Ben Spitz (43/100 improv meals) @DiracDeltaFunk Twitter profile

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Ben Spitz (43/100 improv meals)

1 year

Very pleased to announce that is live! Inspired by resources like and , you can use to query the SmallCategories database for finite categories with certain desiderata.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

I've just published (I think) the first publicly available database of small finite categories. As of writing, it contains all categories with ≤7 morphisms (except the empty category). Please use freely!

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

When a mathematician gets very excited, they are said to be "russelled up", since they cannot contain themselves.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Tfw the prerequisites are elementary

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

11 months

@evanewashington It is actually bizarre how consistently posts like this show chatGPT making mistakes. I'd expect people to want to check the output is correct before posting excitedly about the feature, out of fear of looking silly. But no?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

Please let G be a group. Thank you!

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let G be a group of order 2023 = 7*17². By Sylow, there are unique subgroups A, B ≤ G such that |A| = 7 and |B| = 17². These subgroups are normal and intersect trivially, so AB ≅ A×B and |AB| = |A||B| = |G|, and thus G ≅ A×B. ...

Algebra Etc.

@AlgebraFact

2 years

How many groups have 2023 elements?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

As of today, my age is the unique integer n such that n-1 is a square and n+1 is a cube! How would you try to solve for n?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

🧪 SOLUTIONS 🧪 1. The answer is "no"! We never need AC to pick an element from a single nonempty set, as we discussed above. Good job all.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

✏️✏️✏️ Pop Quiz: How well do you understand the Axiom of Choice? ✏️✏️✏️ I'll give you examples of steps you might want to take in a proof, and you need to decide if such a step requires invoking the axiom of choice. We're working in ZFC, which you should assume is consistent.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

✏️✏️✏️ Pop Quiz: How well do you understand the Axiom of Choice? ✏️✏️✏️ I'll give you examples of steps you might want to take in a proof, and you need to decide if such a step requires invoking the axiom of choice. We're working in ZFC, which you should assume is consistent.

Ready!

1046

Yikes

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Fact: if G is a group in which every element x satisfies x² = 1, then G is abelian. Weird proof: G = Gᵒᵖ iff G is abelian. The canonical iso x ↦ x⁻¹ : G → Gᵒᵖ is the identity map iff x² = 1 for all x ∈ G. Indeed, if idɢ : G → Gᵒᵖ is a homomorphism, then G = Gᵒᵖ!

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Ok, solution time. But first: ☠️ General Discussion ☠️ Let's talk about the Axiom of Choice. What it means for a set X to be empty is that ∀x (x ∉ X). Thus, what it means for a set X to be nonempty is that ¬∀x (x ∉ X). Equivalently, "X is nonempty" means ∃x (x ∈ X).

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

✏️✏️✏️ Pop Quiz: How well do you understand the Axiom of Choice? ✏️✏️✏️ I'll give you examples of steps you might want to take in a proof, and you need to decide if such a step requires invoking the axiom of choice. We're working in ZFC, which you should assume is consistent.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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?

Yes

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No

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Idk / show results

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Classic math joke

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! Cursed.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let f : ℂ → ℂ be an entire holomorphic function. Suppose that, for all θ ∈ ℝ, {f(re^(iθ)) : r ∈ ℝ} is bounded. Must f be constant?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

RP¹ ≈ S¹ A proof by rubber band

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is yes! In fact, ℝ³∖X is simply connected for any countable X ⊆ ℝ³. Proof: let f,g : S¹ → ℝ³∖X be arbitrary. Let H be the space of homotopies h : S¹×[0,1] → ℝ³ from f to g in ℝ³, equipped with the compact-open topology. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Is ℝ³ ∖ℚ³ simply connected?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

Let's imagine we found a solution {a,b,c}, with a+b+c = ... = K. Then we could consider the polynomial p ∈ ℝ[t] with these three roots, i.e. p = (t-a)(t-b)(t-c). By Vieta, we get p = t³ - (a+b+c)t² + (ab+bc+ac)t - abc = t³ - Kt² + (ab+bc+ac)t - abc. ...

