The Tip: I received a hot tip to visit Sawada Coffee, proclaimed (by a facebook friend of my source) as “the best coffee in Chicago.” Here is a review.

First impressions: Somewhat of a skater-hipster vibe. It turned out that Sawada shares its space with Green Street Meats:

The barista didn’t know the difference between a wet and a dry cappuccino, which was not reassuring. Nor was the fact that the smallest sized serving was 8oz rather than 5 or 6 (or 5.5). Still, I ordered away, and was slightly reassured when a different barista actually made my drink, and produced the following very acceptable looking cup:

So at least the milk foaming and pouring skills seemed up to par. (The claim to fame of Sawada seems to be the green tea latte, but a cafe should stand or fall on its espresso and cappuccino.)

The taste: Hmmm. Terrible espresso shot. Probably the most accurate description was that it tasted like the coffee one gets in Germany. This is not a point in its favor.

General Ambience: Terrible music, blaring far too loudly over the base-heavy speakers. (The sound was something like Santana meets rockabilly, but even that description makes it sound better than it was.) Even the St Matthew Passion (full volume with earbuds) wasn’t enough to mask the sound.

Verdict: Erbarm dich, mein Gott!

Posted in cappuccino, Coffee, Food | | 2 Comments

## fivethirtyeight.com doesn’t understand mathematics

I happened to find myself browsing fivethirtyeight.com a few weeks ago, where I came across a column known as the “riddler.” The particular problem of the week (see here) was to answer the following:

Problem: It’s Friday and that means it’s party time! A group of N people are in attendance at your shindig, some of whom are friends with each other. (Let’s assume friendship is symmetric — if person A is friends with person B, then B is friends with A.) Suppose that everyone has at least one friend at the party, and that a person is “proud” if her number of friends is strictly larger than the average number of friends that her own friends have. (A competitive lot, your guests.)

Importantly, more than one person can be proud. How large can the share of proud people at the party be?

If you take a completed graph and omit one edge, then $N - 2$ people know everyone and have $N$ friends, and the remaining two people have $N - 1$ friends. In this case, there are $N - 2$ proud people. On the other hand, at least one person has a smaller or equal number of friends than everybody else, and so they can’t have more friends than any of their other friends let alone their average number. So at least one person is not proud. Hence the real content of the question is to determine whether there can be $N - 1$ proud people. The answer to this question (which is no) is harder than coming up with the example but neither terribly difficult nor particularly interesting. The absolute shocker, however, is that the riddler’s “solution” (scroll to the bottom of the page) to the puzzle is merely to exhibit the example above with $N - 2$ proud people. There’s not any hint that an argument is required to show that $N - 1$ is not possible. I nearly choked on my cappuccino when reading this. (You could try to argue that the formulation of the problem allows for some wiggle room: one is asked to find the highest proportion of people with the indicated property, and so imaging that N is not fixed, one might claim it merely suffices to show that the limit is 1 as N goes to infinity. But I don’t buy this.) Click and Clack would never have made this mistake.

Maybe the author of the riddler was aware of this issue or maybe they weren’t. But the whole point about online media is that it doesn’t require dumbing down the message to reach the right audience. I may well agree that in this (or other) particular cases, the technical details of a correct solution may be a little annoying, but in that case it is OK as long as:

1. One provides a link with the full argument, and crucially:
2. One makes it very clear in the main text that there is something left to do to fully answer the question.

Not mentioning that there is any issue at all represents a failure in what I hoped the website fivethirtyeight.com was supposed to represent. Instead of hiring technical people who can write and training them in journalism, they appear to have simply hired journalists with a fairly mixed level of technical expertise. That’s a missed opportunity.

(I did look up the person responsible for the riddler website, and they appear not to have any scientific training. Rather, they were trained as an economist. At my institution, no less (hmmm)).

## Prime divisors of polynomials

A heuristic model from the last post suggests that the “expected” order of the Galois group associated to a weight one modular form of projective type $A_5$ is infinite. And when one tries to solve the inverse Galois problem for central extensions of this group, one is lead to problems concerning the prime divisors of polynomials and their properties modulo 2. But I don’t know how to answer this type of problems! Here is an analogous question that seems a little tricky to me:

Problem: Show that there are infinitely many integers $n$ such that all the odd prime divisors of $n^2 + 1$ are of the form 5 modulo 8.

To make the problem slightly easier, one can replace (5 modulo 8) by not (1 modulo 2^m) for any fixed m.

