Its summertime so I’ve been trying out a few project Euler problems again.
In the process of doing one of them I realised something about trig integrals that I forgot, or maybe even never knew.
I thought this was cute and wanted to share it.

Consider the integral

this is one where you have to substitute a trig function to work it out, I always found these a little bit magic, so lets look at it in a more elementary way.

Let’s assume to start that $ -1 \le a \le 0$ The function $\sqrt{1-x^2}$ gives us the height of a radius 1 circle in the $xy$ plane, and so computing the integral above is the same as finding the area of enclosed by the $x$-axis, circle and line $y = a$.
However we know what the area of a wedge of a circle is, the full circle (of radius 1) has area $\pi$ and so a wedge of angle $\theta$ has area $\pi\cdot \theta/2\pi = \theta/2$.
Now the area we are looking for is the area of a wedge minus the area of the extra triangle.
What’s the angle of the wedge? We’re stopping at $y=a$ so the wedge has angle $\cos^{-1}(-a)$.
Hence the area of the wedge is $\cos^{-1} (-a) /2 $.

The added triangle will have side length $\mid a\mid$ (i.e. $-a$) and $\sqrt{1-a^2}$ so we get the final formula

Using the fact that $\cos^{-1}(-x) = \pi/2 + \sin^{-1}(x)$ we get a slightly cleaner formula for the indefinite integral

Which before now I would have said was hard to remember, but after seeing it like this, probably ok.

Another interesting note is that if $a \ge 0$ using the above formula is also correct, here adding the approriate triangle to the wedge gives us the integral.

The fact that every finite group is a Galois group is pretty well known (and in fact this post is basically just a transcription of the one in Lang’s Algebra) but I’ve been thinking about it recently and its a really cool result so I figured I’d share it.
Who knows, maybe I’ll post about the extension to profinite groups next time?

The starting point here is the following theorem of Artin, telling us that we can cut out Galois extensions with any group of field automorphisms we like.

Theorem(Artin)

Let \(K\) be a field and \(G\) a finite group of field automorphisms of \(K\text{,}\) then \(K\) is a Galois extension of the fixed field \(K^G\) with galois group \(G\text{,}\) moreover \([K:K^G] = \#G\text{.}\)

Proof

Pick any \(\alpha \in K\) and consider a maximal subset \(\{\sigma_1, \ldots, \sigma_n\}\subseteq G\) for which all \(\sigma_i \alpha\) are distinct. Now any \(\tau \in G\) must permute the \(\sigma_i \alpha\) as it is an automorphism and if some \(\tau\sigma_i \alpha \ne \sigma_j\alpha\) for all \(j\) then we could extend our set of \(\sigma\)s by adding this \(\tau\sigma_i\text{.}\)

So \(\alpha\) is a root of
\begin{equation*}
f_\alpha(X) = \prod_{i=1}^n (X- \sigma_i\alpha)\text{,}
\end{equation*}
note that \(f_\alpha\) is fixed by \(\tau\) by the above. So all the coefficients of \(f_\alpha\) are in \(K^G\text{.}\) By construction \(f_\alpha\) is a separable polynomial as the \(\sigma_i\alpha\) were chosen distinct, note that \(f_\alpha\) also splits into linear factors in \(K\text{.}\)

The above was for arbitrary \(\alpha \in K\) so we have just shown directly that \(K\) is a separable and normal extension of \(K^G\text{,}\) which is the definition of Galois. As every element of \(K^G\) is a root of a polynomial of degree \(n\) we cannot have the extension degree \([K:K^G] \gt n\text{.}\) But we also have a group of \(n\) automorphisms of \(K\) that fix \(K^G\) so \([K : K^G] \ge n\) and hence \([K : K^G] = n\text{.}\)

So now with this in hand we just have to realise our group as a group of field automorphisms of some field.

Corollary

Every finite group is a Galois group.