Matt Enlow

@CmonMattTHINK

10 months

#math #maths #iteachmath #mathchat #mathschat #mathtwitter #mathstwitter

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is ≥50%! Actually, the probability is exactly 75%. Here's why: There are exactly 2 isomorphism classes of groups of order 6: those isomorphic to S₃ (nonabelian), and those isomorphic to ℤ/6 (abelian). Let M be the set of group structures on {1,...,6}. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Pick uniformly at random a group structure • on {1,...,6}. What is the probability that ({1,...,6},•) is abelian?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

Does there exist a group G with Aut(G) ≅ ℤ/7?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

9 months

The answer is: no! But only because of the one case Diff(∅) ≅ Diff(ℝ⁰) lol. Otherwise, if two nonempty smooth manifolds have the same groups of auto-diffeomorphisms, they are necessarily diffeomorphic! See

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

9 months

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

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

#mathfact The classic diagonalization proof that |ℝ| > |ℕ| relies on countable choice (since you have to make a choice for each digit!), and in fact it is consistent with ZF that ℕ is uncountable! 🤯

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

@littmath He should probably be made aware that the nation, in fact, has 30 entire universities! Excellent ones, even. We could greatly improve efficiency by shutting down WVU altogether

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

Cantor chuckled. "You mean the power set?"

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

7 months

The answer is yes! Here's one possible proof: Fix an isomorphism φ : R[x] → ℂ[x], and let p = φ(x). Since φ is surjective, there is some f ∈ ℂ[x] such that f ∘ p = x. Now 1 = deg(x) = deg(f ∘ p) = deg(f)deg(p) implies in particular that deg(p)=1. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

7 months

Let R be a commutative ring (with 1) such that R[x] ≅ ℂ[x]. Must R ≅ ℂ?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The left adjoint to the inclusion Ab → CMon is easily the most overrated idea attributed to Grothendieck. I can't believe we actually use the term "Grothendieck group" -- it's like calling log(ab) = log(a) + log(b) the "Napier formula" or something.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Christian Lawson-Perfect

@christianp

2 years

New meme format

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2K

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let f : ℤ² → ℝ be a function such that its value at any point is the average of its values at the four neighbors of that point. That is: f(x,y) = (f(x-1,y) + f(x+1,y) + f(x,y-1) + f(x,y+1))/4 for all (x,y)∈ℤ². Must f be constant?

Yes

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No

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Idk / show results

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Groups of prime or prime-square order are always abelian. Then by the classification of finite abelian groups, we have G ≅ (ℤ/7)×(ℤ/17²) or G ≅ (ℤ/7)×(ℤ/17)². There are thus two groups of order 2023, up to isomorphism.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is: a finite real number! In particular, 1. First we will show that, for all n≥5, every endomorphism of Sₙ is either an automorphism or factors through the sign homomorphism sgn : Sₙ → ℤ/2. This will give that, for n≥5, hₙ = |Aut(Sₙ)| + |Hom(ℤ/2,Sₙ)|. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

For each positive integer n, let hₙ be the number of homomorphisms Sₙ → Sₙ. What is lim_{n→∞} hₙ/n! ?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Pick uniformly at random a group structure • on {1,...,6}. What is the probability that ({1,...,6},•) is abelian?

≥50%

434

<50%

342

Idk / show results

510

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

11 months

Mathematicians never get to be more smug than when paper-authorship discourse comes around. And rightly so!

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Fun fact: Let G be a group. The subset of G consisting of those elements with finite conjugacy class is a normal subgroup.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is yes! More generally, in any category with an initial object 0, if x ∐ y ≅ 0 then x ≅ y ≅ 0. Proof: Hom(x, –)×Hom(y, –) ≅ Hom(x×y, –) ≅ Hom(0, –) ≅ 1 ⇒ Hom(x, –) ≅ Hom(y, –) ≅ 1 ⇒ x ≅ y ≅ 0 where 1 denotes a constant functor with image a singleton.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let A and B be commutative rings such that A ⊗ B ≅ ℤ. Must A ≅ B ≅ ℤ?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

I've just published (I think) the first publicly available database of small finite categories. As of writing, it contains all categories with ≤7 morphisms (except the empty category). Please use freely!

