Mark J. Penn wrote Microtrends: The Small Forces Changing the World. He argues that the most important trends in the world today are the smallest ones. Such as… declining standards in math education!

What should you do on the educational front if you have a child with an aptitude for numbers, as mine does? Both of you had better get cracking, because American college students are studying less math. As an example, “Microtrends” says Harvard has only 77 math majors out of 6,700 undergraduate students.
The math story is different in China and India, which are graduating as many as 950,000 engineers a year. Granted, both nations are far more populous than the United States, but that is a lot of engineers.
Mr. Penn notes that a 2001 bipartisan commission “said that the greatest threat to American national security - behind only terrorist attacks - was the threat of failing to provide sufficient math and science education in America.”

I haven’t read the book yet but it’s high on my wish-list after reading the NYT-article Why There’s Strength in Small Numbers and the Introduction of the book.

Via PD1, who told me the story on her 23rd birthday, yesterday.

The story starts with Alex Matter, whose father, Herbert, and mother, Mercedes, were artists and friends of Jackson Pollock, famous for his drip-paintings. He discovered a group of small drip paintings in a storage locker in Wainscott, N.Y. which he believed to be authentic Pollocks, and if he is proved right, they would be worth millions of dollars.

Usually such discoveries lead to heated debates among art-critics and Pollock-experts whether one finds proof to authenticate the paintings. Not this time. In steps a mathematician who claims that he can authenticate a Pollock drip-painting by calculating its fractal dimension (??!!)… and claims that these drippings cannot be Pollocks because their dimension is too small… LOL!

This madmatician is Richard Taylor from the University of Oregon in Corvallis.

Taylor took a digital image of a Pollock painting into his lab, broke the image into its separate colors, and computed the fractal dimension of the lines in each color. Each time, he got a number between 1 and 2, confirming his notion that Pollock’s paintings are fractal. “Rather than mimicking nature,” Taylor says, Pollock “adopted its language of fractals to build his own patterns.” In 1999, Taylor reported that the fractal dimension of Pollock’s paintings increased during his life. His early drip paintings have a loose web of lines, mostly at the same scale. Because these paintings show no fractal qualities, their dimension is near 1. But Pollock’s later paintings have a dense network of overlapping lines, ranging from large, bold strokes to delicate threads, Taylor calculated a fractal dimension of 1.72 for these works.

His paper on this “Authenticating Pollock Paintings Using Fractal Geometry” can be found here. Luckily, the story doesn’t end here. In steps a graduate student in astrophysics at Case Western, Katherine Jones-Smith who had to give a seminar talk to her fellow students.

“I was sort of bored with particle astrophysics,” Jones-Smith says, so she looked around for something different. She came across an account of Taylor’s work, and “it sounded really cool,” she recalls. “The obvious check to me was to make sure that not any old scribble would appear to be fractal,” she says. “So, I made some scribbles.” Much to her surprise, when she computed the fractal dimension of her scribbles, they turned out to be greater than 1.

Recently, she arXived her findings in the paper Drip Paintings and Fractal Analysis from which the above doodles are taken, called appropriately “Gross pebbles” and “Mixed stars”.

When Katherine Jones-Smith made some doodles on a page “”pretty ugly” ones, she says”she found that they shared the qualities of a Pollock, according to an analysis that follows Taylor’s approach. “Either Taylor is wrong, or Kate’s drawings are worth $40 million,” says Jones-Smith’s collaborator Harsh Mathur. “We’d be happy either way.“

More on this hilarious story can be found in this science news article, this New-York times story or the Pollocks-bollocks blog entry over at biophemera.

