@Danu I am replying to you. You wondered about a question that I asked @0celo7. Yes, I asked as if I have asked an issue of language, any mature person.
@0celo7 Don't you see that the "bicomplex" $\Omega^{\ast,\ast}$ there is precisely the $\Lambda^{p,q}$, and its elements are all written with the wedge. I suspect writing the symbol joining one complex to the other as a wedge is convention.
That's good. I hate extra work for mods. I was really amazed actually yesterday...I was in one of the CS chat rooms for the first time. and I saw people chatting about whether to protect a question. And I didn't realize people actually posted crap answers. That's so stupid. But they were discussing how it was going on and it just baffles me people have the time to post useless stuff.
@StanShunpike You don't see discussions about protection here because almost everytime I think oh, that should probably be protected, I find that Qmechanic beat me to it ;)
Seeing an interesting title (and post) via the Hot Network Question list leads to plain bad answers
I remember there was someone who asked virtually the same question as a HNQ but with one minor tweak. I VTC'd as a duplicate and the guy got a little offended
It's true. He asked me to provide page numbers in my answer.
@KyleKanos Really? I'm amazed making the HNQ actually makes a difference. I guess though there are a lot of SEs so I don't really appreciate how many people that actually reaches
Wow, thanks for nothing, Physics stack exchange ...
I posted this question on Mathematica stack exchange first because I didn't even realize there was a Physics one. And I didn't have my hopes up because it was kind of off topic there even more so than it could've ever possibly been here, but at ...
lol mobile, ah the disadvantages. but i'm not complaining, i learn a ton. it works great. i actually only got a smart phone 6 months ago and I'm really loving having internet access wherever i go.
@ACuriousMind not really, but I'm trying to learn about them. Wikipedia defines them using free vector spaces and everyone I talked to said that wasnt needed
Ted Shifrin said I should just use matrices in the beginning but I think if I could learn tensor products it would just be more efficient.
I don't know what an ideal is either.
Tensor products and ideals seem like the key ingredients for defining a wedge product
@StanShunpike Understand the tensor product first, then, the wedge is defined out of it. As a slogan, $\wedge$ is the antisymmetrization of $\otimes$.
And, unsurprisingly, I would disagre with the people you talked to, the proper definition of the tensor product is as the free space on a certain basis.
@StanShunpike Now, you can take $k$ $1$-tensors $t_i$ and build a $k$-tensor $t$ out of them as $t(v_1,v_2,\dots,v_k) := t_1(v_1)\cdot t_2(v_2)\cdot\dots\cdot t_k(v_k)$.
@0celo7 I'm not doing your weird complex geometry here :P
@StanShunpike: Now, you have to observe that a) the $1$-tensors are just the dual space of $V$, since they're linear maps $V\to\mathbb{R}$, which is the definition of the dual, and so they have a basis $t_i$ and a vector space and that b) $k$-tensors are also a vector space (in the sense that they satisfy the vector space axioms)
@StanShunpike Good. Now, let's denote the $2$-tensor I can construct out of pairs of $1$-tensors $t_i,t_j$ as $t_{ij}(v_1,v_2) := (t_i\otimes t_j)(v_1,v_2) := t_i(v_1)\cdot t_j(v_2)$.
One can show that all the possible combinations $t_i\otimes t_j$ form a basis of the space of $2$-tensors.
I think so. we started off by constructing $k$-tensors out of $1$-tensors. Now we are saying that this $2$-tensor is simply a construction out of these. Does this mean it forms a vector space?
@StanShunpike This $2$-tensor is just a special case of the earlier construction. You agreed that the $k$-tensors abstractly form a vector space. I now claim that the basis of the space of all $2$-tensors is given by these $t_i\otimes t_j$.
similarily, all possible combinations $t_i\otimes t_j\otimes t_k$ are a basis for the space of $3$-tensors, and so on
So wait, for $g = -t \otimes t + x \otimes x + y \otimes y + z \otimes z$ that works then because we can just add those elements and they form a $2$-tensor?
And every $k$-tensor can be expressed as a sum over the basis $t_{i_1} \otimes \dots \otimes t_{i_k}$ made of the $k$-fold combinations of the basic $1$-tensors.
Haha ohhh @0celo7 consider, say, the sequence of partial sums of (1,2,1,3,1,4,1,5,1,...). ie the sequence 1,1+2,1+2+1,1+2+1+3,... then you can have arbitrarily large gaps. Lim sup gap size =infinity. But after every gap comes a gap of size 1! Nothing more to it than that.
Hmm, amazing. That's awesome. I'm going to have to study the tensor product part to appreciate the wedge part, but that's a really nice way to go about it. Yay! I have a book on exterior algebras coming so this is great. I will hopefully be able to get more out of it now.
@ACuriousMind My uncle gave me a book Topics in Algebra i think it's called
Rings baffle me frankly. I don't know what to do once a group loses it's abelian nature. In fact, I still don't get why rotations aren't abelian. I think that's what I remember someone saying
I think this stuff is fascinating, I just haven't gotten into it yet. I think I need to do some proofs. I haven't yet tried doing the problems in several books.
music was what got me interested in group theory.
Because I always come back to Do.
I always wanted to know
why certain notes are preferred
in certain combinations and if there were ways to describe that through group theory
Anyways, neat stuff
@ACuriousMind When did you first start learning about physics?