GitHub - diracdeltafunk/SmallCategories: A database of small finite categories

A database of small finite categories. Contribute to diracdeltafunk/SmallCategories development by creating an account on GitHub.

github.com

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

@mattecapu I disagree, I think this problem is very hard. And while it's not an "important" problem, I think a lot of people would care if it was resolved.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is yes! First of all, |X²| = |Y²| implies that X is finite iff Y is finite. The function n ↦ n² : ℕ → ℕ is injective, so we can conclude that |X| = |Y| in the finite case. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let X and Y be sets. Suppose |X²| = |Y²|. Must |X| = |Y|?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

5 months

@CmonMattTHINK You can get a good approximation of Stirling's approximation from this picture :)

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

7 months

@eigenknight The coordinates 1,...,n are voting :) Each index i votes either yes (1) or no (0) -- its vote is denoted xᵢ. The final outcome of the vote is either yes (1) or no (0). A voting system is a function f that takes in the votes (x₁, ..., xₙ) and produces a final outcome in {0,1}.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

8 months

I need "intro to rust for people who know what a type system is" Like geez I do not need 400 pages of "Option is kind of like a burrito!", please just tell me what `?` literally actually does Does this exist

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Is ℝ³ ∖ℚ³ simply connected?

Yes

383

No

312

Idk / show results

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! Let V be {measurable functions [0,1] → ℂ}/≈, where ≈ is almost-everywhere equality. Then d([f],[g]) = ∫min(|f-g|,1) is a metric on V, making it into a metric vector space under pointwise addition and scalar multiplication. V* = 0!

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let V be a Hausdorff topological ℝ-vector space such that V* (the vector space of continuous linear maps V → ℝ) is trivial. Must V be trivial?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

Math folks, what's your favorite definition of all time?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

@MathProfPeter @Mrs_Meowmerz Ok let's say we're looking for a polyhedron with 5 vertices and 100 faces. Then by Euler characteristic we need 103 edges. But such a polyhedron has at most 5 choose 2 = 10 edges. Damn.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

More generally: every group of order n is cyclic iff gcd(n, ϕ(n)) = 1!

Algebra Etc.

@AlgebraFact

2 years

Every group of prime order is cyclic.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

In the style of @littmath : At the end of this poll, ranking the options by share of votes received will determine a permutation of {1,2,3,4}. To be precise, 1 ↦ the most popular response, 2 ↦ the second most, etc. How many orbits will this permutation have?

1

103

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265

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148

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

The answer is "yes"! Let f : G × G → G be defined by f(x,y) = xy⁻¹. Then f is continuous and {e} is closed in G, so Δɢ = f⁻¹(e) is closed in G × G. QED

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let G be a T1 topological group. Must G be Hausdorff? (Definition: a topological space is T1 iff every singleton subset is closed)

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

@uberwensch_ Never hurts to hear this a noether time

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

9 months

Curious about this question: let K be the subgroup of ℤ^ℤ consisting of those functions f : ℤ → ℤ with finite image. Is K free abelian? My guess is no, but I don't have a plan for how to prove it! If you have any thoughts please lmk :)

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is ≥50%! Actually, the exact probability is very close to ln(2) ≈ 69%. Here's why: For a fixed k ∈ {101,...,200}, the number of cycles of length k formed from the elements of {1,...,200} is C(200,k)*(k-1)! where C(a,b) denotes a binomial coefficient. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

What is the probability that a random permutation of {1,...,200} contains a cycle of length >100?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

Mathematicians be like: "this is a braid"

Curiositats espais anellats

@AppeleDuNeant

3 years

Mathematicians be like: "this is the complex projective line"

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341

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

7 months

Let R be a commutative ring (with 1) such that R[x] ≅ ℂ[x]. Must R ≅ ℂ?

