we have found the phd student in the room @MikeMiller :P

not that I disagree hahaha

beware!

Mike, I'm sure this stuff is written down in lots of places, but I am too far removed from that and have none of my previous resources.

Fair enough. Don't mean to pressure you in any way.

I left research, I am trying to move to industry move. Research is very intense. Do you guys research in math?

no

@AbhishekBhatia I'm still learning :)

but I'm paid to

lolll

paid to what?

@AbhishekBhatia What industry are you looking to pursue?

Gold

Computer Science with some Machine learning thrown in there.

@AbhishekBhatia that was one sentence: "no, but I'm paid to [research math]"

good good

Can you learn after college?

ill be back in a few. walking home

I am not sure if my college will allow me to take courses after graduating

bye @Antonios

I'm telling you I've never heard that before, in almost 50 years of using/seeing/knowing linear algebra, Demonark. It's nonstandard.

I remember discussing this question with you when I was starting my masters math.stackexchange.com/q/2043548/200649

Huh

It seems pointless to define rank as a definition of something else that already exists

"we say that rank of a collection of vectors is the dimension of their span" -> just say dim(span)

yeah, I vaguely remember that, @AbhishekBhatia.

My professor answered it in the first class of ML.

Linear regression interpreted in terms of projection is super important.

Nice I discussed with you once.

I just find the relation between mathematics and data-science so weird.

There's actually a fair amount of geometry in the linear algebra of statistics, but statisticians tend to ignore it.

Yeah fair

I was working with a phd student from princeton and another one from cambridge. The cambridge gave us a paper to read. I have taken 1 year of classes so I thought I knew something atleast. But I couldn't understand anything in the paper.

I live

i survived the rain

you didn't melt, @Antonios?

lol fortunately not

I brought to the Princeton guy and turns out it was difficult for him to understand.

the temperature is lovely out, but the rain is crappy.

Just made realize we know so little, which is distressing and motivating thing at the same time.

lol...talking about going off a tangent

@TedShifrinI still can't get how he got to T≡r(modC)).

Because $T-r$ is a multiple of $C$.

@Ted So to brief this is the correct way of solving it right.$L(x) = 3x \\ v_1 = x, v_2 =y \\ L(\alpha(v_1)) = L(\alpha(x)) \\ L(\alpha x) = 3 \alpha x \\ \alpha L(v_1) = \alpha(3x) = \alpha 3x \\ L(v_1+v_2) = L(v_1) + L(v_2) \\ L(x+y) = 3(x+y) = 3x+3y \\ L(x)+L(y)= 3x+3y$

@Antonios-AlexandrosRobotis darn

Rain always screws up a nice day

Yes, @Jon, but I would write it as a complete sentence (mathematical or with words).

$L(x+y) = 3(x+y) = 3x + 3y = L(x)+L(y)$.

Similarly, $L(\alpha x) = 3(\alpha x) = (3\alpha)x = \alpha(3x) = \alpha L(x)$.

You'd prefer snow or 95º weather, Demonark?

I did to verify that the LHS = RHS. In the end thats how we know it is a linear transformation.

Its the same thing just better notation

Yours

@TedShifrin snow for sure

Right. I'm just saying it's better form to string things together to show the desired equalities.

Ideal for me is cloudy, 50s, light breeze, dry

Sounds like Berkeley weather, Demonark.

But I doubt there are that many places which are like that all year

Well, it gets into 60s and low 70s.

Oh huh

Eh I'll do 60s. Well

yeah berkeley weather was quite nice

all year round

Makes sense, since you come from Texas.

I guess we know where I'm going to grad school (hopefully)

There's usually a couple of very hot weeks in Berkeley.

Yeah, Demonark, you're obnoxious enough you should do OK even though so many students get ignored there :P

And maybe just live the rest of my life there

Expensive as all hell, and traffic and parking are terrible now. But I'm staying there 4 days in August.

@TedShifrin I think the condition I wanted is just the Leibniz rule, like $d_A [\phi, \psi] = [d_A \phi, \psi] + [\phi, d_A \psi]$. I asked Ciprian, though I guess he probably won't know off the top of his head.

Someone good at this stuff would presumably be able to answer easily.

I am not.

That suggests some sort of Lie algebra invariance, Mike.

I have worked things out like this, but many moons ago, and I have no brains now.

I know the process by which that happens, even if I'm at the start.

The brain-loss. Not the working things out.

LOL

asks self: Self, what is your name?

