Very very happy with this project we ran at the M2 workshop this summer in Madison -- it is now possible to do compute Ext, Tor, etc. of C_p-Mackey functors by computer! The image below shows how you can use the package to compute a free resolution of a C_p-Mackey functor.

... can this be simplified at all? n and m are fixed positive integers, p is a fixed real number between 0 and 1.

An open problem in complex analysis: Let f ∈ ℂ[x] be a polynomial of degree ≥2. Let z ∈ ℂ. Must there exist a critical point w of f such that |(f(z)-f(w))/(z-w)| ≤ |f'(z)|?

An open question in graph theory: Does there exist a finite (simple, undirected) graph which has diameter 2, girth* 5, and is 57-regular? * The girth of a graph G is the smallest length of a cycle in G.

An open problem in number theory: Recall that the totient function ϕ is defined by sending each positive integer n to the number of positive integers k ≤ n which are coprime to n. Conjecture. For all positive integers x, there exists a positive integer y≠x such that ϕ(x)=ϕ(y)

An open problem in order theory: Let L be a finite lattice (i.e. a nonempty finite poset such that any two elements have both an inf and a sup). Must there exist a finite group G with subgroups H, K such that L is isomorphic to the poset {X ≤ G : H ≤ X ≤ K} ordered by ⊆?

Another elementary open problem: Let f : ℕ → ℕ be a bijection. Must there exist a natural number n and a positive integer k such that either f(n) > f(n+k) > f(n+2k) > f(n+3k) or f(n) < f(n+k) < f(n+2k) < f(n+3k)?

An open problem in algebraic geometry: Let n be a positive integer. Let f : ℂⁿ → ℂⁿ be a regular* function such that the determinant of its Jacobian matrix is a nonzero constant. Must f be bijective with regular inverse? *i.e. each component ℂⁿ → ℂ is a polynomial

A crazy open problem in algebra: Conjecture (Poonen): 100% (asymptotic density) of finite commutative rings have characteristic EXACTLY 8, i.e. 4 ≠ 0 and 8 = 0 both hold. From this amazing paper: arxiv.org/abs/math/060...

Another fun open problem: Conjecture (Rota). Let n be a natural number. Let V be an n-dimensional vector space. Let B₁, …, Bₙ be bases of V. Then there exist orderings of these bases Bₖ = (bₖ₁, …, bₖₙ) such that {b₁ₖ, …, bₙₖ} is a basis of V for all 1 ≤ k ≤ n.

My favorite open problem: Conjecture (Frankl). Let X be a finite set, and let S ⊆ P(X) be a collection of subsets of X which is closed under union. If S≠∅ and S≠{∅}, then some element x∈X appears in at least half of the elements of S, i.e. 2|{s ∈ S : x ∈ s}| ≥ |S|.

Let A be an n×n invertible matrix with coefficients in some field F. Must there exist a vector x ∈ Fⁿ such that neither x nor Ax has 0 as one of its entries? ~ spoilers below ~

Fresh off the presses, joint with Scott Balchin! Come check out our fun pictures :) arxiv.org/abs/2507.14068

BABY FROG

Sneak research preview -- fun pictures coming soon In this paper, we also get to take a limit as p (a prime) approaches 1 :^)

Check out our paper for more spooky fun! Maxine and Sam Ginnett worked out this story for cyclic groups a few years ago, and together we realized that the ideas from our ghost paper this October might allow us to push the argument through for all finite groups. "Might" turned out to be "do" :)

I'm in Seattle for JMM! Hmu if you wanna grab a coffee or something :)

I gave a talk this morning on some recent research (joint with Jason Schuchardt and Noah Wisdom). If you like abstract algebra (broadly construed) you might like this one! Title: "Algebraically Closed Tambara Functors" youtu.be/ast_KRMwBOQ?...

Check out our paper if you like Mackey and/or Tambara functors! It's spooky and fun

If someone's diction is such that everyone can understand them, it is clear Conversely, if your hearing is such that you can understand everyone, it is called cochlear

Just made it to Indianapolis for #MathFest24! DM me if you're around and want to grab a coffee :)

A few months ago, Advika Rajapakse, Talon Stark, and I wrote a math parody(?) cover of "Imagine". Last weekend we played it at a party as a singalong! Feat. Talon on guitar, me on piano, Rohan Joshi on the board, and a great crowd of friends on vocals :) youtu.be/xanZ3tAkhRo

I've started recording a short course on the representation theory of finite groups -- the first video is now up! Covering some basic motivation and a sneak peak of some cool facts we'll be able to prove by the end. youtu.be/KvYYpJARP6M

I gave a talk yesterday on algebraic varieties, the Weil conjectures, and a strange application of the Grothendieck-Lefschetz trace formula. My intention was to make the exposition accessible to ~any math grad student; take a look if you're interested! youtu.be/H5PsXY0E7Bw

On my way to JMM! If you're in SF this week and wanna grab a coffee or something, hit me up :)

Let M and N be smooth manifolds such that M and N have isomorphic groups of auto-diffeomorphisms. Must M and N be diffeomorphic?

