.
\end{theorem}

The above Theorem \ref{theorem3} gives us a criterion by which we can detect the representability of $\CH^2(X)_{\hom}$. Precisely when, the Abel-Jacobi kernel is non-trivial, $\CH^2(X)_{\hom}$ is not representable, in the sense that the natural map from the Chow varieties (infact from the symmetric powers of $F(X)$) to $\CH^2(X)_{\hom}$ are not surjective. There are some examples of higher dimensional varieties with non-trivial Abel-Jacobi mappings given in \cite{GG}. So the Theorem \ref{theorem3} forces that such varieties cannot have  :

$\Sym^d F(X)$ surjecting onto $\CH^2(X)_{\hom}$. So for a threefold $X$, the non-triviality of the Abel-Jacobi kernel implies that the $\CH^2(X)_{\hom}$ cannot be generated by integral linear combination of lines on the threefold of any fixed degree.

config: arxiv


TITLE: Angular Velocity cross product with Vorticity
QUESTION [2 upvotes]: I am trying to show that the following identity is true.
$$\Omega\times\omega=(\omega\cdot\nabla)(\Omega\times r)$$
I've tried using vector algebra and index notation to show that this is true with no luck.
I expanded the right side using the identity $$\nabla(A\cdot B)=(A\cdot\nabla)B+(B\cdot\nabla)A+A\times(\nabla\times B)+B\times(\nabla\times A)$$ and I got this.
$$\nabla(\omega\cdot (\Omega\times r))=(\omega\cdot\nabla)(\Omega\times r)+((\Omega\times r)\cdot\nabla)\omega+\omega\times(\nabla\times (\Omega\times r))+(\Omega\times r)\times(\nabla\times \omega)$$
Then I expanded the third term on the right using the identity $$\nabla\times(A\times B)=A(\nabla\cdot B)+(B\cdot\nabla)A-B(\nabla\cdot A)-(A\cdot\nabla)B$$
and I got this.
$$\nabla(\omega\cdot (\Omega\times r))=(\omega\cdot\nabla)(\Omega\times r)+((\Omega\times r)\cdot\nabla)\omega+(\omega\times\Omega)(\nabla\cdot r)+\omega\times(( r\cdot\nabla)\Omega)-(\omega\times r)(\nabla\cdot\Omega)-\omega\times((\Omega\cdot\nabla) r)+(\Omega\times r)\times(\nabla\times \omega)$$
In my problem $\Omega$ is a constant vector so any derivatives of it should go to zero.
$$\nabla(\omega\cdot (\Omega\times r))=(\omega\cdot\nabla)(\Omega\times r)+((\Omega\times r)\cdot\nabla)\omega+(\omega\times\Omega)(\nabla\cdot r)-\omega\times((\Omega\cdot\nabla) r)+(\Omega\times r)\times(\nabla\times \omega)$$
At this point I'm stuck and not sure whether I'm doing something wrong or if I haven't gone far enough. Any help would be appreciated.

REPLY [3 votes]: Since this is a Fluid Dynamics problem, why not just do the following:
$$\Omega\times r=\begin{pmatrix}\Omega_yz-\Omega_z y\\
\Omega_zx-\Omega_x z\\
\Omega_xy-\Omega_y x\\
\end{pmatrix}$$
And: $$\omega \cdot \nabla=\omega_x\frac \partial{\partial x}+\omega_y\frac \partial{\partial y}+\omega_z\frac \partial{\partial z}$$
Since $\Omega_x,\Omega_y,\Omega_z$ are constant, we get: $$(\omega \cdot \nabla)(\Omega\times r)=\begin{pmatrix}\Omega_y\omega_z-\Omega_z \omega_y\\
\Omega_z\omega_x-\Omega_x\omega_ z\\
\Omega_x\omega_y-\Omega_y \omega_x\\
\end{pmatrix}$$
Which is obviously $\Omega \times \omega$
P.S. : I am fully aware that this is a very particular and simplistic way of proving the identity, but given the context I assume this is appropriate...


TITLE: Definition of Matroid in two cases
QUESTION [1 upvotes]: For matroids M = (S, U), the following property applies to U:

If A, B ∈ U and |B| = |A| + 1 , then there must exist an x ∈ {B \ A} such that A∪ {x} ∈ U

I want to prove this property for the following two matroids
a)
S = set of edges of a connected graph G
U = all subsets of S that do not contain circles
My approach: If | B | = | A | +1, then there is at least one edge x in B, which can be added to A. A∪ {x} is a forest and A is a forest, but how do I know that A∪ {x} does not create a circle in A?
b)
S = set of vectors from $
\mathbb{R}^{n}$
U = all sets of linearly independent vectors from S
My approach: If there are |B| linearly independent vectors, and A contains |B|−1 vectors, then there must be a vector that is linearly independent of the vectors in A. Otherwise A would already be a basis of $
\mathbb{R}^{n}$, and B could not have more elements than A.
Would be glad for a scrutinizing look