Proof

Let \(k\) be an arbitrary field, \(G\) any finite group. Now take \(K = k(\overline g:g\in G)\) (i.e. adjoin all elements of \(G\) to \(k\) as indeterminates, denoted by \(\overline g\)). Now we have a natural action of \(G\) on \(K\) defined via \(h\cdot \overline g= \overline {hg}\) and extending \(k\)-linearly. Now \(K\) and \(G\) satisfy the statement of Artin's theorem and hence \(K/K^G\) is a Galois extension with Galois group \(G\text{.}\)

It is interesting to note that we could have started with any field we liked and built a Galois extension with both fields extensions of the base we picked.
They won’t necessarily share a huge amount with it, however it is interesting to note that the characteristic will have to be the same and so we can do this for whatever our favourite characteristic is.

So around a month back I posted the first post in this 2 (or more who knows?) part series on Ribet’s converse to Herbrand’s theorem. This is the sequel, Cuspstruction, it is basically just my personal notes from my STAGE talk with the same name.
We were following Ribet’s paper and this is all about section 3.
The goal is to construct a cusp form with some very specific properties, which we can then take the corresponding Galois representation and use that to obtain the converse to Herbrand.
In this post though we’ll be focussing on constructing the cusp form, hence, Cuspstruction.

Cuspstruction

We will make use the following building blocks, some specific modular forms of weights 2 and type \(\epsilon\)
\begin{align*}
G_{2,\epsilon} &= L(-1,\epsilon)/2 + \sum_{n=1}^\infty \sum_{d|n} d \epsilon(d) q^n\\
s_{2,\epsilon} &= \sum_{n=1}^\infty \sum_{d|n} d \epsilon(n/d) q^n
\end{align*}
the latter is not a cusp form (not cuspidal at the other cusp of \(\Gamma_1(p)\)) we call such forms semi-cusp forms, denote the space of such by \(S^\infty\) (not standard notation) we will also use
\begin{equation*}
G_{1,\epsilon} = L(0,\epsilon) + \sum_{n=1}^\infty \sum_{d|n} \epsilon(d) q^n
\end{equation*}
the Eisenstein series are all hecke eigenfunctions for \(T_n\) \(n\) coprime to \(p\text{.}\)

Fix a prime ideal \(\mathfrak p|p\) of \(\mathbf{Q}(\mu_{p-1})\text{,}\) can think of \(\mu_p\subseteq \mathbf{Q}_p^*\) and take \(\omega\colon (\mathbf{Z}/p\mathbf{Z})^* \xrightarrow\sim \mu_{p-1}\) the unique character with \(\omega(d)\equiv d \pmod{\mathfrak p}\) for all \(d\in \mathbf{Z}\text{.}\)

We start with a key lemma, will use this repeatedly.

Lemma3.1

Let \(k\) be even \(2\le k \le p-3\text{,}\) then \(G_{2,\omega^{k-2}},G_{1,\omega^{k-1}}\) have \(\mathfrak p\)-integral \(q\)-expansions in \(\mathbf{Q}(\mu_{p-1})\) which are congruent mod \(\mathfrak p\) to
\begin{equation*}
-\frac{B_k}{2k} + \sum_{n=1}^\infty \sum_{d|n} d^{k-1} q^n
\end{equation*}
(this is the \(q\)-expansion of \(G_k\)).

By our choice of \(\omega\) we get the desired result for the non-constant terms of the \(q\)-expansion.

