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12:00 AM
so it is monotonely decreasing from some point and bounded, from that follows the convergence
 
Any other options?
 
@anakhro epsilon criteria does not work so i need sandwhich or something
 
@T_01 what do you mean?
 
I have the convergence now, but how do i show that the limit is in fact zero?
 
Oh I see.
Yeah gut feeling is to sandwich it.
 
12:03 AM
No $Mq^k$ for any $M \in \mathbb{R}$ works for sandwhich
since $k$ unbounded
 
I want something with a chapter focused on series
If possible
 
@FuzzyPixelz Lang's book, as mentioned.
 
We're talking about "Real And Functional Analysis"?
 
No I was just thinking undergrad analysis because you never specified anything more.
33 pages of the basics.
 
Oh sweet
 
12:17 AM
It is undergrad stuff, though.
I don't know if it is at all what you want.
 
ok i have it... assume $L$ is the limit of this sequence. But then we can follow that $kq^k < L \Leftrightarrow q < \frac{1}{k^{\frac{1}{k}}} L^{\frac{1}{k}}$ holds from some k since both parts of the product converge to 1
 
@anakhro It is exactly what I want, I'm 2nd year engineering.. yeah
I can't hold a candle to math graduates
 
thanks for your help @anakhro
 
@T_01 I am glad you got something you are happy with.
 
this answer does not sound very convinced XD
 
12:24 AM
I am just working on my own homework and am not motivated enough to check yours.
So that's my disclaimer. ;)
 
yeah no problem
only a joke
 
12:39 AM
:)
 
 
5 hours later…
6:06 AM
Hey
$\ln(x)=\ln(1+x)-\ln(1+1/x)$
 
@AkivaWeinberger hi!
 
so if we do Taylor series
Hi!
$\ln(1+x)=x-\frac12x^2+\frac13x^3-\dotsb$
Substituting $1/x$ gives
$\ln(1+1/x)=\dotsb+\frac1{3x^3}-\frac1{2x^2}+\frac1x$
and subtracting gives
$\ln(x)=\dotsb+\frac1{-3}x^{-3}-\frac1{-2}x^{-2}+\frac1{-1}x^{-1}+\frac11x^1-\frac12x^2+\frac13x^3-\dotsb$
$\ln x=\sum_{n=1}^\infty\frac1n(x^n-x^{-n})$
When does this converge
Does it converge for $x=e^{iz}$, real $z$?
Oh wait this is the Fourier series of the sawtooth wave isn't it
 
 
7 hours later…
12:55 PM
 
 
2 hours later…
2:27 PM
@AkivaWeinberger Try x = i and look at the odd powers.
 
@AkivaWeinberger lol
@AkivaWeinberger How do you do your subtraction...? LHS isn't correct.
 
@TedShifrin your buddy lawson is here
 
Hi
how's it going
 
Quite well, thanks, what about you?
 
2:35 PM
good
 
I've bene looking at some semigroup theory for PDEs in the last few days
 
I'm at a conference
 
What's the topic?
 
differential geometry
@AlessandroCodenotti I thought you weren't interested in that aspect of semigroup theory
 
There's a course on that topic and I need some analysis credits so...
Is it obvious that if $f:[0,t]\to X$ for some Banach space $X$, with $f\in C^1$, has zero derivative then it must be constant?
 
2:45 PM
Good day
 
hi pal
 
Where is that leaky when you need him.
 
@AlessandroCodenotti sure
 
Thanks for the sanity check
 
@AlessandroCodenotti it follows easily from the fact that linear functionals separate points
 
3:00 PM
Hmmm I'm not seeing it
Ah wait
So the map $t\mapsto \langle F,f(t)\rangle$ is constant for all $F\in X^\ast$ because it is a real function with zero derivative
And now I can take $F$ to be the functional which is $1$ on $f(0)$ and zero on all $x$ not in the span of $f(0)$ I think
 
@Secret :O
 
also typo: That remark about intuitionist set theory should be at the finitely enumerable line
 
3:15 PM
@Secret ok
The Cantor ternary set being a compact, nowhere dense, totally disconnected, uncountable set of measure 0.
Existence of the Weierstrass' everywhere continuous, nowhere differentiable function.
Existence of space-filling curves.
Bijectivity between R and the interval (0,1).
The countability of the dense subset Q of the uncountable set R of reals.
Divergence of the series Sigma(1/n).
^They have only one thing in common
Guess?
 
