prep switch to eleventy

_posts/2009/2009-12-09-testing-mathml.md

@@ -10,6 +10,7 @@ tags:
 - mathml
 - tex4ht
 permalink: 0002/
+mathml: true
 ---
 
 One of the tools I want to use in this blog will be MathML. I think MathML is so far the best solution to present mathematical content on the web even though the [discussion](http://terrytao.wordpress.com/2009/10/29/displaying-mathematics-on-the-web/) on Terence Tao’s blog shows that MathML has its own deficits, especially when it comes to accessibility.

_posts/2009/2009-12-15-matrices-vs-idempotent-ultrafilters.md

@@ -12,6 +12,7 @@ tags:
 - Stone–Čech compactification
 published: true
 permalink: 0004/
+latex: true
 ---
 
 Note: as you can see I am not yet in control of how to convert LaTeX to mathml — bear with me, but I thought I should kick myself and start posting…
@@ -20,16 +21,16 @@ The other day I was looking for someone to chat about an interesting example of
 
 So to start: what’s the example?
 
-**Example** The [matrices](http://en.wikipedia.org/wiki/Matrix_(mathematics))  
-$$  
- A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, B=\begin{pmatrix} -1&-2\\ 1&2 \end{pmatrix}  
-$$  
+**Example** The [matrices](http://en.wikipedia.org/wiki/Matrix_(mathematics))
+$$
+ A= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, B=\begin{pmatrix} -1&-2\\ 1&2 \end{pmatrix}
+$$
  generate an 8-element (multiplicative) subsemigroup. Its elements are $A,B,AB, BA, ABA, BAB, ABAB, BABA$.
 
 So what? Well, what is interesting is that although both $A$ and $B$ are [idempotent](http://en.wikipedia.org/wiki/Idempotence) (i.e. $A\cdot A=A, B\cdot B = B$), their product is not, since
 
 $$
- AB = \begin{pmatrix} {-1} & {-2} \\ 0&0 \end{pmatrix}, AB\cdot AB = \begin{pmatrix} 1&2\\ 0&0 \end{pmatrix} = -AB  
+ AB = \begin{pmatrix} {-1} & {-2} \\ 0&0 \end{pmatrix}, AB\cdot AB = \begin{pmatrix} 1&2\\ 0&0 \end{pmatrix} = -AB
 $$
 
 Still, why is it interesting? Well, this example is of interest for people working with [ultrafilters](http://en.wikipedia.org/wiki/Ultrafilter) on semigroups, in particular on $\mathbb{N}$ — one reason following from the following lemma.
@@ -40,12 +41,12 @@ This can be found as Corollary 6.5 in the book ’Algebra in the [Stone–Čech
 
 **Corollary** There are idempotent elements in $\beta \mathbb{N}$ whose sum is not idempotent.
 
-**Proof**  
+**Proof**
 
-* Step 1 Consider the (discrete) finite semigroup generated by $A$ and $B$.  
-* Step 2 By the previous lemma, it is a continuous, homomorphic image of $\mathbb{H}$.  
-* Step 3 The preimage of both $A$ and $B$ is a closed (by continuity) semigroup (by homomorphy) of $\beta \mathbb{N}$.  
-* Step 4 Conversely, the preimage of $AB$ cannot contain an idempotent (or else the image of that idempotent, $AB$, would be idempotent by homomorphy).  
+* Step 1 Consider the (discrete) finite semigroup generated by $A$ and $B$.
+* Step 2 By the previous lemma, it is a continuous, homomorphic image of $\mathbb{H}$.
+* Step 3 The preimage of both $A$ and $B$ is a closed (by continuity) semigroup (by homomorphy) of $\beta \mathbb{N}$.
+* Step 4 Conversely, the preimage of $AB$ cannot contain an idempotent (or else the image of that idempotent, $AB$, would be idempotent by homomorphy).
 * Step 5 In particular, by the Ellis-Numakura Lemma, both preimages contain idempotents $a,b \in \beta \mathbb{N}$.
 * Step 6 But $ab$ is in the preimage of $AB$, hence not idempotent.
 

