@pablo 🇧🇷

Soderbergh’s Solaris: My Love Letter to Soviet Science Fiction

2023-08-02T00:00:00+00:00

This is my review of Steven Soderbergh’s 2002 film adaptation of the 1961 novel Solaris by the polish writer Stanislaw Lem, a story about a psychologist who is sent to assist the crew of an isolated research station in the titular alien planet. I should probably start by saying that Solaris is my very favorite novel — which is to say that this is mostly a book fan complaining about the movie adaptation. Salaris is also the book that made me fall in love with soviet science fiction as a whole. 🌟

A hard copy of the first edition of Solaris

Starring George Clooney and produced by James Cameron, Soderbergh’s Solaris is a full-blown blockbuster. While Soderbergh’s adaptation ultimately disappointed me, I thought I might be able to use this obscure piece of pop culture to discuss some of the aspects I love about the science fiction literature of the Soviet Bloc. These are books that delighted me for their ability to approach deep subjects in an extremely accessible guise, and I hope to convince you to give them a try too.

This article contains spoilers for both the novel and its film adaptation.

We must give to Caesar what is Caesar’s: the movie is quite competent in communicating the story clearly in a short span of time — and in this aspect this adaptation is much more successful than Tarkoviski’s. The film also takes great risks in deviating substantially from the original book. The problem is that these diversions range from simply inconsequential — such as changing the gender of certain characters — to down right mutilations of the source material.

One of the most substantial of these deviations is the fact the movie employs considerably more resources to establish the relationship between the protagonist Kris and his late wife, depicting the events leading up to her suicide in great detail and wasting precious screen time in the process. To me, Solaris is a story about confronting ghosts from our past, and while Soderbergh’s adaptation does explore this aspect of the novel, the movie distracts from the greater themes of the original book by transforming the narrative into a story about Kris' personal pain and regret.

But perhaps the greatest tragedy of this adaptation is its categorical failure to depict the existential horror of the source material, which is one of the landmarks of soviet science fiction. These are profoundly terrifying books, not so much because their characters are placed in immediate danger, but because rationalizing the utterly surreal situations they are confronted with requires concocting explanations of such astronomical complexity that these events threaten the very fabric of reality.

Indeed, perhaps one of the most delightfully frustrating aspects of these novels is the fact we are consistently denied explanations. In Lem’s Solaris we never really find out what is the nature of the visitors, nor why they came to the station in the first place. The experiments at the station lead to no smoking gun and the field of Solaristics never gets its final breakthrough.

A scene from the 2002 movie

These are not mysteries to be solved. Instead, the conflict of these narratives lies precisely in the desperate attempts by their characters to grapple with the irreconcilable nature of the riddles they are faced with. Such conflict is only ever resolved when the characters embrace their lack of understanding and capitulate to the unknowable stone monuments they were rudely forced to climb. As melodramatic as their reactions are, Clooney and the rest of the cast are ultimately unable to impress these emotions on the screen.

This is partly because the absurd nature of the events depicted makes them hard to convincingly enact, but also because the thematic deviations of the adaptation distract from the terrifying character of the source material. This is particularly apparent towards the end of the movie, where the story diverts dramatically from that of the novel. For instance, by revealing at the last minute that the character of Snaut (Snow) was actually a Doppelganger-like visitor who had previously killed him, a development which is entirely absent in the original book, the film replaces the subtle abstract horror of the source material with the physical threat of an alien invader.

All in all, Soderbergh’s Solaris is precisely what one would expect from an adaptation of a soviet masterpiece produced by the likes of James Cameron: a lukewarm, profoundly shallow interpretation of the source material. This is about all I had to say. I really hope my humble words inspire more people to explore soviet science fiction, which is greatly underappreciated in the west. If you’re interested, take a look at the books from the likes of Lem or the Strugatski brothers! 😁🌟

How Likely is a Finite Group to be Hausdorff?

2023-07-31T00:00:00+00:00

This is the third in a series of posts on some fun little toy problems of mine. Today’s question is one I came across while trying to come up with an example of a topological group which is not Hausdorff. Anyone familiar with topological groups knows that such examples are scarce in the literature. This is not without reason: any $\mathrm{T}_0$ group is actually Hausdorff, so non-Hausdorff topological groups are particularly pathological. But how scarce really are the non-Hausdorff topological groups? In other words, how likely is a topological group to be Hausdorff?

See Is Every Set Fieldable? and Classification of Abelian Lie Groups for the previous two posts on this series!

We begin by noting that, as stated, our question is ill-posed. First, we should clarify that we are working with a fixed group $G$ and considering all possible topologies which turn it into a topological group — which we call “ $G$ -topologies”. That is to say, our question is actually: given $G$ and some random $G$ -topology, how likely is it to be Housdorff? However — and this is more important — the meaning of “some random topology” is still unclear.

To make the question precise, we focus on the case where $G$ is finite. This is because a finite set $X$ can only be endowed with a finite number of distinct topologies — indeed, the power set $\mathcal{P}(X)$ is finite. Counting how many topologies there are in terms of the size of $X$ is a famous open problem and these numbers have only being computed for very small $X$ . Here is a plot of (the logarithm of) the first 19 of these numbers:

See the excellent page on these computation at the OEIS website!

What is relevant to us is that this implies a finite group $G$ admits only a finite number of $G$ -topologies. We may thus consider a random variable $\tau \in \{ G \text{-topologies} \}$ with uniform distribution and ask: can we compute $\mathbb{P}(G\ \text{is Hausdorff}) = \mathbb{P}(G\ \text{is Hausdorff under}\ \tau)$ ? This is the question we will attempt to answer in this article. Since any discrete group is Hausdorff, we certainly know

\mathbb{P}(G\ \text{is Hausdorff}) \ge \frac{1}{\#\{G\text{-topologies}\}}.