They are all mathematical propositions
 
3:33 PM
In my ODE book, they work toward a real Fourier Series formula where the bounds of integration can be from $c$ to $c + 2L$, but in the section of complex Fourier Series, they only get as general as a formula integrating from $-L$ to $L$. Is this incidental, or can complex Fourier Series only be taken with $x = 0$ at the center of the interval for some reason?
 
3:45 PM
 
@Secret wooooooooooooooooooooooooooooooooooooooow
 
@AbhasKumarSinha What do you mean?
$\ln(1+x)-\ln(1+1/x)=\ln(x)$
 
@AkivaWeinberger my bad... didn't read correct... :)
 
@LeakyNun Ah - typo - I had the alternating series, but wrote the sigma notation wrong. (Luckily that works also.)
 
Hi @TedShifrin
 
3:57 PM
@Secret The "only subset of the same cardinality is itself" is basically the pigeonhole principle
You know, there's two objects I can think of that are "finite-like" and not finite
Finite-dimensional vector spaces, and compact topological sets
For the former, we have properties like "automorphisms are injective iff they're surjective"
which are incredibly useful in Galois theory
For the latter, the image of a compact set is compact
I feel like half the reason the Jordan Curve Theorem is provable at all (from our basic topological tools anyway) is because compact sets are so restricted in what they can do
 
Also proper maps are nice because it is like mapping points at infinity to points at infinity.
 
And many times I've seen proofs handwave stuff by saying "We can do blah, by compactness"
I'm not actually so familiar with proper maps
Looking it up, it's that the preimages of compact sets are compact
Trying to think of how to understand this. So all inclusión maps are proper
There's a proper map from $\Bbb R\to[0,1]$
Are there any proper maps from $\Bbb R^2\to[0,1]$?
Oh, I think I have it: the magnitude function (distance from origin) is proper from $\Bbb R^2\to\Bbb R$, and then we can compose that with the map $\Bbb R\to[0,1]$
Oh, nope: this doesn't work
 
> For the former, we have properties like "automorphisms are injective iff they're surjective"
> For the latter, the image of a compact set is compact
 
There's no proper map from $\Bbb R\to[0,1]$, I lied
whoops
and inclusion maps aren't always proper
 
I think the key thing that conveys a notion of "finiteness" is when some closure properties are obeyed and something is preserved. But then, you can actually invert the whole thing and said the same for "infiniteness"
 
4:06 PM
but there is a proper map from $\Bbb R^2\to\Bbb R$, yeah?
 
The compact sets in R^2 in the usual topology are always closed disk and unions of them?
 
@Secret Maybe a better way to phrase it to emphasize the finiteness is, "the image of a compact set is contained in a compact set"
@Secret The compact sets are closed bounded sets, so you can draw 1D curves and such
 
ah right, I forgot the curves
 
Hence the connection to JCT
 
Mapping from closed disks to closed line intervals should be proper
or better, curves to lines
 
4:10 PM
Proper maps are the correct maps to look at when you care about "topology at infinity" as in geometric group theory
 
$\Bbb R$ is a line and $\Bbb R\to\Bbb R:x\mapsto e^x$ isn't proper
Inverse image of $[-1,1]$ is $(-\infty,0]$
@AlessandroCodenotti Is it the same as expecting it to lift to a map on the compactifications?
One-point compactifications
 
No, not necessarily
The only maps $X\to\Bbb R$ that extend to the one point compactifications are "constant at infinity"
 
Example?
 
Like a map $\Bbb R\to\Bbb R$ extends to a map $S^1\to\Bbb R$ iff it has both limits at $\pm\infty$ and they agree
 
Extending to $S^1\to S^1$ you mean
What's an example of a proper map that doesn't do this
 
4:18 PM
No, $\Bbb S^1\to\Bbb R$ (I guess you could call it $S^1\to S^1$ by embedding the codomain $\Bbb R$ into $S^1$ if you want!)
 