So it remains to prove that
\begin{equation*}
L(-1,\omega^{k-2})\equiv -\frac{B_k}{k}\pmod{\mathfrak p}
\end{equation*}
\begin{equation*}
L(0,\omega^{k-1})\equiv -\frac{B_k}{k}\pmod{\mathfrak p}
\end{equation*}
we make use of the following expressions (see probably Washington)
\begin{equation*}
L(0,\epsilon) = -\frac 1p \sum_{n=1}^p \epsilon(n) (n- \frac p2)
\end{equation*}
\begin{equation*}
L(-1,\epsilon) = -\frac{1}{2p} \sum_{n=1}^p \epsilon(n) (n^2 - pn + \frac{p^2}{6})
\end{equation*}
\(\omega(n)\equiv n^p \pmod{\mathfrak p^2}\) so
\begin{equation*}
pL(0,\omega^{k-1}) = -\sum_{n=1}^p \omega^{k-1}(n)(n-p/2)
\end{equation*}
\begin{equation*}
\equiv -\sum_{n=1}^p n^{p(k-1) + 1} \pmod{\mathfrak p^2}
\end{equation*}
and
\begin{equation*}
pL(-1,\omega^{k-2}) = -\frac 12\sum_{n=1}^p \omega^{k-2}(n)(n^2 - pn +\frac{p^2}{6})
\end{equation*}
\begin{equation*}
\equiv -\frac 12\sum_{n=1}^p n^{p(k-2) + 2} \pmod{\mathfrak p^2}
\end{equation*}
but for all \(t\gt 0\) even we have the congruence
\begin{equation*}
\sum_{n=1}^{p-1} n^t \equiv pB_t \pmod{p^2}
\end{equation*}
and so
\begin{equation*}
pL(0,\omega^{k-1}) \equiv -pB_{p(k-1)+1} \pmod{p^2}
\end{equation*}
cancelling
\begin{equation*}
L(0,\omega^{k-1}) \equiv -B_{p(k-1)+1} \pmod{p}
\end{equation*}
\begin{equation*}
\equiv -B_{p(k-1)+1} \pmod{p}
\end{equation*}
\begin{equation*}
\equiv -\frac{B_k}{k} \pmod{p}
\end{equation*}
as \(p(k-1) + 1 \equiv k \pmod{p-1}\) using Kummer's congruence. Similarly
\begin{equation*}
L(-1,\omega^{k-2})\equiv -\frac 12 \frac 2k B_k \pmod{p}\text{.}
\end{equation*}

Corollary3.2

Let \(k\) be as before and \(2\le n,m\le p-3\) both even with \(n+m \equiv k \pmod{p-1}\text{.}\) Then the product
\begin{equation*}
G_{1,\omega^{n-1}}G_{1,\omega^{m-1}} \in M_2(\Gamma_1(p), \omega^{k-2})
\end{equation*}
with \(q\)-expansion coefficients still \(\mathfrak p\)-integers in \(\mathbf{Q}(\mu_{p-1})\text{.}\)

Note: the constant term is a \(\mathfrak p\) unit unless \(B_n\) or \(B_m\) is divisible by \(p\text{.}\) We need to remove this condition.

Theorem3.3

Let \(k\) be as before, then there exists \(g\in M_2(\Gamma_1(p), \omega^{k-2})\) whose \(q\)-expansion coefficients are \(\mathfrak p\)-integers and whose constant coefficient is 1.

It suffices to find a form with constant coefficient a \(\mathfrak p\)-unit. If \(p\nmid B_k\) then we can use \(G_{2,\omega^{k-2}}\) by lemma 3.1.

If \(p|B_k\) try the possible products
\begin{equation*}
G_{1,\omega^{m-1}}G_{1,\omega^{n-1}}
\end{equation*}
with \(2\le m,n\le p-3\) even as above with \(m+n \equiv k \pmod{p-1}\text{.}\) We want to claim that at least one of these must work (i.e. we have a pair \(m,n\) with \(p\nmid B_m, p\nmid B_n\)). If this isn't the case if we let
\begin{equation*}
t = \#\{2\le n \text{ even} \le p-3: p|B_n\}\text{,}
\end{equation*}
we must have \(t \ge (p-1)/4\text{,}\) assume otherwise, we will derive a contradiction from this.

Greenberg showed that
\begin{equation*}
\frac{h_p}{h_{\mathbf{Q}(\mu_p)^+}} = h^*_p = 2^? p \prod_{\substack{k=2\\ \text{even}}}^{p-2} L(0,\omega^{k-1})
\end{equation*}
(this is obtained by taking a quotient of the analytic class number formulas for \(\mathbf{Q}(\mu_p),\mathbf{Q}(\mu_p)^+\)) but by lemma 3.1 we know that \(\mathfrak p^t\) will divide the product of \(L\)-values. And so \(p^t|h_p^*\text{,}\) we will get a contradiction if we show
\begin{equation*}
h_p^*\le p^{(p-1)/4}\text{.}
\end{equation*}

