Fractional Euler Characteristic

This talk was given at the student combinatorics seminar by Cihan Bahran. His talk was expository, and although his interest in the topic arose from his research, he did not make any comments about that motivation. This post will be written at a fairly high level; in particular we’re going to need posets and some linear algebra.


Here is a fairly classical story, with a very small twist:

If $P$ is a finite poset, we can choose any linear extension and define the set of matrices $\text{Mat}_P(\Bbb Q)$ whose rows and columns are indexed by the elements of $P$ (according to the linear extension). Note that although this is a set of functions $P\times P\to\Bbb Q$, it is not the incidence algebra: we are in fact not using any of the order structure on $P$ at all.

Define $\zeta$ in the usual way: the $a,b$ entry of $\zeta$ is zero unless $a\leq b$. Note that by definition this means that $\zeta$ is an upper-triangular matrix with ones along the diagonal, and hence is invertible; it’s not hard to convince yourself that the inverse of $\zeta$ is a matrix $\mu$ corresponding to the usual Möbius function.

To every poset $P$ we can associate a simplicial complex called the order complex $BP$, whose faces are the chains of the poset (in particular, the vertex set is the same as the ground set of $P$ and the Hasse diagram exists as a subcomplex of $BP$, but is usually not the entire 1-skeleton). It turns out that the Euler characteristic of $BP$ is $\displaystyle\sum_{a,b\in P} \mu_{a,b}$.


One way to generalize this story is to observe that a poset is a rather simple example of a (finite) category, whose objects are the elements of $P$ and where there is an arrow from $a$ to $b$ if $a\leq b$, and composition of arrows is allowed by transitivity of $P$.

So let’s take a finite category $C$. Nothing stops us from setting up $\text{Mat}_C(\Bbb Q)$ in the same way, except that we have to give up any pretense that there is a “linear extension” and just order the elements arbitrarily.

We can still define a $\zeta$, but in general it will not be invertible. For this talk, Bahran restricted our attention to categories where it is ( “categories with $\mu$”), although he mentioned briefly some more general settings in which the story might go through anyway. In the case that $\mu=\zeta^{-1}$ exists, we define the Euler characteristic of $C$ to be $\chi = \displaystyle\sum_{a,b}\mu_{a,b}$.

It’s possible to make an “order [CW] complex” $BC$ from a category $C$, where the cells are chains of composable non-identity arrows. We note that $BC$ is usually infinite-dimensional and so we can’t define its Euler characteristic. However $BC$ is dimensionally finite: there are only finitely many cells of any fixed dimension. So we can still write down the alternating sum that represents its Euler characteristic, it’s just that this sum will usually diverge.

You may notice that the Euler characteristic doesn’t actually depend on the composition structure of $C$. Philosophically, there is a reason for this. Given the result for posets, it seems like we want to think of $\chi$ as a the “Euler characteristic of $BC$”— that is, a reasonable assignment of a finite value to the divergent series. So since the construction of $BC$ is independent of the composition structure on $C$ we should want $\chi$ to be independent as well.


We have thus obtained the following commutative diagram:

There is one missing arrow, and, well…

If we want to evaluate a divergent alternating series, how do we do that? Here is one possible idea: represent the alternating-ness as a factor of $(-1)^n$ for the $n^\text{th}$ term in the sum, and then replace all the $-1$’s with a variable, say $q$. Now, if the resulting power series has an expression as a rational function and if that rational function is defined at $q=-1$ (or has a removable singularity), then we can plug in $q=-1$ (or take the limit $q\to-1$).

It is a remarkable fact that both of those if-s will be satisfied whenever $C$ is a finite category with $\mu$, and therefore the last arrow in the diagram exists. Moreover, Berger and Leinster proved in 2008 that this diagram commutes.

But it’s not all good news. This $\chi_\text{div}(BC)$ is… not very nice. In particular, it can exist even when $\chi$ does not, and it’s not invariant under homotopy equivalence (!). So there’s no hope that this thing can be related to the Euler characteristic defined by homology groups. 

Moreover: there is a way to define $\chi$ in some cases when $\mu$ does not exist, and in this class of objects it becomes even worse. In this setting, it can fail to exist even when $\chi$ does (!!), or both $\chi$ and $\chi_\text{div}$ can exist but $\chi( C)\neq\chi_\text{div}(BC)$ (!!!). And worst of all, $\chi_\text{div}$ can even fail to be invariant under equivalence of categories (!!!!).

A 15-year-old Pakistani Sikh girl has become the first girl from the Sikh community to top the matriculation exam in the country.

Manbir Kaur, daughter of Giani Prem Singh the head granthi at Gurdwara Sri Nankana Sahib, scored 1,035 marks of 1,100 and emerged on top.

Kaur’s accomplishment is no ordinary feat as Sikhs only comprise 1% of Pakistan’s population.

“She achieved what even boys of the community could not. I think Manbir’s hard work has paid off. She has always been a diligent student,” Kaur’s proud father said, while speaking to the Hindustan Times.

Further, her father said that the 15-year-old had taken admission in a pre-medical course at the Punjab Group of Colleges in Lahore as she wanted to grow up to be a doctor.

