Would $\sum_{i,j=1}^{3} a_{ij}x_ib_{kj}$ be an 3x3 matrix of a multiplied by a 3x1 matrix of x mltiplied by a nx3 matrix of b?

no

it's easier to see what's going on in that example if you reorder things a bit:

$\sum_{i,j=1}^3 x_i a_{ij}b_{kj}$

So from that we first off note that $i$ is the first index, i.e. the row index of $A$

@MatheinBoulomenos, may I ask you if you've ever study Macaulay's inverse systems? It's just curiosity, 'cause my thesis advisor suggested me to grasp this stuff and I was wondering if you know something

that will mean that $x$ is multiplying $A$ from the left, i.e. $xA$ with $x$ being a 1x3 row vector and $A$ a 3-by-3 matrix

next, we note that the second summation index $j$ is the second index of both $a_{ij}$ and $b_{kj}$

But that's not how summation for matrix multiplication works: one should be a column index, the other a row index

to fix that, note that $B^T$ has the same matrix elements as $B$ except with row and column index swapped

so $(B^T){jk}=b_{kj}$

hence that summation can be regarded as $\sum_j a_{ij}b_{kj} = \sum_j (A)_{ij}(B^T)_{jk}=(AB^T)_{ik}$

and combining that with $x$ on the left gives $xAB^T$

that, at least, would be the interpretation of that summation as elements of a row vector

Anyone have a book to recommend on functional analysis?

if you transpose the entire thing, you get $B A^T x^T$ which is a column vector with the same elements

Either interpretation is valid.

Thanks @Semiclassical

@GFauxPas for which level? our teacher suggested R. Bathia's and Brezis' books

I've learned two courses on linear algebra and two on analysis in grad school

First year of grad school

@Semiclassical waht about the case with $x_i a_{ij}$

Is i not the row element for both x and A?

Or is it that you can regard it as being a 1 before the i

so theres only 1 ro but 3 columns

@JakeRose either/or, really

If you reard them both as being rows then is the multiplication really defined?

Are those books good for my level?

the point is more that it can be read as either

there's surely someone who can give you more precise references: those books are pretty classical on functional analysis, but unfortunately I only read some chapters from Bathia's (but I heard Brezis is qite difficult)

in the present context, you interpret $i$ as the column index of $x$ and therefore have $x$ as a row vector

Mhmm a lot of just conventional type stuff in this

but you could equally well write the above as $a_{ij}x_i = (A^T)_{ji}x_i = (A^T x)_j$

in which case you'd read $\sum_i x_i a_{ij}$ as the $j$th element of the product $A^T x$ with $x$ as a column vector

Hm thanks

it's probably smart at this point to pick a convention such as "all vectors are assumed to be column vectors"

since otherwise it gets sorta irritating

at some level you do just have to deal with it, though. i mean, $\sum_i x_i y_i$ can be understood as either $x^T y$ or $y^T x$ for column vectors $x,y$

Just write $\langle x,y \rangle$ if you're afraid of commitment ;)

lol

and $\langle x | y \rangle$ if you're committed to being a heathen physicist :P

@Semiclassical just stop

One of those days I have to teach myself that notation

These

By all vectors being column vectors you mean the index would be the row part

right

Okay cool

And that would mean i have to transpose x

as you did

YEAH GOT IT

gonna stick with that convention I thik

you can still have $x_i$ with $i$ as a column index, of course. It's just that in that case it's the column index of $x^T$

@GFauxPas I found one today, my professor last quarter saw it and was like, shit I wish I knew this book

@0celo7 lol

Functional Analysis, Spectral Theory, and Applications

i'm tempted to put Dirac notation in my presentation now

(I won't, it's not helpful, but still)

Whats dirac notation?

basically, physicists denote vectors in a Hilbert space as $|\psi\rangle$

and co-vectors as $\langle \psi|$

Thanks Dam :D

that's basically the Hilbert space equivalent of "row vectors act on column vectors from the left"

and the action of a covector $\langle \psi|$ on a vector $|\phi\rangle$ is the inner product $\langle \psi |\phi\rangle$

...except we call the first one a 'bra' and the second one a 'ket', which together form a 'bra-ket' (bracket)

Einseidler and ward?