What is perhaps surprising is that $\mathbb{P}(G\ \text{is Hausdorff})$ is precisely $1/\#\{G\text{-topologies}\}$ . In other words, the odds of $G$ being Hausdorff are as slim as they could possibly be! Indeed, it is a well known fact from the theory of finite topological spaces that any a finite Hausdorff space is discrete, so the only finite Hausdorff groups are the discrete ones. All it’s left is to compute $\# \{G\text{-topologies}\}$ . To do so, we investigate the closure $\overline{ \{ 1 \} }^\tau$ of $1 \in G$ under a given $G$ -topology $\tau$ .

Since the translations by elements of $G$ are all homeomorphism, $\overline{\{g\}}^\tau = g \cdot \overline{\{1\}}^\tau$ for all $g \in G$ . It is then easy to see that

n m \in \overline{\{n m\}}^\tau = n \cdot \overline{\{m\}}^\tau \subset n \cdot \overline{\{1\}}^\tau = \overline{\{n\}}^\tau \subset \overline{\{1\}}^\tau

for any $n, m \in \overline{\{1\}}^\tau$ . Similarly, since the inversion $g \mapsto g^{-1}$ is a homeomorphism, we find $n^{-1} \in \overline{\{1^{-1}\}}^\tau = \overline{\{1\}}^\tau$ for any $n \in \overline{\{1\}}^\tau$ . Finally, $g \cdot \overline{\{1\}}^\tau \cdot g^{-1} = \overline{\{g 1 g^{-1}\}}^\tau = \overline{\{1\}}^\tau$ for all $g \in G$ and $n \in \overline{\{1\}}^\tau$ . This goes to show $\overline{\{1\}}^\tau$ is a normal subgroup of $G$ .

There is thus a natural map

\begin{aligned} \{G\text{-topologies}\} & \to \{N \triangleright G\} \\ \tau & \mapsto N_\tau = \overline{\{1\}}^\tau \end{aligned}

between the space of $G$ -topologies and the set of normal subgroups of $G$ . That’s all well and good, but how does any of this helps us to count $G$ -topologies? Well, the point is that, perhaps surprisingly, this map is actually a bijection.

To see that our map is injective, we remark that the collection of subsets of $G$ which are closed under a given $G$ -topology $\tau$ is actually determined by $N_\tau$ . Indeed,

F = \bigcup_{g \in F} \overline{\{g\}}^\tau = \bigcup_{g \in F} g \cdot N_\tau

for any $F \subset G$ which is closed under $\tau$ . Conversely, since $G$ is finite, any union of $N_\tau$ -cosets is closed under $\tau$ — a finite union of closed sets is closed. All in all, the subsets which are closed under $\tau$ are precisely the unions of $N_\tau$ -cosets.

This last equation also goes to show that our map is surjective: given some $N \triangleright G$ , the set $\tau = \left\{ \bigcup_{g \in X} g \cdot N : X \subset G \right\}$ is a natural candidate for a $G$ -topology such that $N_\tau = N$ . It is clear that $\tau$ is a topology and $\overline{\{1\}}^\tau = N$ . With enough effort, one can also show that if $m : G \times G \to G$ and $i : G \to G$ are the group multiplication and inversion, respectively, then

\begin{aligned} m^{-1}(g \cdot N) & = \bigcup_{h \in G} h \cdot N \times h^{-1} g \cdot N \\ i^{-1}(g \cdot N) & = g^{-1} \cdot N \end{aligned},

which goes to show that $\tau$ is indeed a $G$ -topology.

Finally, $\#\{G\text{-topologies}\} = \#\{N \triangleright G\}$ and thus

\mathbb{P}(G\ \text{is Hausdorff}) = \frac{1}{\# \{ N \triangleright G \}}.

We conclude this article with a table of the values of $\mathbb{P}(G\ \text{is Hausdorff})$ for some noteworthy finite groups $G$ , which were computed with the help of the Group Explorer library. I would like to thank my dear friend Eduardo Sodré for his help with the problem. I really hope the ride was as interesting to you as it was us both! 😁

$G$	$\mathbb{P}(G\ \text{is Hausdorff})$
$Q_4$	$1/6$
$K_4$	$1/5$
$S_4$	$1/4$
$S_n$ for $n \ge 5$	$1/3$
$D_n$ for even $n$	$\frac{1}{3 + \tau(n)}$
$D_n$ for odd $n$	$\frac{1}{1 + \tau(n)}$

Moluscos Australes

2023-07-13T00:00:00+00:00

This is a quick post on a lovely little book I recently found on my shelf, named “Moluscos Australes: Una Guia Ilustrada”, which is Spanish for “Austral Mollusks: an Illustrated Guide”.

“Austral Mollusks: an Illustrated Guide. Bivalves and snails from the coast of the southern tip of America”

This book was most likely bought in Punta Arenas, the capital of the Chilean province of Patagonia, when I visited there in late 2012 on my way to the Torres del Paine national park. The book contains a large number of black and white diagrams of seashells, together with detailed descriptions of the species they depict. Upon finding this copy of the book on my shelf more than 10 years later, I was delighted with both the extreme randomness of its contents and the beauty of the illustrations it contains — which where drawn by the book author, Dr. Sandra Gordillo. I thus decided to scan some of these drawings and share them in a blog post. Here are the pictures!

“Mulinia edulis” specimen

“Falsilunatia limbata” specimen

“Trophon geversianus” specimen

“Adelomelon ancilla” specimen

I really don’t have much to add to the drawing themselves. I tried to write detailed descriptions of the diagrams — which are mostly based on the descriptions provided by the book itself — to make the post more accessible. I suppose the only comment I have to add is the following remark: if you ever have the pleasure of visiting Patagonia, don’t miss the chance to try the excellent centolla soup! 🦀🥣

Classification of Abelian Lie Groups

2023-03-10T00:00:00+00:00

As a representation theorist, I’ve always been fascinated by the innocence of classification problems: Now we know a couple of examples of such mathematical structure, how about finding all possible examples? This is the second post in a series on fun little mathematical toy problems I encountered during my studies and, as it turns out, today’s question is a classification problem. Namely: what are all possible examples of Abelian Lie groups?

For the first article on this series see Is Every Set Fieldable?.

