Locally Confused

Separation preorder

last edited around 2019-10-28

$\DeclareMathOperator{\sep}{sep}$Consider Spaces which are $T_1$ but may fail to be $T_2$. For $x,y\in X$ define $x\leq y$ iff $\sep(x) \subseteq \sep(y)$ where $\sep(x)$ refers to the set of all $y$ such that $\{x,y\}$ is a $T_2$ subspace.

Q: Is this preorder respected by continuous maps? By inverse image?

Adding “companion points”

last edited around 2019-02-09

Let $X$ be a topological space, $S\subseteq X$. Define $\DeclareMathOperator{\comp}{Comp}\comp(X, p)$, to be the set $X\sqcup {p}$ equipped with the topology $$ \{U\cup \{p\} \mid U \ \text{Open in}\ X\} \cup \emptyset. $$ It is straightforward to check that this indeed defines a topology. Our new point $p$ drastically introduces new “convergence” behavior: Because $\{p\}$ is open, the only way for a sequence to converge against $p$ is to be constantly $p$ at some $N\in \mathbb N$, but then we have an intriguing side effect: Since every neighborhood around any other point also contains $p$, said sequence also converges against every other point as well!

Now we have some interesting questions:

  1. Can we “add” a point which is a companion point to only a subset of our original space?
  2. If we add two companion points, are they automatically topologically indistinguishable?
  3. Can we construct $\comp{\mathbb R}$ as a pushout or some other “universal” construction?

Companionship: The specialization preorder

The point we added to our topology appears to be a kind of extreme example of what may appear in a weaker fashion: With respect to any different point, every open set containing the latter contains our new point as well – which suggests the following definition:

DEF Let $X$ be a topological space. $c\in X$ is said to “accompany” (or “be a companion to”) $x\in X$ if every open set containing $x$ also contains $c$. Analogously, $x$ is said to “accompany” $S\subseteq X$ if it accompanies every point in $S$. In symbols, we will use $c \sphericalangle x$ or $c \sphericalangle S$ (evaluated pointwise).

The following consequences are immediate:

  1. Companionship is transitive and reflexive, thus amounts to a preorder.
  2. If $x\sphericalangle y$ and $y\sphericalangle x$, those points are topologically indistinguishable (and vice versa). In particular, if our space satisfies the $T_0$ separation axiom, that is, every distinct points are topologically distinguishable, $x\sphericalangle y$ and $y\sphericalangle x$ implies $x=y$, in other words, we have a poset.
  3. If our space is $T_1$, meaning that given two points $x,y$, each of them can be separated by an open set from the other, we directly conclude that neither $x\sphericalangle y$ nor $y\sphericalangle x$ can hold, so any two distinct points are incomparable – meaning our companionship relation is precisely the diagonal.

The latter motivates our order as a nice tool to investigate how “far” a space is from being $T_1$.

For later purposes, we show that it is sufficient to check companionship on a basis (the proof is straightforward):

LEM Let $c, s\in X$, $\mathscr B$ a basis of $X$. If every $B\in \mathscr B$ which contains $s$ also contains $c$, $c$ is a companion to $s$. Furthermore, if $S\subseteq X$ and every $B\in \mathscr B$ which is not disjoint to $S$ also contains $c$, $c$ accompanies $S$.

Proof Let $s\in U = \bigcup_{i\in I}B_i$. This means that for some $i$, $s\in B_i$, and by our condition, we also have $c\in B_i\subseteq U$. For the second part, fix an $s\in S$. We need to show that $c$ is a companion to $s$: Let $s\in U = \bigcup_{i\in I} B_i$, so again, $s\in B_i$ for some $i\in I$. Containing $s$ means in particular not being disjoint to $S$, so by our premise, $c\in B_i\subset U$, and we're finished.