I try again : there is an expert on game theory / nash balance applied in the game of poker(texas hold em)
 
In other news, both finite and infinite can be said to preserve different kind of invariance: Finiteness preserves some notions of "closure", "containment", "size"
Infiniteness preserves some notion of "induction", "recursion", "constancy"
 
1
Q: Game Theory Poker - Nash equilibrium: Choose between bet or raise and range of hands

Z1pp3dI would like to understand how we choose the best action and how we compose the range for it. I have some notion on zero-sum games and minmax strategies. And I know software that does this for texas holdem: I want to understand how they do it This little and simple article Example : -Opponen...

 
@AlessandroCodenotti Are the proper maps $\Bbb R\to\Bbb R$ the ones that extend to continuous maps $S^1\to S^1$?
 
@AkivaWeinberger $f(x)=x$
 
4:21 PM
Consider the Linear Transformation. $T:\mathbb R^m \to \mathbb R^n$. $T(x)=Ax=b$. In the question given that $x$ is unique. So, $T$ should be bijective. right?. When $m\ge n$, we can conclude that $T$ is bijective. For the uniqueness of $x$. $n\ge m$. right?
-2
Q: What could be the relation between $m$ and $n$?

Dbchatto67 Let $A$ be a $n \times m$ matrix and $b$ be a $n \times 1$ vector (with real entries). Suppose the equation $Ax=b, x \in \Bbb R^m$ admits a unique solution. Then we can conclude that $(\mathrm {a})$ $m \ge n$ $(\mathrm {b})$ $n \ge m$ $(\mathrm {c})$ $n=m$ $(\mathrm {d})$ $n

 
@AlessandroCodenotti That lifts to $S^1\to S^1$
Are there any that don't
 
@AkivaWeinberger What's $f(\infty)$?
 
$\infty$
I don't want $S^1\to\Bbb R$, is the point
I'm asking if proper maps are ones that extend to maps of the compactifications, plural
 
Maybe it is true that proper maps can be lifted (the converse is not)
 
It's not?
 
4:28 PM
Constant functions $\Bbb R\to\Bbb R$ are not proper, but you can surely lift them to a constant function $S^1\to S^1$
 
Ah
Thanks
Refinement: maybe proper maps $X\to Y$ are the one that lift to $X\cup\{infty}\to Y\cup{\infty}$ (the one-point compactifications) such that $\infty\mapsto\infty$
 
*$X\cup\{\infty\}\to Y\cup\{\infty\}$
 
If f(X) and k(X) be 2 linear functions in X, what is the nature of sin(f(X))+cos(k(X))?
Sorry for interfering abruptly
 
Probably not much you can say about it
though maybe those^ are relevant
 
4:38 PM
@AkivaWeinberger Umm I know these. And yet I am unable to come to a satisfactory conclusion using this list of formulae.
Oh sorry. Wrong question. So sorry.
 
@Akiva you're familiar with Lie brackets, yes?
 
My question is what would be the nature of Asin(f(X))+Bcos(k(X)).
 
I'm not entirely sure what sort of answer you want, but maybe try graphing some examples? Like go to desmos.com and graph 2sin(20x)+3sin(21x) and see what happens
 
Well I tried, and I gives some sine waves with additional peaks in top of each rise/descent
 
@AlessandroCodenotti Yeah
 
4:46 PM
I'm just trying to get a feel for coordinate computations with some simple examples
 
hmm... I wonder, if infinitesimal can be defined without referencing to infinity...
but that will mean there is something that is kept invariant by an infinitesmal and cannot by infinite and finite
 
Like let's say I work in $\Bbb R^n$ and I have $X=x^1\partial/\partial x^1$ and $Y=x^2\partial/\partial x^2$, what's $[X,Y]$? (maybe let's use $x$ and $y$ instead of $x^1$ and $x^2$)
 
What my actual doubt is, would any linear combination of sines and cosines having the same period represent the displacement of a particle in a SHM (where X axis be assumed the time axis?)
 
If they have the same period you can combine them
 
Or would that denote a CHM? (Complex harmonic motion...?)
 
4:51 PM
For example, $\sin(x)+\cos(x)=\sqrt2\sin(x+\frac\pi4)$
 
@AkivaWeinberger I faced problems in that. As soon as I try combining any two, the period changes.
@AkivaWeinberger Oh well sorry my mistake. It doesn't.
 
