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Timothy Gowers

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We need more of this!
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I don't understand this well enough to know what impact it would have on academic publishing, but maybe it would make Discrete Analysis illegal (since we insist on CC-BY licences for all our papers).
This appears to be a major threat to open-access publishing: a proposed law that forces all copyrighted material to require fees, even for linking to the material rather than re-using it.
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The Abel Prize has just been announced for this year ...

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Yesterday I found myself near the Montparnasse Cemetery in Paris, so I paid it a quick visit. Many famous people are buried there. The grave pictured below is a family grave that contains perhaps the most famous of them, at least to a mathematician. He died at the age of 58 of complications from a prostate operation.
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Cambridge has a system called Symplectic, which automatically trawls the web and lets you know when it thinks it has found a paper of yours, so that you can "claim" it for the purposes of the next Research Excellence Framework -- the controversial exercise that takes place every few years where the research output of university departments is assessed, with huge financial implications.

The one occasion in the past where a paper turned out not to be by me was a mathematical paper by my son, W. J. Gowers (I am W. T. Gowers). It was reasonable for the system to ask, and it gave me a warm glow of parental pride.

But can that feeling compare with my delight at being asked whether

Electroencephalographic assessment of concussive non-penetrative captive bolt stunning of turkeys, by Gibson TJ, Rebelo CB, Gowers TA, and Chancellor NM, BRITISH POULTRY SCIENCE 59(1):13-20 2018

is mine? If I were to accept it, it would be easily the coolest title of any of my papers. But the journal is published by Taylor and Francis, so I can't in all conscience do so. Instead, I wish my namesake T. A. Gowers all the best in his/her study of methods of killing poultry -- a Google search reveals that (s)he has several papers in the area.
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It's been a while since I wrote anything about the mathematical development of my two youngest children, but here's a quick snapshot. The other evening my wife encouraged me to encourage an interest my daughter appears to be showing in mathematics at the moment, so I decided to amuse them by discussing multiples of 37. First I asked them to multiply it by 3. My son (10) got there quickly, but I had told him not to say the answer. A few seconds later my daughter (7) gave the correct answer, so I asked her how she had done it. She said she had multiplied 30 by 3 and got 90. Then she had multiplied 7 by 3 and got 21. Finally she had taken 90, added 10, added another 10, and added 1. I was very pleased by this answer, because it showed that she has internalized many of the important ring axioms, and in particular has a good understanding of distributivity, at least in certain contexts. I then asked them what nine 37s makes, or rather I asked "OK, if three 37s are 111, what are nine 37s?" My son got it quickly again, but my daughter said 300. It was clear that an answer like 300 couldn't be given without at least something right going on in her head, so I asked her to think a little harder. She then said 330. I asked her to think even harder and she said 333. I don't know exactly what was going on in her mind, but I found these answers fascinating. I then asked for fifteen 37s. This time my daughter got it right and my son got the answer 1665. It wasn't hard to reverse-engineer his mistake: he had carelessly worked out 15x111 instead. (By the way, all this was mental calculation.) He also enjoyed 27x37, so as a complement I asked him what 7x11x13 is, which he got, and also enjoyed. (This part of the conversation was clearly beyond what my daughter could do in her head.) Once he'd got that, I could ask him what 5x7x11x13 was, then 3x5x7x11x13, then 2x3x5x7x11x13 and finally 2x3x5x7x11x13x17. It's amusing that if you multiply things in the right order (I recommend 7x13 first, then multiply that by 11, and after you've done that everything is straightforward) then the calculation is easy, but if you try them in other orders then it becomes much harder.

I then spent a couple of days in Cambridge, and when I got back to Paris (where I'm spending this year), my wife told me something that gladdened a proud father's heart. My daughter is gradually getting to know her tables but doesn't know all of them, and while I was away my wife asked her what 7x7 was. She didn't know, but she worked it out in her head. My wife then asked her how she had done it, and got a proper mathematician's answer: she had taken 7, doubled it to get 14, doubled that to get 28, doubled that to get 56, and finally had subtracted 7 to get 49. I've written before about how reluctant I am to tell my children methods (such as long multiplication) for solving classes of problems. It's because I don't want to deny them the opportunity to discover arguments like that one and really develop an intuition for relationships between numbers. Once they've got such an intuition, the methods will codify some of it and save time for some problems. But there will always be problems that are better solved using a bit of thought rather than blindly applying an algorithm.
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Alan Baker 1939-2018

I'm sorry to report that Alan Baker died yesterday after a major stroke a few days ago. I knew him first when I was an undergraduate going to his number theory lectures, and later I was a colleague of his both in the Cambridge mathematics department and at Trinity College.

He became famous when he proved a far-reaching generalization of the Gelfond-Schneider theorem, which answered Hilbert's seventh problem. Hilbert's problem was the following. Suppose that a and b are two real numbers such that a is algebraic and not equal to 0 or 1, and b is irrational and also algebraic. Must a raised to the power b be transcendental (meaning that it can't be a root of a polynomial with integer coefficients)? For example, must 2 raised to the power the square root of 2 be transcendental? If I remember correctly, Hilbert believed that this was one of the hardest of his problems, but in fact it was one of the first to be solved.

One can reformulate the theorem by taking logs. It is equivalent to saying that if a_1 and a_2 are algebraic (and not 0 or 1) and b_1 is algebraic and irrational, then b_1log(a_1)-log(a_2) cannot be zero. We can express this more symmetrically by saying that if a_1 and a_2 are algebraic and b_1 and b_2 are algebraic with an irrational ratio b_1/b_2, then b_1log(a_1)+b_2log(a_2) cannot be zero.

