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September 10, 2023 · 12:35 pm

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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July 5, 2023 · 2:20 pm

Okinawa

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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March 13, 2023 · 11:54 am

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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December 23, 2022 · 3:26 am

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 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 hit a 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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November 15, 2022 · 4:28 am

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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October 20, 2022 · 1:06 pm

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 $\displaystyle{\frac{q-1}{2}}$ .

Origins

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).

Solution

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,

$\frac{q^d-1}{d(q-1)}=\frac{1}{2dm}\sum_{i=1}^d {d \choose i}(2m)^i.$

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

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

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

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

For $i\geq 2$ , we therefore get

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

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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October 13, 2022 · 9:59 am

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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February 25, 2022 · 11:37 am

How to not defend a title (DBNI EOG)

Step 1: Qualification.

Despite winning the DBNI last year, I was not given an automatic qualification to this year’s edition. So I had to go through the qualification slog. A lack of convenient tournaments, given the North American bias of the vFtF scene, and the cancellation of Cascadia meant that I had to rely on some PBEM results together with some players dropping out in order to make it back to the start list this February.

Given the title of this post, I guess I could just say the easy way to not defend my title would be to not qualify, but having qualified, I had to work harder to not defend it.

Round 1: Germany. (backstabbr link)

I “chose” Germany under the auction system in place for the selection of powers. It’s a secret bidding system, hence the quotation marks. I won’t discuss it further except to ask that anyone who discovers the optimal solution please write to me with details.

Evan is in France, with Ben in England. Surely I can work with one of them right? Matt is along for the ride beside me in Russia and one year ago when we were in the same positions, he came much to close to comfort to a solo, an experience I’d rather not repeat. Rounding out the board are Liam in Italy, Katie in Austria and the unstoppable Farren in Turkey. Interesting stat about Farren: Going in to this tournament, in games we’ve played together where we’re not neighbours, her SC counts are 12, 11, 13 and 10. If that continues, it doesn’t leave many centres for me to fight over.

Evan informs me he is ordering PAR-PIC, MAR-BUR so I bounce the latter and get “rewarded” for doing so by being bounced out of HOL by Ben. Not the best of starts, and when Italy walks into MUN in ’02 and BER in ’03, things are looking grim. But with Evan not fully committed to the EF and Ben distracted in Scandinavia, I’m able to hold on until the French move into IRI gives me the diplomatic space to get back in the game.

Now we arrive at a key moment in the game. I’m allied with France and meanwhile a strong Austro-Turkish alliance has blossomed on the other side of the board. I recognise that my chances to top this board depend on not being the next Austrian target. So I negotiate with Katie to give her the space to do absolutely anything except attack me and …

Nope. Farren’s hypnotic powers are total and I am unable to break them.
I try consoling myself with the fact that I am far from the only person unable to break them but it doesn’t work.

The FG goal now becomes to capture the remainder of the English centres while forming a line against AT. All is going according to plan (except, from my point of view, for Evan failing to order ENG-MAO) and then we see this piece of funkiness.

F NTH C LON-YOR

In all my time playing diplomacy, I’ve never been in a game with a kidnapped convoy before, so part of me is happy this happened, even if it seriously jeopardised my chances in the game. I’m also happy that backstabbr allows kidnapped convoys – I’ve seen some fun-hating tournament directors have rules against kidnapping convoys in their tournament rules. The piece de resistance is that I didn’t know that Evan was ordering LON-YOR, nor did I know that Ben would order YOR-LON. In fact Ben and I didn’t even talk that turn!

Some tense negotiation was needed to get from this new position to a draw, but we managed to stalemate the AT and I was left with a 7sc draw second to a Farren’s 10sc Turkey. A nice little secondary score to add to a good score, but not the board top I wanted – I would now have to find a way to pull that off in the next round.

Round 3: Turkey. (backstabbr link)

[Not round 2. The tournament structure is play one of Rounds 1 and 2, and one of rounds 3 and 4].

