We’re hiring (2023 edition)

We’re hiring!

Two permanent positions in pure maths. One in geometry/topology, one open for all areas of pure mathematics. Applications are through mathjobs (link to Job 1, link to Job 2), and are due December 18 (AEDT).

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Ublock origin filters (unimelb/mathoverflow/stackexchange)

Motivated by recent changes to the mathoverflow cookie popup spam, I recently learned something about Ublock origin filters. In particular, a URL X matches *X. Now my custom filters file contains the lines





So the first line above matches every unimelb.edu.au website, independent of the prefix before unimelb. While the last three match every website, so I can use this set of three filters not just on mathoverflow, but on every site in the stackexchange network.

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The Jacobson Density Theorem

I’ve never been at ease with the Jacobson density theorem, every proof that I’d seen felt somewhat weird and never had any stickiness in my mind. So I came up with my own proof, I hope you like it.

The Theorem (Jacobson Density Theorem}
Let k be a field and A a k-algebra (yes unital). Let S be a finite dimensional simple A-module and D=\operatorname{End}_A(S) (which is a division algebra by Schur’s Lemma). Then the canonical homomorphism from A to \operatorname{End}_D(S) is surjective.

First, without loss of generality (replacing A by a quotient), we can assume that S is faithful.

Let us first deal with the case when A is a division algebra. Then S=A. And D=A as well, acting by right multiplication. And we are done.

Since A is not a division algebra it has a non-zero element a which is not invertible.

If a is not nilpotent, then the minimal polynomial m(x)\in k[x] of a factors as x^np(x) where n,\deg(p(x))>0. In particular it has two coprime factors, so by the Chinese remainder theorem k[x]/(m(x)) has a non-trivial idempotent. In particular, A contains a non-trivial idempotent.

If on the other hand a is nilpotent, then we can find v,w\in V with av=0, aw=v and v\neq 0. Since V is irreducible, there exists b\in A with bv=w. Then (ba)v=0 and (ba)w=w. So ba is not nilpotent, and we can run the argument of the previous paragraph again to conclude that A contains a non-trivial idempotent.

Let e be this non-trivial idempotent. We will use the fact that eS is an irreducible eAe-module. This is part of a general fact, that S\mapsto eS is a bijection betwen simple A-modules S with eS\neq 0 and simple eAe-modules. Furthermore, from the nature of the constructions of both directions of this bijection, we can conclude that

    \[\operatorname{End}_A(S)\cong \operatorname{End}_{eAe}(eS).\]

Now we can perform an induction on dimension. The inductive hypothesis tells us that eAe \to \operatorname{End}_{D}(eS) and (1-e)A(1-e)\to \operatorname{End}_{D}((1-e)S) are both surjective. To finish, given the symmetry between e and 1-e, it suffices to show that given any D-hyperplane H with eS\subset H and any D-line \ell\subset eS, there is a nonzero element in A with kernel H and image \ell. (I thought about this step myself in terms of matrices, but actually writing it up like that feels a little gauche).

The inductive hypothesis applied to (1-e)A(1-e) allows us to find a nonzero element c\in A with ker(c)=H. Pick nonzero v\in \im(c) and w\in \ell. Since S is simple there exists d\in A with dv=w. Then dc does the trick and our proof is complete.

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Kanchanaburi, 14/08/23

Wake up to find out what my hotel calls an “American Breakfast”. This turns out to invovle toasted white bread with jam, fried eggs, mango, juice and tea/coffee. The mangos in Thailand are good, the white bread could be improved upon.

Tuk-Tuk to the bus station (80THB). Some words are exchanged between my tuk-tuk driver and someone at the bus stop and I am asked if I want to take a minibus. I agree. I realise now that based on my experience with the bus to Erawan Falls the next day, I am getting a faster and more comfortable ride than the bus would give me, for only 125 THB. Hellfire Pass is my destination this morning, and I am the first passenger dropped off. I’m not sure what the final destination of my minibus is, perhaps the Three Pagodas border crossing?