REPLY [1 votes]: Your approach to (b) is on the right track, but you need to show that there is a vector in $B$ that is linearly independent of the vectors in $A$; you can do this be showing that if $B\subseteq\operatorname{span}A$, then $B$ is not linearly independent.
For (a), suppose that $A$ and $B$ are the edge sets of forests $G_A$ and $G_B$, respectively, and that $|B|=|A|+1$. Let $V_A$ and $V_B$ be the vertex sets of $G_A$ and $G_B$, respectively. Suppose further that for each $e\in B$, $G_A+e$ contains a circuit. This can happen only if the vertices adjacent to $e$ are already in $V_A$, so $V_B\subseteq V_A$. Moreover, the vertices adjacent to $e$ must lie in the same component of $G_A$, so $G_A$ and $G_B$ have the same number $k$ of components.
Since $G_A$ and $G_B$ are forests, $|B|=|V_B|-k$ and $|A|=|V_A|-k$, and therefore
$$|V_B|-k=|V_A|-k+1\,.$$
But then
$$0\le|V_A|-|V_B|=-1\,,$$
which is absurd.


text:
\begin{document}

\title{Forming the Borromean Rings out of arbitrary polygonal unknots}
  \date{\today}
\author[H.N.\ Howards]{Hugh Nelson Howards}
    \subjclass{57M25}

    \keywords{Borromean Rings, Brunnian Links, stick knots, polygonal knots}
\address{Department of Mathematics, Wake Forest University, Winston-Salem, NC 27109, USA}
\maketitle

\newtheorem{definition}{Definition}[section]
\newtheorem{corollary}[definition]{Corollary}
\newtheorem{conjecture}[definition]{Conjecture}
\newtheorem{cor}[definition]{Corollary}
\newtheorem{theorem}[definition]{Theorem}
\newtheorem{lemma}[definition]{Lemma}
\newtheorem{claim}[definition]{Claim}
\newtheorem{ex}[definition]{Example}
\newtheorem{q}[definition]{Question}
\newtheorem{exer}[definition]{Exercise}
\newtheorem{proposition}[definition]{Proposition}

\newcommand{\bi}{\begin{itemize}}
\newcommand{\ei}{\end{itemize}}
\newcommand{\be}{\begin{enumerate}}
\newcommand{\ee}{\end{enumerate}}
\newcommand{\ds}{\displaystyle}
\newcommand{\hs}{Heegaard splitting}
\newcommand{\tp}{thin position}
\newcommand{\sud}{strict upper disk}
\newcommand{\sld}{strict lower disk}
\newcommand{\db}{disk busting}
\newcommand{\si}{strongly irreducible}
\newcommand{\dbc}{double branched cover}
\newcommand{\ul}{\underline}
\newcommand{\po}{\Pi_{\omega}}

\begin{abstract}
We prove the perhaps surprising result that given any three polygonal unknots in $\R^3$, then we may form the Borromean rings out of them through rigid motions of $\R^3$ applied to the individual components together with possible scaling of the components.  We also prove that if at least two of the unknots are planar, then we do not need scaling.  This is true even for a set of three polygonal unknots that are arbitrarily close to three circles, which themselves cannot be used to form the Borromean Rings.
\end{abstract}

\section{Introduction}
\begin{figure}
	\centering
	\includegraphics[width=0.35\textwidth]{borromean.jpg}
	\caption{The Borromean Rings.}
	\label{fig:br}

	\end{figure}

The Borromean Rings in Figure~\ref{fig:br} appear to be made out of circles, but a result of Freedman and Skora shows that this is an optical illusion (see \cite{F} or \cite{H1}).
The Borromean Rings are a special type of Brunninan Link: a link of $n$ components is one which is not an unlink, but for which every sublink of $n-1$ components is an unlink.  There are an infinite number of distinct Brunnian links of $n$ components for $n \geq 3$, but the Borromean Rings are the most famous example.

This fact that the Borromean Rings cannot be formed from three circles often comes as a surprise, but then we come to the contrasting result that although it cannot be built out of circles, the Borromean Rings can be built out of certain sets of convex curves.  For example, one can form it from two circles and an ellipse.  Although it is only one out of an infinite number of Brunnian links of three components, it is the only one which can be built out of convex components \cite{H1}.  The convexity result is, in fact a bit stronger and shows that no 4 component Brunnian link can be made out of convex components and Davis generalizes this result to 5 components in \cite{D}.

While this shows it is in some sense hard to form most Brunnian links out of certain shapes, the Borromean Rings leave some flexibility.  This leads to the question of what shapes can be used to form the Borromean Rings and the following surprising conjecture of Matthew Cook at the California Institute of Technology.

\begin{conjecture} (Cook)
Given any three unknotted simple closed curves in $\R^3$, they can always be arranged to form the Borromean Rings unless they are all circles. \cite{Cook}

\label{conj:cook}
\end{conjecture}

In this paper we show that any three polygonal unknots (consisting of straight edges meeting at a set of vertices) can be used to form the Borromean Rings through rigid transformations  of the components in $\R^3$ together with scaling of $\R^3$ applied to the individual components.  Note that since any knot can be approximated with a polygonal knot that is arbitrarily close to it, any set of three unknots comes arbitrarily close to forming the Borromean rings: even three circles which themselves cannot form the Borromean Rings.