Note that I did not say “connected Abelian Lie groups”. The classification of connected Abelian Lie groups is a classical problem in Lie theory and its solution is well known: every connected Abelian Lie group $G$ is the product of copies of $\mathbb{R}$ and ${\mathbb{S}}^1$ — i.e. $G \cong \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1$ as Lie groups. We are interested in the classification of all Abelian Lie groups, regardless of whether or not they are connected. Nevertheless, we can use the classification of connected Abelian Lie groups to our advantage.

Given any Lie group $G$ , the connected component $G^0$ of $1 \in G$ is itself a closed normal subgroup. In particular, if $\dim G = n$ then $G^0$ is an $n$ -dimensional Lie subgroup of $G$ and $\Gamma = G / G^0$ is a $0$ -dimensional Lie group — i.e. a countable discrete group. We thus have an exact sequence

0 \longrightarrow G^0 \longrightarrow G \longrightarrow \Gamma \longrightarrow 0.

Hence it suffices to classify all exact sequences of Lie groups of the form

0 \longrightarrow G^0 \longrightarrow G \longrightarrow \Gamma \longrightarrow 0,

where $G^0$ is a connected Abelian Lie group — i.e. $G^0 = \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1$ — and $\Gamma$ is a countable discrete group.

We certainly know an example of such a sequence. Namely, given $G^0$ and $\Gamma$ as in the above we may take $G = G^0 \times \Gamma$ . But is this all we got? Surprisingly, the answer to this question is a resounding yes. In other words, we claim that all short exact sequences with $G^0$ and $\Gamma$ in the extremes are isomorphic to

0 \longrightarrow G^0 \longrightarrow G^0 \times \Gamma \longrightarrow \Gamma \longrightarrow 0,

and in particular every Abelian Lie group has the form $G^0 \times \Gamma$ for some $G^0$ and $\Gamma$ as above.

To see this, we introduce the notions of injective groups. An Abelian group $I$ is called injective if given an injective group homomorphism $f : H \rightarrow G$ between Abelian groups $G$ and $H$ and a homorphism $H \rightarrow I$ , there is some group homomorphism $G \rightarrow I$ such that the composition $H \overset{f}{\to} G \to I$ is the same as the map $H \rightarrow I$ .

The reason why we are interested in injective groups is the fact that every short exact sequence

0 \longrightarrow I \overset{f}{\longrightarrow} G \overset{g}{\longrightarrow} K \longrightarrow 0

of Abelian groups with $I$ injective splits. In other words, there is some group homomorphism $s : G \rightarrow I$ such that $s \circ f = \text{id}$ — or, equivalently, there is $s ' : K \rightarrow G$ such that $g \circ s ' = \text{id}$ .

Indeed, by taking $H = I$ and $\text{id} : I \rightarrow I$ for the map $H \rightarrow I$ in the definition of an injective group we get some $s : G \rightarrow I$ as required. As it turns out, $G^0$ is an injective group, but how can we go about proving it? The answer to this question requires us to introduce one more definition, namely the concept of a divisible group: an Abelian group $D$ is called divisible if given, $d \in D$ and $n \in \mathbb{Z}$ , there is some $d ' \in D$ such that $d = n \cdot d '$ .

Examples of divisible groups include $\mathbb{R}$ and ${\mathbb{S}}^1$ . In addition, it is clear that the product of divisible groups is also divisible, so that $G^0 = \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1$ is a divisible groups. It is a well known fact from module theory — recall that Abelian groups are the same as $\mathbb{Z}$ -modules — that every divisible Abelian group is injective. In particular, $G^0$ is injective and our sequence

0 \longrightarrow G^0 \overset{f}{\longrightarrow} G \overset{g}{\longrightarrow} \Gamma \longrightarrow 0

splits in the category of Abelian groups.

We can thus find a group homomorphism $s ' : \Gamma \rightarrow G$ such that $g \circ s ' = \text{id}$ . Since $\Gamma$ is discrete, $s '$ is continuous and therefore our sequence splits in the category of Abelian topological groups. This implies $G \cong G^0 \times \Gamma$ as topological groups. Since every continuous group homomorphism between Lie groups is a smooth map, it follows that $G \cong G^0 \times \Gamma$ as Lie groups.

All in all, we have just seen that every Abelian Lie group is the product of copies of $\mathbb{R}$ and ${\mathbb{S}}^1$ with a countable discrete group $\Gamma$ : $G \cong \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1 \times \Gamma$ . I would now like to conclude this article by saying a feel words on how one may try to generalize our proof to a classification of all Lie groups — regardless of Abelianess.

First and foremost, our proof is heavily reliant on the classification of connected Abelian Lie groups. This classification, in turn, relies on the classification of Abelian Lie algebras: to find all connected Abelian Lie groups it suffices to find all connected Lie groups whose Lie algebra is the Abelian Lie algebra $\mathfrak{g} = \mathbb{R}^n$ , and these are just the quotients of its simply connected form $G = \mathbb{R}^n$ by discrete subgroups.

Since there is no general classification of finite-dimensional Lie algebras, this particular argument cannot be used to obtain a classification of all connected Lie groups. Indeed, the classification of arbitrary connected Lie groups is regarded by most as an intractable problem. Nevertheless, there are classifications of $1$ -dimensional and $2$ -dimensional connected Lie groups, so one could attempt a classification of low-dimensional groups.

Next there is the question of whether or not every short exact sequence of (not necessarily Abelian) Lie groups of the form

1 \longrightarrow G^0 \longrightarrow G \longrightarrow \Gamma \longrightarrow 1,

with $G^0$ connected and $\Gamma$ discrete, splits — which would imply $G \cong G^0 \rtimes \Gamma$ as Lie groups. The issue here is that $G^0$ may not be divisible in the general setting, and even if it is, it is unclear whether or not every divisible group is injective — in the category of arbitrary groups.

That about wraps it up. Hope to see you in the next post of this series! 😛

Minhocão GIF

2023-03-08T00:00:00+00:00

This is just a quick post on a GIF I made during my teenage ears, which I would now like to share with the rest of the world. These pictures where taken at Minhocão (Portuguese for “Big Earthworm”), an infamous elevated highway at my hometown of São Paulo, Brazil.

