Aufenthaltsraum

A topological room.

[aufenthaltsraumspizza]

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New ξ just dropped.

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The previous post made me remember this classic.

[aufenthaltsraumspizza]

The story continues.

[p-locally]

Anyway, all this work was motivated by the fact that apparently given two commuting idempotent endomorphisms L and R of a group G, the homomorphism Δ : G -> G×G, g ↦ (L(g), R(g)) describes a comonoid object in (Grp, ×, {1})

I wonder what happens with the monoidal structure given by the free product (Grp, *, {1})

*loses another three hours*

[aufenthaltsraumspizza]

I disagree.

Let (G, Δ, ε) be a comonoid object in a (cartesian) monoidal category (C, ×, 1). It then follows from the counitality diagram that the composites

   π₁ ∘ Δ ,  π₂ ∘ Δ : G → G

are both the identity morphism of G. The morphism

   Δ : G → G × G

is therefore the diagonal morphism.

There hence exists one, and only one, comonoid structure on every object of C. (This is also what https://ncatlab.org/nlab/show/comonoid states.) This comonoid structure is given by the diagonal morphism Δ : G → G × G and the unique morphism ε : G → 1.

[aufenthaltsraumspizza]

[aufenthaltsraumspizza]

The previous post made me remember this classic.

[aufenthaltsraumspizza]

[p-locally]

Whoever can find a pair of commuting, idempotent endomorphisms L, R of a group* wins a cookie (and a Comonoid object in the category of groups)

mmmh after thinking about it more it seems like we must have L = R after all, what a bummer :(

*non abelian please

[p-locally]

HOW did I not think of that?? I'm really glad I asked this question now!

I still think that's the only L ≠ R example for finite groups, however now I'm fairly sure I can find some infinite groups giving rise to a somewhat less boring L and R generalizing your extreme case.

Details tomorrow as of course late night math is dangerous

[aufenthaltsraumspizza]

Given any two groups G and H we can consider their product P = G × H. Both “internal projections” p : P → P, (g, h) ↦ (g, 1) and q : P → P, (g, h) ↦ (1, h) are group endomorphismn. Both of them are idempotent, and they commute.

[bubbloquacious]

I need a name for a 'semiring-but-it-doesn't-necessarily-have-0-or-1', so what do y'all think of 'distributive bialgebra'.

[p-locally]

Nooooo a bialgebra is already another thing!!

A dark and scary thing B with a multiplication BxB -> B and a comultiplication B -> BxB and other stuff I don't remember

I agree that names like semirng/semirig really suck though, I even think in some texts "Rings-not-necessarily-with-identity" are still called "Rings"

[aufenthaltsraumspizza]

Where × = ⊗.

[p-locally]

LaTeX disaster speedrun

Just hit a replace command "\P -> \PP" on my latex editor (texmaker), I made the mistake of not selecting the text I wanted to replace so it acted on the whole document which is a bad thing by itself I guess

BUT ALSO said editor by default doesn't check for capitalization (you have to manually check this setting) so now all my \pi are now \PPi, all my \prod are now \PProd, my \phi are \PPhi and so on

MOREOVER I just now found out i had already defined another command \PPP so now I have a hellscape of \PP, \PPPP and \PPPPP scattered all across number theory exercises

[aufenthaltsraumspizza]

Hopefully you can just undo your editors replace action.

You could also try to improve the damage done by enabling autosave and having the editor crash while trying to fix the problem. (If the editor refuses to crash, then you can also accidently close it.)

[p-locally]

[7/?] Today I sent the talk slides to the lecturer and he said they were very good! So I can stop worrying about that and focus on the rest of the exam. Two days to revise all the exercises again and the theory (although I know we won’t be asked to know every theorem by heart like in some other exams).

Here’s a very quick/bad proof of a fact that I always struggle to remember:

Thm. Let g be a Lie algebra and h < g be an ideal. If both h and g/h are nilpotent then so is g.  Proof.  A commutator of length m is an expression of the form [a1, [a2, … , [am-1, am]]]. Since the projection g -> g/h onto the quotient is a Lie algebra morphism, commutators in the quotient corresponds to commutators in g. The fact that g/h is nilpotent means that for some k > 0 all commutators in g/h of length > k are null, but this means that all commutators in g of length > k belong to h. This in turn means that for all m > k the m-th derived ideal of g belongs to the m-th derived ideal of h, but since h is nilpotent these are eventually (for some m’ > 0) the trivial ideal, and so g is nilpotent too. \qed

[aufenthaltsraumspizza]

This “theorem” is wrong.

Example. Let g be the two-dimensional Lie algebra with basis x, y and Lie bracket [x, y] = y. The one-dimensional linear subspace h of g spanned by y is an ideal of g. Both h and g/h are one-dimensional and thus nilpotent. But g is not nilpotent.

Exercise. Find the mistake in the above “proof”.

[p-locally]

Whoops! Yes this is wrong, thanks for proof-reading!