While the first few sections prove that many sets of unknots do satisfy Cook's conjecture we conclude in the final section with a possible counter-example to Cook's conjecture.

\section{The main theorems}

To prove our theorems we will need the following lemma about unknots and the disks they bound.  Throughout the paper, $K_i$ will bound a disk $D_i$ and we will abuse notation by preserving the names $K_i$ and $D_i$ even after rigid motions or scaling of the components.   When we refer to a disk or sub-disk as flat or planar, we mean that it is a subset of a flat plane.

For each polygonal unknot $K_i$
 we pick can pick an extremal vertex  (a vertex that is a global maximum with respect to some direction vector) which we will call $v_i$.  The edges adjacent to $v_i$ will be called $e_i$ and $f_i$.

\begin{lemma}
If $v_i$ is a unique global maximum for an unknot $K_i$ (with respect to some direction vector) then we may choose a disk $D_i$ whose boundary is $K_i$ and which also has the point $v_i$ as its unique global maximum in the same direction.
\label{lemma:extremaldisks}
\end{lemma}

This lemma will certainly hold for the special case of polygonal unknots that we study in this paper, but we prove it in general.

\begin{proof}
This can be done with a standard innermost loop argument.  Since there is a plane that intersects $K_i$ in $v_i$ and otherwise contains $K_i$ entirely on one side of it, we can also find
 a sphere tangent to the plane at $v_i$, intersecting the knot only in $v_i$ and which otherwise contains $K_i$ entirely inside of it (as the radius of the spheres tangent to the plane at $v_i$ goes to infinity, the spheres limit on the plane).  Now pick an embedded disk $D_i$ for $K_i$ whose interior intersects $S$ transversally in a minimal number of components.  Since $K_i \cap S$ is a single point there are no arcs of intersection in $S \cap D_i$.  This means all remaining intersections may be assumed to be circles.  If the set of circles is nontrivial take an innermost circle on $S$ (one of the components of $S \cap D_i$ that bounds a disk on $S$ disjoint on its interior from $S \cap D_i$) and
cut and paste $D_i$ replacing the component of $D_i -  (D_i \cap S)$ that is bounded by this circle and does not contain $K_i$ by the corresponding subset of $S$.  Pushing the new disk slightly off of $S$ gives a new disk $D_i'$ that intersects $S$ fewer times than $D_i$ did yielding a contradiction to the minimality assumption and showing that we may assume $D_i \cap S = v_i$.

\end{proof}

\begin{claim}
Given polygonal unknots $K_1$, $K_2$, and $K_3$  and disks $D_1$, $D_2$, and $D_3$ as in Lemma~\ref{lemma:extremaldisks} there exists an $\epsilon >0$ such that we may assume that each $D_i$ is planar in an $\epsilon$ neighborhood of $v_i$ (a subset of the plane containing edges $e_i$ and $f_i$), but such that $v_i$ is still the unique global maximum for $D_i$.
\label{claim:flat}
\end{claim}

\begin{proof} The argument is simple.  We have already shown that there is a plane that intersects $D_i$ only in $v_i$.  We may find a plane parallel to it that intersects $D_i$ only in an arc.  We may need an isotopy of $D_i$ to straighten all such arcs near $v_i$, but $K_i$ stays fixed.

\end{proof}

Translate the three knots so that each $v_i$ is at the origin. Let the closure of the complement of an $\epsilon$ neighborhood of $v_i \subset D_i$ be called $J_i$.  Recall that we have just asserted above that we may assume that $D_i - J_i$ is planar since this is just a small neighborhood of $v_i$.

\begin{lemma}
Given polygonal unknots $K_1$, $K_2$, and $K_3$, disks $D_1$, $D_2$, and $D_3$, and subsets $J_1$, $J_2$, and $J_3$  as above then we may scale $K_3$ and $D_3$ up so that every point in $J_3$ is farther from the origin than the distance of any point in $D_2$ from the origin and we may then scale up $K_1$ and $D_1$ so that every point in $J_1$ is farther from the origin than any point in $D_2$ or $D_3$.
\label{lemma:extremaldisks2}
\end{lemma}

\begin{proof}

For each $i$, let $v_i$ be at the origin and let $r_i$ be the maximum distance from the origin to any point of $D_i$.
We know by the argument above that we may pick an $\epsilon$ such that
an $\epsilon$ neighborhood of the origin in $\R^3$ intersects each $D_i$ only in a flat subset.  Scale $K_3$ up by multiplying by the three by three matrix $\lambda I$, where $\lambda > \frac{r_2}{\epsilon}$.  As mentioned earlier, we will abuse notation and call the scaled up knot and disk $K_3$ and $D_3$.  The scaling will ensure that any point on $D_3$ that is not in the planar portion is farther from the origin than any point in $D_2$.
Now scale $D_1$ and $K_1$ up similarly so that any point on $D_1$ that is not in its planar portion is farther from the origin than any point in $D_2$ or $D_3$.