Sneaking through the beautiful apartment complexes on the winding route of the São João avenue, Minhocão is a symbol of the chaotic and inhumane atmosphere of São Paulo. I would go as far as classifying it as an urbanistic aberration, yet I think it somehow captures the post-apocalyptic beauty hidden behind the rundown façades of São Paulo’s downtown.

Anyway, I would like to thank my friend Caio Almeida for helping me out with this GIF: I was the model and he was the one pulling the trigger.

Is Every Set Fieldable?

2023-01-14T00:00:00+00:00

This is the first in a hopefully long series on some of the most absolutely stupid questions I came across while studying mathematics. The idea is presenting a stupid toy problem, the context behind it and how I came across a solution. My expectation is that none of the questions posed in this series will be of any significance. The first question we will explore is set-theoretic in nature (because of course): can every infinite set be endowed with the structure of a field? In other words, given an infinite cardinal number $\kappa$ , are there fields of cardinality $\kappa$ ?

Notice that the statement is clearly false for finite sets. It is a well known fact that all finite fields are finite extensions of the fields of prime order $\mathbb{F}_p = \mathbb{Z} / p \mathbb{Z}$ . In particular, the order of a finite field must be a prime power.

I first asked myself a similar question while trying to apply a result from Group Theory to an unrelated problem. Specifically, I unfortunately found myself in the position of trying to apply a theorem about groups to a set $X$ which did not posses a natural group structure. The question then becomes natural: does $X$ have any group structure at all? Using Cantor’s normal form I was able to show that $X$ is in bijection with the additive group of a certain commutative ring $A$ . If we fix a bijection $X \rightarrow A$ we can thus endow $X$ with the structure of an Abelian group by declaring that this bijection is actually a group isomorphism.

In fact, the same argument goes to show that every infinite set can be endowed with the structure of a commutative ring! Unfortunately this argument falls short of a proof of our statement about fields, since the ring $A$ I constructed turned out to contain divisors of zero. The question then came: can I somehow improve this proof to account for fields? Coming up with an answer required me to develop a new method, which I would now like to share. Explicitly, we will show that, given a cardinal $\kappa$ , the field

\mathbb{Q}(\kappa) = \left\{ \frac{p(x_1, x_2, \ldots, x_\alpha, \ldots)} {q(x_1, x_2, \ldots, x_\alpha, \ldots)} : \alpha < \kappa, p \text{ and } q \text{ polynomial} \right\}

of rational functions in $\kappa$ variables is in bijection with $\kappa$ .

To be clear, the fact that there are fields of cardinality $\kappa$ is an immediate corollary of the Löwenheim–Skolem theorem, but we will alternatively present a constructive proof. As it turns out, much of the delicacies of this proof come down to one of the many pathologies of cardinal arithmetics. Specifically, throughout this article we will use the the fact that, given two infinite cardinals $\kappa$ and $\lambda$ , $\kappa \cdot \lambda = \text{max} \left ( \kappa , \lambda \right )$ . Our proof has two primary steps, the first of which consists of showing that…

Taking the polynomial ring doesn’t change the cardinality

Denote by $\mathbb{Q} \left [ \kappa \right ]$ the ring of rational polynomials in $\kappa$ variables $x_\alpha$ , $\alpha < \kappa$ . We want to establish $\left | \mathbb{Q} \left [ \kappa \right ] \right | = \kappa$ . Clearly $\kappa \le \left | \mathbb{Q} \left [ \kappa \right ] \right |$ . Indeed, there is a natural injection

\begin{align*} \kappa & \to \mathbb{Q}[\kappa] \\ \alpha & \mapsto x_\alpha \end{align*}

The challenge thus is showing that $\left | \mathbb{Q} \left [ \kappa \right ] \right |$ is no larger than $\kappa$ itself. We start by bounding the cardinality of the set of monomials in the variables $x_\alpha$ . Given a non-negative integer $n$ , there is a canonical surjection

\begin{align*} \kappa^n & \to \{ \text{monomials in } x_\alpha \text{ of degree } n \} \\ (\alpha_1, \ldots, \alpha_n) & \mapsto x_{\alpha_1} \cdots x_{\alpha_n} \end{align*}

Hence $\# \{ \text{monomials in } x_\alpha \text{ of degree } n \} \le |\kappa^n| = \kappa$ . Since the union of countably many sets of size $\kappa$ has size no larger than $\kappa$ , this implies

\# \{ \text{monomials in } x_\alpha\} = \left| \bigcup_{n \ge 0} \{ \text{monomials in } x_\alpha \text{ of degree } n\} \right| = \kappa

By the same token, it follows from the fact $\mathbb{Q}$ is countable that $\mathbb{Q} \left [ \kappa \right ]$ , the set of rational combinations of the monomials in the variables $x_\alpha$ , has cardinality $\kappa$ . Next we show that formally inverting polynomials does not change the cardinality of our rings. In other words…

Taking the field of fractions doesn’t change the cardinality

Let $\mathbb{Q} \left ( \kappa \right ) = \text{Frac} \left ( \mathbb{Q} \left [ \kappa \right ] \right )$ be the field of rational functions in the variables $x_\alpha$ , $\alpha < \kappa$ . We want to establish that $\left | \mathbb{Q} \left ( \kappa \right ) \right | = \left | \mathbb{Q} \left [ \kappa \right ] \right | = \kappa$ . Once again, it is already clear that $\kappa \le \left | \mathbb{Q} \left ( \kappa \right ) \right |$ . Indeed, there is a canonical injection

\begin{align*} \kappa & \to \mathbb{Q}(\kappa) \\ \alpha & \mapsto x_\alpha \end{align*},

given by the composition of the canonical injection $\kappa \rightarrow \mathbb{Q} \left [ \kappa \right ]$ with the inclusion $\mathbb{Q} \left [ \kappa \right ] \rightarrow \mathbb{Q} \left ( \kappa \right )$ .