In WWII, the Japanese used forced labour (both POWs and civilians) to build a rail link to Burma. Hellfire pass is one section of the rail corridor where an interpretive centre and an audioguided hike along the rail corridor has been set up by the Australian government, free of charge. There are no longer any rail tracks along this part of the route, the British ripped out all the rail westward of Nam Tok at the conclusion of the war.

Unlike most visitors, I did the full hike, which is not as difficult as it is made out to be. I’m glad I’m not in a tour group – one arrives after me and the participants don’t get quite enough time to do the full hike and get the full experience. The audioguide has lots of stories told by POWs and the track has some nice views across the valley towards Myanmar. I also see my first monkeys of the trip, and after conversing with a local it seems that monkeys are as common in Thailand as kangaroos are in Australia.

I wholeheartedly recommend visiting this place.

After finishing Hellfire Pass, I bum a ride with a local to Namtok Sai Yok Noi waterfall. This seems to be a place more for children than adults. Though I must say, the chance to cool off under the waterfall is a great relief in this weather. Lunch is had here, from the other side of the main road.

There are two ways now to return by public transport, by bus or by train. I choose the latter (100 THB for foreigners), and walk to Nam Tok station, a walk which is not of any interest whatsoever. The train back to Kanchanaburi is more scenic than the bus, and on the more scenic portion of the trip I’m joined by some tour groups. The highlight is the section around Krasae Cave (you do get to briefly see inside the cave from the train) right next to the river. And the rest of the journey is enjoying riding through the countryside in a civilised manner, with the windows open as much as possible and no A/C in sight.

Back in Kanchanaburi, the train crosses the Khwae Yai along the famous bridge, full of tourists who have to get out of our way. There’s a station immediately after the bridge, which is where I alight.

Next stop is the JEATH war museum. This one got added to the agenda essentially because of the location, and being Thailand, entry is cheap. There’s a section dedicated to the 2nd World War, as you expect, but also some other stuff related to older history of Thailand. As well as some surprising exhibits like a section on the winners of the Miss Thailand beauty pageant. Because why not. The JEATH museum also comes with some nice views of the bridge over the River Kwai and the surrounding area as in the below picture.

Dinner is next, but that is beyond the scope of this blog. And Erawan falls beckons the next morning.

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1. Ta-taki Falls.

This is a short hike along a river to reach a pretty waterfall spot. And when I say along a river, I really mean you are walking through the actual river. There’s a place to hire shoes for the hike at the trailhead carpark. If you’re unsure, you need to hire them. Wear bathers, as you’ll want to have a swim at the pool under the waterfall, as well as giving you more flexibility on your route during the walk. A unique hike, highly recommended.

2. Bitter Melon

Those of you who grew up eating this may wonder what I’m even making a fuss about. There’s this vegetable in Okinawa (and some other places) called bitter melon. When you first taste it, it seems a little strange, but not overly offensive. And then the aftertaste hits you and you wish you had made wiser life choices. Best to avoid this food, although that isn’t always easy, some places are nasty and will do things like hide bitter melon inside tempura.

3. Satsukimaru

We found this lovely little restaurant in Oku, in the northern part of Okinawa. Good cheap food, even if they did do the bitter melon in tempura trick. Small and cozy, you can see the entire restaurant in the picture. Highly recommended.

4. Onishidake (Mt. Onishi)

There is supposed to be a hiking track to the top of Mt. Onishi. We stopped the car and went for a bit of a walk along a side road, but couldn’t find the trailhead. It wouldn’t surprise me if the trail has been neglected and become severly overgrown. If any reader stumbling across this post has more information, please do share it.

5. Mt. Fuenchiji

Skip this. There’s some telecommunications equipment and no view.

6. Katsuudake (Mt. Katsuu)

This one is a short steep hike from the car park to the summit, which is sure to work up a sweat, at least in the humid summer months. You can see that beautiful views will be possible from the below photo, but to see them all, you’ll have to go visit yourself. It is possible to continue hiking beyond the peak but we didn’t look into this. Highly recommended.