\end{proof}

\begin{cor}
We may scale the knots and disks so that if each $v_i$ is sufficiently close to the origin then $D_2$ can only intersect the planar portion of $D_3 \cup D_1$ and such that $D_3 \cap D_1$ is contained in the planar portion of $D_1$.
\label{cor:planar}
\end{cor}

We now state the two main theorems of the paper.

\begin{theorem}
Let $K_1$, $K_2$, and $K_3$ be three polygonal unknots, then we may form the Borromean rings out of them through rigid motions of $\R^3$ applied to the individual components together with scaling of the components.
\label{thm:poly}
\end{theorem}

\begin{theorem}
Let $K_1$, $K_2$, and $K_3$ be three polygonal unknots, at least two of which are planar, then we may form the Borromean rings out of them through rigid motions of $\R^3$ applied to the individual components.
\label{thm:polyplanar}
\end{theorem}

The scaling of Lemma~\ref{lemma:extremaldisks2} and Corollary~\ref{cor:planar} is the only scaling necessary in the proof of Theorem~\ref{thm:poly} and no scaling is necessary in the proof of Theorem~\ref{thm:polyplanar}.  Not using Lemma~\ref{lemma:extremaldisks2} or Corollary~\ref{cor:planar}  in Theorem~\ref{thm:polyplanar} is the only difference between the two proofs.

\bigskip

{\em Proof of Theorems~\ref{thm:poly} and \ref{thm:polyplanar}:} The two proofs are nearly identical, so it will be easy to prove both at the same time.  We start by arguing that we may use rigid transformations of $\R^3$ to position $K_1$, $K_2$, and $K_3$ as they appear in  Figure~\ref{fig:origin} and then use a
translation to arrive at  Figure~\ref{fig:polygonal}.  In the case where two of the components are planar, let them without loss of generality, be $K_1$ and $K_3$.

\begin{figure}
	\centering
	\includegraphics[width=0.45\textwidth]{origin2.jpg}
	\includegraphics[width=0.45\textwidth]{vectors2.jpg}
	\caption{The Knots are initially moved so they have an extremal vertex at the origin.  $\{e_1, f_1\} \subset K_1$ lie in the $xy$-plane,  $\{e_2, f_2\} \subset K_2$ and  $\{e_3, f_3\} \subset K_3$ lie in a plane $P$ containing the $y$ axis.
	 On the right we see a vector ${\vec w}$ with its head at $v_2$ and its tail at the other vertex of $f_2$ and we see the vector
${\vec v}$  with its head at $v_3$ at the origin and its tail between $e_3$ and $f_3$.}
	\label{fig:origin}

	\end{figure}

\begin{figure}
	\centering
	\includegraphics[width=0.65\textwidth]{shifted.jpg}

	\caption{We shift $K_2$ slightly in the direction of the vector ${\vec w}$ and $K_3$ slightly in the direction of ${\vec v}$.}
	\label{fig:polygonal}

	\end{figure}

We initially position $v_1$, $v_2$, and $v_3$  at the origin. To be specific, for $K_1$  place $e_1$
 in the $xy$-plane so that $v_1$ is at the origin and $e_1$ lies on the (negative)  $y$-axis.  Fixing this edge rotate $K_1$ until  $f_1$ lies in the $xy$-plane and has positive values for $x$-coordinates aside from at the point $v_1$ which has $x$ coordinate 0.

We will arrange $K_2$ and $K_3$ so that $e_2$, $f_2$, $e_3$, and $f_3$ all contain the origin, and are all coplanar in a plane $P$ that contains the $y$ axis, such as the $xz$-plane.  This requires a little work since we also care about the intersections of the $D_i$'s.

Place $K_2$ so that $v_2$ is a global maximum for $K_2$ with respect to $z$,  and lies at the origin.  Rotate it around the $z$ axis so that $P_2$, the plane containing $e_2$ and $f_2$ contains the $y$ axis.  $v_2$, of course, remains a global maximum.

Place $K_3$ so that $v_3$ is a global minimum for $K_3$ with respect to $z$,  and lies at the origin.  Rotate it around the $z$ axis so that $P_3$, the plane containing $e_3$ and $f_3$ contains the $y$ axis.

If we may choose $P_2=P_3$ while preserving the above properties, we do so.  If not, and $P_3$ is steeper than $P_2$
rotate them around the $z$ axis so that the upper half plane of each $P_i$ (the portion above the $xy$-plane) has non-positive $x$-coordinates.
If $P_2$ is steeper than $P_3$ rotate them so that the upper half plane of each $P_i$ has non-negative $x$-coordinates.
 $v_2$ and $v_3$ remain extremum for $K_2$ and $K_3$ respectively
and each $P_i$ still contains the $y$-axis.