For me since it is the induced connection on $\text{End}(E)$ of a connection on $E$ I imagine I have all the invariance I want.

It's like you're assuming the connection is parallel with regard to the Lie group.

I would think it's just assuming it's compatible with the Lie group. Like, it's a connection for $E \otimes E^*$ with the structure that is induces from $E$, which is a subset of all connections on $E \otimes E^*$.

Any recommendations of a good math chat site for subjective questions/open ended ones?

potentially right here @frogeyedpeas :P

I guess my point is that you're assuming the Lie bracket operation is parallel, i.e., has covariant derivative 0.

Kasmir! Did you go to the doctor?

@TedShifrin Yeah, I see your point. I think it should be. I don't think that restricts the space of connections too much.

Should still be rather beefy.

Presumably, a $G$-invariant connection should give you that. I'd have to ponder.

@Antonios-AlexandrosRobotis, alright here goes: has anyone looked at matrix multiplication and asked the question of "in what sense is the definition of matrix multiplication natural for rectangles". To sort of highlight what I mean, suppose I make a trianglular collection of numbers, and wish to somehow define a notion of multiplication between them such that this notion reduces to classical matrix multiplication.

how could one go about doing this

Put the triangle inside a rectangle?

I see. But (for frame of reference) I am thinking of this as living over some random 3-manifold, not on the bundle $\mathfrak g$ over $G$.

Yeah, but $G$ still acts on the bundle, @Mike, so what I said makes sense.

@frogeyedpeas what ted said makes a good bit of sense.

Ah, by adjoint action. I see.

In your case, yeah.

the reason the matrix multiplication is meaningful as it's written is because it encapsulates the notion of linear combination

Well, adjoint just meaning the action of $G$ on its Lie algebra.

Maybe you prefer 'conjugation'.

Well, you're doing End, so, yeah, ...

So to follow @TedShifrin's construction n some sense a triangle or any shape, is just a collection of numbers that are non-zero, then we can make the remaining elements 0, until it fills up to a rectangular array, and we have a definition of matrix multiplication to recall from here

If you try to multiply with other "shapes" I think the question you should answer first is "why."

Upper- (or lower-) triangular matrices is a thing ...

@Antonios-AlexandrosRobotis just meandering thoughts, which I didn't feel were qualified enough for a proper math.se post

@Antonios: Welcome to teaching linear algebra :P

@TedShifrin I really care about Aut(E) < End(E) and I am cheating by expanding the sections of interest to make a nonlinear problem into a linear one.

@frogeyedpeas no judgement, interesting question at least.

@TedShifrin :P i ought to get used to it

I thought for a bit that $\ker(\Delta) \cap \text{Aut}(E)$ was a subgroup but that's probably not true because harmonic times harmonic is usually not harmonic... but whence, the desire for calculation.

@MikeM: Or are you differentiating?

Right, I have no idea what structure that'll have.

Nah. I can solve things after differentiating.

Oh, I see your point. For me E is equipped with eg unitary structure so Aut(E) is meant to preserve that structure. Its tangent bundle is skew-Hermitian endomorphisms, but it sits inside arbitrary endomorphisms.

Since Laplacian is linear, sums are good, but products are yuck.

Yup.

Yeah, but still maybe you should look at the skew-Hermitian ones first.

Presumably the harmonic ones form a subspace and you should exponentiate?

Yeah. I understand the story in that case.

But I don't understand the story away from the identity.

Oh, no, but exponential and Laplacian won't commute.

I'm confident that this is well-known to the right people.

Yup. Maybe I should just ask an MO question.

Terry Tao would know. Robert Bryant would know.

It's kind of fun to get back into the functional analysis but it's one of those things where I knew what I was doing a year ago and have to claw my way back into that.

Not that this question is functional analysis. I just got here in that process.

Hello everybody

What is a complex number raised to a complex number?

geocalc, do you know the definition of a number raised to an irrational number

no

okay, let's start with positive real numbers

agree or disagree: $5^9 = \exp \ln(5^9) = \exp (9 \ln 5)$

In this answer math.stackexchange.com/a/913529/200649 , I don't understand what does (modC) to r mean?