_posts/2010/2010-01-07-matrices-vs-idempotent-ultrafilters-part-2.md

@@ -13,6 +13,7 @@ tags:
 - Stone–Čech compactification
 published: true
 permalink: 0006/
+latex: true
 ---
 
 In [an earlier post](/0004/) I gave a short introduction to an interesting finite semigroup. This semigroup could be found in the $2\times 2$ matrices over $\mathbb{Q}$.
@@ -33,19 +34,19 @@ So, as a first approach we thought about the following question.
 
 Thinking about the problem a little and experimenting with [Macaulay 2](http://en.wikipedia.org/wiki/Macaulay2) we ended up with the following classification
 
-**Proposition** For $$A = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$ the solutions for $B$ being of rank one consist of four one – dimensional families, namely (for $x\in \mathbb{Q}$)  
- $$  
- F_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix},  
- F_2(x) = \begin{pmatrix} 1 & 0 \\ x & 0 \end{pmatrix},  
- F_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix},  
- F_4(x) = \begin{pmatrix} 0 & 0 \\ x & 1 \end{pmatrix}.  
- $$  
- Additionally, we have four special solutions  
- $$  
- G_1 = \begin{pmatrix} – 1 & 1 \\ – 2 & 2 \end{pmatrix},  
- G_2 = \begin{pmatrix} – 1 & – 1 \\ 2 & 2 \end{pmatrix},  
- G_3 = \begin{pmatrix} – 1 & 2 \\ – 1 & 2 \end{pmatrix},  
- G_4 = \begin{pmatrix} – 1 & – 2 \\ 1 & 2 \end{pmatrix}.  
+**Proposition** For $$A = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$ the solutions for $B$ being of rank one consist of four one – dimensional families, namely (for $x\in \mathbb{Q}$)
+ $$
+ F_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix},
+ F_2(x) = \begin{pmatrix} 1 & 0 \\ x & 0 \end{pmatrix},
+ F_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix},
+ F_4(x) = \begin{pmatrix} 0 & 0 \\ x & 1 \end{pmatrix}.
+ $$
+ Additionally, we have four special solutions
+ $$
+ G_1 = \begin{pmatrix} – 1 & 1 \\ – 2 & 2 \end{pmatrix},
+ G_2 = \begin{pmatrix} – 1 & – 1 \\ 2 & 2 \end{pmatrix},
+ G_3 = \begin{pmatrix} – 1 & 2 \\ – 1 & 2 \end{pmatrix},
+ G_4 = \begin{pmatrix} – 1 & – 2 \\ 1 & 2 \end{pmatrix}.
  $$
 
 Note: due to technical problems, this post continues [here](http://thelazyscience.blogspot.com/2010/01/testing.html) .
@@ -61,15 +62,15 @@ Luckily enough, we get something very similar from our alternative for $A$.
 **Proposition** In case$$
 A = \begin{pmatrix} 1 & 1
  \\ 0 & 0
- \end{pmatrix}$$ the solutions for $B$ being of rank one consist of five one – dimensional families namely (for $x\in \mathbb{Q}$)  
- $$  
- H_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix},  
- H_2(x) = \begin{pmatrix} x+1 & x \\ ( – x – 1) & – x \end{pmatrix},  
- H_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix},  
- H_4(x) = \begin{pmatrix} ( – x+1) & ( – x+1) \\ x & x \end{pmatrix},  
+ \end{pmatrix}$$ the solutions for $B$ being of rank one consist of five one – dimensional families namely (for $x\in \mathbb{Q}$)
+ $$
+ H_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix},
+ H_2(x) = \begin{pmatrix} x+1 & x \\ ( – x – 1) & – x \end{pmatrix},
+ H_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix},
+ H_4(x) = \begin{pmatrix} ( – x+1) & ( – x+1) \\ x & x \end{pmatrix},
  $$
  $$
- H_5(x) = \begin{pmatrix} ( – x+1) & ( – x – 1 – \frac{2}{x – 2}) \\ x – 2 & x \end{pmatrix} , x \neq 2.  
+ H_5(x) = \begin{pmatrix} ( – x+1) & ( – x – 1 – \frac{2}{x – 2}) \\ x – 2 & x \end{pmatrix} , x \neq 2.
  $$
 
 As before we can describe size and structure.