If two of the knots are planar, recall that we chose them to be $K_1$ and $K_3$.  Since $D_3$ is a flat disk totally contained in $P_3$, and neither $D_1$ nor $D_2$ have points with positive $z$ coordinates, we may rotate $K_3$, $D_3$, and $P_3$ around the $y$ axis until $P_3 = P_2$ without introducing any new intersections of the disks.  If the knots are not planar then we now apply Lemma~\ref{lemma:extremaldisks2} to scale the knots.

Now a rotation of $K_3$, $D_3$, and $P_3$ around the $y$-axis through the acute angle between $P_2$ and $P_3$ keeps $v_3$ at the origin and preserves the properties of  Corollary~\ref{cor:planar}.  We have set up the planes so that no matter whether the upper half plane of $P_3$ was above or below the upper half-plane of $P_2$, the rotation goes in the same direction.  It is a clockwise rotation around the $y$-axis if looking towards the origin from a point on the positive $y$-axis.  Equivalently a point of the form $(x,y,z)$ with $z>0$ will see its $x$ coordinate decrease as a result of the rotation.

As we rotate $P_3$ around the $y$-axis onto $P_2$, the total rotation is less than 90 degrees and any points other than $v_3$ where $D_3$ now intersects the $xy$-plane have negative $x$ values since they all had positive $z$ values before the rotation.  Thus they are disjoint from the planar portion of $D_1$, which had only non-negative $x$ values.  We have, of course, asserted that this is the only portion of $D_1$ which $D_3$ can intersect.  Similarly by Corollary~\ref{cor:planar} the only points of $D_3$ that could intersect $D_2$ are in the planar portion of $D_3$ and aside from $v_3$ this remains above the $xy$-plane.  Since $D_2$ is strictly below the $xy$-plane aside from $v_2$, we know that even after the rotation the only intersection of $D_2$ and $D_3$ must be at the origin.

Thus we may now assume that $P_2=P_3$ and we rename the new plane $P$.
  Thus $D_i \cap D_j$ is the origin for $i \neq j$ and $e_2$, $f_2$, $e_3$ and $f_3$ all lie in the same plane $P$.  Note that $v_3$ may no longer be a global minimum with respect to $z$, but this is not a problem since we now completely understand the intersection patterns of the disks.

We do not want any of $e_2$, $f_2$, $e_3$ and $f_3$ to be collinear, but by general position, this may be ensured by rotating one of the knots by $\epsilon$ around the line perpendicular to $P$ and through the origin without creating any new intersections.

 Note that $J_s \cap D_t = \empty set$ for $s \neq t$.  Since each $D_s$ and $J_t$ is  a pair of disjoint compact sets we may find a minimum distance $d$ from $J_s$ to $D_t$ over all $s \neq t$.
For the rest of the paper we will make sure that no $D_s$ is moved more than $d/2$ and thus this disjoint property will be preserved.  {\em Thus from here on out $D_s \cap D_t$ for $s \neq t$ will only occur in the flat triangular portions of the disks that lie in an $\epsilon$ neighborhood of the origin.}

\medskip

We have completed the only scaling we need in the proof of Theorem~\ref{thm:poly} and no scaling is needed in the proof of
Theorem~\ref{thm:polyplanar}.   Otherwise the proofs of the two theorems are identical.  Although $P$ may not be the $yz$-plane, we will picture it in this manner in the figures since that will not impact the future arguments (we only use the fact that $P$ exists and contains the $y$-axis, not the specific angle it makes with the $xy$-plane).

Let $f_2$ be the least steep edge from the collection $\{e_2, f_2, e_3, f_3 \}$ (the absolute value of the slope of $f_2$
in $P$  is less than the absolute value of the slopes of $e_2$, $e_3$ and $f_3$ in the same plane).  If we chose the knots wisely from the start this is a safe assumption, but if not this may require returning to the start of the argument and relabeling of $K_2$ and $K_3$ and going through the above steps, all of which will still work fine and lead to this desired steepness result.

We now fix $K_1$ for the rest of the proof and move the other two knots slightly starting with $K_2$.

\begin{claim}
Given  disks $\{D_1, D_2, D_3\}$ intersecting only at the origin and bounded by knots $\{K_1, K_2, K_3\}$ as above, then for any $\epsilon >0$, let $\{J_1, J_2, J_3\}$ be equal to $\{D_1, D_2, D_3\}$ minus the portion of the $D_i$'s in an open $\epsilon$ ball around the origin, and given any translation of $\R^3$ acting on a given $D_i$ or any rotation of that $D_i$ about a fixed axis $l$, then there exists an $\epsilon'$ such that any translation of distance less than $\epsilon'$ leaves the $J_i$'s pairwise disjoint.
Similarly there exists an $\alpha > 0$ such that any rotation about $l$ of angle less than $\alpha$ leaves the $J_i$'s pairwise disjoint.
\label{claim:simpint}

\end{claim}

{\em Proof of Claim~\ref{claim:simpint}:}
Since $J_i \cap J_j = \empty set$ and they are both compact, there is a positive minimal distance $s$ between any point in $J_i$ and $J_j$.  Setting $\epsilon' < s$  will ensure that translating one of the disks less than $\epsilon'$ cannot create an intersection.  Similarly after fixing an axis of rotation we can pick a small enough angle $\alpha$ such that rotating $J_i$ will move no point of $J_i$ more than $s/2$ completing the proof of the claim.