$\exp(t)$ is the same thing as $e^t$

yeah i agree with that

@AbhishekBhatia: You never learned mod? I already told you. $x\equiv y\pmod C$ means precisely that $x-y$ is divisible by $C$. This means they give the same remainder when divided by $C$.

do you know what the limit of a sequence is geocalc?

yes

let $r$ be an irrational number

@TedShifrin yeah, I am not sure what does "to r" mean here?

in theory, you can define $5^r$ to be the limit of $5^{q_n}$ where $(q_n)$ is a sequence of rational numbers converging to $r$

but it turns out that while that works, its kinda hard to prove things or calculate things that way

@AbhishekBhatia: If $17=2\times 7 + 3$, then $17\equiv 3\pmod 7$.

its easier to say that $5^\pi = \exp(\pi \ln 5)$

$17$ is equivalent to $3$ mod 7.

it gives you the same number as taking the limit of a sequence but it's easier to work with. you can think about the process of raising $5$ to $\pi$ to be in terms of a limit or in terms of $\exp(\pi \ln 5)$, both intuitions are correcrt

following so far?

yep

now let $z,w$ be complex numbers

we would like to say that $z^w = \exp(w \log z)$, as long as $z \ne 0$, and that's what we do

but unlike the natural logarithm $\ln$ on positive real numbers, the complex logarithm on complex numbers isn't a function because it's multivalued

$e^{2 \pi i} = e^0 = 1$ but $2\pi i \ne 0$

but we can still write $i^i = \exp( i \log i)$ but the problem is that "equal" sign is ambiguous

you with me?

Nah, it's an equality of sets ;)

true! if you want to do it like ted said, you can write something like

so at periodic intervals there are places where we get two values for the complex logarithm

$\{i^i\} = \{\exp(i \log i)\}$ which is an equality of sets

No, most places you have infinitely many values, @geocalc

Ugh, don't do that, @GFauxPas.

okay I wont

better?

LOL, I was fine with just saying $i^i$ represented a set of values. No set symbol. But whatever ...

okay I wasn't sure what you meant :)

so in some contexts leaving it as ambiguous is good enough. In other contexts, you need a convention to pick a "best" value. This choice, called a branch of the multivalued function, is more of a human thing than something intrinsic to the nature of the universe

it turns out that any two logarithms of a (non-zero) number differ by an integer multiply of $\pi i$

$2\pi i$

sorry $2 \pi i$

oops

okay

as you already showed because of $e^0 = e^{2\pi i}$.

so you take the logarithm modulo $2\pi i$ and you narrow down the possibilities

in the sense that

like

$2 \equiv 4 \equiv 0 \equiv -2 \pmod 2$ but

But the original question expects the infinite list ...

So don't pick any branches.

I didn't know htere was an original question

I thoguht it was just "what does $z^w$ mean when $z$ and $w$ are complex"

@geocalc: What did you ask at the beginning of this?

Well, that's multivalued.

I suppose. I've seen places where the intent of the writer is "for some suitable choice of branch" and other places where it means "the set of values satisfying"

Is a complex number raised to another complex number...

My interpretation is always that it's the infinite set, unless there's a specific reference to a particular branch of logarithm.

...another complex number

okay, I'll go with that

it might be real or it might be complex

No, @geocalc. It's an infinite list.

Almost always.

but it - yes

so a complex number raised to another complex number is a multitude of values

the definition is $z^w = \exp( w \log z)$ where $\log z$ is an infinite list of values, but the difference between any two values in $\log z$ is an integer multiple of $2 \pi i$

interesting

you can write that as $i^i = e^{i \log i} = e^{-\pi/2 + n2\pi)}, n \in \mathbb Z$, for a particular example

Double check that, @GFauxPas. You lost an $i$.

oh youre right

I really like the function 1/ln(z)

but its not really a function because it is multivalued right

@GFauxPas: It's interesting because you get a bunch of real multiples of $i$.

No, still wrong.

oops lost the window of time where i could edity it

i was trying :(

$e^{-\pi/2 + 2n\pi}$

wait not done

No minus sign there.

NO.

there

It should be $e^{i\log i} = e^{i(\log 1 + i(\pi/2+2\pi i n))} = e^{i\pi/2} e^{-2\pi n} = ie^{-2\pi n}$.

So that set for all integers $n$ ...

but the principle value is $e^{-\pi /2}$ and I'm trying to get that as a possible value with yours

$n=0$, of course ...

No, that's not right.

Oh, hell, wait. I have it wrong, too.

math.stackexchange.com/questions/191572/… I'm looking at this

It's too hard doing this typing, especially while I'm busy elsewhere too.

You're right. I messed up.

we all make mistakes. double triple quadruple checking, the answer is $e^{- \pi/2 + 2 \pi n}$ for $n \in \mathbb Z$

:)

$e^{i(i(\pi/2+2\pi n))} = e^{-\pi/2 -2\pi n} = e^{-\pi/2}e^{-2\pi n}$.