_posts/2010/2010-01-11-matrices-vs-idempotent-ultrafilters-part-2-5.md

@@ -13,6 +13,7 @@ tags:
 - Stone–Čech compactification
 published: true
 permalink: 0007/
+latex: true
 ---
 
 Note: there seems to be some problematic interaction between the javascripts I use and blogspot’s javascripts which prevents longer posts from being displayed correctly. As long as I don’t understand how to fix this, I will simply split the posts.
@@ -25,13 +26,13 @@ We can also describe size and the algebraic structure.
 
 Luckily enough, we get something very similar from our alternative for $A$.
 
-**Proposition** In case $A = \begin{pmatrix} 1 & 1 \\\ 0 & 0 \end{pmatrix}$ the solutions for $B$ being of rank one consist of five one – dimensional families namely (for $x\in \mathbb{Q}$)  
- $$  
+**Proposition** In case $A = \begin{pmatrix} 1 & 1 \\\ 0 & 0 \end{pmatrix}$ the solutions for $B$ being of rank one consist of five one – dimensional families namely (for $x\in \mathbb{Q}$)
+ $$
  H_1(x) = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix},  \\
  H_2(x) = \begin{pmatrix} x+1 & x \\ ( – x – 1) & – x \end{pmatrix},  \\
  H_3(x) = \begin{pmatrix} 0 & x \\ 0 & 1 \end{pmatrix},  \\
  H_4(x) = \begin{pmatrix} ( – x+1) & ( – x+1) \\ x & x \end{pmatrix},  \\
- H_5(x) = \begin{pmatrix} ( – x+1) & ( – x – 1 – \frac{2}{x – 2}) \\ x – 2 & x \end{pmatrix} , x \neq 2.  
+ H_5(x) = \begin{pmatrix} ( – x+1) & ( – x – 1 – \frac{2}{x – 2}) \\ x – 2 & x \end{pmatrix} , x \neq 2.
  $$
 
 As before we can describe size and structure.

_posts/2010/2010-04-15-welcome-again.md

@@ -14,6 +14,7 @@ tags:
 - textile
 published: true
 permalink: 0014/
+latex: true
 ---
 
 Welcome to the new home of thelazyscience. I hope the new setup makes everything better.

_posts/2010/2010-04-18-understanding-the-central-sets-theorem.md

@@ -17,6 +17,7 @@ tags:
 - Stone–Čech compactification
 published: true
 permalink: 0015/
+latex: true
 ---
 
 To write the first post on the new domain I thought I might just write a little about what I’ve been studying recently — the Central Sets Theorem.
@@ -43,8 +44,8 @@ I might write about the history (and applications) of the Central Sets Theorem s
 
 So, what does the usual formulation look like?
 
-**Central Sets Theorem**  
- Imagine you are given finitely many sequences in a commutative semigroup $(S,+)$, say $\mathbf{y^0}, \ldots, \mathbf{y^\alpha}$ as well as a central set $C \subseteq S$.  
+**Central Sets Theorem**
+ Imagine you are given finitely many sequences in a commutative semigroup $(S,+)$, say $\mathbf{y^0}, \ldots, \mathbf{y^\alpha}$ as well as a central set $C \subseteq S$.
  Then you can find a sequence $\mathbf{a}$ in $S$ as well as a sequence $\mathbf{h}$ of non-empty, disjoint and finite subsets of $\mathbb{N}$ such that for $\beta \leq \alpha$
 
  \\[ FS ( {a_n} + {\sum_{i \in h_n} y_i^\beta} ) \subseteq C. \\]
@@ -53,8 +54,8 @@ Complicated, no? I mean, a random bunch of sequences, some strange set and you f
 
 Let’s cut it down a little and just consider the case $\alpha = 0$.
 