Suffice to say that I did not want to play Turkey. The fundamental problem with playing Turkey is that your three neighbours covet your corner position, which creates a bias towards early attacks on Turkey.

At the start of the game as Turkey, the clock is ticking. You have three game years to make soemthing happen, otherwise you’re dead.

1901 came and went with no progress, only with Christophe in Austria lying to me about wanting an AT.

In 1902 I was able to hold Bulgaria due to Russian neutrality, but Greg wasn’t intereted in actively working with me and turned down my offer of Serbia. The neutrality did buy me an extra year though.

1903 and there’s still no progress. For some unknown reason Austria tries to take CON with the wrong unit so I keep it.

Fall 1904, with an Italian fleet already docked in Smyrna and finally we see a crack with an Austrian swing at RUM. We’ve survived the onslaught and are back in the game. Let’s go! Interestingly once this happened, I started getting much more nervous, believing I had a real chance again.

Meanwhile, on the other side of the board, after some initial indications of an FG, Karthik (E) and Farren (F) had entered into a strong alliance, which quickly swept aside Timothy in Germany and was looking to roll the board. Not wanting to sit in the corner on 3 for the rest of the game, I played an aggressive game, cooperating with Greg to pick up a couple from Italy in a single year, before turning on Greg himself, all while the EF marched onwards.

This changed in Fall 1910, when Russ explained to the rest of the southern powers how we could form a 14 centre stalemate line in two turns, assuming we made a couple of good guesses. I went for it, looking for something to break up the EF. One nervous wait for adjudication later to find out if my convoy to APU would be disrupted and we made it.

Stalemate

And went straight into the final phase of the game. Where nothing moved. I figured Karthik would stab, given he needed a strong board top to make it through to the final. And yet the stab never came. Meanwhile I was busy holding my line, with EF (together or separately) not offering me any inducements to stab that I felt I could take seriously.

However despite nothing happening, Karthik would not agree to a draw. Now most tournaments have a rule along the lines of the Tournament Director being able to force a draw if there is no significant change in a certain amount of time. But the DBNI did not have such a rule, though now thanks to us, it does. At some point Zach (our TD) came in and told us that he was instigating this rule for this game, starting from when he announced it to us.

Still nothing of substance changed, with some turns going by quickly due to everyone clicking to adjudicate early and others being taken up with frantic negotiation between myself, Farren and Karthik. And so with Karthik taking Kiel on the last turn for an 11sc board top, we were force drawn with me on 8 centres, ending my bid to defend my title.

I am pleased I got to fight in some good-spirited and well-fought tough games. I specifically enjoyed getting to play against two quality players I had never played before in Christophe and Greg in my last game and I hope to cross swords with them on a board again soon. Special mention and congratulations must go to Farren, who dominated both my games and thoroughly deserves her place on the top board.

Speaking of the top board, it is being played on the weekend, and you can watch all the action at the DBN Channel with the stream scheduled to start at 00:30 GMT on Sunday 27 February. I look forward to seeing how the final chapter of this season unfolds.

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June 16, 2021 · 2:07 pm

Jucys-Murphy elements and induction

This post concerns the representation theory of the symmetric group over the complex numbers. Recall that the irreducible representations of the symmetric group $S_n$ are indexed by partitions of $n$ . Let $S^\lambda$ be the irreducible representation indexed by $\lambda$ . I want to say some words about the theorem that the decomposition of the induced module $\operatorname{Ind}_{S_{n}}^{S_{n+1}}S^\lambda$ is given by the decomposition into eigenspaces under the action of the Jucys-Murphy element.