@konoa I haven't studied them

so a bra-ket product is an inner product, whereas a ket-bra is an outer product

and something like bra-bra only makes sense if you interpret it as a Kronecker product

What's that

@MatheinBoulomenos ahahahahah no problem, just to know

Nm looked it up

wikipedia: "In mathematics, the Kronecker product, denoted by ⊗, is an operation on two matrices of arbitrary size resulting in a block matrix. It is a generalization of the outer product (which is denoted by the same symbol) from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis."

physicists would just say "tensor product" but they mean it in that kind of way

Never really saw the point of bra-ket tbh. I can work with the canonical isomorphism from a Hilbert space to its dual and evaluate a functional at a vector without that just fine. (But I'm also not doing physics)

I feel two ways about it

on the one hand, I feel like it makes mysterious something which is, as you say, rather simple

on the other hand, it is nice to be able to write something like $|\uparrow \uparrow\rangle$ and not have lots of annoying stuff in the subscripts

I have no problem with calling the Kronecker product "tensor product" as a mathematician. That's exactly what the tensor product functor does on morphisms

Semi does this notation only work in spaces where $V^{**} \equiv V$

pretty much, yeah

@Semiclassical Thats cool, what year do you think Id do tha tin a degree?

It's often true

What's the name of that property

depends. the place you'd be likely to see it is QM

but depending on the level of the course, you may only see it mentioned

I'd say you also need some nondegenerate bilinear form or equivalently an isomorphism from V to its dual

a grad school course would definitely use it

@GFauxPas reflexive

Thx

Im doing physics at uni now

First year

I'll note here that, if you go to ncatlab, you find stuff like this:

"Large parts of quantum mechanics and quantum computation are naturally formulated as the theory of †-categories that are also compact closed categories in a compatible way – dagger compact categories."

I heard physicists say that it's even useful for math, but I really don't see any benefit

@MatheinBoulomenos i think the most generous reading of that is how bra-ket notation is a prototype for diagrammatic representations of such calculations

which in turn gets into category theory stuff

I believe the physicists that it's nice for physics

but, uh, I don't do category theory

But you can work with adjoints without bra-ket

yeah

I know we did in our LA course

i mean, as a visual way of keeping track of a vector and its adjoint, fine

I am a big hater of bra-ket because it leads to many immoral abuses

lol

it should also be noted that physicists tend to use it glibly regardless of whether you're doing a finite-dimensional Hilbert space or an infinite-dimensional one

in the former case, there's no issue

do physicists ever specify which Hilbert space they work in?

looooooooooooooooooooooool

they're like, "there's more than one?"

Not explicitly, no.

eh, that's not fair

a physicist knows that the basis you use for a particle in a box depends on the boundary conditiosn

Oh please, 95% of physicists couldn't define what a Hilbert space is

Lol

The percentage is probably higher in Europe

Physicists here take at least 3 rigorous math courses

complete inner product space? I forget the precise definition myself.

You're one of the 5%

meh

L2 is the only Hilbert space for L.p., right?

If the average physicist can write down what a Cauchy sequence is, I'd be shocked honestly

I'd also point out that physicists are really concerned not with vectors in Hilbert space so much as rays in Hilbert space

Here's where Mathein says that's the projective Hilbert space

well, sure

Lol

Sure

I think the main thing is that when a physicist hears the phrase "hilbert space" they don't think of the definition of hilbert space

they think of the examples of such which actually occur in their work

I can't say what they think, I'm not a physicist

Anyone here try Spivaks Physics for Mathematicians

But can the average physicist define L^2...probably not.

L^2 the norm or L^2 the space

The obstacle is of course saying what the measure is

the L^2 norm they do all the time. if you do a QM, you do use the phrase 'square-integrable' a lot

Besides finite-dimensional hilbert spaces, the main go-to example for a physicist are various examples of Fourier analysis

In any case, the person who asked...the Hilbert space is usually some $\Bbb C^N$ or $L^2(\Bbb R^n)$, or tensor products of these

One physicist told me that he encountered group cohomology in some advanced QM because of something something phases don't matter you actually care about projective representations, but linear representations are easier to work with, so ask about lifting projective representations yo linear ones. That was actually the historic motivation for group cohomology, predating homological algebra

Or $L^2$ on a cube or whatever

@MatheinBoulomenos I know there's physics people who care about what K-theory is

doesn't mean most of them do, though

That's the 5%. I wonder how accurate that figure is

I think the fraction of physics people who have heard of K-theory is about 5%.