-**simple Central Sets Theorem**  
- Imagine you are given a sequence $\mathbf{y}$ in a commutative semigroup $(S,+)$ as well as a central set $C \subseteq S$.  
+**simple Central Sets Theorem**
+ Imagine you are given a sequence $\mathbf{y}$ in a commutative semigroup $(S,+)$ as well as a central set $C \subseteq S$.
  Then you can find a sequence $\mathbf{a}$ in $S$ as well as a sequence $\mathbf{h}$ of non-empty, disjoint and finite subsets of $\mathbb{N}$ such that
 
  \\[ FS ( {a_n} + {\sum_{i \in h_n} y_i} ) \subseteq C. \\]
@@ -65,8 +66,8 @@ Even this special case of the standard formulation somehow focuses on aspects th
 
 Now, the theorem says all kinds of complicated things about the existence of a sequence of disjoint finite subsets of $\mathbb{N}$. Can I get around this? I thought I should be able to. Let’s start with a much weaker version of the theorem.
 
-**A weak simple Central Sets Theorem**  
- Imagine you are given a subsemigroup $T \subseteq \mathbb{N}$ as well as a central set $C \subseteq \mathbb{N}$.  
+**A weak simple Central Sets Theorem**
+ Imagine you are given a subsemigroup $T \subseteq \mathbb{N}$ as well as a central set $C \subseteq \mathbb{N}$.
  Then you can find a sequence $\mathbf{a}$ in $\mathbb{N}$ as well as a sequence $\mathbf{b}$ in $T$ so that
 
  \\[ FS ( {a_n} + {b_n} ) \subseteq C. \\]
@@ -79,35 +80,35 @@ This is much weaker than the statement before. Of course, given a sequence $\mat
 
 So where does this leave us? Well, when I hear finite subsets of $\mathbb{N}$ I think of my favourite structure — in fact the favourite structure for a lot of algebra in the [Stone–Čech compactification](http://en.wikipedia.org/wiki/Stone%E2%80%93%C4%8Cech_compactification#Addition_on_the_Stone.E2.80.93.C4.8Cech_compactification_of_the_naturals) on $\mathbb{N}$, the semigroup $\delta \mathbb{F}$. But let’s step back a little. The best way to think about $\delta \mathbb{F}$ is in terms of partial semigroups.
 
-**(Adequate) Partial Semigroups**  
- A _partial semigroup_ operation on a set $S$ is a map $\cdot: S \times S \rightarrow S$ such that associativity $s \cdot (t \cdot u) = (s \cdot t) \cdot u$ holds in the sense that if one side is defined so is the other and they are equal. A partial semigroup is _adequate_ if the sets  
+**(Adequate) Partial Semigroups**
+ A _partial semigroup_ operation on a set $S$ is a map $\cdot: S \times S \rightarrow S$ such that associativity $s \cdot (t \cdot u) = (s \cdot t) \cdot u$ holds in the sense that if one side is defined so is the other and they are equal. A partial semigroup is _adequate_ if the sets
 
- \\[  
- \sigma(s) := \left\\{ t\in S : {s \cdot t} \mbox{ is defined} \right\\}  
+ \\[
+ \sigma(s) := \left\\{ t\in S : {s \cdot t} \mbox{ is defined} \right\\}
  \\]   generate a filter, i.e., finitely many elements have a common compatible element.
 
 This notion was introduced by [Bergelson, Blass and Hindman](http://www.math.lsa.umich.edu/~ablass/bbh.pdf) ([DOI](http://doi.org/10.1112/plms/s3-68.3.449)) in the 90s. It tells us that the operation, although partial, is associative in a strong way. Additionally, it makes sure the operation is not just empty but defined for many elements (well, ok it could be just one for all, but that’s not the point).
 
 For ultrafilters the critical point is the following.
 