@AlessandroCodenotti I think zero?
 
Hmm I should work more before coming out with incorrect nonsense...
 
ok, maybe this will work: Infinitesimal are those process that continues without interruption, except becoming smaller but will never become empty
 
Continues without interruption... Till?
 
4:53 PM
@AkivaWeinberger I think that too, phew
 
@BaiduryaMathaddict In general, $A\sin x+B\cos x$ will equal a new wave with amplitude $\sqrt{A^2+B^2}$. (I'd encourage you to try to find a geometric explanation that connects with Pythagoras)
 
And therein comes infinity, I guess
@AkivaWeinberger I'll look up and let you know. Thanks a lot
 
you just don't specify it, if it is finite, there is an interruption in the form of switching from start to stop
 
(Hint: imagine a rectangle rotating about one of its vertices)
(and think about the x-coordinates of each of the vertices as they rotate)
 
@Secret Hmm that's okay, but not specifying is like running away from an obvious thing that's supposed to be there
@AkivaWeinberger ah wow. Wow. That is fantastic
 
4:57 PM
:)
 
Well, I am trying to decouple infinity, infintesimal and finite from each other by define them in terms on what they kept invariant
 
@AkivaWeinberger On a slightly off the track note, are you a researcher/professor?
 
they are all related, but they kept different invariants unchanged
 
No, I'm a freshman college student
 
@AkivaWeinberger Aha. Where?
 
4:59 PM
Yale
 
* nods of appreciation *
 
@AlessandroCodenotti Yeah geometrically, imagine you start on the point $(x,y)$
Move along $X$ an infinitesimal amount gets you to $(x+x\epsilon,y)$
Move along $Y$ an infinitesimal amount from there gets you to $(x+x\epsilon,y+y\epsilon)$
and then move backwards along $X$ and then move backwards along $Y$
You get back to $(x,y)$, so it's zero
 
Makes sense
 
5:22 PM
More generally if $X$ and $Y$ have constant coefficents, then $[X,Y]=0$
 
hi demonic @Alessandro, DogAteMy
 
Question: let $\Gamma \subset \Bbb R^n$ be complete (i.e. $\dim_{\Bbb R}\Gamma = n$).. then $\Bbb R^n/\Gamma$ is gonna have dimension $0$ right?
(uhh $\Gamma$ a lattice)
 
TED.
 
@ÍgjøgnumMeg: Not to beat a dead horse, but it seemed to me that you should review the statement of the Weierstrass M-test. It applies to series, not sequences.
yo @anakhro
 
5:34 PM
Did you know, Ted gave the first TED talk?
 
I think Ted talks preceded this Ted.
 
@Ted I know, I was trying to use it with $f_n = $ partial sums (incorrectly), but I managed to reach the conclusion without using the $M$-test
 
@ÍgjøgnumMeg: No, not dimension $0$. Compact of dimension $n$ !!
 
HmHMMMMhmmmM
 
Start with $\Bbb R^2$ and $\Bbb Z^2$. What is the quotient?
Or $\Bbb R/\Bbb Z$, for that matter.
 
5:39 PM
ah
"a circle"
 
Yup.
 
and then a sphere
yeah
nice
 
No, not a sphere.
You have to identify the entire boundary of the rectangle to get a sphere.
 
okay a circle zips mouth
 
Hmm, I wonder what the product of a circle with a circle might be.
 
5:40 PM
@ÍgjøgnumMeg Didn't you want to get something two dimensional?
 
A torus?
 
A torus — that sounds promising :)
 
@Ted for the other problem, I was able to show the uniform convergence on compact subsets directly
 
OK, I can't think of any way to do that without the M-test (or just reproving it in this context, which is easy anyhow).
Uniformly Cauchy is fine, too.
 
by setting $f_k(z) = \sum_{n=1}^k e^{\pi i n^2 z}$ you can bound $\rvert f_k(z) - f(z)\rvert$ above by a tail of a geometric series
and with something else
that I've written down
 
5:48 PM
Hmm, it certainly doesn't look geometric.
Oh, but if you separate positive and negative $n$, then I guess that works.
 
If I just identify $\Gamma$ with $\Bbb Z^n$ then I can just get a finite product of compact spaces
which is itself compact
right
trembling
 
Yup.
 