I’m missing an hypothesis: h must be a central ideal, and I guess the mistake is in the last lines (and also when I say “derived ideal” I mean the lower central series of course, I don’t know why I called it that way).

It’s been quite some time since I thought of this stuff but I think it can be fixed like so: we have that [g, … m times … [g, g]] + h = 0 in g/h so [g, … m times … [g, g]] si in h. Now since h is central [g, … (m+1) times … [g, g]] = [g, [g, … m times … [g, g]]] = [g, h] = 0 so g is nilpotent.

[aufenthaltsraumspizza]

Yes, if h is central in g then the argument works. It sufficies more generally that h is “relatively” nilpotent in g in the sense that the decreasing sequence    h,   [g, h],   [g, [g, h]],   [g, [g, [g, h]]],   … eventually reaches zero. (†)

I like to think about this in term of representations:

If M is a representation of a Lie algebra g, then we can consider its suprepresentation gM, which is spanned (as a vector space) by all elements xm with x in g and m in M. We say that the representation M is nilpotent, or that g acts nilpotently on M, if the decreasing sequence    M,   gM,   g(gM),   g(g(gM)),   … eventually reaches zero.

We then have the following:    • The Lie algebra g is nilpotent if and only if its adjoint representation is nilpotent.    • An ideal h of g in relatively nilpotent (in the above sense) if and only if the adjoint action of g on h is nilpotent.    • If h is an ideal of g and M is a representation of the quotient Lie algebra g/h, then we can regard M as a representation of g, and have (g/h)M = gM.

Your original “proof” can be modified to give the following observation:

Proposition. Let g be a Lie algebra, let M be a representation of g and let N be a subrepresentation of M. The representation M is nilpotent if and of if both the subrepresentation N and the quotient representation M/N are nilpotent.

The above proposition and three statements give (†). The key point here is that everything is done relative to the fixed Lie algebra g.

[twocubes]

ways to end proofs

QED

which was what we wanted (abbreviated w⁵, or u¹⁰, or even v¹⁰)

■ (or □ for lemmas)

♥

So there.

The rest is obvious.

[aufenthaltsraumspizza]

Proof by contradiction has led to one of my favorites:

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And another one, this time from Bredon.

[aufenthaltsraumspizza]

This time from Halmos.

[draculasdaughter]

This is about Sci-Hub. yeah we get it.. gatekeep knowledge and protect the interests of capital…

[three-blogs-in-a-trenchcoat]

Listen, this is serious.

Do not use the website called Sci-Hub!

It lets people access scientific articles for free. This is dangerous. It helps the free flow of knowledge and reduces the competitive edge of all the people who worked really hard to have been born into a wealth.

Like, it’s literally a website where you can type in the DOI of an article and read it, without ever having to pay the publisher who exploited the author.

So, again, do not, under any circumstance, use Sci-Hub. I mean, can you imagine a world where knowledge is free and easily accessible to everyone? Even, y'know, poor people?

[whateverthebeeswant]

Libgen also has many books online, including textbooks, searchable by name, author, and ISBN. Can you imagine textbook companies not getting their hard-earned income from poor college students? Here is the link just so you make sure that you never accidentally stumble across this horrible, unethical website.

[three-blogs-in-a-trenchcoat]

Oh, and while we’re talking about books, if you’ve managed to stay clear from Libgen, definitely don’t go to zlibrary, where you can also find a lot of textbooks, but unfortunately they’re completely free.

[trzpiotka]

nooo don’t use sites that make academic papers accessible to everyone, why would you do such a terrible thing

[p-locally]

[7/?] Today I sent the talk slides to the lecturer and he said they were very good! So I can stop worrying about that and focus on the rest of the exam. Two days to revise all the exercises again and the theory (although I know we won’t be asked to know every theorem by heart like in some other exams).

Here’s a very quick/bad proof of a fact that I always struggle to remember:

Thm. Let g be a Lie algebra and h < g be an ideal. If both h and g/h are nilpotent then so is g.  Proof.  A commutator of length m is an expression of the form [a1, [a2, … , [am-1, am]]]. Since the projection g -> g/h onto the quotient is a Lie algebra morphism, commutators in the quotient corresponds to commutators in g. The fact that g/h is nilpotent means that for some k > 0 all commutators in g/h of length > k are null, but this means that all commutators in g of length > k belong to h. This in turn means that for all m > k the m-th derived ideal of g belongs to the m-th derived ideal of h, but since h is nilpotent these are eventually (for some m’ > 0) the trivial ideal, and so g is nilpotent too. \qed

[aufenthaltsraumspizza]

This “theorem” is wrong.

Example. Let g be the two-dimensional Lie algebra with basis x, y and Lie bracket [x, y] = y. The one-dimensional linear subspace h of g spanned by y is an ideal of g. Both h and g/h are one-dimensional and thus nilpotent. But g is not nilpotent.

Exercise. Find the mistake in the above “proof”.

[aufenthaltsraumspizza]

???

How is this site still alive?!

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The two genders

[aufenthaltsraumspizza]

Homotopy is the pumpkin spice of mathematics.