If $V$ is an inner product space and $W$ a subspace of $V$, it is always true that $V = W \oplus W^{\perp}$?

@user193319 No

@Narcissusjewel Shoot...Do you know of any conditions on $W$ that will guarantee that such a conclusion holds?

Well it is guaranteed if $W$ is a closed linear subspace*

Closed with respect to the metric topology induced by the inner product?

Yep

Hi (and Merry Christmas! to y'all), sorry for interrupting – Random newbie question: Why is $\sqrt[-2]{4}$ equal to $0.5$? Just because $4^{\frac{1}{-2}}=0.5$?

@Narcissusjewel Sweet! Thanks!

@user193319 No problem

@Mr.Xcoder Why is $1/\sqrt{4}=1/2$?

Because $\sqrt{4}=2$

Oh, I was rewriting your question.

$4^{-1/2}=1/4^{1/2}$

Ah thanks

facepalm

@Narcissusjewel Sorry to bother you again. Would you happen to know of a link to the proof of that theorem? I tried google, but I couldn't find much; perhaps I am not using the right search terms.

Hey @Mr.Xcoder

@SimplyBeautifulArt hi

:D I wrote a derivative calculator

tio.run/##RcyxCoMwEADQX3FM1FMjOIXLj0gGxcgFzFFsa6@I3552c3vT29/…

Learning how to pass around functions/lambdas

nice

(6 pages)

@Narcissusjewel Listening to Wolves in the Throne Room album right now.

I hope it's legal to listen to black metal on Christmas :p

@BalarkaSen Hahaha, as long as the lyrics aren't satanic it is. Enjoying it?

Love it so far. It's beautiful.

@BalarkaSen I put my analysis question about Ted's book into an MSE question

+1ed. I still think the answer lies in bad counterexamples like the positive measure Cantor set.

@Narcissusjewel Oh. So $V$ has to be a Hilbert space in order for that direct sum result to hold?

I only have that $V$ is an inner product space.

@KasmirKhaan you clearly didn't read anything I said

@Mr.Xcoder define $\sqrt[-2]4$.

Someone answered that question already

@AkivaWeinberger sure

$\sqrt[r]b=b^{1/r}$

but it may still be helpful for $\Huge{\rm him}$ to define it

btw $M/N$ is defined even when $N \not\le M$ :O

It's nice that $\sqrt[2]2=\sqrt[4]4$

I certainly learnt something new today

Mm, @Akiva. If $g$ is $C^1$, the set of critical values of $g$ is measure $0$, isn't it? Sard's theorem.

So it seems you still cancel out those anyway. That's strange.

Critical values. Codomain.

Like write the right hand side as an integral over $\Omega - K$ plus integral over $K$ where $K$ is the critical set of $g$ (where $\det Dg = 0$). The integral over $K$ vanishes because the determinant of Jacobian vanishes.

The way Ted wrote it, it seems like it's talking about measure $0$ in $\Omega$ rather than ${\bf g}(\Omega)$

On the left hand side the integral cancels over $g(K)$ because that's measure 0.

So it seems you're still fine?

Oh. Hm.

@Akiva Yeah I'm channeling my confusion here.

Wait, is $\Omega-K$ open? Do we need that?

Oh wait

No, $U$ is open, $\Omega$ can be whatever

I am not sure why openness is relevant, no.

So we don't need that, so your thing still looks like it works

Yea

I don't get it

::confused::

@LeakyNun $4$ raised to the power of $\frac{1}{-2}$

Gtg now

1 plus 2 equals pi minus 1 that's e quick engineering

cc @Daminark

@BalarkaSen I thought pi and e are equal

good meme

Nice

hi

HIIIIII!

HIIIIIIIIII!!!!_!_!_!_!

Hallow

Hollow.

I m not doing mathematics today just onna hang out with you guys :P

@MikeMiller the way you say that reminds me of how a friend of mine links [this](niceme.me)

@Daminark That is genius. I shall have to use and abuse this.

Fucking hell

No hyperlink for some reason

@Balarka yeah it's great

link

what the shit

just put http

ah

I should put on my christmas hat

@Daminark youtube.com/watch?v=ZlVFegYA0fo

I don't know if your familiar with those songs :P

Not particularly, I'll check it out!

I grew up with space toon

I don't know if your familiar with it

Nope

Oh hell yes

Best random click of my life

How'd... You know I'm not questioning it

Youtube recommendation works sometimes

Also ffs that "I'm blue" song is stuck in my head

I need to listen to the full album, Atma

@Daminark lmao

Also @Adeek the intro to space toon is pretty nice for sure

yeah

@BalarkaSen All numbers are finite and thus can be rounded down to 0

all numbers are quasi-isometric to 0

ohyeah

@BalarkaSen here's my random song rec:

@AkivaWeinberger I don't think that $-1$ rounds down to zero...