On the other hand, there is a surjection

\begin{align*} \mathbb{Q}[\kappa] \times (\mathbb{Q}[\kappa] \setminus \{0\}) & \to \mathbb{Q}(\kappa) \\ ( p(x_1, \ldots, x_\alpha, \ldots), q(x_1, \ldots, x_\alpha, \ldots) ) & \mapsto \frac{p(x_1, \ldots, x_\alpha, \ldots)} {q(x_1, \ldots, x_\alpha, \ldots)} \end{align*}

and therefore $\left | \mathbb{Q} \left ( \kappa \right ) \right | \le \left | \mathbb{Q} \left [ \kappa \right ] \times \left ( \mathbb{Q} \left [ \kappa \right ] \setminus \{ 0 \} \right ) \right | = \kappa^2 = \kappa$ . It thus follows from the Cantor-Bernstein theorem that $\left | \mathbb{Q} \left ( \kappa \right ) \right |$ is precisely $\kappa$ .

This concludes our proof. I would like to point out that this proof also works for constructing fields of positive characteristic. Indeed, if we take ${\mathbb{F}}_p \left ( \kappa \right )$ instead of $\mathbb{Q} \left ( \kappa \right )$ , the exact same argument show that $\left | {\mathbb{F}}_p \left ( \kappa \right ) \right | = \kappa$ . In addition, I am strongly convinced that similar methods can be used to show that every infinite set can be endowed with the structure of a algebraically closed field — of arbitrary characteristic!

On the other hand, I should note that most of what we discussed is nonsense unless we work under the assumption of the axiom of choice, for if the axiom of choice does not hold we can find a set $X$ which does not have a cardinal. I’m not entirely sure of whether or not I expect our result holds without the assumption of the axiom of choice, but I am inclined towards trying to disprove it. I’ll leave that for another post though. See you then!

How to Convince People to Stop Smoking

2022-08-18T00:00:00+00:00

As illustrated by the following picture, I used to smoke cigarettes. In fact, I used to smoke a lot of cigarettes, I was a heavy smoker by most standards: about a pack and a half a day. Anyway, as indicated by the use of the past tense, as well as the “🚭” emoji featured in the navigation bar, I’m past smoking! It’s been 111 days since I last smoked a cigarette. This is a short essay on my experience as an ex-smoker and what ultimately pushed me to quit.

I should point out that the opinions expressed in here are entirely based on my personal experience, not on serious research of any kind.

Me simultaneously smoking 4 cigarettes I found on the ground at FAU-USP

I started smoking when I was 16 years old. Anyone who says they enjoyed the first time they smoked a cigarette is lying, but at the time I was very uncomfortable with the fact that I looked younger than I was, and I (unconsciously) felt like cigarettes would help me look older — as they say in Cidade de Deus: “Eu fumo. Eu chero. Já matei. Já roubei. Sou sujeito homem.” Over time I started smoking more and more, menthol cigarettes gave way to red filter cigarettes, until “being a heavy smoker” became part of my identity.

I distinctly remember an episode, I was about 13 years old, when I though to myself: “Why on god’s green earth would anyone on their right mind smoke a cigarette knowing how harmful it is?”. Nevertheless, three years later I found myself smoking 20 cigarettes a day. This is because, of course, smoking as a habit is not a conscious decision. Becoming a heavy smoker is usually a very gradual process, the product of an uncountable number of tiny decisions. No one wakes up one morning screaming “I will start smoking today! This is a great idea!”

This is what I think most anti-tobacco propaganda gets wrong: no one chooses to be a smoker, so aggressively telling people how harmful of a habit smoking is is not an effective way of preventing them from doing so. In fact, I used to feel a lot of hostility towards anti-tobacco propaganda because of this reason. I was distinctly aware of the harmful effects of tobacco consumption, so the numerous attempts to shock me with this information sounded like an insult to me. Did people really think I didn’t already know cigarettes are bad?

The grotesque imagery displayed on the back of cigarette packs — if I had to pick a favorite, I would certainly go for the picture of a bleeding red fetus inside a white porcelain toilet — is a great example of this: the aggressiveness of it made me feel attacked. Instead, I argue that we should treat smokers — a.k.a. potential ex-smokers — with compassion, as this is the strategy that ultimately pushed me to quit after 6 years of smoking.

Brazilian anti-tobacco propaganda: “YOU SUFFER: this product causes miscarriages and premature birth”

More precisely, what convinced me to consider quitting was the Ciência Suja podcast episode on the tobacco industry, entitled “Cigarette: the father of modern science denial” — if you happen to speak Portuguese, I strongly recommend listening to this episode. There are a number of reasons why this particular episode managed to cut through my hostility towards anti-tobacco propaganda. First of all, the episode heavily features the testimony a laryngeal cancer survivor named Ricardo Gama.

I can’t really grasp why, but for some reason I was much more sympathetic towards the voice of an ex-smoker. This is perhaps the primary reason why this specific piece of propaganda was so effective to me: the fact that the cautionary tale of a cancer survivor was told by the survivor himself made it feel like the warm words of a friend who understood me. Ricardo described how he started smoking when he was about 17 years old — which was, of course, very relatable to me — and how he quit when he was 26 years old, after about 10 years.

He then described how he was diagnosed with laryngeal cancer when he was 41, and how the subsequent removal of his larynx — a procedure known as larygectomy — affected his everyday life. This ties up into the next reason why I was convinced: I used to fool myself with the though that smoking wasn’t that risky if I were to quit early on. Unfortunately for him, Ricardo proved this is clearly not the case. I was also profoundly moved by the segment at the start of the episode on the Sua Voz choir, a choir formed by larygectomy patients such as Ricardo himself.

I guess what I’m trying to get at is I think that switching the perspective from “you, the smoker” to “we, the smokers/ex-smokers” may lead to an increase in the effectiveness of anti-tobacco propaganda. Again, this opinion is entirely based on my personal experience, which is anecdotal evidence. I’m actually very curious to know whether or not there are rigorous studies that investigate this hypothesis, please let me know if you know some! Either way, I’m hoping this article can provide some insight on anti-tobacco propaganda from the perspective of an ex-smoker.