Yes

299

No

266

Idk / show results

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! For example, S³ is a non-contractible space on which every vector bundle is trivial. That's because a k-vector bundle (k ∈ {ℝ, ℂ}) on S³ is determined by the homotopy class of its clutching function S² → GLₙ(k), and π₂(GLₙ(k)) is trivial for all n.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let X be a topological space. Suppose every vector bundle on X is trivial. Must X be contractible?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

11 months

@CihanPostsThms Had this as part of a homework problem once!

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

@frobunnius @littmath For the non-expert: "quasi often" means "induces often on homology"

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Found a pretty elegant proof for a statement that I thought was too strong to be true. Cautiously pogging.

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

8 months

The answer is yes! Consider the composite A → B → A, which we will call f. f is injective, and thus by invariance of domain it is open. Now let A' be an arbitrary connected component of A, and note that f|_A' is an open embedding with connected closed image. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

8 months

Let A and B be closed topological manifolds, and suppose there exist embeddings A → B and B → A. Must A and B be homeomorphic?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! We can take C to be the empty set :) Also, Roberts gave a more interesting counterexample in 1977:

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let V be a Hausdorff topological ℝ-vector space. Let C be a compact convex subset of V. Must C have an extreme point?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let A and B be abelian groups such that A⊗B ≅ ℤ. Must it be the case that A ≅ B ≅ ℤ?

Yes

204

No

320

Idk / show results

569

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

@grassmannian Surface integrals from calc 3 classes

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! In fact, every finite simplicial complex is weak homtopy equivalent to a finite space (basically, the space of barycenters), and in particular there is a finite space with the same homotopy type as S². But S² has infinitely many nontrivial homotopy groups!

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let X be a finite topological space. Must there exist a natural number N such that, for all n > N and all x ∈ X, πₙ(X,x) is trivial?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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?

Yes

131

No

282

Idk / show results

251

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

What's your favorite example of a space with finite non-abelian fundamental group?

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Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

For each positive integer n, let hₙ be the number of homomorphisms Sₙ → Sₙ. What is lim_{n→∞} hₙ/n! ?

0

119

A positive real number

211

∞ or does not exist

148

Idk / show results

266

13

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39

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! This was first proven by Olshanskii in the 80's, who showed that for any prime p > 10^75, there are continuum-many isomorphism classes of such groups. In 1992, Adyan and Lysënok improved this bound, showing that such groups exist for any prime p > 1003.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let p be a prime. Let G be a group such that every nontrivial proper subgroup of G is cyclic of order p. Must G be finite?

6

2

8

2

12

42

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

7 months

The answer is no! For example, let G be an infinite product of copies of ℤ/2 (with the discrete topology). Then G is a compact Hausdorff topological group (by Tychonoff) which is 2-torsion. Also G is infinite, and thus not discrete.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

8 months

Let G be a locally compact Hausdorff topological group. Suppose G is torsion, i.e. for all g ∈ G there exists a positive integer n such that gⁿ = e. Must G be discrete?

4

1

12

2

1

38

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

@SamiDouba My feeling is that "most" "interesting" infinite groups are better understood with a background in topology. There's lots to say about SL(n, R), for example, but mostly through the lens of Lie theory!

1

40

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

So, a solution is {1/2, (1+√(3/2))/2, (1-√(3/2))/2} with common power-sums K = 3/2.

0

41

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

This is the ideal definition. You may not like it, but this is what peak performance looks like.

Artem Chernikov

@archernikov

3 years

@AndresECaicedo1

2

1

16

1

2

40

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

8 months

The answer is no! For example, let A be [0,1] and let B be [0,1] ⨿ [0,1].

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

8 months

Let A and B be compact Hausdorff spaces, and suppose there exist embeddings A → B and B → A. Must A and B be homeomorphic?

