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21:00
@Ted I'm aware.
Past the security checkpoint the airport looks like a mall.
the first time I saw the tensor product was in linear algebra but the prof just copied the construction for modules and then talked about kronecker products for a day so I didn't really see the point
it gives Vect(k) a symmetric monoidal structure, so obviously it's interesting
in geometry/manifolds, it shows up first in the dual setting most naturally, @Samuel
waits for smacks
@BalarkaSen why is that interesting
21:02
ignores Balarka and sends him to bed without his math
@TedShifrin the dual setting?
Hi
it was a joke on bourbaki-ish math, nevermind me.
i have been telling that one over and over, but it ever gets funnier.
Tensor products of $V^*$ show up more naturally than those of $V$, @Samuel.
21:03
E.g., inner products
but yeah the second time I saw tensors was in commutative algebra and the universal property confused the hell out of me for like, months
but I got it eventually
afternoon chat
right, but if you're used to thinking of bilinear linear maps as linear maps on the tensor product, it's not so crazy, @Samuel
hi @Semiclassic
"pullbacks are better than pushforwards"?
21:05
true but the general problem at that point was that I wasn't really used to anything so everything was hard
bilinear linear maps as linear maps...now that's multilinear
We learn about inner products and differential forms before we learn about multivectors, typically, @MikeM
I don't have any particular objection to that
@TedShifrin it'd be interesting to compare where those concepts emerge in the physicist line of math (supposing a lack of an explicit diff-geo course, to be sure)
and, with regard to your pullback comment, interesting how we can pull back forms but can't always push forward tangent vectors
21:07
though of course I'm fond of or Clifford algebras
I think physicists deal with tensors (aside from the moment of inertia) mostly in GR, @Semiclassic, but you should know way better than I.
How could I simplify $H_d(d,k)/k!$ if $$H_d(d,k) = (k-1)!\ \sum_{j=0}^{k-1} \frac{(-1)^j}{(k-1-j)!} \sum_d H_d(n/d, k-1-j)$$
I stupidly tried to replace $(k-1)!$ with $1/k$ but it's clearly not the way to do it...
that's definitely where it's developed the most deeply, yeah. riemann tensor etc.
stress-energy tensor
though when i hear 'tensor product' i actually think of quantum mechanics rather than GR
21:09
sure, that's because then it's representation theory
right, right
you do see tensors in places when you need to represent anisotropic responses---say, an electric field giving rise to an electric current which isn't parallel with the field
well, obviously, for $m = n = 1$, the map $k \otimes_k k \to k$ is $x \otimes y \mapsto xy$. the generalization is $k^m \otimes k^n \to k^{mn}$ by $(x_1, \cdots, x_n) \otimes (y_1, \cdots, y_m) \mapsto (x_1y_1, \cdots, x_ny_m)$
seems like it should be an isomorphism. shouldn't be hard to verify. let me see
yeah, well, duh. of course.
(Recall, @Balarka, I also said you should try to prove it as many ways as possible. :) But your 30 minutes are up and it's past your bedtime. You need to be awake to see mr. prof.
alternatively use the definition of $k^n$ and the fact that tensors commute with direct sums so you don't have to do this
i don't think tensor product commutes with direct product.
or am i just darn wrong?
21:17
er, you have the thing that looks like distributivity
oh, yeah. br.
been some time since i did these.
$(M \oplus N) \otimes S \cong (M \otimes S) \oplus (N \otimes S)$
I can't get it
right, right.
every argument I think of involves dividing the interval
into some pieces
21:19
Reread the hint I gave you, @Karim.
multivectors, though---not sure how to pin down where that'd emerge in physics
@TedShifrin yeah, i guess i should go and sleep
Gnight, mr @Balarka
that's what pseudovectors are, @Semiclassic
21:22
i thought you guys were big on geometric algebra/calculus/whatever it's called
with clifford algebras and whatnot
There's a guy, whose name I always forget, who tries to do everything with that on here, @MikeM ... Some of it is quite nice.
Muphrid
Right.
I've never seen the point in what he does
Sometimes he's pedagogically totally naive.
21:23
I know uses for Clifford algebras but they go above and beyond "well, it's like differential forms, but vectors!"
I haven't learned it well enough, @MikeM, but my impression is that it unifies things a bit, but with a ton of baggage.
@TedShifrin Hey man do you have any knowledge on how I can start this problem? Let n and k be integers such that n > k ≥ 0. Show that nCk + nCk + 1 = n + 1Ck + 1? I'm completely lost and I need somewhere to start.
@Deathslice: Are you sure you typed that right?
yes. How are you reading it?
Your left hand side has two of the same thing.
21:29
read it as n choose k
parentheses were invented for a reason
Ohhhhh
1+1*2 is not the same as (1+1)*2, Deathslice
Thanks, @anon.
Indeed anon
Alright I'll rewrite it
21:30
There's no way I would have guessed what it was supposed to say. I'm dumb.