I guess this is all I had to say. See you later 😛

Errata on Tensors

2022-06-02T00:00:00+00:00

February this year I posted some comments on how tensors show up in geometry. However, having enrolled in Claudio Gorodski's course in Riemannian Geometry at IME-USP, I now know a thing or two about tensors I didn’t knew at the time. I’m proud to say that most of what I said stands up to the test of time, but I would still like to make some corrections.

If you’re interested in Riemannian geometry, I strongly recommend Claudio’s course notes.

First and foremost, in the previous post I explicitly stated that there are vector bundles $E_i \rightarrow M$ and $F \rightarrow M$ such that the map

i : \Gamma \left ( \text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \right ) \rightarrow \text{Hom}_{C^\infty ( M )} \left ( \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) , \Gamma ( F ) \right )

that takes a tensor $T \in \Gamma \left ( \text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \right )$ to the homomorphism $i T : \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) \rightarrow \Gamma ( F )$ given by $(i T)(\xi^1, \ldots, \xi^n)_p = T_p(\xi_p^1, \ldots, \xi_p^n)$ is not surjective. This is not the case: every $C^\infty ( M )$ -multilinear map $T : \Gamma ( E_1 ) \times \cdots \times \Gamma ( E_n ) \rightarrow \Gamma ( F )$ is the image of some tensor, which is to say, the value of $T(\xi^1, \ldots, \xi^n)_p$ depends only on the values of $\xi_p^i$ .

This result is known as the tensor field characterization lemma, and a proof can be found in here. This particular proof is specific to the case where each $E_i = T M$ and $F = M \times \mathbb{R}$ — which is to say, the proof is specific to the case where $T : \mathfrak{X} ( M ) \times \cdots \times \mathfrak{X} ( M ) \rightarrow C^\infty ( M )$ — but I believe the same argument should work for sections of arbitrary smooth vector bundles. The proof, however, is heavily reliant on the existence of smooth partitions of unity, which hints at the fact that perhaps the lemma does not hold for holomorphic vector fields over complex manifolds — i.e. perhaps there some complex manifold $X$ with holomorphic vector fields $E_i \rightarrow X$ and an $\mathcal{O}_X$ -multilinear map $T : \Gamma ( E_1 ) \times \cdots \times \Gamma ( E_n ) \rightarrow \Gamma ( F )$ which is not the image of a holomorphic tensor.

The fact that this proof is restricted to the particular case where $E_i = T M$ and $F = M \times \mathbb{R}$ is not a coincidence: in most cases we are interested in tensors with the signature $\mathfrak{X} ( M ) \times \cdots \times \mathfrak{X} ( M ) \times \Omega^1 ( M ) \times \cdots \times \Omega^1 ( M ) \rightarrow C^\infty ( M )$ . The space of tensors which take $s$ fields and $r$ 1-forms is usually referred to as $\mathcal{T}^{(r, s)} M$ . In particular, if $M$ is a Riemannian manifold with metric $\mathrm{g}$ , the canonical isomorphism $T_p M \rightarrow T_p^* M$ given by the metric ${\mathrm{g}}_p$ induces an isomorphism of $C^\infty ( M )$ -modules $\mathfrak{X} ( M ) \rightarrow \Omega^1 ( M )$ .

Another way to put it is to say that $\mathrm{g}$ induces an isomorphism $\mathcal{T}^{(0, 1)} M \to \mathcal{T}^{(1, 0)} M$ , which in turn induces the so called musical isomorphisms $\sharp : \mathcal{T}^{(s, r + 1)} M \to \mathcal{T}^{(s + 1, r)} M$ and $\flat : \mathcal{T}^{(s + 1, r)} M \to \mathcal{T}^{(s, r + 1)} M$ . In particular, $\mathcal{T}^{ (r, s) } M$ is canonically isomorphic to $\mathcal{T}^{ (0, r + s) } M$ via $\sharp^s$ , which is the space of tensors with signature $\mathfrak{X} ( M ) \times \cdots \times \mathfrak{X} ( M ) \rightarrow C^\infty ( M )$ — so that the proof of the lemma for this particular case is sufficient. That about wraps it up. Hope this helped someone 😛

How to Export Playlists from NewPipe

2022-05-21T00:00:00+00:00

I am, like many other free software advocates, an enthusiastic user of the alternative YouTube client NewPipe, and I keep all of my playlists in there. While working on a page for my playlist on mathematical lectures the need to export playlists from NewPipe naturally appeared. I figured that with enough digging I would find an obscure button somewhere to export my playlists to JSON or something, but apparently NewPipe has no such feature. This is a tutorial on how to hack your way around this limitation.

As such, this tutorial relies on internal implementation details of the NewPipe app, and may brake with future updates. Since there is no official feature to extract this data from the app, we have to filter this data from all of the app data. To get this data, you should follow the official instructions on how to export the data and you will get a ZIP file containing a SQLite database file newpipe.db. Among other things, this file contains information on the videos inside each playlist, but the issue is that the information on videos from all playlists is stored in a single table named streams.

To find out from which playlist each video in stream is we need to look at the playlists and playlist_stream_join tables. First and foremost, we need to find out the so called “playlist id” of our playlist. This is a code which represents our playlist in NewPipe’s internal system. The playlist id can be obtained by inspecting the playlist_id column of the row corresponding to our playlist in the playlists table, as in

select playlist_id, title from playlists;

Next we need get the information (url + title) of the videos inside our playlist. Again, there is no single table containing this information. NewPipe stores the information on which videos are contained in which playlists at the playlist_stream_join table, while the information on each individual video is stored at the streams table. The order the videos are stored in the streams table is essentially arbitrary, so to get get the videos in the same order as in the playlist we are trying to export we need sort our query by the join_index column of the playlist_stream_join table. This can be done via the following query.

select url, title from streams
where uid in
  (select stream_id from playlist_stream_join where playlist_id = <PLAYLIST ID>)
order by
  (select join_index from playlist_stream_join where stream_id = uid and playlist_id = <PLAYLIST ID>);

You should replace “<PLAYLIST ID>” from the last query with the id obtained by inspecting the entries of the playlists table. After that, you can convert the result of this last query to your favorite data format, such as CSV or JSON.