Is this an open problem?

Posted in Mathematics | | 4 Comments

## Central Extensions and Weight One Forms

As mentioned in the comments to the last post, Kevin Buzzard and Alan Lauder have made an extensive computation of weight one modular forms in characteristic zero (see also here). Thinking about what that data might contain, I wondered about the following question: what are the images of the Galois representations associated to the weight one forms of type $A_5$?

Let us take a step back. Consider a projective representation

$\psi: G_{{\mathbf{Q}}} \rightarrow {\mathrm{PGL}}_2({\mathbf{C}})$

with image $A_5$, and assume that it is odd. (That is, complex conjugation has order $2.$) According to Tate, there exists a lift

$\rho: G_{{\mathbf{Q}}} \rightarrow {\mathrm{GL}}_2({\mathbf{C}}).$

This lift is unique up to twisting. Since the Schur multiplier $H_2(A_5,{\mathbf{Z}})$ of $A_5$ is ${\mathbf{Z}}/2{\mathbf{Z}}$, there is a unique minimal lift up to twist whose image is a central extension ${\widetilde{A}_5}$ by a cyclic group $\Delta$ of $2$-power order. Note that $\Delta$ is not trivial, since $A_5$ does not have any two-dimensional representations. If $|\Delta| = 2$, then the determinant of the corresponding 2-dimensional representation of ${\widetilde{A}_5}$ is trivial, which contradicts the assumption that $\psi$ is odd. (Equivalently, there is an obstruction at $\infty$ to lifting to the central extension by ${\mathbf{Z}}/2{\mathbf{Z}}.$) Hence $4$ divides $|\Delta|.$ What is the expected distribution of $\Delta$ as one runs over all odd $A_5$-extensions?

My first guess (without any prior thought or computation) was that this might obey some form of Cohen–Lenstra heuristic, suitably interpreted.

Note that the image of the determinant has order $|\Delta|/2.$ The corresponding determinant representation is a character of ${\mathbf{Q}}$ of $2$-power order. Since ${\mathbf{Q}}$ has trivial class number, the order $|\Delta|/2$ is equal to the maximal ramification degree $e_p$ of this representation over all primes $p.$

Over ${\mathbf{Q}}$, Tate’s lifting theorem has the following stronger form: one may choose a lift $\rho_p$ of $\psi_p := \psi|_{D_p}$ and insist that $\rho|_{I_p} = \rho_p | I_p$; that is, they agree on inertia. This is essentially a consequence of the fact that ${\mathbf{Q}}$ has trivial class group. For convenience, suppose that $\psi$ is unramified at $2$ and $3.$ Suppose that $\psi$ is ramified at $p.$ There are three possibilities:

1. The image of $\psi_p$ at a ramified prime $p$ is cyclic of order 2, 3, or 5.
2. The image of $\psi_p$ at a ramified prime $p$ is $S_3.$
3. The image of $\psi_p$ at a ramified prime $p$ is ${\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2{\mathbf{Z}}.$

For a fixed $p$, let $\epsilon$ denote the Teichmuller lift of the mod-$p$ cyclotomic character. (Fix an isomorphism of $\mathbf{C}$ with $\overline{\mathbf{Q}}_p$ for all $p.$)

Let us consider the three cases in turn.

In the first case, the image factors through a cyclic quotient. One may thus take $\rho_p$ to be a direct sum which, on inertia, has the shape $\chi \oplus 1$ up to twist. By comparing this to the projective representation, we see that $\chi$ has order 2, 3, or 5, and so, after finding the twist such that the determinant has $2$-power order, we see that $e_p = 1$ or $e_p = 2.$

In the second case, the lift on inertia is (up to twist) of the form:

$\omega^{r}_2 \oplus \omega^{pr}_2$

for some $r.$ Since the order of $\omega_2$ is $p^2 - 1$, the order of the ratio is

$\displaystyle{\frac{p+1}{\gcd(r,p+1)}}.$

which must be equal to 3. It follows that $r$ is even. Yet the determinant is equal to

$\displaystyle{ \omega^{(p+1)r}_2 = \epsilon^{r}},$

Since $r$ is even, we see that, after twisting, we may take $e_p = 1.$

Finally, in the third case, the lift is of the form:

$\omega^{r}_2 \oplus \omega^{pr}_2$

for some $r.$ We now find that the order of the ratio of these characters is

$\displaystyle{\frac{p+1}{\gcd(r,p+1)}}.$

which must be equal to 2, and the determinant is $\epsilon^r.$ If $r$ is even, then, as above, we may twist so that $e_p = 1.$ Hence, the only way that the image after minimal twist does not have $|\Delta| = 4$ is if we are in this third situation with $p \equiv 1 \mod 4$, with $r$ odd, and then (after twisting) we find that $e_p$ is the largest power of $2$ dividing $p - 1.$

(I confess that I originally forgot the fact that the third possibility could occur, and was only after noticing that this seemed to imply the inverse Galois problem was false thought a little bit more about the possibilities.)