@StanShunpike I'm not sure. I soaked up what my father told me, but my parents are both chemists, so that was only rarely physics. My contact with physics is only through school, I think.
Probably 6th or 7th grade, but I was never someone who thought much about these things. It was good to know that there was knowledge out there, but it could wait.
@ACuriousMind So I'm still struggling to get a grasp on what the current working framework is for understanding blackholes. Is the Schwarzchild metric a tool used for defining blackholes?
I just mean
from what I have heard, GR fails to describe blackholes to some extent.
@StanShunpike It conceptually sounds like something that would interest me, but I think I've been deterred by a bad lecture on it, and now I'm so immersed in quantum theories that my interests don't go in that direction anymore.
@StanShunpike Yes. If there is no path to null infinity inside of the horizon, then light from inside the black hole can't join light from the outside off in the distance.
user54412
I'd argue that escape velocity is a misleading way to think about things. In particular, in a Newtonian universe with an arbitrary speed limit of c, you could still escape from a black hole.
user54412
After all, with continuous thrust, you can escape at any velocity
@0celo7 Oh, I'm probably "advanced", but not a specialist. I'll call myself a specialist when I understand everything in, for example, here.
user54412
Moreover, even without continued thrust, launching yourself at less than the escape velocity moves you some distance off the surface, which is decidedly not the case with GR and black holes
My English teacher at Caltech said she was the first person to take Feynman to Huntington Library where they keep a copy of Newton's principia. She said Feynman took the Principa and started reading it. He came to a page where he saw some problem. He thought through the proof and turned the page, only to find Newton had done the same proof.
@NeuroFuzzy Uh...they're things you can divide out of rings, since they're the kernels of ring morphisms, and it often happens that you want to consider the equalities in some algebra (which is essentially a ring with additional structure) "up to" something, and the way to make that precise is to divide out the ideal these somethings lie in...I don't have a physical explanation of what they are, and I don't know whether there is one.
@ACuriousMind What does the phrase "up to" mean in mathematics? I hear it all the time. Like "up to isomorphism" or something like that. The only association I make with it is "up to no good" lol
@StanShunpike The prime example is "clock arithmetics", i.e. $\mathbb{Z}/n\mathbb{Z}$. Take some number n, for example 8, and say that two numbers are equal if their difference is a multiple of 8, that is $4 = 44$ "up to $8\mathbb{Z}$".
"up to" means stuff differs only in respects you don't care about, generally
@NeuroFuzzy Yeah, that's a bit difficult to see, and you can live a happy physicists' life without ever knowing what an ideal or a ring is, I think.
You can't live a happy mathematical physicists' life, though ;)
@StanShunpike The precise topological definition of my earlier black hole definition is $B:=M-J^-(\mathcal{J}^+)$ where $J^-$ is the causal past and $\mathcal{J}^+$ is future null infinity.
I just had an interesting idea. Since there are points in our universe which appear to be moving away from us at a speed faster than that of light, does our universe satisfy the Newtonian definition of a black hole?
After all, to escape our universe in the Newtonian sense, we would have to go faster than light.
@ACuriousMind Does this sound sufficiently crackpot?
@ACuriousMind I just went back over the notes I took on tensor products. And then the definition of the wedge product you gave. Here's what I don't understand.
@ACuriousMind the tensor product was defined as $t_i,t_j$ as $t_{ij}(v_1,v_2) := (t_i\otimes t_j)(v_1,v_2) := t_i(v_1)\cdot t_j(v_2)$
But dot product is commutative
And the wedge product definition you gave was
$s\wedge t := s\otimes t - t\otimes s$
So if I let $a$ be the tensors and I write $s = a_i(v_1) \cdot a_j(v_2)$ and $t = a_k(v_3) \cdot a_l(v_4)$, then
Damn sorry let me rewrite that
What I'm trying to say is that
$s$ and $t$ are the tensors so we just write them as a dot product
So that would mean
$s\wedge t = s \cdot t - t \cdot s = 0$
@ACuriousMind there we go, I got it. What I just wrote. And that doesn't make any sense. I must be misunderstanding something conceptually
Forget the part involving $a$ that was a mistake.
user54412
7:17 AM
@StanShunpike I think you just need to keep track of all your products and what's acting on what.
user54412
start with $s \wedge t = s \otimes t - t \otimes s$
user54412
then $(s \wedge t)(u,v) = (s \otimes t)(u,v) - (t \otimes s)(u,v) = s(u) t(v) - t(u) s(v)$
user54412
In other words, there's no dot product between $s$ and $t$. Your "dot" product is just the multiplication in the base field, whatever it is. $s(u)$, $t(v)$, etc. are scalars in this field and so can be multiplied in this way.
@0celo7 Oh, you think $c$ is your ally, but you merely adopted $c$. I was born in it, molded by it. I didn’t see Newtonian mechanics until I was already a man.
@StanShunpike Dude, $s(u)\neq s(v)$, and the same holds for $t$
Why would it be zero?
Also hah caught up on chat
Mornings/early afternoon in EU is the perfect time for that---nice 'n' quiet.
@Danu dude, (1) you said dude a lot and (2) dude I didn't see that the inside of the vectors dude flipped dude. So dude no wonder its not zero dude. Dude, I think that like dude that answers the question dude.