I think people underestimate how many physicists are doing materials science or experimental stuff.

The fraction of people who actually know about it is roughly 5% of that, I wager

Only in Germany would those people be subjected to k theory

lol

main thing i know about K-theory is that it's something to do with classifying clifford algebras

I know a physics undergrad who skipped physics lectures si he could attend slgebraic number theory

Look, Mathein, we've established that Germans are crazy

You all may have a warped impression from Germany based on taking me as representative

somehow K-theory is related to random matrix theory

i was born and raised there

Germans are 1% of the world population by my calculations

I mean in terms of university education

So there's 4% of physicists who are not German who have heard of Cauchy sequences

I am assuming that a.e. German physicist knows functional analysis...seems fair

eh. i think that underestimates the number of physicists who also did a math degree in undergrad

if you control for physicists who only took physics, then I find that number more plausible

That's not accurate. They will know measure theory and a bit of Lp spaces, but not taken a whole course on functional analysis with the major theorems

Who only took physics? Have you actually heard that term in a real physics class?

not really, no

but I mean people who, while they may have had calculus and stats, never went further towards a math degree

If you do a math degree, you probably have had to take real analysis, and therefore will have seen Cauchy sequences

So in that respect I think that the fraction of physicists who have seen them is higher than 5%.

@Semiclassical Hopefully...

(what fraction of them remember what a Cauchy sequence is, on the other hand...)

@Semiclassical Ah, but the original claim was who can actually write it out correctly.

For Hilbert spaces, yes

oh, hmm

This is what we should be polling, not presidential approval ratings.

lol

One funny story a friend of mine told me from when he was teaching intro PDE (proof based, but no prior knowledge of functional assumed) he asked the students if they're familiar with convolution. The guys who took the regular analysis sequence saw them for R^n. The prof who taught the alternative "math for physics" sequence apparently did convolutions wrt to a general Haar measure, but the only thing one student could remember was "something with groups"

I don't mean teaching, I mean TAing

A Cauchy sequence is, obviously, a sequence that converges in a Hilbert space

@GFauxPas A topological space is a space of open sets

^

@MatheinBoulomenos What is with German math professors doing things in the most difficult possible setting for no reason

also "wrt to" hehe

The prof who did that was trained as a physicist

That way you only do a proof once

@MatheinBoulomenos Ok, all Germans then

Hi Demonark

Why don't you just define Cauchy nets in uniform spaces? Don't want to do the same thing again for topological vector spaces

Though I think representation theory is supposed to be really useful in quantum for some reason, so it wouldn't surprise me that someone takes the opportunity to do that kinda business

It's excessive but... It's not a surprise

@TedShifrin hello Ted!

Having a Haar measure can also be useful if you want to do averaging arguments on a compact (say Lie) group

Hello @Ted

I know what a Haar measure does, but clearly the guy didn't get anything across if the students remembered "something with groups"

tbf that was only one student

Hi Mathein. I only know about Haar measure for Lie groups (never thought about topological groups) and so I have much easier ways of having volume forms on manifolds.

Haar = integration measure for a Lie group?

for a topological group ... invariant measure/volume-form.

I’ve run into it in the context of coherent state quantization

@Semiclassical left invariant measure

Mmkay

Invariance is obviously crucial if one's on a group (or has a group acting effectively).

You can do cool things with that in algebra

Are Lie groups necessarily orientable?

Also important in geometry.

Yes, Demonark. Why?

I was wondering about your mention of the volume form

No, my "why?" was for you to explain why it should be to me.