-**The semigroup $\delta S$**  
- Given an adequate partial semigroup and $p,q$ ultrafilters containing all $\sigma(s)$. Then the operation  
+**The semigroup $\delta S$**
+ Given an adequate partial semigroup and $p,q$ ultrafilters containing all $\sigma(s)$. Then the operation
 
- \\[  
- p \cdot q = \left\\{ A \subseteq S : \left\\{ s : \left\\{ t : s \cdot t \in A \right\\} \in q \right\\} \in p \right\\}  
- \\]  
+ \\[
+ p \cdot q = \left\\{ A \subseteq S : \left\\{ s : \left\\{ t : s \cdot t \in A \right\\} \in q \right\\} \in p \right\\}
+ \\]
 
 is well-defined and associative and semi-continuous. In other words, $\delta S$ is a closed semi-continuous semigroup.
 
 Now this is somewhat surprising. Even though our operation is partial, these ultrafilters are a full semigroup! With all the bells and whistles it takes to do algebra in the Stone–Čech compactification.
 
 What does this have to do with the Central Sets Theorem?
 
-Denote the non-empty, finite subsets of $\mathbb{N}$ by $\mathbb{F}$. Consider the restriction of $\cup$ on $\mathbb{F}$ defined by  
+Denote the non-empty, finite subsets of $\mathbb{N}$ by $\mathbb{F}$. Consider the restriction of $\cup$ on $\mathbb{F}$ defined by
 
- \\[  
- s + t \mbox{ defined } \Longleftrightarrow \max(s) \cap \min(t) = \emptyset.  
- \\]  
+ \\[
+ s + t \mbox{ defined } \Longleftrightarrow \max(s) \cap \min(t) = \emptyset.
+ \\]
 
 Then in fact this constitutes a partial semigroup, adequate at that.
 
@@ -117,11 +118,11 @@ This partial semigroup structure could be called the free partial semigroup in t
 
 To come back to the weak version of the Central sets theorem — partial semigroups are exactly what it talks about. So let us reformulate,
 
-**simple Central Sets Theorem**  
- Imagine we are given a partial subsemigroup $T$ of $(S,+)$ as well as a central set $C \subseteq \mathbb{N}$. Then we find sequences $\mathbf{a}$ in $\mathbb{N}$ and $\mathbf{t} \in T$ such that $FS ( {t_n} ) \subseteq T$ and  
+**simple Central Sets Theorem**
+ Imagine we are given a partial subsemigroup $T$ of $(S,+)$ as well as a central set $C \subseteq \mathbb{N}$. Then we find sequences $\mathbf{a}$ in $\mathbb{N}$ and $\mathbf{t} \in T$ such that $FS ( {t_n} ) \subseteq T$ and
 
- \\[  
-  {FS( a_ {n} + t_{n}) \in C.}  
+ \\[
+  {FS( a_ {n} + t_{n}) \in C.}
  \\]
 
 Now this sounds much closer to the original theorem. Since any sequence generates a partial semigroup on its $FS$-set (isomorphic to $\mathbb{F}$), this is in fact the Central Sets Theorem for just one sequence.
@@ -130,15 +131,15 @@ Now this sounds much closer to the original theorem. Since any sequence generate
 
 However, the actual theorem is more than just some kind of induction on the above version. It is considerably stronger and here it is time to let go of the simplifications of partial semigroups again. For the theorem really does talk about $FS$-sets, i.e., partial semigroups isomorphic to $\mathbb{F}$. The strength lies in the fact that the infinite sequences can be chosen uniformly in the sense that we pick from the different partial semigroups in the same prescribed way.
 
-**Central Sets Theorem**  
- Imagine you are given finitely many $FS$-sets in a commutative semigroup $(S,+)$, say ${FS( {\mathbf{y^0}} )}, {\ldots}, {FS( {\mathbf{y^\alpha}} )}$ as well as a central set $C \subseteq S$.  
+**Central Sets Theorem**
+ Imagine you are given finitely many $FS$-sets in a commutative semigroup $(S,+)$, say ${FS( {\mathbf{y^0}} )}, {\ldots}, {FS( {\mathbf{y^\alpha}} )}$ as well as a central set $C \subseteq S$.
  Then you can find a sequence $\mathbf{a}$ in $S$ as well as one disjoint sequence $\mathbf{h}$ in $\mathbb{F}$ such that for all $\beta \leq \alpha$
 
  \\[ FS ( {a_n} + {\sum_{i \in h_n} y_i^\beta} ) \subseteq C. \\]
 
 To see this strength at work it is time to look at the classical application.
 