\medskip

We may pick any $\epsilon > 0$ and be certain that no point on any disk will be moved more than $\epsilon$ for the rest of the proof.
The virtue of Claim~\ref{claim:simpint} is that we now know that during all our remaining manipulations of the disks and knots the existing components of intersection may change in size and shape (or even go away), but no new intersections will be introduced.  Also all intersections will remain in an $\epsilon$ ball neighborhood of the origin and on the flat triangular pieces of the $D_i$'s running between $e_i$ and $f_i$ (i.e. the complement to $J_i$ in $D_i$) .

Let $ {\vec w}$ be the vector with its head at $v_2$ and parallel to $f_2$ (its tail may be thought of as lying on the other vertex of $f_2$) as in Figure~\ref{fig:origin}.  Translate $K_2$ by adding $\epsilon * {\vec w}$ to every point on $K_2$ for a sufficiently small $\epsilon$ in order to translate $K_2$  (and $D_2$) minimally up in a direction parallel to $f_2$. We want to be certain that $e_2 \cap e_1$ remains nontrivial and that no new intersections are introduced outside of a neighborhood of the origin.
By Claim~\ref{claim:simpint} choosing a sufficiently small $\epsilon$ will ensure all of these properties, as would any positive translation
smaller than $\epsilon$.
Now
$v_2$ is very close to, but above $v_1$, $f_2$ intersects the origin ($v_1$), and $e_2$ intersects $e_1$ in some point other than $v_1$.  $K_1$ and $K_2$ look as they do in Figure~\ref{fig:polygonal} and we need to reposition $K_3$ to match the figure.

The fact that $f_2$ is not as steep as $e_3$ and $f_3$ ensures that  both $e_3$ and $f_3$ are on the same side of the line containing $f_2$ in $P$.  $f_2$ being less steep than $e_2$ and on its right in $P$ ensures that the points of $f_2$ all have non-negative $y$-value.
Let ${\vec v}$ be a vector with its head on $v_3$ at the origin and its tail between $e_3$ and $f_3$ as in Figure~\ref{fig:origin}.
Translate $K_3$
by $\delta * {\vec v}$ for a small $\delta$.  This
keeps $e_3$ and $f_3$ in $P$.
Choosing a sufficiently small $\delta$  again makes sure that all changes in the intersections of the disks occur in a neighborhood of the origin, that
$K_2 \cap K_3$ consists of exactly two points, $e_3 \cap f_2$ and $f_3 \cap f_2$.
Finally to complete the figure pick $l$, a line in $P$ parallel to $f_2$, but separating $f_2$ from $v_3$.  Let $l \cap e_3$ be called $p_e$ and let $l \cap f_3$ be called $p_f$.  We must pick $l$ close enough to $f_2$ so that $p_e$ is above $e_1$.  This is easy to do since  $e_1$ lies on the $y$-axis and we need only make sure that $p_e$ has positive
$z$-coordinate.  The point $e_3 \cap f_2$ has positive $z$ coordinate so if $l$ is sufficiently close to $f_2$ then $p_e$ will, too.  On the other hand, $f_2 \cap f_3$  has negative $z$-coordinate, and $p_f$ is below this point, $p_f$ will have negative $z$-coordinate.

Recall that we have already established  that in Figure~\ref{fig:polygonal} we may assume the $D_i$ are all disjoint from the neighborhood of the origin depicted except in the obvious flat triangular  sub-disks and that the $D_i$ are disjoint from each other outside of the figure.

\begin{figure}
	\centering
	\includegraphics[width=0.45\textwidth]{k12.jpg}
	\includegraphics[width=0.45\textwidth]{k13.jpg}
	\includegraphics[width=0.45\textwidth]{k23.jpg}
	\caption{The initial intersections of the disks $D_1 \cap D_2$, $D_1 \cap D_3$, and $D_2 \cap D_3$
	respectively are shown in gold. They are not yet in general position.}
	\label{fig:brstick4}
	\label{fig:originint}

	\end{figure}

Now we want to put the knots and disks in general position. This process will take us from Figure~\ref{fig:polygonal} to Figure~\ref{fig:brstick}.  Because general position is always easy to attain with infinitesimally small transformations we can, as mentioned earlier, pick a small number $\epsilon$ and no point of the knots or disks will move more than $\epsilon$ over the rest of the proof. This ensures that the only new intersection patterns between the disks will be the result of local changes in the current intersection patterns.

\begin{figure}
	\centering
	\includegraphics[width=0.65\textwidth]{rotated.jpg}
	\caption{$K_3$ has been rotated and the Borromean rings will be formed once we translate $K_2$ slightly in the direction of horizontal vector  ${\vec u}$ which lies in the $xy$-plane with its tail at the origin and its head between $e_1$ and $f_1$.}
	\label{fig:rotated}