Here comes the shameful part where I realize that the order I defined not only already exists, but was already known to me. It is called the “specialization preorder” – however, usually it is defined by the following, equivalent characterization:

LEM $\DeclareMathOperator{cl}{cl}$Let $x, y\in X$. $\cl(x)\subseteq \cl(y) \Leftrightarrow y\sphericalangle x$.

Proof First, we will reformulate the right hand side to deal with closed, and not open sets: \begin{align} y\sphericalangle x \Leftrightarrow&\forall U \:\text{open}\colon\: x\in U\implies y\in U\\ \Leftrightarrow&\forall U \:\text{open}\colon\: y\not\in U\implies x\not\in U\\ \Leftrightarrow&\forall U \:\text{open}\colon\: y\in U^c\implies x\in U^c\\ \Leftrightarrow&\forall C \:\text{closed}\colon\: y\in C\implies x\in C\\ \end{align} Now, we will remember that $\cl$ is a closure operator and satisfies extensitivity, monotony, and idempotence, so if $x\in C$, writing this as $\{x\}\subseteq C$ and applying $\cl$ to both sides, we get $\cl(x)\subseteq \cl(C)=C$, and on the other hand, because $x\in \cl(x)$, the latter implies $x\in C$ as well, so we have an equivalence: \begin{align} &\forall C \text{ closed}\colon\: y\in C\implies \cl(x)\subseteq C\\ \Leftrightarrow&\forall C \text{ closed}, y\in C \colon\: \cl(x)\subseteq C\\ \Leftrightarrow& \cl(x)\subseteq \bigcap_{\substack{y\in C \\ C\:\text{closed}}} C = \cl(y)\\ \end{align} Which completes the proof.

In most scenarios, I find it more intuitive to describe topologies with respect to their open sets, but that is just the personal preference of a layman.

Note that we can make the following statements about points:

  1. Let $\{x\}$ be open. Then $y\sphericalangle x$ implies in particular that $y\in \{x\}$, i.e. $y=x$ – in other words, open points are minimal.
  2. Let $\{x\}$ be closed. With the dual characterization, $x\sphericalangle y \Leftrightarrow \cl(y)\subseteq \cl(x) = \{x\}$, and since $y\in \cl(y)$, we also have that $y=x$: closed points are maximal!

A useful thing to keep in mind is that when two are maximal (respectively, minimal), they are either equal or incomparable.

Other ideas (WIP)

DEF Let $X$ be a topological space. Define $\operatorname{Comp}(X, S, p)$ to be the set $X\sqcup {p}$ equipped with the topology generated by the following set: \begin{align} \mathscr T &:= \langle \mathscr A \cup \mathscr A^\prime \cup \{\{p\}\} \rangle, \end{align} where \begin{align} \mathscr A &:= \{ O\cup \{p\} \mid O\cap S \neq \emptyset, O \ \text{open in }\ X \} \\ \mathscr A^\prime &:= \{ O \mid O\cap S = \emptyset, O \ \text{open in }\ X \} \end{align}

DEF Let $(X,\mathscr T_X), (Y, \mathscr T_Y)$ be a topological spaces, $S\subseteq Y$. Define a topological space $$ \comp(X,S,Y) := \{ X\sqcup Y, \mathscr T_X \cup \Sigma \cup \mathscr R \} $$ where \begin{align} &\Sigma := \{ X \cup O \mid O\in \mathscr T_Y,\,O\cap S\neq \emptyset \},\\ &\mathscr R := \{ O \mid O\in \mathscr T_Y,\,O\cap S=\emptyset \} \end{align}

LEM The generating set given above is a basis.

Proof We wil verify that pairwise intersections can be represented as unions of our generator. To do that, consider the following cases separately:

LEM I

Proof We will verify that the generating set given in the above definition forms a basis.

Q What do companions to the whole space mean for the preorder / depth associated to a finite topology?

Q Is a point that's the only minimum in its connected component necessarily open?

