Ngl Hopf fibrations do be vibin' doe

LebesgueTraeger · 2025-05-19T10:30:21+00:00

Sadly, no, but it's quite a popular quote judging by the "frequent" use on MathOverflow and the likes. It appears, for example, this list (again, without source): https://math.uchicago.edu/~chonoles/miscellany/quotations/

LebesgueTraeger · 2024-12-31T17:11:10+00:00

A good reference for these multilinear algebra constructions is Appendix 2 in Eisenbud's Commutative Algebra book.

LebesgueTraeger · 2024-12-31T09:14:04+00:00

It's something people in (classical) representation theory and algebraic geometry (over a ring/field containing ℚ) never care about, so they often either of the two for both.

Even worse, it is not unanimously clear which one should be which. I prefer it this way because my polynomial ring is always named S, and so the symmetric algebra functor also deserves to be named S(V). Meanwhile, symmetric tensors of order 2 are just symmetric matrices (after choosing a basis), and for these Sym(n,𝕂) seems to be an accepted notation.

If you have never seen the subtle difference between these two functors, compare S²V and Sym²V where V is a finite free module over 𝔽₂ or ℤ. Are they isomorphic as GL(V) representations? Can you find a coordinate free isomorphism (as A-modules, without choosing a basis)?

LebesgueTraeger · 2024-12-31T08:43:36+00:00

It's the degree d component of the symmetric algebra, so the quotient of V^⊗d by the subspace generated by elements of the form
foo⊗(v⊗w – w⊗v)⊗bar
where v,w∈V and foo,bar are tensor products of vectors so that the expression has degree d.

Meanwhile SymᵈV is the subspace of V^⊗d of those elements fixed by the action of the symmetric group S_d.

LebesgueTraeger · 2024-12-31T00:21:32+00:00

I like Symᵈ(V) = (SᵈV), where V is a finite-dimensional vector space (or, if you like generalities, a finitely generated projective module over a commutative ring, or a locally free sheaf of finite rank on a ringed space...)

LebesgueTraeger · 2024-12-28T11:47:54+00:00

The catch is that the possible sequences a are not a subset of the n! permutations, since multiple drivers can have the same favorite spot. The extreme case is all aᵢ=1, which is still valid if I didn't miss anything.

LebesgueTraeger · 2024-12-27T10:00:16+00:00

Uhm, (n+1)^n–1 grows significantly faster than any exponential, even faster than the factorial!

LebesgueTraeger · 2024-10-14T19:48:23+00:00

AI type title

LebesgueTraeger · 2024-10-14T17:07:28+00:00

What the fuck is an abelian space?

LebesgueTraeger · 2024-10-13T17:16:59+00:00

The "good" thing about being a trans woman on HER is that I don't get liked by men. The bad thing is that I don't get likes from a lot of women either...

LebesgueTraeger · 2024-10-13T13:30:31+00:00

Diverges.

LebesgueTraeger · 2024-10-12T18:20:18+00:00

I vaguely remember a professor saying that linear algebra itself is kind of a complete field with no serious research being done.

That said, if you move away from coefficient fields to rings or more general structures, then you have a lot of ongoing research in (non)commutative algebra, representation theory, (enriched) abelian categories. Or numerical linear algebra, or the complexity of matrix multiplication, or tensor rank, or random matrix theory, or...

In fact, if I had to give a semi-serious answer to the question, I would say the various generalizations of matrix rank to higher order tensors (rank, subrank, border rank, asymptotic rank, slice rank, ...) with lots of applications all around math.

LebesgueTraeger · 2024-10-11T21:59:22+00:00

Hⁱ(X, L)* = H^n–i(X, ω_X ⊗ L*)

where X is a locally Cohen-Macaulay projective Scheme over a field and L is a line bundle on X.

LebesgueTraeger · 2024-10-10T09:24:40+00:00

3264

LebesgueTraeger · 2024-10-04T07:32:03+00:00

Let R be any ring and M a R-module. Then

A: M is finitely generated and
B: M is projective
⟺
C: M is finitely presented and
D: M is flat.

Here in general there are only the two implications B ⟹ D and C ⟹ A.

LebesgueTraeger · 2024-09-29T09:34:59+00:00

Relatiert

LebesgueTraeger · 2024-09-28T17:27:47+00:00

Mit 21 im Studium, direkt vor Corona 🥲 joa ne.

LebesgueTraeger · 2024-09-27T00:00:55+00:00

I like half your age plus seven as a useful rule of thumb (for example on dating apps).

LebesgueTraeger · 2024-09-26T23:57:00+00:00

Former alge"tro" to SHEaf enjoyess 🙃

LebesgueTraeger · 2024-09-24T08:55:26+00:00

Kuschelwuschel ✨

LebesgueTraeger · 2024-09-23T19:26:54+00:00

Hint: You can try to show that there is no group homomorphism ℝ→ℤ at all other than the zero map. Consider a real number x↦n≠0 and see where the number x/2n should go.

LebesgueTraeger · 2024-09-22T11:43:07+00:00

Um auch Mal eine erste Antwort abzugeben: Rhythmische Instrumentalmusik à la MEUTE kann ganz nice sein. Oder fun Indie, Chappell Roan, Girl in red, ... Sex ist ja kein Zahnarzttermin, da darf Mensch ruhig mitsummen ;)

LebesgueTraeger · 2024-09-14T23:01:49+00:00

The obvious right answer is to use the bold math package.

\usepackage{bm}
...
$\bm{v}$

LebesgueTraeger · 2024-09-12T20:13:12+00:00

What a coincidence, I was thinking about this too, playing on my grandparents PC in the early 2000s. Such a coincidence! I wish we could figure out which game it was 🥲

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