Yeah, that's right.

@geocalc33 it's not a function unless you choose a suitable convention to make it a function

it's a theorem that there is no way to choose a convention to make $\ln z$ continuous everywhere, even if you exclude $z = 0$

sadface

so what is a complex number raised to 1/ln(z) where z is a complex number

choosing a suitable branch cut

well I've given you enough information that we can do this together

Do you mean $z^{1/\ln (z)}$?

$z^w = \exp( w \log z)$, so....

That's sorta cool.

let's figure this out

it equals e

Hi, @EricSilva, slob.

yoyoyo

by the way the convention is to use the word $\log$ instead of $\ln$ when dealing with the multi-valued logarithm defined on complex numbers

how goes it

:) About to disappear, Eric.

why is it cool ted?

sad

Sad, why, Eric?

so if $z^w = \exp(w \log z)$ then ... help me out geocalc33 I don't know the answer

I mean I know how to do it but i havent done it

cause I just appeared!

Oh.

z^w

$z^{1/\log z}$

yea that's 2.718

$\exp(\log z \frac 1 {\log z}) = \exp(1)$ as long as both $\log$ expressions are choosen to have the same value at the same time, which is a reasonable expectation

:)

what if we didn't insist that the two $\log$ expressions have to have the same value at the same time?

what's $\dfrac{\log_1(z)}{\log_2(z)}$ where $\log_1 \ne \log_2$?

you are using a lot of symbols

do you have mathjax turned out, the code that makes things look like formulas

no

where can i turn it on

instructions are in jthe upper right corner

LaTeX in chat

I'm saying that $\exp\left({\dfrac{\log z}{\log z}}\right) = e^1$ if we insist that you choose the values of $\log$ in the numerator and the denominator to be same, but I'm investigating what happens if we allow the numerator and denominator to be different.

okay

great :)

so let's consider $\exp \dfrac {\log_1 z}{\log_2 z}$ where you choose different values for the $\log$ in the numerator and the denominator. Remember that any logarithm value differs from the other values by an integer multiple of $2 \pi i$

okay

scratch that

so write the numerator as $r + 2\pi in$ and the denominator as $r + 2\pi i m$ but maybe $m \ne n$

you get $\exp \left({\dfrac {r + 2\pi i n}{r + 2\pi i m}}\right)$

not so pretty

don't think you can simply it

hope that explanation helped you

yes thank you

come back any time

I have another question

sure

so if you have a function that spits out a complex number and you raise that complex number to 1/log(z)

okay

how would you solve for the roots

basically im just asking what

well you set up the equation $f(z) = 0$ but in general it's hard to get solutions to that

it's more complicated for complex analysis

?

but the exponential function $e^z$ has no roots

$\exp(-|z|) = 0$ if you consider $z = \infty$ to be allowed, but otherwise $e^z$ has no roots

but for example

say you have a complex function that does have roots, and you want to transform those roots by raising that complex function to 1/log(z)

does that seem reasonable

$\log 0$ is not defined

does that help

so raising $z \mapsto z^{1/\log z}$ will give you interesting things at the roots of $z$

what interesting things?

well actually

we saw before that $z^{1/\log z} = e^1$ if you insist that the $\log$ takes the same value in th enumerator and denominator, which is a reasonable assumption

so then it would just be a constant map and you'd lose all information

:(

:(

but you can take functions and raise them to the power of other functions such that the new function has discontinuities

but i don't want to map it to z^(1/log z)

say you have a function $f(z)$ and you sent it to $\csc{f(z)}$

oh, then I dont understand your question sorry

(function with zeros)^(1/log Z)

oh, $f(z) \mapsto f(z)^{1 / \log z}$?