Work of Carlitz-Olson '55, Maillet's determinant shows that
\begin{equation*}
h_p^* = \pm\frac{D}{p^{(p-3)/2}}
\end{equation*}
where \(D\) is the determinant of a \((p-1)/2 \times (p-1)/2\) matrix with entries in \([1,p-1]\text{.}\) So recalling Hadamard's inequality
\begin{equation*}
|\det(v_1\cdots v_n)| \le \prod_{i=1}^n ||v_i||\text{,}
\end{equation*}
or the simpler corollary
\begin{equation*}
|A_{ij}| \le B \implies |\det(A)| \le n^{n/2}B^{n}
\end{equation*}
and applying with \(B = p, n=(p-1)/2\) gives
\begin{equation*}
|D| \le \left(\frac{p-1}{2}\right)^{(p-1)/4} p^{(p-1)/2} \lt 2^{-(p-1)/2} p^{(3p-3)/4}
\end{equation*}
so
\begin{equation*}
h_p^* \lt p^{(p+3)/4} 2^{-(p-1)/4}\text{.}
\end{equation*}
And we are done as \(h_p^* = 1\) for \(p\le 19\) and as \(p\le 2^{(p-1)/4}\) for \(p\gt 19\text{.}\)

Now we fix \(2\le k\le p-3\) even with \(p|B_k\) and let \(\epsilon = \omega^{k-2}\text{,}\) \(k\) must really be at least 4 (or even 10) so \(\omega\) is a non-trivial even character, we will work in weight 2, type \(\epsilon\) from now on.

Proposition3.4

There exists
\begin{equation*}
f= \sum_{n=1}^\infty a_nq^n\in S_2^\infty(\Gamma_1(p), \epsilon)
\end{equation*}
with \(\mathfrak p\)-integer \(a_p\in \mathbf{Q}(\mu_{p-1})\) with
\begin{equation*}
f\equiv G_k \equiv G_{2,\epsilon}\pmod{\mathfrak p}\text{.}
\end{equation*}

Let
\begin{equation*}
f= G_{2,\epsilon} -cg
\end{equation*}
with \(c = L(-1,\epsilon)/2\text{.}\) This is a semi-cusp form by construction. We get \(f\equiv G_{2,\epsilon}\pmod{\mathfrak p}\) because
\begin{equation*}
c \equiv -B_k/2k \equiv 0 \pmod{\mathfrak p}
\end{equation*}
by lemma 3.1 (again!) as we assume \(p|B_k\) now. Additionally by the same lemma \(G_k \equiv G_{2,\epsilon}\pmod{\mathfrak p}\text{.}\)

Let's take stock, we have a semi-cuspidal form which mod \(\mathfrak p\) looks like \(G_k\) and is hence an eigenform mod \(\mathfrak p\) but we want an actual eigenform, bro do you even lift?

Proposition3.5

There exists
\begin{equation*}
0\ne f'\in S_2(\Gamma_1(p), \epsilon)
\end{equation*}
which is an eigenform for all \(T_n\) with \((n,p) =1\text{.}\) With the eigenvalue \(\lambda(\ell)\) for \(T_\ell\) (\(\ell \ne p\)) satisfying
\begin{equation*}
\lambda(\ell) \equiv 1+\epsilon(\ell)\ell\equiv 1 +\ell^{k-1}\pmod{\mathfrak P}
\end{equation*}
for some prime \(\mathfrak P| \mathfrak p\) in \(\mathbf{Q}(\mu_{p-1},\lambda(n):(n,p)=1)\text{.}\)

We start with \(f\) from the proposition above it's a mod \(\mathfrak p\) eigenform and so we can use Deligne-Serre lifting lemma (6.11 in Formes modulaires de poids 1) to obtain a semi-cusp form \(f'\text{,}\) that is an eigenvalue for the Hecke operators stated.

To promote the semi-cusp form to a full blown cusp form we observe that the space \(S_2^\infty(\Gamma_1(p),\epsilon)\) is generated by the cusp forms and \(s_{2,\epsilon}\) which is also an eigenform we only have to check that \(f'\) isn't \(s_{2,\epsilon}\) (or it's scalar multiple). So we check the eigenvalues mod \(\mathfrak p\text{.}\)
\begin{equation*}
\epsilon(\ell) + \ell \equiv 1 + \ell\epsilon(\ell)\pmod{\mathfrak p}
\end{equation*}
implies \(\epsilon(\ell) = 1\text{,}\) but \(\epsilon\) is non-trivial!