2

1

12

3

1

40

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

We know Aut(ℤ/6) ≅ (ℤ/6)* ≅ ℤ/2 and Aut(S₃) ≅ S₃, so there are 6!/2 = 360 abelian group structures and 6!/6 = 120 nonabelian group structures on {1,...,6}. That's a total of 480 group structures, 360 of which are abelian, giving a probability of 360/480 = 75%. QED

0

40

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

Claim: Every bijection {1} → {1} has a fixed point. Proof: Consider {1} as a 0-manifold and apply the Lefschetz trace formula.

0

3

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Thanks for playing! Check back tomorrow for the answers :)

0

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is yes! By the existence of Jordan canonical form, write X = P(D+N)P⁻¹ for some invertible matrix P, diagonal matrix D, and strictly upper-triangular matrix N such that DN=ND. Then I = exp(X) = exp(P(D+N)P⁻¹) = P(exp(D+N))P⁻¹ ⇒ exp(D)exp(N) = exp(D+N) = I. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let X be a square matrix with coefficients in ℂ. Suppose exp(X) is the identity matrix. Must X be diagonalizable?

6

1

11

2

5

38

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is yes! Proof idea: Let μ be a Haar measure on (the additive group of) F. By inner regularity, pick a compact set S₀ ⊆ F such that μ(S₀) > 0. Then let S = S₀ ∪ -S₀, so that S is compact, μ(S) > 0, and S = -S. Now define n : F → [0,∞) by n(x) = μ(xS)/μ(S). ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let F be a locally compact Hausdorff topological field. Must F be metrizable?

2

0

17

1

6

40

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

How many isomorphism classes of rings are there with underlying additive group ℤ²?

2

19

6

8

38

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

What is the probability that a random permutation of {1,...,200} contains a cycle of length >100?

≥50%

308

<50%

457

Idk / show results

231

12

5

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

The answer is that this is equivalent to CH! In fact, ZFC proves that every uncountable compact metric space has cardinality 2^ℵ₀. Let X be an uncountable compact metric space. X is separable, so it injects (via its metric) into ℝ^ℕ, hence |X| ≤ |ℝ^ℕ| = 2^ℵ₀. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Does there exist a compact metric space of cardinality ℵ₁?

9

4

29

1

5

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let X be a set with 2023 elements. Let * be a binary operation on X such that a*(b*c) = (a*b)*(a*c) for all a,b,c ∈ X (i.e. * distributes over itself). Must * be constant? In other words, must it be the case that a*b=c*d for all a,b,c,d ∈ X?

Yes

65

No

120

Idk / show results

269

7

1

39

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let f : ℂ → ℂ be an entire holomorphic function. Suppose that, for all θ ∈ ℝ, {f(re^(iθ)) : r ∈ ℝ} is bounded. Must f be constant?

Yes (in ZFC)

131

No (in ZFC)

52

Independent of ZFC

46

Idk / show results

122

9

5

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

When you localize your category

0

3

36

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

@QuantaMagazine Hi Quanta, this is misleading in a bad way and should be revised. A genus-g surface has H¹ of rank 2g. In particular, a genus-2 surface has H¹ of rank 4, not 6.

2

1

36

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Bringing this back

0

6

36

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

@dzackgarza Right! Although in my mind ℚ(√{-d}) means ℚ[x]/(x²+d) while ℚ(i√d) means "the smallest subfield of ℂ which contains i√d". In particular, writing ℚ(i√d) says to me that we've already fixed a choice of the field ℂ and the element i in it. Same field, lil different vibes.

0

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

9 months

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

Yes

144

No

259

Idk / show results

263

18

4

36

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is no! @schala163 provided a nice counterexample: Let (X, Ω) = (ℕ, 2^ℕ), where 0 ∉ ℕ (😥). Let μ(S) = ∑_{s∈S} 2^(-s), and let ν be the same but with the measures of {1} and {2} swapped. For any p∈[0,1], its binary expansion is a set S such that μ(S) = p. ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let μ and ν be probability measures on a measurable space (X, Ω) such that, for all p ∈ [0,1], there exist S, S' ∈ Ω such that μ(S) = p = ν(S'). Must there exist some T ∈ Ω such that μ(T) = 0.5 = ν(T)?