Practicalities: I needed to produce my international drivers licence to rent a car in Japan. 40 AUD for someone to look at my license and write down details on some document + 12 AUD for a photo. I should muscle my way into this industry, it seems like a nice profit making machine.

Acknowledgements: Thanks to Iva Halacheva for the photos.

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How to log in to Backstabbr on Firefox

You can’t log in to Backstabbr on Firefox without disabling enhanced tracking protection.

The good news is you can turn it off for this site only, log in, then turn the protection back on and you’ll remain logged in.

To turn off the enhanced tracking protection, look to the left of the address bar, and click on the shield. You’ll then be given a switch you can toggle.

I’m recording this here for future me, and/or anyone else who stumbles across this problem.

Thanks to the pseudonymous user shinythebald for pointing out this solution to me.

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The $140,000 Hit, by Neil Marks

Back in the ’90s, there was a monthly cricket magazine published in Australia, called “Inside Edge”, which as of the time of writing, doesn’t even have a wikipedia page.

In the middle of the magazine was always a double sided poster, each side featuring a picture of a player. Below we see an example, chosen for inclusion here because of the accompanying poem (which certainly was a one-off)

I don’t know which issue this poster came from. Steve Waugh hit the Mercantile Mutual Cup sign on 22 October 1995 (video (youtube), scoreboard (cricinfo)), and the only other reference I can find online for the poem is this Barker College magazine (pdf), dated April 1996, which puts bounds on when the poster was released. Perhaps some kind reader will know the exact issue, or I will stumble across my archive of Inside Edge magazines and update this post.

I now include the text of the poem.

In Perth, the WACA Cricket Ground
Has seen great feats unfurled,
And all those famous cricket deeds
Are known throughout the world.
Greg Chappell made his maiden ton
And Lockie showed his tricks,
The Windies fell to brave Mery Hughes
And Douggie hit that six.

It was here that Garth McKenzie
First bowled himself to fame,
Where the mighty Dennis Lillee
First learned to play the game.
Though their deeds will last forever,
The greatest deed, for mine,
Was a day in late October
When Tugga hit the sign.

The Mercantile Insurance Co.
Had offered lots of cash
To hit a little piece of tin,
So all would have a bash,
Around Australia’s major grounds
From straight hit down to fine,
In eight strategic places stood
A little painted sign.

Big hitters round the wide brown land
Had tried to no avail,
Jones, Ponting, Bevan, Border, Boon
Showed even champs can fail.
For two years unsuccessfully
Men slogged for all they’re worth
But never hit a bloody sign
Until that day in Perth.

The WACA’s not an easy ground,
On which to score a win,
But let me state it’s harder still
To hit a piece of tin.
The ground is fast and long and wide,
(I saw one hit score nine),
You’d give long odds to Bradman that
He couldn’t hits sign.

This day the Western Warriors
Were playing men in blue,
And Joey Angel made the break,
Knocked Slater’s stumps askew.
Then from the darkened locker room,
With countenance benign,
Out strode the great Steve (Tugga) Waugh,
His eyes fixed on the sign.

He took his guard and settled in,
For Reid was bowling well,
And with his nose down on the pitch
Survived a hostile spell.
Waugh then unleashed his finest shots,
Still playing down the line,
He smashed the ball all round the field,
But nowhere near the sign.

Tom Moody bowled to keep it tight
And sent a slow one in,
But Tugga saw it quickly and
He aimed it at the tin.
The clouds stood still, the strong wind dropped,
The sun began to shine.
The ball flew like a tracer shell
And crashed against the sign.

The crowd’s cheers rent the western skies,
The Blues let out a roar,
(No doubt the greatest hitting since
McDougall topped the score).
Tug’ clenched his fist and held it high,
Emotion to the fore,
Not much for ‘Slats’ or ‘Mo’ perhaps,
But quite a lot for Waugh.

All the rest was anti-climax,
Of that there is no doubt.
Soon after Waugh’s dramatic hit,
Young Stewart got him out
The skipper seemed quite happy, though
with Tayls’ it’s hard to tell.
But he asked the ump a Question:
“Did we win the game as well?”