To be honest, I don’t blame the NewPipe team for not implementing this functionality: it is such an obscure feature. I would like to be able to somehow share my playlists with friends in some sort of text format, but I don’t really see how exporting playlists to machine readable formats such as JSON would help. Anyway, at least we geeky people can access this information. 😛

What does the curvature of a surface have to do with tensor products?

2022-02-01T00:00:00+00:00

The question I would like to address in this article is: what is a tensor? This question has two answers. If you ask an algebraist, he (she) will tell you it is an element of the tensor product of two modules. If you ask a geometer, she (he) will ramble about “global constructions that only depend on point-wise values” for hours on end. We should note that we will primarily focus on what a tensor is from the perspective of a geometer, the intuition behind it and how we get from that to the usual formalism.

Frankly, I feel like there isn’t much to explain, yet I never had this explained to me and I always felt it was difficult reconcile my intuition with the formalism most commonly adopted. This is the primary reason I wrote this article: I would have loved to read it in the past.

In differential geometry and related fields, information can often be obtained by passing from the non-linear to the linear via infinitesimal approximations. Often times this comes in form of $C^\infty$ -linear functions between the spaces of smooth sections of two fiber bundles. Specifically, if $M$ is a smooth manifold and $E \rightarrow M$ , $F \rightarrow M$ are vector bundles over $M$ , then the sets $\Gamma ( E )$ and $\Gamma ( F )$ of smooth (global) sections of $E$ and $F$ , respectively, have a natural structure of a $C^\infty ( M )$ -modules, and sometimes linear maps $\Gamma ( E ) \rightarrow \Gamma ( F )$ show up. If such a map $\tau : \Gamma ( E ) \rightarrow \Gamma ( F )$ satisfies the condition that $\tau {\left ( \xi \right )}_p$ depends only on $\xi_p$ — and not $\xi$ on its entirety — then $\tau$ is called a tensor.

Often times it is convenient to also consider multilinear maps $\tau : \Gamma ( E_1 ) \times \Gamma ( E_2 ) \times \cdots \times \Gamma ( E_n ) \rightarrow F$ — i.e. maps that are linear in each coordinate. Again, if $\tau(\xi^1, \xi^2, \ldots, \xi^n)_p$ is determined by $\xi_p^i$ then $\tau$ is called a tensor. The classical examples of tensors are differential forms. A perhaps more interesting example is a Riemannian metric: for each point $p \in M$ we fix a positive-definite bilinear form $\text{g}_p : T_p M \times T_p M \rightarrow \mathbb{R}$ which “varies smoothly with $p$ ”. This construction induces a tensor

\mathrm{g} : \mathfrak{X}(M) \times \mathfrak{X}(M) \to C^\infty(M) \cong \Gamma(M \times \mathbb{R})

where $\left ( \text{g} ( V^1 , V^2 ) \right ) ( p ) = \text{g}_p ( V_p^1 , V_p^2 )$ .

This is what a tensor is supposed to be: for each $p \in M$ we fix some multilinear function between the $p$ -fibers of some vector bundles that “varies smoothly with $p$ ”. The meaning of “varies smoothly with $p$ ” is still imprecise, dare I not say unclear. We should point out that often times it is more convenient to define tensors in terms of global sections rather than defining the fiber-wise transformations, such as in the case of the curvature tensor $R ( X , Y ) Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z - \nabla_{[ X , Y ]} Z$ of a connection $\nabla$ or the Nijenhuis tensor $N ( X , Y ) = [ X , Y ] + J [ J X , Y ] + J [ X , J Y ] - [ J X , J Y ]$ of an almost complex structure $J$ .

Hence the need to consider tensors in geometry. Working with multilinear maps can be a bit of an annoyance, however. It would be convenient if we could somehow look at a tensor as a straight linear map — instead of a multilinear map. This brings us to the algebraic answer to our initial question. Given a ring $R$ and two $R$ -modules $M$ and $N$ , their tensor product $M \otimes_R N$ is the $R$ -module which enjoys the universal property that

\text{Hom}_R \left ( M \otimes_R N , L \right ) \cong \text{Bil}_R \left ( M \times N , L \right ) ,

where $\text{Bil}_R \left ( M \times N , L \right )$ is the module of $R$ -bilinear maps $M \times N \rightarrow L$ .

In other words, $R$ -multilinear maps $M_1 \times M_2 \times \cdots \times M_n \rightarrow N$ naturally correspond to $R$ -linear maps $M_1 \otimes M_2 \otimes \cdots \otimes M_n \rightarrow N$ . We should point out that the tensor product of modules can always be shown to exist by means of an explicit construction — whose elements are usually called tensors. If we fix $R = \mathbb{R}$ , this construction induces a construction in the category of vector bundles over some fixed manifold $M$ : if $E_i \rightarrow M$ are bundles over $M$ , there is a vector bundle $E_1 \otimes E_2 \otimes \cdots \otimes E_n \rightarrow M$ whose fibers are

\left ( E_1 \otimes E_2 \otimes \cdots \otimes E_n \right ) \text{|}_p = E_1 \text{|}_p \otimes E_2 \text{|}_p \otimes \cdots \otimes E_n \text{|}_p

The relationship between these two notions of tensor should now be clear: tensors $\Gamma ( E_1 ) \times \Gamma ( E_2 ) \times \cdots \times \Gamma ( E_n ) \rightarrow \Gamma ( F )$ are called tensors because they correspond to $C^\infty ( M )$ -linear maps