Now that I know, you can do it by algebra, or you can do it by thinking.
just say nCk+nC(k+1)=(n+1)C(k+1). what's your definition of nCk?
the best definition is the number of ways to choose k things out of n. (i.e. k-subsets of {1,...,n})
<--- yields the floor to expert @anon :)
but sources can also define it using a formula involving factorials, or as coefficients appearing in expansions of binomials raised to powers (a la binomial theorem)
i mostly know about clifford algebras in connection with systems of interacting fermions
I concur with @anon. Or you could use the formula $\binom nk = \dfrac {n!}{k!(n-k)!}$.
21:35
Oh I see what you mean
Well know I can do it using simple algebra
@Semiclassical: can you say that in math words
But the number of ways to choose ... definition gives a much more satisfying argument.
@MikeM: When does your plane leave?
eh, probably not. it's not something with which i'm very well acquianted
three hours from now
oh ugh
21:37
meh, airport is a pretty little place
was that choice purposeful or the only flight you could get?
and i'll be able to eat here instead of the more expensive US airports
cheapest flight
i figured it would be painful but this is actually a pretty pleasant airport
I decided to take a 1-stop flight on American from ATL to SAN ... even though I hate that. But it wasn't worth $125 to get a nonstop.
@Semiclassical: the only place i know it showing up in physics is seiberg-witten theory, where you need a clifford multiplication to define the equations
but that probably speaks more to my taste and ignoramosity than it does to the applications of clifford algebras
They're all over Lawson/Michelsohn, @MikeM
21:41
which is why they're in SW theory
@TedShifrin: 1/2 of the SW eqns is $D^A\varphi = 0$, where $D^A$ is the Dirac operator assoc'd to a connection $A$ and $\varphi$ is a section of the bundle in question. the correct statement there is that you need a Clifford multiplication to define $D^A$, and of course L/M are big fans of Dirac operators
they show up in the study of topological insulators in physics, with kitaev providing a classification in terms of K-theory (real and complex). (i don't know that story so well, since there are other ways to formulate that same classification)
sure, @MikeM ... makes sense.
in the SW case you're working with a spin^c structure on M instead of a spin structure, but the story is similar
21:46
I've never really grokked spin^c
here's kitaev's paper that got the ball rolling on that: arxiv.org/abs/0901.2686
just another real vs complex thing... I think the real reason it's introduced is because you can't actually put a spin structure on every 4-manifold, but they all have spin^c structures
which even just from the abstract you can see the prominence of clifford algebra stuff
on every 4-manifold with $H_1(M;\mathbb Z_2) = 0$, I mean; if you have 2-torsion you can fail to have a spin-c structure IIRC
22:02
I'm feeling a migraine coming on... :(
Eating might help, @MikeM
that's the plan... I brought my excedrin migraine with me
 
1 hour later…
23:09
hm
@robjohn Hi. Are you familiar with epsilon-delta proofs? Let's say I want to prove $\displaystyle\lim_{x\to 0}e^x=1$. We have that if $|x|<\delta$ then $|e^x-1|<\varepsilon$. Manipulating this last inequality I got $\ln(-\varepsilon+1)<x<\ln(\varepsilon+1)$ but I'm stuck there.
@Cristopher So you need to find a range around $0$ for $x$. If you set $\delta=\log(1+\varepsilon)$, then if $|x-0|\lt\delta$, $|e^x-1|\lt\varepsilon$
@Cristopher Note that $\log(1+\varepsilon)\le-\log(1-\varepsilon)$
23:32
@robjohn Hm. if $\delta=-\ln(1-\varepsilon)$ then $\ln(1-\varepsilon)<x<-\ln(1-\varepsilon)$ ?
@Cristopher Well, yes, but more importantly, $\log(1-\varepsilon)\lt-\log(1+\varepsilon)\lt x\lt\log(1+\varepsilon)$
Oops, that's backwards.... you want the smaller range. I have fixed the preceeding.
Oh you changed the delta...
@Cristopher yes, you want the smaller of the two bounds, and $\log(1+\varepsilon)\le-\log(1-\varepsilon)$ since $\log(1-\varepsilon^2)\le0$.
23:50
@robjohn So $|x|<\ln(1+\varepsilon)$ is a smaller bound than $\ln(1-\varepsilon)<x<\ln(\varepsilon+1)$ ?
@Cristopher yes, since $\log(1-\varepsilon)\le-\log(1+\varepsilon)$ (smaller = more restrictive)
Pretty general question here, but does anyone know of interesting functions like the Gamma function that I could read more on?
@KhallilBenyattou look into "special function" theory
@robjohn Okay, I see it now. Thank you
Do you know of any good sources for that, @anon or am I a quick Google away from it?
23:56
@KhallilBenyattou find some keywords and branch out from there. hyperbolic trig functions, riemann zeta function, L-functions, modular forms, weierstrass p functions, elliptic functions, hypergeometric functions, etc.
(many of those concepts are probably beyond your current level, but the basics in others might be approachable)
Thanks, @anon! I'll do some digging! :-)

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