The answer is: yes! For each d, you can take any finite coproduct of K_{d+1}'s. However, here's an interesting result of Biggs and Smith: there are exactly 12 finite connected 3-regular distance-transitive graphs, up to isomorphism.

A graph G is said to be distance-transitive if, for all vertices v,w,v',w', if d(v,w) = d(v',w'), then there is an automorphism φ of G such that φ(v) = v' and φ(w) = w'. Are there infinitely many finite d-regular distance-transitive graphs for all natural numbers d?

The answer is: yes! Such a set S is called a Steiner (n,k,m)-system. It turns out there is a unique Steiner (4,5,11)-system up to isomorphism! It has 66 elements, and can be found here: mathoverflow.net/a/452571/54637

A positive integer n is said to be "organizable" if there exist integers m > k > n and a set S ⊆ 2^{1,...,m} such that (i) each element of S has cardinality k (ii) for each X ⊆ {1,...,m} of cardinality n, there is a unique Y ∈ S such that X ⊆ Y Is 4 organizable?

The answer is: yes! First, let p be an arbitrary prime ideal of R. Then R/p is a finite domain, and hence a field. Thus, p is maximal. Also R is finite, so it has finitely many ideals. Thus, Spec(R) consists of finitely many closed points, so Spec(R) is finite and discrete.

Let R be a finite commutative ring. Must R be a product of local rings?

The answer is: no! I originally saw this posed on twitter by @chmonke, which sniped my friend Clark and I for a few days. You can read about our solution here twitter.com/DiracDeltaFunk… Quick counterexample: R = ℤ M = A⊕B with A, B the Prüfer 2- and 3-groups, resp. f = (a,b)↦(0,b)

Let R be a commutative ring, and let M be an R-module. Suppose f is an endomorphism of M with the following property: For all m ∈ M, there exists r ∈ R such that f(m) = rm. Must there exist r ∈ R such that for all m ∈ M, f(m) = rm?

The answer is: no! Proof: Let G be a group with Aut(G) ≅ ℤ/7. Then Inn(G) ≅ G/Z(G) is cyclic, so G is abelian. Now x ↦ -x : G → G is an automorphism of order ≤2. The only element of ℤ/7 of order ≤2 is the identity, so in fact x = -x for all x ∈ G. ...

Puzzle of the Day Does there exist a group G with Aut(G) ≅ ℤ/7?

The answer is: yes! This was asked by "mathlander" on math.stackexchange in 2022; I wrote a proof here: math.stackexchange.com/a/4587423/19006

Let C denote the ℝ-algebra of smooth functions ℝ → ℝ. Let d : C → C be a nonzero linear operator s.t. (i) d(f*g) = d(f)*g + f*d(g) (ii) d(f∘g) = (d(f)∘g)*d(g) i.e. d≠0 satisfies the product rule and the chain rule. Must d(f) = f' for all f ∈ C?

The answer is no! For example, we can take Y to be a the one-point space, and X any nondiscrete space. Then Hom(X,Y) is a singleton, which must be discrete.

Let X and Y be nonempty topological spaces such that Hom(X,Y) (with the compact-open topology) is discrete. Must X and Y be discrete?

The answer is: yes! Let χ be a character of a finite group G. Then ker(χ) := {g ∈ G : χ(g) = χ(1)} is a normal subgroup of G (equal to the kernel of the corresponding representation of G). Conversely, if N is a normal subgroup of G, then G acts on ℂ[G/N] with kernel N.

Suppose G and H are finite groups with the same character table. Suppose G is simple. Must H be simple?

The answer is: no! For example, take the ring of upper-triangular 2×2 matrices with integer coefficients. This ring is not commutative; in particular [2 0] [0 1] and [1 1] [0 1] do not commute.

Let R be a ring (unital, associative, not necessarily commutative) with underlying additive group ℤ³. Must R be commutative?

The answer is: 0! More generally: If x₁, ..., xₙ are real numbers that x₁² + … + xₙ² = x₁³ + … + xₙ³ = x₁⁴ + … + xₙ⁴, then x₁, ..., xₙ ∈ {0,1}ⁿ. So if n>2, the xᵢ's cannot be distinct.

Inspired by this question from Matt Enlow on twitter: twitter.com/CmonMattTHIN... How many solutions are there to w + x + y + z = w² + x² + y² + z² = w³ + x³ + y³ + z³ = w⁴ + x⁴ + y⁴ + z⁴ with w,x,y,z being distinct real numbers?

The answer is: ℵ₀! It turns out that any such ring is isomorphic to ℤ[t]/(f) for some monic quadratic polynomial f ∈ ℤ[t]. There are ℵ₀ such f, so there are ≤ℵ₀ such isomorphism classes. Moreover, it turns out that the rings ℤ[t]/(t²-p) are pairwise non-isomorphic as p ranges over the primes. ...

If you followed me on twitter, you know my thing is math polls. But there are no polls on bluesky :( so let's make it open-ended! How many isomorphism classes of rings are there with underlying additive group ℤ²? Finitely many? ℵ₀? More? Please use rot13.com to avoid spoiling others :)

Sheefs