Comparing more things than points

As mentioned, as soon as we have $T_1$, the specialization preorder is discrete, thus does not convey any more information. But what if we compare subsets of a partition in a similar way? Take $\mathbb R = (-\infty, 0) \cup [0,\infty)$. We can “represent” this partition by the homomorphic image onto a sierpinski space. There, the specialization preorder reflects that it is not discrete, i.e. that partition is not a partition into closed sets. Questions:

n-fold transitivity of topological spaces

last edited by Lukas around 2019-01-09

DEF. Let $X, C$ be topological spaces. $X$ is said to be $n$-fold transitive with respect to $C$ if for every collection of $n$ injective continuous maps $\iota_i\colon C \to X$ with disjoint image and every permutation $\sigma\in S_n$ we find an automorphism $h\colon X\to X$ satisfying $h\circ \iota_i = \iota_{\sigma(i)}$.

Here, the $C$ is meant to stand for “characteristic”. Note that this is quite a strong statement: Even if we take $C$ to be the simplest topological space, namely the point, even only 2-fold transitivity appears pretty strong: every point can be transformed into every other, implying there must be a lot of automorphisms (read: symmetry) intrinsic to $X$. However, it might be of interest to study maximal n-fold subspaces. Also, keep in mind that n-fold transitivity implies $n-1$-fold transitivity by considering a permutation that leaves one image fixed – which matches the suspicion that transitivity for higher $n$ represents a stronger property.

Q. Can a space be 2-fold with respect to $S$, but not a subspace $T\subseteq S$? A. Yes. Consider the “converging sequence” space $$\newcommand{\C}{\mathscr C} \left\{\frac{1}{n}\mid n\in \mathbb N\right\}\cup \{0\} =: \C, $$ and take two copies of that: $X := \C \sqcup \C$. This is 2-transitive w.r.t. $\C$ (check that!), but not w.r.t. the one-point space $\ast$, since the two points $(0, 0)$ and $(0, \frac{1}{2})$ (the first component denotes the coloring introduced by our disjoint union) cannot be interchanged by a homeomorphism, since one point is isolated, and the other is not.

This appears weird, but is actually quite interesting in the following way: being n-fold transitive with respect to the one-point space is a pretty harsh restriction already, but perhaps by choosing „wisely“, we can still draw information about the symmetry of other spaces we wouldn't have been able to “see” by just considering the continuous permutability of points.

From here on, if we make no mention of the embedded space, we mean to analyze the one-point case.

Ex. $\mathbb R$ is 2-transitive, but not 3-transitive: There is no homeomorphism taking $(-1,0,1)$ to $(-1,1,0)$, as every homeo must preserve the usual order (wait, is that true?)

Ex. $\DeclareMathOperator{cofin}{CoFin}X:=\cofin(S)$ for an infinite set $S$ is n-transitive for any $n$ and any $C$, because every bijection $S\to S$ is a homeomorphism: For bijections, preimages of finite sets are finite, therefore, preimages of cofinite sets must be cofinite. Given a permutation of images of $C$ in $S$, the map that permutes the images and acts as the identity on everything else constitutes a valid choice of such an automorphism.

Q. Is the cofinite space the only one with this property (fixing $C$, demanding arbitrary high $n$)?

Note. When constructing examples by taking coproducts, keep in mind that once we can partition our space into (finitely many) connected components, every automorphism must be a permutation of those connected components, implying by restriction that they may only mediate between connected components which are homeomorphic.

Conjecture. If $X$ is not n-transitive w.r.t. $C$, it is probably not $n$-transitive w.r.t. $C\sqcup C$.

Example. Given any $n\geq 2$, a subpace of $\C$ is $n$-transitive ($n\geq 2$ implies it must contain at least two elements) if and only if it does not contain $0$. Therefore, there is no maximal $n$-transitive subspace. So let's try to make the characteristic space more complicated: finite or larger discrete spaces probably won't behave any differently, as they're essentially just a bunch of points, but let's look at $\C$ itself, which is the “next more complicated“ example (it has only one non-isolated point). Unfortunately though, we can only inject one copy of $\C$ into $\C$ at a time, given that the non-isolatedness of $0$ must be preserved. So $n\geq 2$-transitivity is impossible as well.