Nice, that's coo'
thanks
Trying not to be a vampire here lol
 
@TedShifrin I can't think up an elegant way of showing $\mathbb Z[i]$ mod 5 has no non-zero nilpotent elements.
Binomial theorem didn't seem helpful.
And I can't think of a way to exclude each element from a prime ideal.
So checking the multiplication table was the only way I have shown it so far. :P
 
Can you write that as a direct product of simpler rings?
 
6:04 PM
I think it is allegedly Z[i]/(1 + 2i) x Z[i]/(1 - 2i), but that seems more complicated.
 
Right, or (2+i)(2-i).
 
Yeah.
These are prime ideals though
So the components are fields?
 
Yup.
 
Thus no nilpotent elements in either component.
 
Maximal ideals, in fact.
 
6:05 PM
Same thing!
 
So is it easy to show this direct product... hmm
 
Well, but to say the quotient is a field ...
 
Well I know that any element in both is zero.
 
This is the same reason $\Bbb Z/6 \cong \Bbb Z/3 \times \Bbb Z/2$. Do you know how to prove such things?
 
Yes. I think I have it now.
Thanks Ted, that helped.
 
6:06 PM
Okey dokey.
 
What are you up to today, @TedShifrin?
 
Being a lazy bum, watching the ATP finals from London.
 
@TedShifrin no lazy .. lazy kill the man.
is good to see the winner of Thiem
but this post wait u
1
Q: Game Theory Poker - Nash equilibrium: Choose between bet or raise and range of hands

Z1pp3dI would like to understand how we choose the best action and how we compose the range for it. I have some notion on zero-sum games and minmax strategies. And I know software that does this for texas holdem: I want to understand how they do it This little and simple article Example : -Opponen...

lol
:D
 
I'm probably rooting for Stephanos over Dominic, but ... Anyhow, it's on now, so bye.
 
bye
 
6:34 PM
-1
Q: What function does $P(2^{2^{2^s}},2^{2^{2^{-s}}})$ trace out?

UltradarkThis is related: Do these points trace out a function? $ P(2^{2^s},2^{2^{-s}}) $ What function does $P(2^{2^{2^s}},2^{2^{2^{-s}}})$ trace out? I tried going through the answer that was given in the previous post but could not figure it out for this extension of the problem. I don't understand h...

 
What is ATP?
 
tennis^
 
Oh, I see.
 
7:00 PM
How is the Gauss-Newton method related to Newton's method on the sum of squares?
 
7:38 PM
I have a question.
If we do the tensor product of two abelian groups, say M and N.
And we are given a right $A$-module structure on $M$ while a left module structure on $N$.
Defining the multiplication $a(m\otimes n)=am\otimes n$. Does this define a left $A$-module structure on the tensor product?
I cannot deduce the distributivity for the tensors.
So the question is: construct the tensor product two $A$-modules as an abelian group, can we naturally define an $A$-module structure on the tensor product?
 
@WilliamSun the tensor product of an (R,S)-bimodule and an (S,T)-bimodule is an (R,T)-bimodule
so if M is just a right A-module (i.e. an (Z,A)-bimodule) and N is just a left A-module (i.e. an (A,Z)-bimodule) then $M \otimes_A N$ is just a (Z,Z)-bimodule, i.e. just an abelian group
 
7:55 PM
Thank you for the answer, but I don’t quite understand how this relates to if one can define a left A-module structure on the tensor product by $a(m\otimes n)=a\otimes an$.
This failed in general I guess.
 
@WilliamSun I don't know what's your map, you wrote it differently both times
 
Oh my bad! It should be $a(m\otimes n)=m\otimes an$
So recall that M is a right module and N left module
 
so given a fixed $a \in A$, you're claiming that $\varphi: M \times N \to M \otimes_A N$ given by $(m, n) \mapsto m \otimes an$ is bilinear
 
I am claiming that this defines a left A-module structure on M\otimes N.
Nvm if that’s the same thing.
 
and that means, among other things, that $\varphi(m,bn) = \varphi(mb,n)$; but $\varphi(m,bn) = m \otimes abn$ and $\varphi(mb,n) = mb \otimes an$
@WilliamSun well I'm talking about well-definedness
the only way you can get a map from the tensor product is from a bilinear map
and I'm arguing that your map isn't well-defined, since $a(mb \otimes n) = mb \otimes an$ and $a(m \otimes bn) = m \otimes abn$
that's what you get if you unfold my proof
 
8:07 PM
Let $W = \Lambda^2 \Bbb{C}^4$. If $w \wedge y = 0$ for all $y \in W$, does it follow that $w =0$? If so, why? I'm not too comfortable with wedge products.
 