@Semiclassical tbh it wasn't that random :) I was listening to a ton of black metal and doom metal

lol

what you linked sounds EDMish

@Akiva even infinity can be rounded down tbh

Just turn it to the side a bit and then it's finite

i think it came up on pandora on an explosions in the sky station

If $n> k$ and $1 \leq k \leq n$ is it trivial that $\binom{n-k}{k} \leq \binom{n-1}{k}$? just verifying that this late hour isn't buzzing my brain

Hi @Semiclassical

hi

I solved that other q we were talking about

neat

was damn exhausting

@Semiclassical I dunno why but Explosions in the Sky doesn't ring with me

I really like post-rock, and have tried a couple of their albums, but I don't like their style

to each their own

first one is (n-k)(n-k-1)...(n-2k+1)/k!, second is (n-1)...(n-k+1)/k!

so you're trying to compare (n-k)(n-k-1)(n-2k+1) with (n-1)(n-2)...(n-k+1)

so it is trivially true

looks like, yeah

its also logical

yeah, especially if you think in terms of Pascal's triangle

choosing k elements out of n-1 should have many more options than choosing k elements out of n-1

Oh, how so? I am not familiar

please do explain

well, do you know Pascal's triangle?

BTW. I am a computer science masters student. I deal with math alot, but not deeply. Mostly I end up writing algorithms and publishing papers about them. That's why I come here often... to learn deeper stuff and different "points of views"

Yes, that I do, Pascal's triangle.

How can I think "in terms of it"?

well, $\binom{m}{k}$ corresponds to a particular line of elements in Pascal's triangle

more precisely, if you write the triangle as

Yes, that I know @Semiclassical

it's the kth column

mth row, kth column

Right

So you're saying that n-1 is deeper than n-k

right.

So it must be either equal to it or greater than it in terms of choosing k elements

and you're definitely increasing along a column.

yup :)

great!

and only way to be equal is if k=1

oh, wait

?

it's also true for k=0, but you excluded that

Yup.

i.e. it doesn't work for the very first column

For $k=0$ its true since there is equality

sure, I meant my 'increasing' comment

Ah. sure. its not strictly greater than.

Erk. I don't think I have proved that the stable/unstable distributions of an Anosov diffeomorphism are integrable

In engineering, $\ln\ln x=1$

@Semiclassical here is an interesting one. Show that $p(n)$, the number of partitions of a natural number $n$ is $\geq 2^{\sqrt{n}}$. I thought about showing that for any subset of $\{1,...,\sqrt{n}\}$ we can create a partition of $n$ that contain the elements of that subset and numbers outside of $\{1,...,\sqrt{n}\}$

@Akiva kek, one time a prof of mine made the joke that $\ln(\ln(x))$ was basically bounded by 5, then he just waved his hands innocently and was like "Don't tell mathematicians I said that" in a room of mathematicians

@Daminark omg

@Daminark how do i prove that the stable/unstable subspaces of $T_xM$ are closed under Lie bracket

Lie bracket is a strange objecto

@TheNotMe What's $\sum\{1,\dots,\sqrt n\}$? (The sum of the things in the set) I feel like it should be less than $n$…

We need it to be less than $n-\sqrt n$, actually, don't we

@AkivaWeinberger indeed it is. it is $\frac{n+\sqrt{n}}{2}$

which is strictly less than $n$ for $n \geq 1$

Let me assume we take each element into the partition (except for $1$, lets say) only once, if any. So you're saying, fix a subset $A$ of $\{1,...,\sqrt{n}\}$. At the very maximum it could sum up to $n- \sqrt{n}$ and in order to complete it we must either only add $1$'s or stuff that are greater than $\sqrt{n}$?

kinda foggy still

Oh blah. $df[X, Y] = [df(X), df(Y)]$.

should do it

@BalarkaSen I don't know the Lie bracket

@AkivaWeinberger ?

Er, I don't know how the norm behaves with the bracket

@TheNotMe Oh, wait, I forget. Are partitions allowed to have repeated elements?

@Daminark all the better

Yeah :/

Oh. So my $n-\sqrt n$ thing is probably inaccurate

But it is something along those lines

I mean, I don't know the answer, I'm trying to figure it out

@Balarka is the Lie bracket painful to work with?

@AkivaWeinberger take a look: math.stackexchange.com/questions/2497506/…

I found this

but I don't really understand it in depth

Whenever I see a partitions problem, my first instinct is to think about (Ferrers) diagrams

same here

@Daminark I'll tell you the definition. If $X$ and $Y$ are two vectors fields on a smooth manifold $M$, the Lie bracket $[X, Y]$ is defined to be the vector field on $M$ which, after you take directional derivative of $f$ wrt $[X, Y]$ at each point, spits the function $[X, Y](f) := X Y(f) - Y X(f)$. ($Y(f)$ and $X(f)$ are directional derivatives of $f$ on the direction of $X$ and $Y$ at each point, and $XY(f)$ is directional derivative of $Y(f)$ in the direction of $X$ and similar for $YX(f)$.)