	\end{figure}

\begin{figure}
	\centering
	\includegraphics[width=0.65\textwidth]{brs.jpg}
	\caption{The Borromean Rings.}
	\label{fig:brstick}

	\end{figure}

Rotate $K_3$ around $l$ so that the $x$ coordinate of $v_3$ decreases and so that $D_3$ is in general position with respect to both $D_1$ and $D_2$ (although $D_1$ and $D_2$ are still not in general position with respect to each other).  Rotating by a small enough angle will ensure that no point on $K_3$ or $D_3$ moves more than $\epsilon$.
The rotation will fix $p_e$ and $p_f$, will cause all the points on the same side of $l$ as $v_3$ to have decreasing $x$ coordinates and all the  points of $e_3 \cup f_3$ on the other side to have increasing $x$ coordinates.
Before rotating, $D_3$ intersected $D_1$ in a single arc, a subset of $e_1$  running from $e_3 \cap e_1$ to $v_1$, including the single point
$l \cap D_1$.  After the rotation $D_1 \cap D_3$ will remain an arc, $l \cap D_1$ will be one endpoint, and the arc of intersection will rotate about this point.  The other end point will move away from the origin ($v_1$) to a point on $f_1$ with positive $x$ coordinate as in Figure~\ref{fig:brstick3}.

Before rotating $D_3$, $D_2 \cap D_3$ was a triangular subset of $P$ formed by intersecting the triangle subset of $D_2$ running from $e_2$ to $f_2$ and the analogous triangle on $D_3$ from $e_3$ to $f_3$.  The two triangles and thus the intersection contained the portion of $l$ running from $p_e$ to $p_f$.  After rotating this portion of $l$ will be the only portion of $D_3$ near the origin contained in $P$.  Since we have already established that all intersections will occur near the origin this means that $D_2 \cap D_3$ is exactly the arc of $l$ running from $p_e$ to $p_f$.   Now $D_3$ is in general position with respect to both $D_1$ and $D_2$.

Finally we must translate $K_2$ and $D_2$ slightly  so that $D_1 \cap D_2$ is in general position.  This will move $D_2 \cap D_3$ infinitesimally, but since these two disks are already in general position and the move will be minimal it will not be enough to change the intersection pattern of those two topologically so for our purposes we may think of it as essentially unchanged.  Before translating $D_2 \cap D_1$ is the subset of the edge $e_1$ running from $e_1 \cap e_2$ to $v_1 = e_1 \cap f_2$.
  Let ${\vec u}$ be a vector in the $xy$-plane with tail at the origin ($v_1$) and head on the triangular portion of $D_1$ between $e_1$ and $f_1$ as in Figure~\ref{fig:rotated}.  Translate $K_2$ by $\rho * {\vec u}$
where $|\rho * {\vec u}|$ is small enough to satisfy Claim~\ref{claim:simpint}.
Since $D_2$ and $D_3$ were in general position and our translation was minimal,
 $D_2 \cap D_3$ remains an arc as before (although it is no longer a subset of $l$).  $D_1$ and $D_2$ now are in general position and $D_1 \cap D_2$ becomes an arc from $e_2 \cap D_1$ to $f_2 \cap D_1$ that is parallel to, but now disjoint from $e_1$.

   All the disks are now in general position and the link looks locally like Figure~\ref{fig:brstick}.   The disks now each intersect the union of the other two in a cross as in Figure~\ref{fig:brstick5}.  It is not hard to show that this intersection pattern can only result from the Borromean Rings.  See, for example, \cite{H2}.  This concludes the proof of the two main theorems.

\bigskip

Although no set of three circles can be used to form the Borromean rings, Theorem~\ref{thm:polyplanar} has the following interesting Corollary.

\begin{cor}
Given any three circles, $C_1$, $C_2$, and $C_3$ and any $\epsilon >0$ there exists unknots
$K_1$, $K_2$, and $K_3$ with each $K_i$ contained in an $\epsilon$ tubular neighborhood of $C_i$  and isotopic to $C_i$ in that neighborhood, such that the Borromean Rings may be formed from $K_1 \cup K_2 \cup K_3$ through rigid motions of the components in $\R^3$.
\end{cor}

This follows immediately by picking planar polygonal unknots arbitrarily close to each $C_i$.  It then shows that while the Borromean Rings can't be formed out of three circles, they can in some sense come arbitrarily close to forming them.

\begin{figure}
	\centering
	\includegraphics[width=0.45\textwidth]{k1k2.jpg}
	\includegraphics[width=0.45\textwidth]{k2k3.jpg}
\includegraphics[width=0.45\textwidth]{k1k3.jpg}
	\caption{The Intersections of the disks in the final format $D_1 \cap D_2$, $D_1 \cap D_3$, and $D_2 \cap D_3$
	respectively are shown in black. Note that each time the black arc has end points on one of the knots (in the top left $K_1$, top right $K_2$, and bottom $K_3$) and on the interior of the disk bounded by the other knot.
 If all three arcs were drawn in the same picture we would see that the top two form a cross in the $xy$-plane intersecting in a single point on the interior of both arcs.  The third arc intersects the $xy$-plane
they would cross in a single point on its interior, which also is the unique point where it intersects the other two arcs. }
	\label{fig:brstick3}
	\end{figure}

\begin{figure}
	\centering
	\includegraphics[width=0.25\textwidth]{cross.jpg}
	\caption{$D_r \cap (D_s \cup D_t)$ looks like the figure above
	for any distinct choices of $r,s,t \in \{1,2,3\}$.  This can only happen in the case of the
	Borromean rings.}
	\label{fig:brstick5}