The final challenge is to ensure that \(f'\) is also an eigenform for \(T_{p^i}\text{.}\)

Proposition3.6

\(f'\) is an eigenform for all Hecke operators, so we can normalise as
\begin{equation*}
f' = \sum_{n=1}^\infty \lambda(n)q^n\text{.}
\end{equation*}

Use the theory of newforms. There are no oldforms for \(\Gamma_1(p)\) as
\begin{equation*}
M_2(\operatorname{SL}_2(\mathbf{Z})) = 0\text{.}
\end{equation*}
A newform that is an eigenform for all hecke operators coprime to the level \(p\) is also an eigenform for the remaining Hecke operators.

So in conclusion:

Theorem3.7

Assume \(p|B_k\) then there exists
\begin{equation*}
f =\sum_{n=1}^\infty a_nq^n\in S_2(\Gamma_1(p),\epsilon)
\end{equation*}
which is an eigenform for all \(T_n\) and \(\mathfrak p|p\) an ideal of \(\mathbf{Q}(a_n)\) such that
\begin{equation*}
a_\ell \equiv 1+\ell^{k-1}\equiv 1+\epsilon(\ell)\ell \pmod{\mathfrak p}
\end{equation*}
for all \(\ell \ne p\text{.}\)

Remark

Word on the internet is that Mazur, Mazur-Wiles' proof of the Main conjecture of Iwasawa theory is modelled on this.

That's all for now, in the remainder of Ribet's paper he constructs a Galois representation from this and use it to prove the theorem.

Tomorrow I’m giving the STAGE talk on Ribet’s converse to Herbrand’s theorem, after I’ll try and post more notes, but for now here’s a little intro to get us thinking about the problem.

Ribet's converse to Herbrand

We are interested in the class groups of cyclotomic fields
\begin{equation*}
h_p = h_{\mathbf{Q}(\mu_p)}\text{.}
\end{equation*}
Lets list the first few of these

\(p\)

2

3

5

7

11

13

17

19

23

29

31

37

41

43

47

53

59

61

67

\(h_p\)

1

1

1

1

1

1

1

1

3

8

9

37

121

211

695

4889

41241

76301

853513

\(p|h_p\)

no

no

no

no

no

no

no

no

no

no

no

yes

no

no

no

no

yes

no

yes

DefinitionRegular primes

We'll call primes for which \(p\nmid h_p\) regular primes. Otherwise irregular primes.

Why is this important from a number theory perspective?

TheoremKummer 1850

Fermat's last theorem is true for regular prime exponents.

It's hard to tell when a prime is a regular prime, you'd have to compute the class group.

DefinitionBernoulli numbers

The Bernoulli numbers are the sequence of integers given by the exponential generating function
\begin{equation*}
\frac{x}{e^x - 1} + \frac x2 - 1 = \sum_{n\ge 2}^\infty B_k\frac{x^k}{k!}\text{.}
\end{equation*}

These have a number of cool properties, such as:

TheoremKummer's congruence

If \(h\equiv k \pmod {p-1}\) then
\begin{equation*}
\frac{B_k}{k}\equiv \frac{B_h}{h} \pmod{p}\text{.}
\end{equation*}

But most important for us is the relation to class numbers:

TheoremKummer's Criterion

\(p\) is a irregular prime if and only if there exists some \(2\le k \le p-3\text{,}\) even with \(p\) dividing the numerator of \(B_k\text{.}\)

This is a great theorem relating class numbers to the Bernoulli numbers, but can we do better? What if I know a specific \(k\) so that \(p|B_k\text{,}\) can I say anything more specific about the class group? Yes; there is a strengthening of this theorem due in this form to Herbrand (in one direction) and Ribet (later, in the other direction).

