Welcome!

This is a mathematics blog of Colin Tan. I work on positivity and sums of squares with my advisor To Wing Keung at the math department of the National University of Singapore.

A description of my current research can be found  at my wiki.

I am recently married to Gao Man.

Sequentially Resolved Pointsets

Def. A sequentially resolved pointset is a set together with a set of sequences in , declared as convergent sequences, that is closed under subsequences.

Def. A subset of a sequentially resolved pointset is sequentially closed if the following holds: If a sequence in is converges and is equivconvergent with a constant sequence, then this constant sequence lies in .

Manifold

n. A manifold is a space patched up from local constructions. It is an object of the (a?) topos containing these local patches.

Space

n. An object in a topos.

What is the area of a circle?

In spite of differences of culture, practice, and theological expression, each faith leads to the ultimate reality we call God. The lessons we need to learn. . . are to practice religious tolerance and to sincerely follow your chosen deity.

Quoted from “Eastern Wisdom for Western Minds” by Parachin and Dy.

Meditating on any mathematical problem should lead to infinity, the central discourse of mathematics.  The choice of problem should not matter.

In the following blog post, I will meditate on the area of the circle. So what is the area of a circle? I know, and you know, that the area of a circle is , where is the radius of the circle. Now, what is ? Mathematicians declare as the ratio of the circumference to the diameter. Ah. The question I’m really after is “What is the area of a circle in terms of its circumference (and radius)?”

Piecewise Linear Curve

To answer this question, I need to articulate what “Area of a circle” and “length of the circumference” is. How should I define the length of a curve? In the first place, what is the length of a line segment? Let me draw a line segment.  Then I lay the straight segment out. 

Straightened out

I will identify a piecewise linear curve  with its straightened version . Let me say this in better English by using “straighten” as a verb. I straighten piecewise linear curve out into a line segment. Okay. How do I add in the ”identification” part? I declare straigtening as an equivalence.

Let us explore the degenerate case. Among the piecewise linear curves, is the curve with no kinks — a line segment. 

A line segment -- no kinks!

A line segment  is a degenerate piecewise linear curve. It is already straight so straightening out does nothing: .

Suppose  and are as in the following diagram.

Compare Line Segments

I declare that if I can rigidly move to be totally contained in .

In this way I can compare line segments.

My identification imposes the relation too, allowing me to compare line segments.

Definition. The length of a piecewise linear curve  is the equivalence class of . 

Interesting Research Problems

A list of interesting research articles.

1. The authors construct exceptional collections in the derived category of coherent sheaves on a compact homogeneous space whose symmetries are semisimple algebraic of types B, C and D. I could extend their construtions to the exceptional groups.

Fundamental Theorem of Algebra

Thm. The image of a complex polynomial function is either a point or the entire complex plane.

Lem 1. If a continuous map can be extended asymptotically to the circle at infinity , then is null-homotopic.

Symbolically, the map for .

Pf of Lem 1. Fold the plane into a cone whose origin as vertex. This decomposes the plane into a family of circles parametrized by its radius.

Adding the circle of infinite radius compactifies the plane as . But is homeomorphic to , so homotopes to , a constant map. QED.

Lem 2. If the maps converge asmptotocially at the circle at infinity, then converge asymptotically at the circle at infinity. Here is the homotopy equivalence .

Lem 3. If the power map is null-homotopic, then .

Pf of Thm. Suppose that is a complex polynomial whose image is not the entire complex plane. Thus there must be a point excluded from the image. By a suitable translation (that is consider instead of ), I can assume that this point is the origin. Therefore, I have a polynomial map to the punctured complex plane. To prove the assertion of the theorem, I have to show that the image of is a point.

Let . Then . Applying Lem 2 tells me that extends to the power map asmyptotically at the circle of infinity. Applying Lem 1 tells me that this power map is null-homotopic. Applying Lem 3 tells me that . Therefore is a constant polynomial and has a single point as image. QED.

Coro. The complex field is algebraically closed.

Algebraically Closed

adj. The free monoid on linear polynomials quantifies the polynomial ring. 

Eg. An algebraically closed field enables us to zero locus a polynomial into a mutli-subset of the affine line distinctly.