I'm wondering what happens to the zeros

when you apply the exponent

like that?

yeah i suppose

$f(z)^{1/\log z} = \exp\left({\frac{\log f(z)}{\log z}}\right)$

gross

okay

yeah it's not pretty

then if $f(z) = 0$ and $\log z \ne 0$ that won't make a lot of sense

so it would basically be undefined at places where $f(z) = 0$

why not

oh

because $\log 0$ is undefined

an interesting question would be

would it be defined on a neighborhood around any of the roots of $f(z)$

and what kind of discontinuities are they

I'm really interested in where the zeros of f(z) end up after applying the exponent

like where do they move to

well as I said, they wouldn't be defined, so the question is, what happens nearby to those points

yeah

would that be a solid research question?

oh I'm not the one to ask questions like that to

I'm not good at that kind of advice sorry

I think i actually know how to get rid of the discontinuities

I consider $f(z) = \sin z$ and I asked Wolfram alpha what $f(z)^{1/\log z}$ does to the real axis

tinyurl.com/y8fjtjub

@LeakyNun Thanks for reminding me! Very helpful discussion

A very important question which has received little attention so far:

hi chat

Consider $L \subset \mathbb{P}^n$ a codimension 3 subspace. What's the meaning of "$L$ imposes $(n-2)$ conditions to hyperplane? I know it's a definition I could google, but I can't find anything satisfying :/

Are there planar graph layout algorithms which produce a "nice" layout on the surface of a sphere?

All algorithms I found that work for every planar graph tend to produce unreadable and ugly output. E.g. Schnyder's algorithm.

Tutte embeddings work for some planar graphs and often produce pleasing results, but sometimes squeeze the vertices together too much in the middle.

I am imagining something like the latter, but on the surface of a sphere, which should help avoid the squeezing.

We have created an essential singularity on 12 June 2018. This is going to be very interesting

12/6/2018 = 563 days

Hi, if I know that $A \cap B$ is open and also that $B$ is open, can I conclude that $A$ is open too?

Not if $A \subsetneq B$

okay... I'm trying to prove that an $n$-dimensional submanifold of $\Bbb R^n$ is an open set.

The weird set notation is only there because a $n$-dim submanifold is only a special case of the more general definition $k$-dimensional submanifold.

I tried to reason that since $V_p$ is open and $\phi^{-1}$ is continuous we have $\phi^{-1}(V_p) = U_p \cap M$ also open.

And from this I wanted to conclude that $M$ is open.

Hello, guys. How is the tangent space to a point in the boundary of a manifold defined?

I don't think you have enough information to conclude $M$ has to be open. E.g. if M is the closed disk and $U_p$ is an open ball center at $p$ for example, then $U_p \cap M$ can be open

It is an excercise so it should be possible...

Quick question: $e_n$ does not converge weakly to $(0,0,...)$ in $\ell^1$, because otherwise we'd have $\lim_{n \to \infty} \sum_{k=1}^\infty \delta_{nk} b_k = 0$ for every $(b_k){k=1}^\infty \in \ell^\infty$, which amounts to $\lim_{n \to \infty} b_n = 0$ for every $(b_k)_{k=1}^\infty \in \ell^\infty$, which is certainly not true.

Does this sound right?

A $2\times 2$ matrix with real entries can't have eigenvalues $w$ and $w^2$, where $w$ and $w^2$ are cube roots of unity other than 1. Is this reason correct: characteristic polynomial of that matrix has degree 2, but $w$ and $w^2$ can't be roots to a real-coefficient polynomial with degree less than 3. @0celo7

@philmcole you can just repeat this argument for every $p$ in $M$. Then you have $M = \bigcup_p U_p\cap M$ which is open in $\mathbb{R}^n$.

Thanks @loch !

@Silent looks right

Suppose it's infinite instead and argue by contradiction

ok thx will try

here's a very stupid question - what if you just take an infinite number of (isolated) points in $\mathbb{R}^n$?

Ah wait I missed that, are you sure it's not a compact submanifold?

The excercise was only the direction finite set $\implies 0$-dim. submanifold. I kind of optimistically assumed that I could prove the converse too like for the "$n$-dim. submanifold iff open" proof

No you can't. What about a compact 0-dim submanifold though?

What's the definition of a compact submanifold?

just the additional premise that $M \subseteq \Bbb R^n$ is also compact?

Also the case $\phi_p(U_p \cap M) = \{\}$ can't happen right? Since $0 \neq U_p \cap M = \phi_p^{-1}(\phi_p(U_p \cap M))$ and if $\phi_p(U_p \cap M) = \{\}$ so would be $U_p \cap M$.

Context is homology. Hatcher uses the term $\Delta$-complex without saying how to pronounce it. I'm guessing "simplicial complex" but it could be "Delta-complex"?

"Triangle complex"?

Delta complex. Some discussion from Hatcher himself here: mathoverflow.net/a/6302/55904

Thanks

Well that's a decent source :P

lolyes

google-fu to the rescue

Can I get a hint for the proof of the above?

I don't see how I involve the finite/infinite assumption in this

or the compactness