Regarding girls’ education in Pakistan, Kaur’s father said, “I think women should be treated no differently from men as they have as much a right to pursue their dreams. If they can run homes so efficiently, why shouldn’t they be allowed to conquer the world outside?”

Claiming that Kaur’s cousins had also done really well in the Class 10 exams, Giani Prem Singh said that the 200 Sikh families living in Nankana Sahib were encouraging their daughters to excel in every field.

After the result was announced, the Pakistan Sikh Gurdwara Parbandhak Committee honoured Kaur with a siropa (robe of honour).

The Boards of Intermediate and Secondary Education announced results of the secondary school certificate examination across Punjab last week.

Graduate Level Mathematics Appears in Gravity Falls... Holy Shit

If you’re a viewer of Gravity Falls, you’ve likely paused the opening theme near the end in order to get a glimpse of the lovely triangle thingy that appears for a few frames.  If you manage to pause correctly, what you will see is this:

While the image in the center, and the symbols in the circle are clearly the thing that draws the eye of the viewer most, take a look in the upper right corner.  What do you see?

Matrices.  The bane of every high school student who went past the first year of algebra.  I was looking at this today (since my life is apparently now consumed by this show), and noticed that the symbols and patterns of the matrices were familiar for some reason.

I looked up how well the page had been interpreted/translated, and the only explanation I found for the matrices were: Geometric Transformation.  That’s all well and good, but there was something still not clicking.  I determined that I needed a better look at the full matrix equation (you can tell there’s some missing from the top of the previous image, if you look closely).

So, I found an image of the page (from S1E19, taken from the Gravity Falls Wiki) which showed the other portions of the matrices:

Here we have the full matrix equation.  It turns out that what we can see in the title card is the final form of the equation itself, which is an expanded form of a column vector with components [hx, hy_z, h, 1].  Suddenly, it hit me:

This is something I learned about in a 6000 level Stat course last semester!

This is not a random “geometric transformation.”  This is a projective transformation from 2D to 3D! These transformations preserve the connections between points, but can stretch and warp the connective lines beyond recognition.

If I were to extrapolate this identification into an implication about Bill Cipher, the lovely triangle in the center, I would think it implies three things, one of which is confirmed:

1) Bill can project into minds, and possibly into 3D in general. [Confirmed by S1E19]

2) Bill is trapped in 2D space in general.  The summoning described in the ring (which is not fully deciphered/understood in the context of the show) will permanently project him into 3D space/the “real” world (of the show).

3) Bill can alter his form, but will overall stay contained by the form of a triangle (projective transformations can stretch and warp the lines that connect the three points).

I am super, super excited to see this, and was a bit shocked to find a graduate level mathematics reference/clue hidden in a 1-frame image in the opening theme of a children’s television show.

tl;dr - There’s another clue in the title card of Gravity Falls, and it depends on an understanding of a particular topic of graduate-level mathematics.

You hit my car? Come back here, I want to show you something.

Today, I was parked on a busy road that’s pretty narrow. The parallel parking spots are pretty limited, so you have to get up tight with the other people. It was afternoon, and hey, just my luck, the last available space pulled out.

The parallel parking spots are set up in a little enclave beside the road. Once all the spots are taken up, there remains enough space behind the last vehicle to maybe squeeze in a Smart or a motorcycle.

Being in a rather senior part of town, this 65+ year old never-had-to-work-a-day-in-my-life grandma decided to try to squeeze in her Lexus behind the last car (my car). I was waiting in my car and could tell right away that she was going to hit. You know that moment, that moment right before you know you’re about to win something and you’re just waiting for the sparks to fly? That was me. As her car slowly creaked in beside, the inevitable was about to happen. She’s gonna park her car, hit me, and leave. I know it, I know it.

And that’s exactly what happened. She hit me not once, but twice during her parallel parking escapade. At this point, I make my sortie. She’s buried in her phone, texting away or something. Pretending that nothing happened. Ahh, I love it. I check the back of my bumper, not enough damage to call her out on. Damn, well. I can do something petty at least.

Miss self-entitled exits the car and just starts to leave the scene, I give her a snarly stink eye, trying to not argue or anything, as I know what I’m gonna do. I head behind her car, and take down her license.

She then starts yelling at me. In the most annoying Californian high class prick accent you can think of, she goes off.

“Hey, what’s your problem?”

“How about the fact that you hit me, twice.”

“It’s NORMAL! What do you expect?”

“It’s normal for you to hit cars? That must be fun.”

She scoffs, and leaves.

Now, my petty revenge plan begins. I eyeball her down all the way down the street, about 2 blocks away she turns into some furniture store. She’s old enough that you can tell it’s difficult for her to walk far. I care not.

What this lovely bourgeois example of a lady neglected to consider is that I am now impossibly pinned between two cars. I’m going to have to tell her to get over here and move her car. HOWEVER, she is also parked completed illegally on the road. There was a officer at the street across, so I flag him down and get him to get a tow.

In the mean time, I go back to that furniture store, and complain at her that she needs to move her car. I get her to come back with me–just in time to see her own car getting towed.

Ahh, that was sweet.

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