To summarize:

Lemma Assume that $\psi$ is unramified at $2$ and $3$ and has projective image $A_5$, and a lift with image $\widetilde{A}_5$ with minimal kernel. Then order of $\Delta$ is $4$ unless there exists a prime $p \equiv 1 \mod 4$ such that the image of the decomposition group at $p$ under $\psi$ is ${\mathbf{Z}}/2{\mathbf{Z}} \oplus {\mathbf{Z}}/2 {\mathbf{Z}}.$ In this case, we have $\Delta$ to be twice the largest power of $2$ dividing $p -1$ for all such primes $p.$

Let $\Delta(\psi)$ denote the corresponding power of $2.$

We see that $\Delta(\psi)$ is determined by purely local phenomena. This still doesn’t quite answer what the distribution of the extension $\widetilde{A}_5$ will be. However, I imagine that Bhargava style heuristics should certainly be able to predict the ratio of $A_5$ with $\Delta(\psi) = 2^n.$ Does anyone have a sense of how easy this might be to prove? Or, much more modestly, how easy it would be to compute from these quite precise heuristics the exact predicted distribution of the central extensions of $A_5$ coming from weight one modular forms?

(I confess, it is not even obvious to me from this construction how to prove that all central extensions $\widetilde{A}_5$ occur as Galois groups — but I presume this is known, and hopefully one of my readers can provide a reference.)

(According to KB, BTW, all the $A_5$ representations with $N \le 1500$ have $\Delta = 4.$)

## LMFDB!?

The LMFDB has gone live!

I previously expressed on this blog a somewhat muted opinion about certain aspects of the website’s functionality, and it seems that my complaints have mainly been addressed in the latest version. On the other hand, this live version has been released with a certain amount of fanfare, not to mention press releases. (Is AIM involved? yes it is.) First of all, there’s something about press releases in science (or mathematics) which I find deeply troubling. What is the point of such a press release? To drum up future funding? To generate mainstream press articles and thus communicate a sense of wonder and amazement to the public, who rarely get to glimpse the excitement of modern mathematics in action? Hopefully it’s the later which is true. But if so, does it really require that we stretch the truth about what we are doing in order to generate such excitement? (To see that the answer to the latter question is no, one need only look back on Scientific American articles on mathematics and physics from the ‘60s and ‘70s.) To be fair, I should also link to a more modest description of the project here. But then again, I dare you to click on the following website.

But back to the topic at hand. I have asked the opinion of at least three mathematicians about either the LMFDB or on computational aspects of the Langlands Programme more generally. For reasons of anonymity, I will not mention their names here. The first comment addresses a widely circulated quote from John Voight in the press release: “Our project is akin to the first periodic table of elements.” One source offered the following take on this (literally copied from my email and modified only by adjusting the spelling of the Langlands Programme to the preferred Canadian spelling):

The periodic table was a fantastic synthesis of decades, maybe even centuries, of empirical investigation, that led to profound new theoretical insights into chemistry and the physical world. The LMFDB is a rag-tag assortment of empirical facts in the Langlands Programme which lag far behind the theoretical advances of the past decade.