Last time we saw that the algebra $(\OmegaV~C Q,Circ)$ of relative differential forms and equipped with the Fedosov product is again the path algebra of a quiver $\tilde{Q}$ obtained by doubling up the arrows of $Q$. In our basic example the algebra map $C \tilde{Q} \rightarrow \OmegaV~C Q$ is clarified by the following picture of $\tilde{Q}$
(which generalizes in the obvious way to arbitrary quivers). But what about the other direction $\OmegaV~C Q \rightarrow C \tilde{Q}$ ? There are two embeddings $i,j : C Q \rightarrow C \tilde{Q}$ defined by $i : (u,v) \rightarrow (a,x)$ and $j : (u,v) \rightarrow (b,y)$ giving maps
$\forall a \in C Q~:~p(a) = \frac{1}{2}(i(a)+j(a))~\quad~q(a) = \frac{1}{2}(i(a)-j(a))$ Using these maps, the isomorphism $\OmegaV~C Q \rightarrow C \tilde{Q}$ is determined by $ a0 da1 \hdots dan \rightarrow p(a0)q(a1) \hdots q(an)$ In particular, $p$ gives the natural embedding (with the ordinary multiplication on differential forms) $C Q \rightarrow \OmegaV~C Q$ of functions as degree zero differential forms. However, $p$ is no longer an algebra map for the Fedosov product on $\OmegaV~C Q$ as $p(ab) = p(a)Circ p(b) + q(a) Circ q(b)$. In Cuntz-Quillen terminology, $\omega(a,b) = q(a) Circ q(b)$ is the _curvature of the based linear map $p$. I\’d better define this a bit more formal for any algebra $A$ and then say what is special for formally smooth algebras (non-commutative manifolds). If $A,B$ are $V = C \times \hdots \times C$-algebras, then a $V$-linear map $A \rightarrow^l B$ is said to be a based linear map if $ l | V = idV$. The _curvature of $l$ measures the obstruction to $l$ being an algebra map, that is $\forall a,b \in A~:~\omega(a,b) = l(ab)-l(a)l(b)$ and the curvature is said to be nilpotent if there is an integer $n$ such that all possible products $\omega(a1,b1)\omega(a2,b2) \hdots \omega(an,bn) = 0$ For any algebra $A$ there is a universal algebra $R(A)$ turning based linear maps into algebra maps. That is, there is a fixed based linear map $A \rightarrow^p R(A)$ such that for every based linear map $A \rightarrow^l B$ there is an algebra map $R(A) \rightarrow B$ making the diagram commute In fact, Cuntz and Quillen show that $R(A) \simeq (\OmegaV^{ev}~A,Circ)$ the algebra of even differential forms equipped with the Fedosov product and that $p$ is the natural inclusion of $A$ as degree zero forms (as above). Recall that $A$ is said to be _formally smooth if every $V$-algebra map $A \rightarrow^f B/I$ where $I$ is a nilpotent ideal, can be lifted to an algebra morphism $A \rightarrow B$. We can always lift $f$ as a based linear map, say $\tilde{f}$ and because $I$ is nilpotent, the curvature of $\tilde{f}$ is also nilpotent. To get a uniform way to construct algebra lifts modulo nilpotent ideals it would therefore suffice for a formally smooth algebra to have an algebra map $A \rightarrow \hat{R}(A)$ where $\hat{R}(A)$ is the $\mathfrak{m}$-adic completion of $R(A)$ for the ideal $\mathfrak{m}$ which is the kernel of the algebra map $R(A) \rightarrow A$ corresponding to the based linear map $id_A : A \rightarrow A$. Indeed, there is an algebra map $R(A) \rightarrow B$ determined by $\tilde{f}$ and hence also an algebra map $\hat{R}(A) \rightarrow B$ and composing this with the (yet to be constructed) algebra map $A \rightarrow \hat{R}(A)$ this would give the required lift $A \rightarrow B$. In order to construct the algebra map $A \rightarrow \hat{R}(A)$ (say in the case of path algebras of quivers) we will need the Yang-Mills derivation and its associated flow.

The previous post in this sequence was moduli spaces. Why did we spend time explaining the connection of the quiver

to moduli spaces of vectorbundles on curves and moduli spaces of linear control systems? At the start I said we would concentrate on its double quiver Clearly, this already gives away the answer : if the path algebra $C Q$ determines a (non-commutative) manifold $M$, then the path algebra $C \tilde{Q}$ determines the cotangent bundle of $M$. Recall that for a commutative manifold $M$, the cotangent bundle is the vectorbundle having at the point $p \in M$ as fiber the linear dual $(Tp M)^$ of the tangent space. So, why do we claim that $C \tilde{Q}$ corresponds to the cotangent bundle of $C Q$? Fix a dimension vector $\alpha = (m,n)$ then the representation space
$\mathbf{rep}{\alpha}~Q = M{n \times m}(C) \oplus Mn(C)$ is just an affine space so in its point the tangent space is the representation space itself. To define its linear dual use the non-degeneracy of the _trace pairings $M{n \times m}(C) \times M{m \times n}(C) \rightarrow C~:~(A,B) \mapsto tr(AB)$ $Mn(C) \times Mn(C) \rightarrow C~:~(C,D) \mapsto tr(CD)$ and therefore the linear dual $\mathbf{rep}_{\alpha}~Q^ = M{m \times n}(C) \oplus Mn(C)$ which is the representation space $\mathbf{rep}{\alpha}~Q^s$ of the quiver