1

4

26

1

3

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

The answer is "no"! For example, let τ be the Zariski topology on ℂ, and let T be the Zariski topology on ℂ². T ≠ τ×τ, because Δ_ℂ is a closed subset of (ℂ², T), while (ℂ, τ) is not Hausdorff. However, it is true that Zariski closure commutes with cartesian product! ...

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let (X, τ) be a topological space. Let T be a topology on X×X with the following property: For all subsets A,B ⊆ X, the closure of A×B in (X×X, T) is equal to A̅×B̅ (where A̅ and B̅ denote the closures of A and B, respectively, in (X, τ)). Must T be the product topology τ×τ?

1

2

8

2

7

34

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

@littmath Great minds :)

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

When a mathematician gets very excited, they are said to be "russelled up", since they cannot contain themselves.

9

104

597

0

1

37

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

Let A and B be nonzero, torsion-free abelian groups. Must A⊗B be nonzero?

Yes

263

No

188

Idk / show results

479

9

2

34

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The idea is that you can cook up an entire function which is bounded except on a strip {z : a<Im(z)<b} with a>0. Then each line {re^(iθ) : r∈ℝ} has compact intersection with the strip, so f is bounded on the line. Thanks to Zach Baugher for making me aware of this horror.

2

1

32

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

2 years

📣📣 Stream Announcement 📣📣 I have a teeny amount of experience with Lean, but for a while I've wanted to actually get good and maybe try to contribute to mathlib. Well, I now have no excuse: this Friday, @grassmannian will be giving me a crash course on Writing Serious Lean.

3

1

34

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

3 years

@mathemensch Concepts are closed under finite disjunction, so we conclude that any finite number of concepts can be explained to a 6yo. Then by compactness there exists a 6yo to whom every concept has been explained.

1

0

35

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Note that H is nonempty and completely metrizable! For each x∈X, let Hₓ = {h∈H : x∉img(h)}, i.e. Hₓ is the set of homotopies which avoid x. For all x∈X, Hₓ is open and dense in H. By the Baire category theorem, ⋂_{x∈X} Hₓ is dense in H, and in particular nonempty. ...

4

2

34

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

A square matrix X ∈ Mₙ(ℝ) is said to be "funky" if: 1. Each entry in X is +1 or -1 2. The columns of X are pairwise orthogonal Do there exist funky matrices in all dimensions?

Yes

204

No

346

Idk / show results

287

19

3

36

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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.

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

10 months

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

1

2

16

5

4

35

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

@Anthony_Bonato .... it would be MUCH more surprising to see chatGPT beat any skilled human at integration, let alone Cleo in particular

2

0

33

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

There's a basic problem with this though: a proof may not have infinitely many steps! This is exactly what the axiom of choice is for. AC says precisely that whenever we have *any* set of nonempty sets, we may pick an element from each simultaneously, in a single proof step!

1

0

34

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

@taz_chu ↦ is already this but better :)

0

1

34

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

The answer is 2! B is corepresented by the ring ℤ[x]/(x²-x) ≅ ℤ². By the Yoneda lemma, Aut(B) ≅ Aut(ℤ²). You can check directly that there are exactly two automorphisms of ℤ²: the identity and (x,y) ↦ (y,x).

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

Let B : CRing → Set be the functor sending each commutative ring to its set of idempotents, and sending a morphism X → Y to its restriction B(X) → B(Y). How many natural automorphisms does B have?

2

3

14

2

32

Ben Spitz (43/100 improv meals)

@DiracDeltaFunk

1 year

For each natural number n, let Gₙ be the number of isomorphism classes of (simple, undirected) graphs with n vertices. Does ∑_{n=2}^∞ 1/log(Gₙ) converge?

Yes

176

No

61

Idk / show results

165

12

3

34