-**Central sets in $( \mathbb{N},+)$ contain arbitrarily long arithmetic progressions**  
+**Central sets in $( \mathbb{N},+)$ contain arbitrarily long arithmetic progressions**
  Take $\mathbf{y^\beta}$ to be the multiples of $\beta$ (for $\beta \leq \alpha$). Then the central set theorem guarantees we find $a_1, h_1$ such that for all $\beta \leq \alpha$
 
  \\[ (a_1 + \beta \cdot \sum_{i\in h_1} i) \in C.\\]

_posts/2010/2010-04-25-van-douwen-spaces.md

@@ -15,6 +15,7 @@ tags:
 - van Douwen space
 published: true
 permalink: 0016/
+latex: true
 ---
 
 At the [winterschool](http://www.winterschool.eu) [Alan Dow](http://math.uncc.edu/~adow/) gave quite challenging tutorials. He also mentioned something about van Douwen spaces.
@@ -27,7 +28,7 @@ As formulated [here](http://math.uncc.edu/~adow/vDspace.pdf)
 
 What caught my interest was that there is an example that has something to do with [idempotent](http://en.wikipedia.org/wiki/Idempotency) [ultrafilters](http://en.wikipedia.org/wiki/Ultrafilter). Let me introduce something first.
 
-**A partial order** On the idempotent ultrafilters (on $\mathbb{N}$) define a partial ordering by  
+**A partial order** On the idempotent ultrafilters (on $\mathbb{N}$) define a partial ordering by
 
  $$ p \leq_r q \Leftrightarrow q + p =p. $$
 
@@ -39,7 +40,7 @@ This partial order (as well as its left counterpart and their intersection) is q
 
 For van Douwen spaces it is useful to go in the other direction. There exist many right-maximal elements in this order, but even more can be said.
 
-**Strongly right maximal idempotents** An idempotent ultrafilters $p \in \beta \mathbb{N}$ is _strongly right maximal_ if  
+**Strongly right maximal idempotents** An idempotent ultrafilters $p \in \beta \mathbb{N}$ is _strongly right maximal_ if
 
  $$ q+ p =p \Rightarrow q= p. $$
 
@@ -53,7 +54,7 @@ Let $p$ be strongly right maximal. Then $\mathbb{N} + p$ is a van Douwen space.
 
 And this is what Alan Dow mentioned. Ignoring the crowdedness, this is really easy for in fact more holds in this case.
 
-If $p$ is strongly right maximal, then  
+If $p$ is strongly right maximal, then
 
 $$ \rho_p: \mathbb{N} \rightarrow \beta \mathbb{N}, n \mapsto n+ p $$
 

_posts/2010/2010-04-28-new-what-is-videos.md

@@ -14,6 +14,7 @@ tags:
 - what is seminar
 published: true
 permalink: 0017/
+latex: true
 ---
 
 ### Backlog
@@ -26,7 +27,7 @@ For now, I’m happy to say that two new videos are up. First there is Joscha Di
 
 Second there is the first ‘tag team’ talk by [Dror Atariah](https://twitter.com/drorata) and [Tobias Pfeiffer](http://www.mi.fu-berlin.de/en/math/groups/ag-geom/people/former-members/pfeiffer.html) on [‘What is … a geodesic on a Riemannian Manifold’](http://vimeo.com/11253670) .
 
-I also must say I love [Felix Breuer’s](http://www.felixbreuer.net) beautiful video [Der Kleine Gauß](http://vimeo.com/10014698) which was made for a general (German speaking) audience explaining the equation  
+I also must say I love [Felix Breuer’s](http://www.felixbreuer.net) beautiful video [Der Kleine Gauß](http://vimeo.com/10014698) which was made for a general (German speaking) audience explaining the equation
 
 $$ \sum_{i \leq n} i = \frac{n\cdot (n+1)}{2}$$