A second source, speaking more broadly on the question of computations in the Langlands programme, wondered if there had been any fundamental discoveries made via numerical computations since the BSD conjecture. While these are certainly pretty strong statements, I feel comfortable agreeing at least that, in our field, the theory is way in advance of the computations. We can prove potential modularity theorems for self-dual representations of any dimension, and yet it’s barely possible to compute even weight zero forms for U(3) (David Loeffler did some computations once). In part, I think this actually provides justification for effort into understanding how to actually compute these objects. After all, a really nice aspect of number theory is having a collection of beautiful concrete examples, from X_0(11) to Q(sqrt(-23)) (take that, Geometric Langlands!). Yet these statements also provide an alternate framework with which to view the LMFDB, which is less glorious than the press releases suggest. To continue with the periodic table analogy, the LMDFB is less a construction of a periodic table, but more a collection of sample elements from the periodic table neatly contained in small glass vials. Let me make the following point clear: some of the samples took a great deal of effort and ingenuity to extract. But it’s not entirely obvious to me how easy it will be to actually use the data to do fundamental new mathematics. This was the main point of my third commentator: the problem is that, if one actually wants to undertake a fairly serious computation, the complete set of data one has to compute will either not be available on the LMFDB (however big the LMDFB grows), or not available in a format which is at all practical to extract for actually doing computations. So, in the end, if you need some serious data, you are probably going to have to compute it again yourself. In some cases you may be able to do this, and in other cases not.

This leads to probably the most frustrating thing about the website from my perspective. I thought quite a bit about what the most useful format for some of the data might be (for me). Here is a typical thing that I might want to do: find a Hecke eigenform of some particular weight and level, and then compute information about the mod-p representations for various primes p (as well as congruences between forms). This is a little tricky to do at the moment, in part because the data for the coefficients is given in terms of a primitive element in the Hecke field (often the eigenvalue of T_2), and then the order generated by this element has (almost always) huge index (divisible by many small primes) inside the ring of integers, which makes computing the reductions slightly painful. There are certainly ways to address this specific problem and incorporate such functionality into the website. But it ultimately would be silly to customize the LMFDB for my particular needs. Instead, in the end, what I think would actually be most useful would be if the webpages on modular forms included enough pari/gp/sage/magma code to allow me to go away and compute the q-expansion myself. This is why I think that, even within computational number theory, the impact of programs like pari/sage/mamga will be far greater than the LMFDB.

If I think of the three most serious computations I have been involved with recently, they include partial weight one Hilbert modular forms, non-liftable classes of low weight Siegel modular forms, and Artin L-functions of S_5 extensions. The first required customized programs in magma (some written in part by John Voight), and the tables of higher weight HMFs in the LMFDB would not have been of any use. The computations of low weight Siegel modular forms in finite characteristic, which are absolutely terrifying, require completely custom computations (which, by the way, are completely beyond my capability of doing and involve getting Cris Poor and David Yuen to do them). Finally, the computation which would be the closest to an off the shelf computation was proving that the Artin L function $L(\rho_5,s)$ of some $S_5$ extension provably had no zero in $[0,1].$ However, the LMFDB tables only go up to degree four representations. Fortunately I knew that Andy Booker was an expert in this sort of thing and he did it for me. (Then again, even if the data in this case was included, it’s totally unclear to me the extent to which the data in the tables has been “proved.”) And my point here is not to complain about the lack of certain computations being in the LMFDB, just to caution against the idea that its existence will be particularly useful for future computations.

The bigger point, however, is surely this: Is hype in science or mathematics a necessary evil to generate public enthusiasm, or is it an ultimately corrosive influence?

Posted in Mathematics, Rant | | 31 Comments

## Report From Berkeley

My recent trip to Berkeley did not result in a chance to test whether the Cheeseboard pizza maintained its ranking, but did give me the opportunity to attend the latest Bay Area Number Theory and Algebraic Geometry day, on a (somewhat disappointingly) rainy Saturday in Evans Hall. The weather was somewhat better on Sunday, however, allowing myself to make the trip to Mint Plaza for the following cup, which should bear some resemblance to the banner picture on this site. (Unfortunately, they were no longer serving their mini-Brioche buns.)

But now on to the good stuff, a report on some of the talks:

Jaclyn Lang gave a talk on her work concerning the image of big Galois representions. The setup is roughly as follows. Let

$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{F}_p)$

be an absolutely irreducible odd Galois representation over a finite field (hence modular). Suppose this Galois representation was the residual representation associated to an ordinary modular form that lived inside a Hida family that was smooth over weight space. Then one might expect that the corresponding representation