Back in the rowdy locker room
We quaffed the beer and wine,
And we raised our glass to Tugga —
Our mate who hit the sign.
Still it makes you stop and wonder
Why Destiny’s design,
Made sure the greatest player was
The first to hit the sign.

And when the World’s Great Umpire gives
This faithful servant out,
I’ll journey to Elysian Fields
With not a care or doubt.
For I have seen the champions play
This golden game divine,
And was watching at the WACA
When Tugga hit the sign.

Neil Marks

For more cricket poetry, you might want to learn about How McDougall Topped the Score.

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Tension in teaching

The following quote is by Matt Emerton (in a comment on MathOverflow)

I think there is a genuine tension between proofs that a professional will like (where professional here may mean professional algebraist!) and ones that are elementary. For professionals, reductions and devissages are easy, natural, and we don’t even think of them as real landmarks in the proof; they are just serve as passages between the key points and ideas. But in writing things out, they can take a lot of words, and seem (as you wrote) mysterious and difficult. I don’t know the best way to deal with this tension.

Interestingly, Matt posted it as part of a discussion about exactly what I wanted to talk about in this post, the teaching of the structure theorem for finitely generated abelian groups, or more generally, of finitely generated modules over a PID.

My personal connection is that I taught this as part of our third-year algebra course this year at the University of Melbourne, and am slated to do so again next year. I think that I did not do a particularly good job of teaching it in 2022, primarily because I got distracted by the reductions and devissages and tried to proceed along those lines as much as possible, when what I have learned is more appropriate for one of these courses is the more prosaic approach involving matrix manipulations. It is with the matrix manipulations (directly proving Smith Normal Form) that I plan to teach this part of the course in 2023 (and beyond, if necessary).

For completeness, allow me to state the professionals’ proof: Split off the quotient by the torsion subgroup to reduce to the torsion case. Then canonically decompose the module into a direct sum of its p-primary components. Then use the fact that R/(p^e) is injective over itself to manually split the remaining short exact sequences needed to complete the classification.

While it may not be reasonable to expect a third-year student to follow this proof, I think it is fair to expect any PhD student of mine to be able to understand and execute this proof.

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SMMC 2022 A4

The Simon Marais Mathematics Competition happened last weekend. It is a maths competition for undergraduate students across Europe, Asia, Africa and Oceania. This post is about problem A4, which I submitted. I’ll talk a bit about where the problem came from, a generalisation, a conjecture and also provide a solution. The entire paper is available on the Marais website, and solutions should be put up there at some time in the near future.

Problem (SMMC 2022 A4)
Let n be a positive integer, and let q\geq 3 be an odd integer such that every prime factor of q is larger than n. Prove that

    \[ \frac{1}{n!(q-1)^n}\prod_{i=1}^n (q^i-1) \]

is an integer that has no prime factor in common with \displaystyle{\frac{q-1}{2}}.


Let G=GL_n(\mathbb{F}_q) and let N be the subgroup of monomial matrices (a matrix is a monomial matrix if and only if it has exactly one nonzero entry in each row and column). I show below in my solution that this question is equivalent to the fact that the integer |G|/|N| is coprime to (q-1)/2. Now why would I ever care about that?

This coprimality fact implies that the cohomology of G with mod (q-1)/2 coefficients is isomorphic to the cohomology of N with mod (q-1)/2 coefficients. And I was interested in these cohomology groups because the second cohomology group classifies central extensions, which is what I used to think about back in my PhD days. The group N feels somewhat more “combinatorial” than G, so it is nice to be able to pass information from N to G for free.

Generalisations (known and conjectural)

Let G be a split reductive group over \mathbb{F}_q, which I conflate with its \mathbb{F}_q-points below in an abuse of notation. Let T be a maximal split torus and N its normaliser in G. Then

    \[ \frac{|G|}{|N|}=q^{|\Phi^+|}\prod_i \frac{q^{d_i}-1}{d_i(q-1)}. \]

Here \Phi^+ is the set of positive roots and the collection of integers \{d_i\} are the exponents of the Weyl group. Then the same argument as in my proof below shows that this fraction is an integer, relatively prime to \frac{q-1}{2}.