Sorry just a minute let me write it down.
Oh but should we assume that mb\otimes n=m\otimes bn then?
I mean this does not follow directly from what have been defined right!
(“?” instead of “!”)
 
@WilliamSun how did you define your tensor product
 
That’s not correct let me try again
It is the quotient of the free abelian group generated by the product of M and N by the subgroup generated by the relations.
 
what relations
 
((m_1+m_2), n) ~ (m_1,n)+(m_2, n)
And same for n_1, n_2 on the right.
 
8:17 PM
one more
 
(m, n_1+n_2) ~ (m, n_1)+(m, n_2)
 
that's the previous one
 
@user193319 Yes
 
There are only two for abelian groups I guess?
 
40 mins ago, by William Sun
If we do the tensor product of two abelian groups, say M and N.
39 mins ago, by William Sun
And we are given a right $A$-module structure on $M$ while a left module structure on $N$.
so you're taking $M \otimes_\Bbb Z N$?
 
8:19 PM
Yes that’s what I mean by “the tensor product of abelian groups”
Sorry for causing confusion...
 
Write $w$ out as $a_1 e_1\wedge e_2 + a_2e_1\wedge e_3+ \cdots + a_6e_3\wedge e_4$, and then since $w\wedge y =0 $ for every $y\in W$ you can consider the four vectors taking the place of $y$: $y_1e_1\wedge e_2,\cdots y_6 e_3\wedge e_4$, and wedge these with your original thing, taking $y_i$ to be whatever you want.

In each case you'll get one term that doesn't vanish, which will be $(-1)^{\sigma(i,j)} a_jy_i e_1\wedge e_2\wedge e_3\wedge e_4$
 
@WilliamSun in that case I think it does define an A-module structure
 
@tigre So then you conclude that those scalars are all $0$, hence $w=0$. This works for more general wedge products, right?
 
Yes but I am trying to verify that the scalar distributes over two tensors.
It does not seem too obvious to me.
 
@user193319 Well, you'd have to be more precise with the 'more general wedge products'
 
8:23 PM
@WilliamSun what do you mean
 
@user193319 If you mean $w,y$ for $W=\bigwedge^{n}\Bbb C^{2n}$, then sure
 
Let me reformulate the question at this stage.
 
@tigre Oh, that's weird, so the dimension of the vector space being wedged together has to be right.
 
@user193319 Well, I just mean that if you are taking $w,y\in \bigwedge^3\Bbb C^4$, then of course $w\wedge y=0$ for every $y$, for any $w\ne 0$
Since it's alternating, you'd always have some $e_i\wedge e_i$ appear
 
Define for an element $a$ in the ring $A$ and $m\otimes n$ in the tensor product of $M$(right A-module), $N$(left A-module) as $\mathbb Z$-modules a product $a(m\otimes n)=m\otimes an$, why do we have $a(m_1\otimes n_1+m_2\otimes n_2)=a(m_1\otimes n_1)+a(m_2\otimes n_2)$?
 
8:28 PM
You extend the action by linearity
 
I cannot see how we can “combine” the two tensors in order to apply the definition of the product defined above.
 
@WilliamSun use the universal property
 
@tigre Okay. I'm trying to show that the bilinear form $B$ on $W$ defined by $w_1 \wedge w_2 = B(w_1,w_2) e_1 \wedge ... \wedge e_4$ is a symmetric non-degenerate form....But why is it symmetric? It seems like it should be antisymmetric.
 
I cannot see how one can apply universal property to this problem.
 
@WilliamSun Well, by construction, $m_1\otimes n_1+m_2\otimes n_2$ is a sum of generators in the tensor product, and can't be reduced to a pure tensor just using the defining relations (unless of the the $m_i,n_j$ is zero say). So $a(m_1\otimes n_1+m_2\otimes n_2)$ only makes sense after you've defined the $A$-action on the generators, and extended it by bilinearity right?
 