Q. Are there non-discrete characteristical spaces which give rise to non-trivial transitivity of $\C$? A. Let's look at $\operatorname{Aut}(\C)$ explicitly: Every such automorphism $h$ must preserve $0$, ergo be a permutation n the rest. But which permutations are allowed? Well, every open set containing 0 is of the form $[0,1/k]\cup I$, where $I$ is an arbitrary subset of points in $C$ outside of said interval (the interval is to be interpreted not as a subset of $\mathbb R$, but of the poset $\C$). Put another way, they are precisely the cofinite sets containing 0. The property of being cofinite is clearly preserved if $h$ is a bijection, and as it also preserves 0, the preimage is open without need for further demands. So indeed, any permutation leaving 0 constant constitutes a valid automorphism of $\C$. However, a pretty basic problem cannot be resolved: Anything that's not countable does not admit an injection into $\C$, and nor does anything with more than one non-isolated point. The only countable space with one isolated point is $\C$(?), so lets pick the space with no isolated points: the discrete space $\mathscr P (\mathbb N)$. However, injecting this space does not even allow for 2-transitivity, because again, one injection might containg $0$ in its image and another might not, and both are not translatable by an automorphism. So $\C$ does not admit any transitivity, and may rightfully be proclaimed totally asymmetric.

Pre-Proof (the only ctbl space with one isolated point is $\C$): idea: embed $\C$ injectively into this space (possible, as the poset of open sets containing the non-isolated points does not have a minimal element, apply choice to this chain of decreasing open sets), and note that the complement consists of isolated points and is thus discrete. The complement can be probably made finite by choosing a maximal chain (note that every open set can “constructively” be extended by gluing a point), so we have hopefully our image is an open set

Homo-more-or-less

last edited around 2018-12-31

$\require{AMScd}\newcommand{\C}{\mathscr C}\newcommand{\D}{\mathscr D}\DeclareMathOperator{\obj}{Obj}\DeclareMathOperator{\id}{id}$

Motivation

Let $L\leq \mathfrak{gl}(n,k)$ be a matrix lie algebra. The transpose has an interesting interplay with the commutator:$$ [A^\intercal, B^\intercal] = A^\intercal B^\intercal - B^\intercal A^\intercal = -(AB - BA)^\intercal = -{[A,B]}^\intercal. $$ Thereby, it is not formally a lie-Algebra homomorphism, but it still does not seem to change very much: For instance, The transpose maps lie subalgebras to themselves. What would happen if we would allow for more things to be a “Lie-algebra homomorphism” in a way that we can consider homs “up to slight correction”?

Construction

Such a correction can be described as a natural transformation from some functor to itself, because what they need to fulfil is precisely that the family of morphisms can be „dragged“ through any function just like the minus sign does. In fact, in the case of vector spaces, the NT needs to leave every 1D subspace invariant, which means we are restricted to scalar multiples of the identity. TODO REWORD

DEF Let $U\colon \C\to \D$ be a faithful functor. Define $\C_\sim$ to be the category with objects $\obj(C)$ and morphisms $$ \C(X,Y) := \{Uf\,\alpha_Y \mid f\colon X\to Y, \alpha: U\to U\}, $$ where composition is inherited from $\D$.