Yet another way of expressing the result is to say that if a_1 and a_2 are algebraic numbers and their logarithms are linearly independent over the rationals, then these logarithms must in fact be linearly independent over the algebraic numbers.

Alan Baker extended this to an arbitrary number of algebraic numbers. That is, if a_1,...,a_k are algebraic and log(a_1),...,log(a_k) are linearly independent over the rationals (meaning that no non-trivial rational combination gives zero), then they are linearly independent over the algebraic numbers (meaning that no non-trivial linear combination with algebraic coefficients gives zero). This immediately implies the transcendence of all sorts of other numbers.

In fact, Baker did more: he gave a lower bound for how far away an algebraic combination of the logs of the a_i had to be from zero (in terms of various "heights", which tell you how complicated a polynomial you need to demonstrate that a number is algebraic). This had important applications to Diophantine equations.

One result I like of Baker's is an effective version of a consequence of Roth's theorem. I won't go into details, but a famous theorem of Roth implies that for every c>2 there is a constant delta>0 such that if p and q are positive integers and a is the cube root of 2, then |a - p/q| must be at least delta/q^c. Liouville's theorem, which is much easier, shows that it must be at least 1/q^3 or something similar. A problem with Roth's theorem is that it is ineffective: it does not tell you how small the constant delta needs to be. Baker managed to prove that if c=2.995 then you can take delta to be 0.000001. I like it because it is apparently such a negligible improvement over what you get from Liouville's theorem (indeed, q has to be very large before it gives any improvement at all), but that of course reflects how difficult the problem is.
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As a bit of background to the story about Elsevier giving £1 million to Oxford mathematics, here are three links. The main one below is to a blog post that gives some idea of how Elsevier doesn't give a ... about its lower-tier journals.

The other two are to information about Big Deal negotiations in South Korea ( and Finland ( The South Koreans appear to have capitulated UK-style when they got worried that access to journals would be cut off (ignoring what has happened in Germany, where Elsevier has maintained access despite not being paid for it), while the Finns have gone for a disappointing Dutch-style deal -- it looks to me as though it's probably roughly what they had before but with some discounts on APCs thrown in. That's why Elsevier can afford to splash out the odd million here and there.
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A Faustian pact?

Elsevier has given £1 million to Oxford to support postdocs and workshops in mathematics. Perhaps it's worth remembering that a few years ago Oxford was paying Elsevier £1 million or so per year (it's probably slightly more than that now) for its Big Deal, and many other universities round the UK were paying similar amounts, and still are after those who negotiated for us last time round managed so successfully to demonstrate to Germany how not to do things.

What was the right thing for Oxford to do here? Turning down the money seems tough on the potential postdocs who would benefit from it (and other Oxford mathematicians who would benefit from their presence). But accepting it without mentioning that Elsevier taketh away with the other hand is difficult for others to stomach.

Whatever one thinks of it, if we did away with Big Deals, we'd all be much better off, even if we stopped receiving gifts like this.
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Would you pay £500 in order to avoid a bill for £50 and therefore have full control over how you spent that £50? No? Thought not.

To summarize Johnson's arguments.

1. By the time the UK leaves the EU, the gross, and net contributions will have increased.

2. It is reasonable to use the gross figure, because this is the amount that we will have full control of afterwards.

3. But even the net figure is a substantial amount of money, which could be used to increase NHS funding.

The business about taking back control is very important to Johnson, because it is what allows him to claim afterwards that he isn't flat-out lying. Of course, he is being intentionally misleading, but it seems that that's OK in politics.

It's hardly necessary, but let's address one or two of the arguments. First, there are two net figures that matter here. One is the amount we pay in net of the rebate that we get straight back. Johnson seems to think it is terribly important to have control of the difference here -- I think because in theory the EU could decide to end the rebate. But it hadn't shown any sign of doing so, so the situation up to now has been that we send the EU X-Y instead of a notional gross figure of X, meaning that we have been 100% free to spend Y as we like. So even if for some weird reason you want to take a principled stand about the possibility that at some point in the future the EU would charge X instead, you cannot claim that we have not up to now had full control of Y.

The second net figure that matters is what we pay, net of what the EU gives us back in the form of things like agricultural subsidies, regional development funds, scientific grants, and so on. Here it is true that we do not have full control of how the money is spent. But why does that matter? For Johnson to make the case that it does, he has to say how he would spend the money differently, and that means telling us who would get less as well as who would get more, something he has notably failed to do. If he proposes to spend the same amount of money on regional development, science and agricultural subsidies, then the loss of control is hardly a serious problem. (It might be for his ego and feeling of power, but the rest of us just care about how the money is spent, and some of us don't want to give Boris Johnson more power.)

That leaves the figure net of both the rebate and what we get from the EU. And now we come to the most important part of all (though one that is insufficiently emphasized). OK, we won't be spending that money any more, but that money buys us access to the single market, losing which will cost us far more than we currently pay in. So what Johnson basically wants is for the government to have full control of a pot of money worth X, even if that comes at the cost of a much bigger pot of money, namely a large chunk of the government's tax take. If the government wants to spend as much as possible on the NHS, then it should make sure that its tax take is as big as possible, and the way to do that is to pay into the EU and rake in far more than that as a result of the benefits of the single market.
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