$\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_2(\mathbf{Z}_p[[T]])$

to have image which was as big as possible, namely, containing $\mathrm{SL}_2(\mathbf{Z}_p[[T]]).$ This can’t always be the case, however; for example, the residual representation (or the entire family) could be dihedral. However, if the residual representation contains $\mathrm{SL}_2(\mathbf{F}_p),$ and one additionally assumes that the image of inertia at p is sufficiently large, then this is indeed the case (probably this assumes that the residue characteristic is at least 5). There have been a number of generalizations of this result due to Hida and others which improves the result by weaken the various hypotheses; for example, allowing coefficients, allowing the residual representation to be dihedral, and weakening the ramification hypothesis at p. For these results, one can’t expect that the image is full, but rather that the image contains an appropriate congruence subgroup of $\mathrm{SL}_2.$ I like to think of this as follows: at classical specializations, one knows that the image (if it is not CM or weight one) will have open image; the results of this talk and of previous work show that index can be controlled in families. Actually, this is not quite true, because another obstruction to having open image even classically is the existence of inner twists. The main result of the talk was to deal with this issue of inner twists, and hence also allow for a generalization of the results not only to smooth Hida families but to any irreducible component of any Hida family. (More details to be found here.)

A natural question: one output of Lang’s result is to give an ideal $\mathfrak{b}$ of the Hida family for which the image of these Galois representations contains the $\mathfrak{b}$-congruence subgroup (after accounting for inner twists). In characteristic zero, my impression from the talk was that one can identify the support of this ideal as coming from CM points and classical weight one modular forms. On the other hand, apparently there is also a version of this result in the reducible case (due to Hida and with extra hypotheses); in that case the zeros should correspond to the reducible locus, or equivalently, the zeroes of the p-adic zeta function. However, a stronger result is true, namely, that $\mathfrak{b}$ can essentially be identified with this p-adic zeta function. So, returning back to the residually irreducible case, the natural question is: can the support of $\mathfrak{b}$ contain the prime p?

Kestutis Cesnavicius gave a talk on the Manin-Stevens and Manin constants for elliptic curves, with emphasis on the prime p=2. He raised the following question: Suppose that $N$ is odd. Is there a surjection from the space of weight 2 classical modular cusp forms of level $\Gamma_0(N)$ with coefficients in $\mathbf{Z}_2$ to the space of weight 2 Katz cusp forms of the same level over $\mathbf{F}_2?$ The issue here is that the latter space is really the cohomology of the associated stack, not the course moduli space. Unfortunately, this question distracted me a little as I tried to find a counter-example (I failed). A result of Serre and Carayol basically implies that the result can only fail after localizing at a non-Eisenstein maximal ideal $\mathfrak{m}$ of the Hecke algebra $\mathbf{T}$ if the corresponding representation $\overline{\rho}_{\mathfrak{m}}$ is induced from $\mathbf{Q}(\sqrt{-1}).$ (Analogously, for $p =3,$ when the representation is induced from $\mathbf{Q}(\sqrt{-3}).$) This is related to the classic failure of the first version of Serre’s conjecture for $p =3$ at level $\Gamma_1(13).$ However, as Serre quickly realized, this failure ultimately comes from a failure to lift mod-p forms as above, except in this case from the intermediate curve $X_H(13),$ not from $X_0(13).$ I ultimately convinced myself that lifting was always possible unless $\mathfrak{m}$ was not only Eisenstein but also the ideal containing $T_{p}$ for all odd p not dividing N. I think this must be related to Ken’s result on component groups of Neron models, and how the non-Eisenstein parts arise for $X_H(N)$ but not for $X_0(N)$ or $X_1(N).$ (More details here.)

The final speaker of the day was Daquin Wan. The key question that arose in his talk was the following. Suppose that $D(k,T)$ is the characteristic power series of the $U$ operator on the space of overconvergent p-adic modular forms in integral weight k. Can one show that

$L(k,T) = \displaystyle{\frac{D(k+2,T)}{D(k,pT)}}$

has infinitely many zeros and infinitely many poles? One actually has to assume that $k \ne 0$ here, since otherwise the result is false, as this will be a polynomial of dimension the space of weight two forms. One feels that p-adic Langlands should be able to say enough about slopes in these weights to obtain a contradiction, but I don’t unfortunately see how to do it. The main point of the talk was two-fold. There is an argument of Coleman that shows that $D(k,T)$ is not itself a polynomial. This argument can be generalized to prove that $L(k,T)$ is not a rational function. Second, the product $L(k,T) L(-k,p^k T)$ is actually a rational function because of the properties of the theta operator. So one deduces that at least one of these functions had infinitely many poles and the other had infinitely many zeroes. This also relies on a previous result of Wan that these functions are meromorphic. (Oh, I should mention that this was joint work with … and here I didn’t take notes for a talk two weeks ago … Liang Xiao? Please correct me if I’m wrong)

(Romyar Sharifi also talked, but since I am actively trying to understand something about that talk on a more technical level, so I will have to return to a discussion of it later.)