First we need to recall the mod \(p\) cyclotomic character \(\chi\colon \operatorname{Gal}(\overline{\mathbf{Q}}/\mathbf{Q}) \to \mathbf{F}_p^*\) defined by
\begin{equation*}
\zeta_p^{\chi(\sigma)} = \sigma (\zeta_p)\text{.}
\end{equation*}

TheoremHerbrand-Ribet

Write \(C = \operatorname{cl}(\mathbf{Q}(\mu_p))/\operatorname{cl}(\mathbf{Q}(\mu_p))^p\) this is an \(\mathbf{F}_p\) Galois representation which decomposes as a sum of eigenspaces
\begin{equation*}
C = \bigoplus_{i=0}^{p-1} C(\chi^i)\text{.}
\end{equation*}
Then for \(2\le k\le p-3\) even we have
\begin{equation*}
p| B_k \iff C(\chi^{1-k}) \ne 0\text{.}
\end{equation*}

The \(\Leftarrow\) direction was proved by Herbrand in 1932. And the \(\Rightarrow \) direction by Ribet in 1974.

Now for completeness here is a table of factorisations of Bernoulli number numerators.

Some vague thoughts about a weird category me and my housemate got to thinking about recently, unfortunately I’m a little too sleepy to write anything more coherent right now, but beeminder demands tribute.

When doing non-abelian group cohomology (and many other things) you end up dealing with the category of pointed sets, that is, sets with a point specified, and where morphisms must map specified points to each other.
This category is actually fairly nice (or at least nicer than plain $\mathrm{Set}$ is anyway) insomuch as it has a zero object (i.e. an object that is both initial and terminal), the one element set, this allows us to make sense of kernels etc. which is a nice thing to be able to talk about.
So why do we get a 0-object here? Well the one element set is already terminal in $\mathrm{Set}$ so we get terminalness for free, as for why it is initial it’s revealing to describe the category of pointed sets in a different way:
We can equivalently describe it as the coslice category $\{* \} \downarrow \mathrm{Set}$, where the objects are morphisms in $\mathrm{Set}$ from the one element set (giving you the specified point) and the new morphisms are commuting triangles in $\mathrm{Set}$.
When we do this we of course get the object we cosliced at (or more specifically its identity morphism) as an initial object, and as we already had a unique map from any object to this object we get that its identity map to itself is terminal also.

This got me thinking about $\mathrm{CRing}$, which infamously doesn’t have kernels, so what if we follow the recipe above or at least it’s dual and consider the slice category over the initial object $\mathbf{Z}$.
By the dual of the above this should now have a 0-object (the identity map on $\mathbf{Z}$) and so we could form kernels, indeed the kernel of a morphism $A \to B$ (where both $A$ and $B$ have an associated map to $\mathbf{Z}$) should be the preimage of $\mathbf{Z}$ I suppose, and this is probably a subring?

Before we get to that though be should ask what does this category $\mathrm{CRing}\downarrow \mathbf{Z}$ even look like?! It’s not absolutely trivial as far as I can tell as we get maps from polynomial rings into $\mathbf{Z}$ from evaluating at integer vectors.
We can also add in nilpotents and other stuff that just ends up mapping to zero but is still fine.
However on the other hand we immediately rule out positive characteristic rings, and anything which inverts an element of $\mathbf{Z}$.

I really have no conclusions about what this category of commutative rings with a map to $\mathbf{Z}$ actually is, but it is quite fun to play with.

As part of a discussion in our Galois representations course John Bergdall challenged us to come up with a representation that is irreducible but not absolutely semi-simple.
I found this a pretty fun thing to think about so I thought I’d write up my progress and what the next steps are.

First things first a reminder of the definitions: an irreducible representation is one with no fixed subspace, and a semisimple representation is one which can be written as a direct sum of irreducible representations.
Adding the word absolutely just means that we require the property to be true for the representation acting on the vector spaces over algebraic closure of the base field.
By allowing more general coefficients things that are irreducible to start with can easily become reducible.

To start our search lets work with the most simple type of potentially interesting representations I can think of, these look something like

which are entirely determined by the image of 1, lets try and find an appropriate matrix to make the example we want work.

In my head non-semi-simple things look something like this

So in order to find an irreducible non absolutely semisimple representation we want to find a matrix over a field which has no eigenvalues, but which over the algebraic closure has repeated eigenvalues.
This is not possible over $\mathbf{C}$ for example, as the trace would have to be twice the eigenvalue.
This would then give that the eigenvalues were themselves in the ground field, and therefore the eigenvalues were defined over $K$ in the first place.