and therefore we have that the cotangent bundle to the representation space $\mathbf{rep}{\alpha}~Q$ $T^* \mathbf{rep}{\alpha}~Q = \mathbf{rep}{\alpha}~\tilde{Q}$ Important for us will be that any cotangent bundle has a natural symplectic structure. For a good introduction to this see the course notes “Symplectic geometry and quivers” by Geert Van de Weyer. As a consequence $C \tilde{Q}$ can be viewed as a non-commutative symplectic manifold with the symplectic structure determined by the non-commutative 2-form
$\omega = da^* da + dx^* dx$ but before we can define all this we will have to recall some facts on non-commutative differential forms. Maybe next time. For the impatient : have a look at the paper by Victor Ginzburg Non-commutative Symplectic Geometry, Quiver varieties, and Operads or my paper with Raf Bocklandt Necklace Lie algebras and noncommutative symplectic geometry. Now that we have a cotangent bundle of $C Q$ is there also a tangent bundle and does it again correspond to a new quiver? Well yes, here it is
and the labeling of the arrows may help you to work through some sections of the Cuntz-Quillen paper…

In the previous part we saw that moduli spaces of suitable representations of the quiver locally determine the moduli spaces of vectorbundles over smooth projective curves. There is yet another classical problem related to this quiver (which also illustrates the idea of looking at families of moduli spaces rather than individual ones) : linear control systems. Such a system with an $n$ dimensional state space and $m$ controls (or inputs) is determined by the following system of linear differential equations $ \frac{d x}{d t} = A.x + B.u$ where $x(t) \in \C^n$ is the state of the system at time $t$, $u(t) \in \C^m$ is the control-vector at time $t$ and $A \in Mn(\C), B \in M{n \times m}(\C)$ are the matrices describing the evolution of the system $\Sigma$ (after fixing bases in the state- and control-space). That is, $\Sigma$ determines a representation of the above quiver of dimension-vector $\alpha = (m,n)$

Whereas in control theory (see for example Allen Tannenbaum\’s Lecture Notes in Mathematics 845 for a mathematical introduction) it is natural to call two systems equivalent when they only differ up to base change in the state-space, one usually fixes the control knobs so it is not natural to allow for base change in the control-space. So, at first sight the control theoretic problem of classifying equivalent systems is not the same problem as classifying representations of the quiver up to isomorphism. Fortunately, there is an elegant way round this which is called deframing. That is, for a fixed number $m$ of controls one considers the quiver $Qf$ having precisely $m$ arrows from the first to the second vertex and the system $\Sigma$ does determine a representation of this new quiver of dimension vector $\beta=(1,n)$ by assigning to the arrows the different columns of the matrix $B$. Isomorphism classes of these quiver-representations do correspond precisely to equivalence classes of linear control systems. In part 4 we introduced stable and semi-stable representations. In this framed-quiver setting call a representation $(A,B1,\hdots,Bm)$ stable if there is no proper subrepresentation of dimension vector $(1,p)$ for some $p \lneq n$. Perhaps remarkable this algebraic notion has a counterpart in system-theory : the systems corresponding to stable quiver-representations are precisely the completely controllable systems. That is, those which can be brought to any wanted state by varying the controls. Hence, the moduli space
$M^s{(1,n)}(Qf,\theta)$ classifying stable representations is exactly the moduli space of completely controllable linear systems studied in control theory. For an excellent account of this moduli space one can read the paper [Introduction to moduli spaces associated to quivers by Christof Geiss. Fixing the number $m$ of controls but varying the dimensions of teh state-spaces one would like to take all the moduli spaces $ \bigsqcupn~M^s{(1,n)}(Qf,\theta)$
together as they are all determined by the same formally smooth algebra $\C Qf$. This was done in a joint paper with Markus Reineke called Canonical systems and non-commutative geometry in which we prove that this disjoint union can be identified with the infinite Grassmannian $ \bigsqcupn~M^s{(1,n)}(Qf,\theta) = \mathbf{Gras}m(\infty)$ of $m$-dimensional subspaces of an infinite dimensional space. This result can be seen as a baby-version of George Wilson\’s result relating the disjoint union of Calogero-Moser spaces to the adelic Grassmannian. But why do we stress this particular quiver so much? This will be partly explained next time.