\Gamma ( E_1 ) \otimes_{C^\infty ( M )} \Gamma ( E_2 ) \otimes_{C^\infty ( M )} \cdots \otimes_{C^\infty ( M )} \Gamma ( E_n ) \rightarrow \Gamma ( F ) ,

which are in turn canonically identified with $C^\infty ( M )$ -linear maps

\Gamma \left ( E_1 \otimes E_2 \otimes \cdots \otimes E_n \right ) \rightarrow \Gamma ( F )

In fact, there’s a natural isomorphism of sheaves of $C^\infty$ -modules $\Gamma(-, E_1) \otimes_{C^\infty} \Gamma(-, E_2) \otimes_{C^\infty} \cdots \otimes_{C^\infty} \Gamma(-, E_n) \cong \Gamma(-, E_1 \otimes E_2 \otimes \cdots \otimes E_n)$ 🤡

To recap: we’ve just shown that a tensor $\tau : \Gamma ( E_1 ) \times \cdots \Gamma ( E_n ) \rightarrow \Gamma ( F )$ can be naturally identified with some $\tau \in \text{Hom}_{C^\infty ( M )} \left ( \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) , \Gamma ( F ) \right )$ . A natural question to ask ourselves at this point is: does $\tau$ correspond to some $\tau \in \Gamma \left ( \text{Hom} \left ( E_1 \otimes \cdots E_n , F \right ) \right )$ ? First of all, why does this make sense? Recall that given two vector spaces $V$ and $W$ , the set $\text{Hom} ( V , W )$ of linear transformations $V \rightarrow W$ is again a vector space. Hence we can consider the vector bundle $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \rightarrow M$ whose fibers are

\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \text{|}_p = \text{Hom} \left ( E_1 \text{|}_p \otimes \cdots \otimes E_n \text{|}_p , F \text{|}_p \right )

The previously mentioned example of Riemannian metrics does hint at an inclusion

i : \Gamma \left ( \text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \right ) \rightarrow \text{Hom}_{C^\infty ( M )} \left ( \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) , \Gamma ( F ) \right ) ,

which takes $\eta \in \Gamma \left ( \text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \right )$ to $i \eta : \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) \rightarrow \Gamma ( F )$ with $i \eta {\left ( \xi \right )}_p = \eta_p \left ( \xi_p \right )$ — notice this is precisely what we did to get from “a bilinear form in $T_p M$ for each $p \in M$ ” to a Riemannian metric seen as a tensor. The meaning of “a transformation at each fiber $p$ that varies smoothly with $p$ ” is now much clearer too: this is a smooth section of $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right )$ . The inclusion $i$ is not surjective. This is because in general if $\varphi : \Gamma \left ( E_1 \otimes \cdots \otimes E_n \right ) \rightarrow \Gamma ( F )$ is a homomorphism the value of $\varphi(\xi^1, \ldots, \xi^n)_p$ may very well depend on $\xi^i$ in their entirety, and not only on $\xi_p^i$ .

This last statement is actually false! See the errata on this post.

We claim, however, that the image of $i$ consists precisely of the multilinear functions $E_1 \times \cdots \times E_n \rightarrow F$ that are tensors — i.e. such that $\tau(\xi^1, \ldots, \xi^n)_p$ is determined by $\xi_p^i$ . Indeed, if we consider the map

s : \mathcal{T}(E_1 \times \cdots \times E_n, F) \to \Gamma(\operatorname{Hom}(E_1 \otimes \cdots \otimes E_n, F))

given by $s \tau_p (v_1, \ldots, v_n) = \tau(\xi^1, \ldots, \xi^n)_p$ , where $\mathcal{T}(E_1 \times \cdots \times E_n, F) \subset \operatorname{Hom}_{C^\infty(M)}(E_1 \otimes \cdots \otimes E_n, F)$ is the subspace of tensors and $\xi^i \in \Gamma ( E_i )$ are such that $\xi_p^i = v_i$ , we can very quickly check that $i = s^{- 1}$ , establishing an isomorphism of $C^\infty ( M )$ -modules

\Gamma(\operatorname{Hom}(E_1 \otimes \cdots \otimes E_n, F)) \cong \mathcal{T}(E_1 \times \cdots \times E_n, F)

The definition of $s \tau_p (v_1, \ldots, v_n)$ does not depend on the choice of $\xi^i$ precisely because the value of $\tau(\xi^1, \ldots, \xi^n)_p$ depends only on $\xi_p^i = v_i$ ! In conclusion, a tensor $\tau : E_1 \times \cdots \times E_n \rightarrow F$ is called a tensor because it corresponds to a smooth section of $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right )$ . To finish things of, I would like to conclude our discussion by explaining a small notational quirk the reader will probably encounter in the literature: most people refer to $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right )$ as $E_1^* \otimes \cdots \otimes E_n^* \otimes F$ .

This is because given two vector spaces $V$ and $W$ , the space $\text{Hom} ( V , W )$ is canonically isomorphic to $V^* \otimes W$ . Taking $V = E_1 \text{|}_p \otimes \cdots \otimes E_n \text{|}_p$ and $W = F \text{|}_p$ , this translates to an isomorphism of vector bundles $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right ) \rightarrow E_1^* \otimes \cdots \otimes E_n^* \otimes F$ . In fact, usually the differential structure of $\text{Hom} \left ( E_1 \otimes \cdots \otimes E_n , F \right )$ is defined via the identification with $E_1^* \otimes \cdots \otimes E_n^* \otimes F$ . This is the formalism generally adopted, which is to say, when a geometer says “a tensor” in a formal sense he most likely means “some $\tau \in \Gamma \left ( E_1^* \otimes \cdots \otimes E_n^* \otimes F \right )$ ”.

Also, if $F \rightarrow M$ is the trivial line bundle $M \times \mathbb{R}$ , one usually refers to $E_1^* \otimes \cdots \otimes E_n^* \otimes F$ by simply $E_1^* \otimes \cdots \otimes E_n^*$ , because tensoring by $M \times \mathbb{R}$ is the same as doing nothing. For instance, a Riemmanian metric is most often defined as a tensor $\text{g} \in \Gamma \left ( T^* M \otimes T^* M \right )$ satisfying special conditions. That about wraps it up. I hope this helped someone 😛