This sort of weird multiple roots of irreducible polynomials stuff happens only for non-perfect fields (by definition), of which the most quoted example is $\mathbf{F}_p((t))$.
Here the polynomial $x^p - t$ is irreducible but has repeated roots over the algebraic closure as it factors as $(x-\sqrt[p]{t})^p$.
As we are dealing with 2-dimensional representations here we should look for a matrix over $\mathbf{F}_2((t))$ with characteristic polynomial $x^2 - t$, one simple example of this is the matrix

So we have a matrix that has no eigenvalues over the base field but repeated eigenvalues over the algebraic closure.
We now need to check that it only has a one dimensional eigenspace, to be sure that we have a non absolutely semisimple representation.
This eigenspace is given by the kernel of

which is indeed one dimensional and so we are done.

One property of this example is that if we consider the restriction of the representation to the subgroup $2\mathbf{Z}$ we get something semisimple as the square of our matrix $M$ is diagonal.
The new improved challenge is therefore to find a representation for which this doesn’t happen either; more explicitly to find an irreducible not absolutely semisimple representation which remains that way on all finite index subgroups.
I don’t think the example I have right now can be extended to this case and it might be necessary to look at representations of more exotic groups than simply $\mathbf{Z}$ for this.

Just a quick post to explain my recent absence (and get beeminder off my back).

Recently I haven’t posted that much, the reason being that I just moved to the US to start a PhD in maths at Boston University and things have been kinda busy with the move.
I’m really excited about starting doing maths full time again, and I’m sure there will be a lot more to post about in the very near future.
I just moved into an apartment today after a brief stay in a hotel while I attended a week long international orientation, not so much maths but I did get the chance to give a 10 minute talk on the 4-colour theorem.

So I’m currently on my way back from the 7th European Congress of Mathematics, which took place in Berlin over the course of the last week.
While everything is still fresh(ish) in my memory I wanted to record some of the bits that I (personally) found most interesting so that I’ll be able to look back when I invariably do forget what I did for a whole week.
If you like similar things to me then maybe you’ll find something interesting here too (if you don’t like similar things to me then I question your choice to read this blog).
Some of these topics I’d like to revisit in more detail (possibly in a future post?!) but for now these short snippets will have to do.
So in approximately no particular order and undoubtedly with some gaps.

Peter Scholze

This is the most obvious one for me, Scholze talked about Perfectoid spaces, which since he introduced them in his thesis have become a hot topic in number theory.
Scholze himself has been awarded various prizes and honours for developing this theory (including another at the ECM).

The motivation is to transfer problems involving the $p$-adics (which can be problematic as they are mixed characteristic) over to the similar looking ring $\mathbf{F}_p((t))$.
Scholze said he wants to fight for the freedom of $p$.
The way he does this is via a technique called tilting, which was originally used by Fontaine and Wintenberger to prove a result relating the Galois theories of these fields.
Scholze (and independently Kedlaya-Liu) takes this result and distils out the definition of a field where this works and calls it a perfectoid field.
He then does some natural looking things (which I’m sure are actually very complicated to do properly) and constructs perfectoid algebras and perfectoid spaces.
He shows that various things work out nicely and that one can tilt the spaces too in a way that matches what you’d hope for simple spaces like projective space.

In the last part of the talk Scholze got increasingly high level so rather than embarrass myself trying to replicate it I’ll just say that it definitely looks like he has more impressive things forthcoming.

One thing that was especially interesting for me in this talk was mention of work by Jared Weinstein who works at Boston University (where I will also be from September).
I was aware that Weinstein was doing work in this area so it was really interesting to have a bit of it put into context.

Karin Vogtmann

Vogtmann first gave a nice introduction to outer automorphisms of the free group and why one might care, and got me interested with the observation that abelianising gives automorphisms of $\mathbf{ZZ}^d$.
She also described various related “outer space”s which are spaces which the outer automorphisms of free groups act on.
The construction is really cool and involves deforming certain metrized graphs using the interpretation of $\operatorname{Out}(F_n)$ as the automorphisms of the rose graph.
(One thing she mentioned in passing here that I’d like to learn more about is Gromov’s topology on the space of all metric spaces, so meta.)
The gist was that the various slightly different outer spaces were all very related to the main one.
The pictures were very good and extremely helpful and for me this talk definitely wins the Oscar for best visual effects.