E.g. you can recover the norm of a local field from its topology

The argument is really neat and also works for R and C

Ted there was a fun question on my final btu I don't know if it's right and the professor is going to France for a few months so I can't ask him :(, it was this

@Mathein: Other than having espied them in Kelley's Topology Book, I've never in my mathematical life encountered a uniform space in that generality.

@GFauxPas: I'd like to go to France, too.

let $u \in L^1(\mathbb R)$. Prove $\lim_{n \to +\infty} \int \sin(nx)u(x) d\lambda = 0$

Oh, the Riemann-Lebesgue lemma.

Hmm, so I know S^3 is parallelizable and this has to do with its being unit quaternions

That tells you that Fourier coefficients have to die off.

Demonark, it's true for all orientable 3-manifolds, but you're right.

So maybe that's a thing in Lie groups, I'll think about this

chat.stackexchange.com/transcript/message/44554098#44554098 chat.stackexchange.com/transcript/message/44554831#44554831 chat.stackexchange.com/transcript/message/44554786#44554786 @TedShifrin Do you know if there's some less messy way of doing this?

Yes. It is.

Think back to basic stuff you learned in Neves's course, Demonark.

we didn't do Fourier coefficients in the class, he skipped that so hed have more time for sighned measures :/

@Ted Me neither. Apparently a lot of results on topological groups carry over, so that's neat (topological groups come up a lot in what I do)

Well, each of the Fourier (sine) coefficients is, up to an overall constant, just the $nth such integral

I've never seen a question like that @B.Mehta, so I have nothing to contribute. You obviously have to play off finite interval versus infinite interval.

The integrand is dominated by $u$ so I wanted to use the DCT

But there's no pointwise limit, @GFauxPas.

@Semiclassical physicist proof: the fast oscillations kill the integral

Hey everyone!

So “the Fourier sine coefficients go to zero as n->infty” is literally what that means

Hey @TedShifrin, @Daminark

Hi @Perturbative.

@TedShifrin Okay thanks!

Yes that is a problem Ted :/

Hey @Perturbative!

What happens if you approximate $L^1$ by $C^1$?

Then I can IBP

?

That's certainly where the intuition comes from.

Yes.

K let me try that

Hmm so the natural thing to do is to say okay, map $G\times T_eG \to TG$ by pushing forward using left multiplication

Right, Demonark.

So the tangent bundle is trivial.

I should go through the details of this somewhat (we never really talked about Lie groups) but from here, the task would be to show that manifolds with trivial tangent bundle are orientable

@Daminark Now construct a left-invariant volume form.

Demonark: More generally, any trivial vector bundle is orientable.

$\int \sin nx \tilde{u} = -\cos(nx)/n \tilde{u}(x) + \int \frac{\cos nx}n \tilde{u}'(x)$?

But you guys didn't talk about vector bundles in general. (Nor did I in the G&P course.)

Not at a ,computer so idk if the mathjax rendered

@Daminark First show how to construct a left-invariant Riemannian metric on $G$.

He doesn't know all this stuff, 0celo.

A vector bundle is orientable iff the top exterior power is trivial

I think you need a Riemannian metric in here somewhere.

Yeah I'm not all too familiar with Riemannian business, my idea here was to try to take a nowhere vanishing vector bundle and work with that

No, 0celo, you don't.

Huh? Demonark

Nowhere vanishing what?

Field, sorry

You mean a global frame of the vb.

No, no, you need triviality, not a single nowhere zero section (that just gives you a trivial line subbundle).

Then take a Riemannian metric and make the volume form. Dunno what could be easier.

That doesn't help understanding general vector bundles, however.

You just said he doesn't know that

But he doesn't know what you're saying, either.

Okay Ted its definitely true is $u$ is C1. And that's it, by a density argument?

Trying Euler characteristic was probably misguided, if it's a trivial bundle then just choosing a basis for the fiber and copying it over should just do it

@Daminark So you have a frame $e_1,\dotsc, e_n$ of $TG$. Now assume you can find a coframe (do you know what that is?) $\theta^1,\dotsc, \theta^n$ such that $\theta^i(e_j)=\delta^i_j$. Then the volume form is $\omega=\theta^1\wedge\cdots\wedge\theta^n$

To construct $\theta^i$ you can use a Riemannian metric

Pretty much, @GFauxPas. I'm a bit worried about the density, though. Is that right?