LEM $\C_\sim$ is well-defined as a category, and contains $\C$ as a subcategory by means of the embedding of morphisms $$ f\mapsto Uf;\id_X $$

Proof Identities are given by $\id_{UX} = U(\id_X\id_X) = U\id_X\,U\id_X$ which is of the required form because $X\mapsto U\id_X$ is a natural transformation from $U$ to itself. To see that our morphisms remain valid under composition, note that given $f: X\to Y, g: Y\to Z$ and two natural transformations $\alpha, \beta: U\to U$, we have $$ Uf\,\alpha_Y\,Ug\,\beta_Z = UfUg \alpha_Z\beta_Z = U(fg) (\alpha\beta)_Z, $$ which is a valid representation of a morphism in $\C_\sim$ because $\alpha\beta$ is still natural. To practise drawing a commutative diagram, this is the flip we've done.$$\tag*{$\blacksquare$}$$

\begin{CD} UX @>Uf>> UY @>Ug>> UZ \\ @. @V{\alpha_Y}VV @V{\alpha_Y}VV \\ @. UY @>Ug>> UZ \\ @. @. @V{\beta_Z}VV \\ @. @. UZ \\ \end{CD}

What does it preserve?

In this section, we will examine the introductory example of $k$-lie algebras, $\operatorname{char} k=0$, equipped with the forgetful functor $U$ to $\underline{\operatorname{k-Vec}}$. Remember that in this case, the only natural transformations $U\to U$ are scalar multiples. Greek letters like $\lambda, \mu$ will denote elements of our field.

Zero objects, kernels

It is straightforward to check that $0$ is indeed a zero object.

Products

Let $K, L_1, L_2$ be lie algebras, where $K$ is associated with two morhisms $\lambda_i f_i\colon K\to L_i$. Can we find a product object $P$?

Generalizing 0 and 1: Graded semigroups

last edited around

For some reason, I assumed that for any ring $R$, the subset $R_{\neq 0}$ is a monoid under multiplication. However, that is only true if there are no zero divisors, i.e. in a domain! On the other hand, including the zero just gives us a semigroup, since $1=0$ implies that 1 cannot be an identity. What a pity, I was hoping to deal with a monoid… Pretend that the last sentence has never been written, and continue with me.

However, on second sight, multiplication by $1$ and $0$ is not too different, after all:

This motivates the following definition:

DEF A semigroup $S$ equipped with a system of subsets $S_0\subsetneq \cdots \subsetneq S_n\subseteq S$ is called an $n$-graded semigroup if the following holds: Let $T_0 := S_0$ and $T_{i+1}:=S_{i+1}\setminus S_i$. For every $0\leq i \leq n$ there is an element $e_i\in T_i$ such that

  1. $e_i$ is a left and right identity for all elements of $T_i$
  2. Left and right multiplication with $e_i$ have image in $S_i$. We call ${(S_i)}_i$ a grading on $S$, and ${(e_i)}_i$ the identities of the grading.

Note that the requirement to be an identity on $T_i$ does not imply that the latter is a subsemigroup – for that, we'll see an example soon. However, It is not necessary to ensure uniqueness of the identities:

LEM Let $S$ be a semigroup and $S_0\subsetneq \cdots \subsetneq S_n\subseteq S$. If Elements $e_i$ satisfying the identity axiom (1) exist, they are unique.

Proof. Assume there were identities $e_i and f_i\in T_i$. Then $e_i=e_i f_i = f_i$.

LEM $S_i\leq S$

Proof Let $s,t\in S_i$ and let $j\leq i$ such that $t\in T_j$. Then $st = s(te_j)=(st)e_j\in S_j\subseteq S_i$.

LEM Let $(S, {(S_i)}_i$ be a graded semigroup with identities $e_i$, and $\lambda_x,\rho_x$ the left and right multiplication with $x\in S$, respectively. Then $S_i = \bigcup_{j\leq i} \mathrm{Im} \lambda_{e_i} = \bigcup_{j\leq i} \mathrm{Im} \lambda_{e_i}$.

Proof.

EX Consider $S={e_0,e_1,e_2}$ with the multiplications necessary to make it into a grading, i.e. $e_0\cdot - = - \cdot e_0 = e_0$, $e_1^2 = e_1$, $e_2^2 = e_2$. To see what other restrictions we need, note that $bc,cb\in {a,b}$