Don Zagier

I found Don’s style of presenting (rapidly moving through hand written slides while barely breathing) engaging and super impressive, multiple times he switched to a slide and corrected something before I’d even had time to read the first sentence.
The content itself was also really fascinating, he started with a fairly random looking recurrently defined sequence that Apery used to prove that $\zeta(3)$ is irrational and gave multiple interpretations of where such a sequence comes from (though in fact there is still something very special going on).
He ended up giving some motivation for the concept motives (pun not intended) and presented us with a few results concerning various special values that were conjectured by a coauthor of his based on pure thought on the level of motives and proved by Zagier and others on the concrete level.

Unfortunately I can’t remember quite as much of what went on as I’d like due to the above-mentioned speed and Don’s recommendation at the start that we “don’t even bother trying to take notes”.
I do hope a video of the lecture will appear at some point as out of all the talks I saw I think this is the one I’d most benefit from rewatching in its entirety.

Another fun coincidence for me personally was that he mentioned some work of Irene Bouw, who I met only days beforehand in Ulm.

Tommy Hofman

Tommy talked about an algorithm he and Claus Fieker developed to compute Hermite normal forms over Dedekind domains.
This was another exceptional example of the law of large numbers (or whichever law it is that explains funny coincidences) as previously I worked on implementing algorithms for computing HNFs and doing this work was in some way the original cause of me coming to Kaiserslautern, which is where Tommy works!
The main takeaway from this talk (other than a nice new algorithm) is that quotients of rings of integers by powers of prime ideals are always Euclidean rings.

Other nice talks

Fedor Pakovich found a relationship between Davenport-Zannier pairs (generalisations of pairs of polynomials $(f,g)$ such that $f^3-g^2$ is of minimal possible degree, excluding trivial things) and Dessins d’Enfants.
He used this to obtain a classification of these pairs (roughly 10 famillies and 11 sporadic pairs) by couting the corresponding graphs (which ended up being interpretable as certain weighted trees or something like this).
One of the reasons I found this so interesting was that it’s the first time I think I’ve seen Dessins used in anger, I started reading Girondo and Gonzailez-Diaz’s book on compact Riemann surfaces and Dessins about year ago but got distracted before I got to the juicy bit.

Thomas Willwacher talked about cohomology of graph complexes which it turns out is really hard to compute but relates many other areas of maths (including automorphisms of free groups!).

Maryna Viazovska gave a nice talk about her (and others) recent proof of sphere packing in dimensions 8 and 24, which got a lot of popular attention at the time so was nice to hear about.

Meinolf Geck spoke about his work computing with groups, remarked that finite groups of Lie type $G(\mathbf{F}_q)$ make up a large portion of the classification of finite simple groups. He wants to construct generic character tables as $q$ varies for different algebraic groups and possibly different primes - and this actually looks to work! That this is even possible totally blew my mind.

Uli Wagner spoke about the topological Tverberg conjecture, Gunter Ziegler gave a talk that touched on this in Warwick in 2014 which I really enjoyed, so it was cool to hear about recent progress regarding it.

Jeremy Grey spoke about Poincare and Weyl and it was a lot more philosophical than I was expecting, contrasting their views on philosophy of science over time, but it was fascinating stuff.

There were certainly many many other interesting talks that I saw (and indeed many that I didn’t) but I’m tired and my memory is failing now and I have to make a post or Beeminder will take my money.

Other thoughts

There were a few political statements made during the conference for example by Pavel Exner condemning the situation in Turkey (where academics were recalled/banned from travelling) and from Timothy Gowers regarding the Brexit (short version: 48% of us voted to stay).
This is something that I think is appropriate for large I/ECM like conferences, which represent the whole mathematical community.

Non-maths

Berlin is a really awesome city so even outside of the conference I did a lot of nice things, and I’m really glad I got a good excuse to visit before I finish my time in Germany.
I took quite a lot of photographs (especially the Sunday before the conference when it was really nicely overcast), maybe I’ll upload some to Flickr and put a link here when I’m back in the UK and have some free time.