It's a purely algebraic/analytic fact that this is a vector field. $XY$ and $YX$ are not individually vector fields; the have second order terms in them. But they cancel out because $\partial^2/\partial x \partial y = \partial^2/\partial y \partial x$

I don't have much intuition for this object.

If we take the elements of any subset $A$ of $S = \{1,..,\sqrt{n}\}$ we will still need at least $\frac{n-\sqrt{n}}{2}$ of value to complete the partition into $n$

($f$ there is a smooth function $M \to \Bbb R$, I should have emphasized but forgot to)

I've never dealt with directional derivatives in terms of vector fields at all

Yeah that's the algebraists's definition of a vector, as an operator which eats a function and spits a scalar at each point (in a way that it satisfies linearity + Leibniz's rule)

Also called derivations

the other route that I think about is the generating function approach:

Alright so let's say $X$ is a vector field on $M$ and $f:M\to\mathbb{R}$. How do you define $X(f)$?

$\sum_{n=0}^\infty p(n)x^n=\prod_{k=1}^\infty (1-x^k)^{-1}$

Instead of being a disgusting analyst, let $G$ be an algebraic group (i.e. abstract group with Zariski topology), and let $A=K[G]$. Let $G$ act on $A$ by left translations $(\lambda_xf)(y)=f(x^{-1}y)$. Then consider the algebra $\text{Der}_K(A)$ of $K$-derivations of $A$, which is a Lie algebra under the commutator bracket. Then one can consider the nice Lie algebra of the algebraic group, given by left invariant derivations.

@Daminark Locally. Pick a point $p \in M$ and take a chart $U$ around $p$. Then $f|U : \Bbb R^n \to \Bbb R$ is your plain vanilla multivariable function.

$X(p)$ now descends down to a vector $v$ in $\Bbb R^n$ by the chart

hmm, that doesn't really help here though. so I'll leave that aside.

Define $X(f)_p = D_v f(p)$ - the directional derivative of $f$ at $p$ in the direction of $v$.

No it doesn't.

@Narcissusjewel rages

What I'd start with, I guess, is a special case like $n=m^2$.

rolls all 105.3 eyes

Hey @Narcissus!

Alright lemme process this real quick

@Balarka so... why does this not depend on choice of chart?

ex qawlasi qawad labit

@Narcissus so a $K$-linear derivation of $A$ is a map $D:A\to A$ which is $K$-linear (I'm guessing $K$ is a field here)

And which satisfies the product rule for derivatives basically

@Daminark Good question. Exercise :)

We want to show that the commutator is a Lie bracket

To figure out the exercise you'd have to unwrap everything and avoid my abuse of notation, of course, just as a note

@BalarkaSen You saw the comments on the question on Ted's thing?

Oh I didn't. So we were right?

Seems like it

Maybe Ted didn't want to state the theorem in such greater generality

It's always nice when I can get 15 points of reputation off of Ted's mistake

Well, it's technically not a mistake :)

You can actually drop the injectivity hypothesis too, sometimes

Is $\delta(\mathbf{x}) = \frac{1}{(2 \pi)^n} \int_{\mathbf{k} \in \mathbb{R}^n} \exp i \mathbf{k} \cdot \mathbf{x}$ a correct representation of the $n$-dimensional Dirac delta function?

Like in the case of covering spaces

You get a factor of the degree of the cover instead

So at the cost of that, you turn out to be all good

Turns out I'm dumb. The integrability is easy because the stable/unstable distribution is $f$-invariant and expands when iterated in the backwards/forward direction respectively.

@BalarkaSen Is this a response to me?

Nah, just a general comment on the thing I was stuck on before.

Probably nobody cares

Anyway, the important point is that the proof is - Proof: $f$, $f$, $f$. I N F L A T E. Done.

Mm… infinite lattes

"Infinite Latte" - David Foster Wallace

hi

can some one help me with automata language and grammar?

There is a language {a^n b^m c^k | n=m OR m<< k}

<<?

what js this OR in here? what does it mean?

that << is less or equal

It's the set of all things of the form a^n b^m c^k such that either n=m or m<=k

(People usually write <= and >= for less-than-or-equal and greater-than-or-equal on computers)

I know Iwas in a rush :(

But it's the set of things such that one or both of the conditions are satisfied

what happens to the third if either of them are satisfied?

Third?

if m <=k , how about n?

If m<=k, then n can be anything for it to be in the set

tnx

So, Elon Musk is joking about selling flamethrowers to help fund his company

At least, I think he's joking. It's hard to tell.

@Akiva it's not a joke, I bought one earlier this morningd

How else would you fund a trip to Mars?

:P

@Alessandro Sell a lot of copies of "Astrophysics for people in a hurry"?

guys I'm back

@Balarka i have a better plan

(The author also happens to be a musician playing with Vulfpeck, a greag funk group)

lol