@Daminark Now, if you choose $e_i$ to be left-invariant as you did above, and additionally take the metric to be left-invariant (which can be done in a similar manner), then this gives you a Haar measure

I feel like this approximation only works for compactly supported functions

@GFauxPas C^1_c is dense in L^1, so...

Oh, then okay

But Ted is telling me to think about a possible flaw in the argument

Hm

We wants you to justify taking the limit here

No, Ted is rusty. I was thinking of Lusin's Theorem and not seeing how to get around the set of small, but positive measure.

There's two limits, right? $n\to\infty$ and $j\to\infty$ in $u_j\to u$

So just make sure that everything works out

But, yes, then you have to make sure that you can get things OK on all of $\Bbb R$ ... If you have $L^1$ close, then the fact that $|\sin nx|\le 1$ should do it.

But, @GFauxPas, that's the key idea (and using ibp, which you said instantaneously) you were missing.

@0celo7 not sure, does frame = basis and coframe = dual basis?

Suppose that $x, y, z$ and $x', y', z'$ are all real numbers and $(x-x')^2 + (y-y')^2 + (z-z')^2 < 1$ and I know that $x^2 + y^2 < 1$ is there any quick way that I can show that $(x')^2 + (y')^2 < 1$? All I can conclude is that $-2(xx' + yy' +zz') + x^2 + y^2 < 1$

Can't you just take a chart on U containing the identity element and translate that to make an atlas? The transition maps between these charts are all trivial unless I messed up somehow

Right, thanks Ted

Well I mean, smoothly varying

I don't think it's quite right, @Perturbative, but it's close.

Think about the first condition geometrically (and forget about $z,z'$).

@Ted also what does orientation entail for bundles in general? The only things I know about orientability involve smoothness in some form

Well, remember that you learned what an orientation on a vector space is.

@TedShifrin Geometrically this condition gives an "open" cylinder in $\mathbb{R}^3$

@Perturbative: Just stick to $\Bbb R^2$. If $(x',y')$ is within distance $1$ of $(x,y)$, what can you conclude?

Demonark: For vector bundles, you want that orientation to be consistent on (sufficiently small) open sets (just like triviality).

You can do stuff with exterior powers, to be fancy.

@TedShifrin Well by definition we'd get $(x-x')^2 + (y-y')^2 < 1$

Alright, and what that should mean is that if you take a small enough open set around a point where $\pi^{-1}(U) \cong U\times \mathbb{R}^n$, that isomorphism should be orientation-preserving on the fibers?

Which is what you started with, @Perturbative. But do it by thinking about distances, not by algebra.

Right, Demonark.

Oh well I could also say that $(x', y') \in B((x, y), 1)$

I just heard this result: If you choose a matrix randomly (meaning via the Haar probability measure) from SU(2), and it has eigenvalues a and b, the average of $a^k+b^k$ will be 0 for all $k\geq1$, EXCEPT for k=2; then the average is -1.

Okay that makes sense, I think. I should probably at some point revisit manifolds but more formally, along with perhaps learning about vector bundles in general

I'm sure you will at some point, Demonark.

anyone have any insight, it sounds unbelievable to me

But you see that if you have a global trivialization, you have a global orientation.

Yeah this definitely makes sense now

Cool, Demonark.

@TedShifrin Those are the two things I can conclude I believe

@B.Mehta: Well, you know that the eigenvalues are complex conjugate and have norm 1.

Sure

I should say I'm not looking for an actual proof; I'm pretty sure it's beyond me - just some intuition for why that should make sense

or rather, why it doesn't not make sense

Right @Perturbative, so that says $x'^2+y'^2<2$.

I've heard something like this before, @B.Mehta, and I know someone to ask (tomorrow), but I don't think I believe it quite like that.

Much appreciated!

Ohhhhh (mini mind blown as that fact blew right over head)

Thanks @TedShifrin :)

Yup.

Isn't Ted great to have around helping us? :D

thank you Ted

in general

tiptoes out of the room

NO COME BACK

thank you Ted

Thank you Ted