## Tensor Products

Let $W$ be an irreducible representation of a finite group $G.$ Say that $W$ is tensor indecomposable if any isomorphism $W = U \otimes V$ implies that either $U$ or $V$ is a character.
In conversations with Matt and Toby (which permeate the rest of this post as well), the following problem came up:

Question: Let $G$ be a finite group. Let $V$ be an irreducible representation of $G.$ Is there a unique decomposition

$V = V_1 \otimes V_2 \ldots \otimes V_k$

of $V$ as a tensor product of tensor indecomposable representations (up to re-ordering and twist)? I don’t think this can be too hard, but I confess I don’t see how to do it. (Since we didn’t really need this, we didn’t think about it too hard.)
(edit: when I say I don’t think this can be too hard, I don’t mean to imply that I think it is true; just that I think either a proof or counterexample should not be too hard to find — hopefully not both.)

One can ask an analogous problem for Lie groups. Actually, the problem for Lie algebras is actually quite simple (and the answer is positive). It is related to the following:

Lemma:
Let $V$ and $W$ be irreducible non-trivial representations of a simple Lie group $\mathfrak{g}.$ Then $V \otimes W$ is reducible.

Proof: Assume that $V \otimes W$ is irreducible. In particular, it is determined by its highest weight. Let the highest weights of $V$ and $W$ be $\lambda$ and $\mu$ respectively. Then the highest weight of $V \otimes W$ must be $\lambda + \mu.$ But now, by the Weyl character formula, we deduce that

$\displaystyle{1 = \frac{\dim(V) \dim(W)}{\dim(V \otimes W)} = \prod_{\Phi^{+}} \frac{ \langle \rho,\alpha \rangle \langle \rho + \lambda + \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.$

The product term can also be written as:

$\displaystyle{1 + \frac{ \langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle} { \langle \rho + \lambda,\alpha \rangle \langle \rho + \mu, \alpha \rangle}}.$

In particular, since the pairing is non-negative between positive roots and highest weights, we deduce a contradiction unless

$\langle \lambda,\alpha \rangle \langle \mu, \alpha \rangle = 0$

for all $\alpha \in \Phi^{+}.$ The assumption that $\mathfrak{g}$ is irreducible, however, is equivalent to saying that $\Phi^{+}$ has a maximal root $\beta,$ and for such a maximal root, we have

$\langle \lambda, \beta \rangle = 0 \Rightarrow \langle \lambda, \alpha \rangle = 0, \ \forall \alpha \in \Phi^{+} \Leftrightarrow \lambda = 0.$

Note that this lemma is actually a special case of a theorem of Rajan, who proved that, for simple $\mathfrak{g},$ the factors of a (not necessarily irreducible) tensor product are determined by the representation. In particular, the tensor product of two non-trivial irreducible representations cannot be irreducible.

The problem with the initial question is that it’s hard to construct tensor products of irreducible representations which are irreducible. Or rather, it is easy, simply by taking $U \otimes V$ where $U$ is an irreducible representation of $H$ and $V$ is an irreducible representation of $G$ and the tensor is irreducible for $H \times G.$ Yet the interesting case is something closer to assuming that $U$ and $V$ are faithful. Actually, this motivates the following question:

Question: Do there exist irreducible non-trivial representations $U$ and $V$ of a finite simple non-abelian group $G$ such that $U \otimes V$ is irreducible?

The argument above for Lie groups suggests that this may not happen for Chevelley groups (although it certainly doesn’t prove this). It also suggests (relating the representation theory of $A_n$ and $S_n$ to $\mathrm{GL}_n$) that it doesn’t happen for the alternating groups either. It almost surely doesn’t happen for the sporadic groups either. So my guess that the answer to the problem above is no, and that this is probably known, and probably requires classification. (Please comment if you know the answer.) Actually, this also reminds me of a similar problem which I think is open.

Question: Fix N. Does there exist a non-trivial representation $V$ of a finite group $G$ of dimension N such that $\mathrm{Hom}^0(V,V)$ (of dimension $N^2 - 1$) is irreducible?

This question came up in my paper with Barry, where I was surprised to find very few examples. I seem to remember that the Mathieu group $M_{12}$ has an $11$-dimensional representation whose corresponding 120 dimensional adjoint is irreducible. Can one classify all such examples coming from simple groups?