	\end{figure}

\section{Conjectures, open questions, and a possible counter-example to Conjecture~\ref{conj:cook}}

In spite of our evidence in {\em partial} support of Cook's conjecture, we now state a conjecture of our own that would contradict it.

\begin{conjecture}
It is possible to find three unknotted curves, one of which is not a circle that cannot be used to form the Borromean rings. (Here scaling is not allowed.)
\end{conjecture}

Jason Cantarella suggests the following example as likely to satisfy this new conjecture (and arguably a counter-example to Cook's original conjecture).  Let $T$, shown in Figure~\ref{fig:torus}, be a torus that can be parameterized as follows:

$x= 5.1*\cos(\theta)+5*\cos(\psi)*\cos(\theta), y= 5.1*\sin(\theta)+5*\cos(\psi)*sin(\theta), z=5*\sin(\psi)], 0 \leq \psi \leq  2*\pi,  0 \leq \theta \leq  2*\pi$.

Let $L$ consist of two circles $K_1$ and $K_2$ of radius 10 together with an unknot $K_3$ that is isotopic {\em on $T$} to an $(n,1)$ torus knot, but consisting of $n$ arcs of meridian circles outside of a small neighborhood of the origin and then short arcs on the torus connecting adjacent arcs to complete a single knot.  An example of such a knot with $n=18$ is shown in Figure~\ref{fig:untorus}.  The knot consists of meridian curves like those in in Figure~\ref{fig:torus} away from a neighborhood of the origin.  Within a neighborhood of the origin there will be short, non-circular arcs.   These are in the dense central portion of the bottom picture in  Figure~\ref{fig:torus}.

Now because of the relatively large radii of $K_1$ and $K_2$ compared to the size of $T$, from their perspective the exposed portions of $K_3$ consist exclusively of arcs of circles.  The interactions in the construction of Theorem~\ref{thm:poly} are local and cannot work in this situation for the same reasons it would not work with three circles.  It is not as clear that the construction from Theorem~\ref{thm:polyplanar}, however, could not work, where scaling was allowed.  If we are allowed to scale the components, then if we scale $K_3$ up enough, then the non-circular portions become exposed to the other knots.  This, therefore, might be a counter-example to  Conjecture~\ref{conj:cook} if  scaling is not permitted, but fails to produce a counterexample if it is.   The exact phrasing of Cook's conjecture is ambiguous since the word ``arranged" could be interpreted to allow scaling or it could be interpreted not to, but it seems more likely to exclude scaling.    It is highly possible that this nuance could be the difference between the conjecture being true or false!

It is also worth noting that while from afar Figure~\ref{fig:untorus} appears to consist of arcs of circles as desired, due to the nature of computer generated images, if you truly look closely the knot pictured consists of straight segments that are very, very short.  No matter how short they are, this means that two computer generated ``circles," which also would consist of short straight segments, but would look like circles, together with the knot pictured would in truth be able to form the Borromean Rings without scaling by Theorem~\ref{thm:polyplanar}.  This shows how subtle the line is between knots that can form the Borromean Rings and those that cannot.

\begin{figure}
	\centering
	\includegraphics[width=0.75\textwidth]{torus.jpg}
	\caption{A torus.}
	\label{fig:torus}

	\end{figure}

\begin{figure}
	\centering
	\includegraphics[width=0.65\textwidth]{torusunknot1.jpg}
	\includegraphics[width=0.65\textwidth]{torusunknot2.jpg}
	\includegraphics[width=0.65\textwidth]{torusunknot3.jpg}
	\caption{Various perspectives of an unknot embedded on a torus.}
	\label{fig:untorus}

	\end{figure}

We conclude with a few more open questions and conjectures.

\begin{conjecture}
Any three planar curves can be used to form the Borromean rings as long as at least one is not a circle.
\end{conjecture}

Planar was convenient and was necessary at times for the proofs in \cite{H3}, where it is shown that any three planar unknots (here they need not by polygonal) can always be used to form the Borromean rings through rigid transformations and scaling as long as one of them is not convex, but it is not clear that the theorem fails without it even if this proof does.

\begin{q}
Can any three unknots can be used to form the Borromean rings through rigid transformations and scaling applied to the individual components as long as at least one is not a circle?
\end{q}

Thanks to Jason Cantarella for the idea behind the link represented in Figure~\ref{fig:untorus} and to Matt Mastin for generating the images in Figure~\ref{fig:untorus}.

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\centerline{\textsc{References cited}}

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