8:36 PM
@WilliamSun basically for each $a \in A$ you're defining a group homomorphism $M \otimes N \to M \otimes N$
use the universal property to do that
 
@tigre My reasoning is as follows. Let $v,w \in W$. Then $B(v,w)$ is that scalar such that $v \wedge w = B(v,w) e_1 \wedge ... \wedge e_4$, and $B(w,v)$ is that scalar such that $w \wedge v = B(w,v) e_1 \wedge ... \wedge e_4$ or $v \wedge w = -B(w,v) e_1 \wedge ... \wedge e_4$. By uniqueness of the scalars, $B(v,w) = - B(w,v)$.
 
Oh I understand! Thank you! @tigre @LeakyNun
 
@user193319 You should see why your logic fails by considering $(e_1\wedge e_2)\wedge(e_3\wedge e_4)$ and $(e_3\wedge e_4)\wedge(e_1\wedge e_2)$
It all occurs when you reorder all the $e_a\wedge e_b\wedge e_c\wedge e_d$ to $e_1\wedge e_2\wedge e_3\wedge e_4$
@WilliamSun No problem :)
 
...But isn't $w \wedge v =- v \wedge w$?
 
@user193319 $e_3\wedge e_4\wedge e_1\wedge e_2 = -e_3\wedge e_1\wedge e_4\wedge e_2 = e_1\wedge e_3\wedge e_4\wedge e_2 = -e_1\wedge e_3\wedge e_2\wedge e_4=e_1\wedge e_2\wedge e_3\wedge e_4$
 
8:44 PM
@user193319 only for degree-1 wedges
 
^
It's also not true that $\omega\wedge\omega = 0$ for the record (for an arbitrary $\omega$)
Both of those are common misconceptions when you first encounter the exterior algebra
 
$\alpha \land \beta = (-1)^{kp} \beta \land \alpha$
 
Hmm...what is a degree-1 wedge? @tigre Why I want to rearrange the factors in the wedge that way?
 
what's a space that is homeomorphic to $\Bbb R^2$
 
@user193319 Because your bilinear form has $e_1\wedge e_2\wedge e_3\wedge e_4$ at the end, which yields a scalar, and in this case showed you it was symmetric, not anti-symmetric
@user193319 Degree one means it is some $ae_1 +be_2+ce_3+de_4$
Degree $3$ means its some $ae_1\wedge e_2\wedge e_3 + b e_1\wedge e_2\wedge e_4 + c e_1\wedge e_3\wedge e_4 + d e_2\wedge e_3\wedge e_4$ for example (in $\bigwedge^3\Bbb C^4$)
Degree $k$ wedges live in $\bigwedge^k\Bbb C^n$ ($n\geq k$) or over the appropriate ring
@Ultradark Pick your favourite open disk
 
8:50 PM
@tigre and it's homeomorphic because if you increase it's radius?
 
So, with respect to the formula Leaky Nun cited, we have $p = k = 2$, so the negative vanishes?
 
@Ultradark It's a homeomorphism because its a homeomorphism
@user193319 negative doesn't vanish it's just $(-1)^{\text{even}}=1$
but yes up to being pedantic (sorry)
 
oh @tigre you mean you can contract $\Bbb R^2$ to a disk?
 
@Ultradark I mean that there exists a homeomorphism between $\Bbb R^2$ and an open disk, so they are homeomorphic
You'll have to find the map yourself :-)
 
That's what i mean by vanish.
 
8:55 PM
@tigre do you think i can define a coordinate system that is not homeomorphic to a subspace of $\Bbb R^2$
 
@user193319 I did say I was being pedantic :P. Typically I would always reserve 'vanish' for meaning it because $0$, rather than it not being written down anymore
 
Oh, haha...didn't see that. Sorry!
 
@Ultradark A coordinate system should be a choice of charts to give manifold structure to $\Bbb R^2$. Is this the sense that you want to pick a coordinate system?
If not, I have no idea what your question means
Since a 'coordinate system' doesn't sound like something that can be homeomorphic to anything (it doesn't sound like a topological space)
 
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