If we remove the assumption that G is split, then I suspect the same conclusion is satisfied, but there is an additional argument needed as the formula for the quotient has additional factors. I have not worked out this argument and really don’t want to resort to case by case arguments, so there is your conjecture (I expect we now need to say T is a maximal torus containing a maximal split torus).


First we show that the fraction in the question is an integer. Since q-1 divides q^d-1 as a polynomial for all d, the statement only depends on the residue class of q modulo n!. Since every prime factor of q is greater than n, q is relatively prime to n!. So by Dirichlet’s theorem on primes in arithmetic progressions, we may assume without loss of generality that q is prime.

Let G=GL_n(\mathbb{F}_q) and let N be the subgroup of monomial matrices. Then

    \[|G|=q^{\frac{n(n-1)}{2}}\prod_{d=1}^n q^d-1 \qquad \mbox{and}\qquad |N|=n!(q-1)^n.\]

By Lagrange’s theorem |G|/|N| is an integer. Since q is relatively prime to |N|, we can further divide by the largest power of q in |G| and deduce that

    \[ \frac{|G|}{q^{\frac{n(n-1)}{2}}|N|}=\frac{1}{n!(q-1)^n}\prod_{d=1}^n (q^d-1) \]

is an integer.

Now let p be a prime dividing \frac{q-1}{2} and let d be a positive integer. To conclude, it suffices to show that the fraction

    \[ \frac{q^d-1}{d(q-1)} \]

has zero p-adic valuation. Write q=1+2m, then by the binomial theorem,

    \[ \frac{q^d-1}{d(q-1)}=\frac{1}{2dm}\sum_{i=1}^d {d \choose i}(2n)^i. \]

Let a=v_p(d) and b=v_p(m). Since v_p(i!)=\lfloor\frac{i}{p}\rfloor+\lfloor\frac{i}{p^2}\rfloor+\lfloor\frac{i}{p^3}\rfloor+\cdots <\frac{i}{p-1}, we get

    \[ v_p\left( {d \choose i}(2n)^i \right)\geq v_p\left(\frac{d(2n)^i}{i!}\right)>a-\frac{i}{p-1}+i(b+v_p(2)). \]

We have the inequality v_p(2)-\frac{1}{p-1}\geq \frac{v_p(2)-1}{2}, so

    \[ v_p\left( {d \choose i}(2n)^i \right)>a+i\left(b+\frac{v_p(2)-1}{2}\right). \]

For i\geq 2, we therefore get

    \[ v_p\left( {d \choose i}(2n)^i \right)>a+2(b-\frac{1}{2})=a+2b+v_p(2)-1\geq a+b+v_p(2)=v_p(2dm) \]

as b\geq 1 from our assumption that p divides m.
Thus in our sum, the term with i=1 has a strictly smaller p-adic valuation than every other term, so determines the p-adic valuation of the sum, and we get

    \[ v_p \left( \frac{q^d-1}{d(q-1)} \right)=v_p\left(\frac{2md}{2md}\right)=0, \]

completing the proof.

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Infinite dimensional vector spaces

This is a quick note to prove that two bases of an infinite dimensional vector space have the same cardinality. We freely use the axiom of choice and other standard facts about cardinalities of infinite sets. We will in fact prove the following:

Theorem: Let V be a vector space with basis \{v_i\}_{i\in I} with I an infinite set. Let \{w_j\}_{j\in J} be a linearly independent subset of V. (e.g. a basis of a subspace). Then |J|\leq |I|.

To prove this, WLOG J is a basis of V (by extending \{w_j\}_{j\in J} to a basis of V if necessary). For all i\in I, write

    \[v_i=\sum_j c_{ij}w_j.\]

Let E\subset I\times J be the set of pairs (i,j) with c_{ij}\neq 0. Then E\to I has finite fibres, since the sum above is finite, and E\to J is surjective, since the v_i lie in the span of the w_j with j in the image of V, but also the v_i generate V. Since I is assumed infinite, this is enough to prove that |I|\geq |J|, as required.

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