First, the relevant Jucys-Murphy element is

$X:=(1,n+1)+(2,n+1)+\cdots+(n,n+1)\in \mathbb{C}[S_{n+1}].$

The way it acts on $\operatorname{Ind}_{S_{n}}^{S_{n+1}}S^\lambda=\mathbb{C}[S_{n+1}]\otimes_{\mathbb{C}[S_n]}S^\la$ is not as an element of $\mathbb{C}[S_{n+1}]$ but by $X\cdot (a\otimes v)=aX\otimes v.$ This is well-defined since $X$ commutes with $\mathbb{C}[S_n]$ .

What this action defines is a natural transformation from the functor $\operatorname{Ind}_{S_{n}}^{S_{n+1}}$ to itself. The induction functor is (bi)-adjoint to the restriction functor and this natural transformation is even simpler to construct on the adjoint side. Recall that if $\mathcal{F}$ and $\mathcal{G}$ are adjoint functors, then there is an isomorphism

$\operatorname{End}\mathcal{F}\cong\operatorname{End}\mathcal{G}.$

Here $\operatorname{End}\mathcal{F}$ refers to the natural transformations from $\mathcal{F}$ to itself, and the map in this isomorphism is given by pre- and post-composition by the unit and counit of the adjunction.

And the way that $X$ yields a natural transformation from $\operatorname{Res}_{S_n}^{S_{n+1}}$ to itself is very simple, it’s just by its usual action as an element of $\mathbb{C}[S_{n+1}]$ . If you transport this natural transformation to a natural transformation of the induction functor via the method I just mentioned, then you get the formula mentioned above.

Now given a pair of adjoint functors $\mathcal{F}$ and $\mathcal{G}$ , a natural transformation $X$ from $\mathcal{F}$ to $\mathcal{F}$ (and hence from $\mathcal{G}$ to $\mathcal{G}$ ) and a complex number $a$ , we can define a functor $\mathcal{F}_a$ by

$\mathcal{F}_a(V)=\{w\in \mathcal{F}(V)\mid Xw=aw\}.$

and similarly for $\mathcal{G}$ (this requires some linearity assumptions, but they’re satisfied here. Also you could take generalised eigenspaces if you wanted to, but in our application there is no difference).

When you do this, the functors $\mathcal{F}_a$ and $\mathcal{G}_a$ are adjoint:

Proof: Both $\operatorname{Hom}(\mathcal{F}_aV,W)$ and $\operatorname{Hom}(V,\mathcal{G}_aW)$ are the $a$ -eigenspace of the action of $X$ on $\operatorname{Hom}(\mathcal{F}V,W)\cong\operatorname{Hom}(V,\mathcal{G}W)$ .

Now apply this to our situation. We also use the following standard fact about the action of the Jucys-Murphy element (as developed e.g. in the Vershik-Okounkov approach):

Consider the decomposition

$\operatorname{Res}_{S_n}^{S_{n+1}}S^\mu=\bigoplus_{\lambda\to\mu}S^\lambda.$

Then the Jucys-Murphy element $X$ acts by the scalar $c(\alpha)$ on $S^\lambda$ , where $c(\alpha)$ is the content of the box $\alpha$ added to $\lambda$ to get $\mu$ .

Now translating this statement via the above yoga onto the adjoint side, we get

In the decomposition

$\operatorname{Ind}_{S_{n}}^{S_{n+1}}S^\lambda=\bigoplus_{\lambda\to\mu}S^\mu,$

the Jucys-Murphy element $X$ acts by the scalar $c(\alpha)$ on $S^\mu$ , where $c(\alpha)$ is the content of the box $\alpha$ added to $\lambda$ to get $\mu$ .

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June 14, 2021 · 5:17 am

Oaxaca Photos (December 2019)

I picked up my old phone and decided to try turning it on again. And after charging it, I was surprised, it turned on for the first time in 18 months. I guess the remedy for fixing a water damaged phone is to just wait a long long time.

This means I got access to some photos that I thought were previously lost, and I’ll present some of them here today.

Our trip begins in Oaxaca City, where I was visiting Banff in Mexico. First up, we have a visit to Monte Albán, an archaeological site on top of a hill right next to the city itself.