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12:01 AM
@Daminark Lol late response: I'm beginning to compose music... That's pretty new lol. You?
 
heya @Dair
 
@Ted Hey.
 
Not too much, just relaxing a bit. Gotta decide classes for next quarter soon
 
@Daminark the two body problem is a primary concern for me but it’s looking like my so will likely end up on the east coast
 
Well, if this grad school thing doesn't work out (fingers crossed), I guess I'll just have to put the CS and math on hold and turn over to the world of musical composition... :P
 
12:07 AM
Hello!
 
Ah I see. Yeah I think my choice will probably just come down to comparing faculty. One thing I've become more aware of is how previous students of professors have fared
 
@Daminark it turns out that the differential distance squared is non-zero in the hyperreals, but it is in other frameworks
 
Not a perfect metric since you students might decide on their own that academia isn't worth it, or maybe they just weren't as strong even if they have a good adviser, but it does help a bit to know that a professor has at least something of a decent record at getting students into good postdocs and whatnot
 
@Dair: Since my dad was a composer, I can tell you that's a great way to make tons of money ...
 
@Ted You haven't seen compositions like this before.
 
12:10 AM
Uh huh.
 
@TedShifrin Hey! How are you?
 
hi @topologicalmagician
 
I know that's what they all say, but they don't have what I have: a math degree.
 
@topologicalmagician perhaps, again I'm not really familiar with hyperreals or infinitesimal distance in general, but most working mathematicians who have to deal with the content either work with infinitesimals just as a general heuristic or, if they wanna be rigorous/conceptual, just do delta-epsilon and not bother with them at all
 
I don't believe what you're saying, @topologicalmagician.
 
12:13 AM
@TedShifrin Which part?
 
@Daminark neves told me to basically go on math genealogy to see ppls students and where they go so this is prob a good thing to do generally
 
being zero "in other frameworks" ...
then again, I am not sure I have any idea what you're actually talking about.
 
@Daminark Have you tried looking at the profs erdos number? I know it is an inaccurate metric, but it still may be useful.
 
ridiculous, @Dair
 
@TedShifrin I've been told that it might be the case in synthetic differential geometry, because nilpotent infinitesimals are found in synthetic diff geo
 
12:14 AM
@Ted kidding.
 
I've seen some business about working in $\mathbb{R}[\epsilon]/(\epsilon^2)$
 
that's a scheme-theoretic way of getting the cotangent space, Demonark.
A morphism to that ring of dual numbers.
 
There was actually a thing about this where Fermat or somebody had the first primitive notion of a derivative by discarding squared (or higher) terms...
maybe it was decarte..
lmao. brain is mush.
 
My guess is that this might've been what topologicalmagician saw wrt other frameworks?
 
I am still quite skeptical, @topologicalmagician.
 
12:16 AM
Lmao, I don't think I have too many results that have been misattributed to Fermat
 
What are you calling synthetic differential geometry?
Fermat did figure out the integrals of $x^n$ by using a geometric partition of $[1,b]$. Neat computation.
 
OK, differential geometry in arbitrary characteristic is not my cup of tea, so I'm going to remove myself.
 
Okay I think my mentioning hyperreals earlier probably led you too far down a rabbit hole. I mostly did it to do justice to infinitesimals since they can technically be made rigorous, while in reality the takeaway from what I said should've been closer to "forget about infinitesimals forever"
 
many of the computations to find derivatives before the infinitesimal definition by newton (and leibniz) were really interesting...
 
12:19 AM
Or like, while you're in a class that works with them be able to work with them however you must, but to eventually supplant it completely with limits done right
 
@Daminark It is a legitimate field of study, too bad the most popular intro book is kind confusing about it...
 
i'm annoyed because I'm unable to submit to heuristic arguments easily, and in my notes, a lot of derivations were based on the idea that (dx)^n is 0, like in the case of deriving formula for the gradient in general coordinates
 
What course is doing that crap?
 
Is this taught by an engineer or physicist?
 
12:20 AM
Keisler's book.
 
It's legitimate, sure, but unless you have a particular reason to go down that direction it's a diversion at best. Most working analysts don't gain from this perspective, I imagine
 
I tried giving it a go.
 
I think physicist
 
Just ended up being more confused after.
 
You can do chain rule and transformation laws for differential forms quite rigorously.
There's no need for this nonsense. But it is important to use the right notion of product with differentials ... sometimes symmetric, sometimes wedge product.
 
12:22 AM
@TedShifrin Oh, you weren't talking to me.
 
If you try to do $dx\,dy = (\cos\theta\,dr - r\sin\theta\,d\theta)(\sin\theta\,dr + r\cos\theta\,d\theta)$ you're going to get garbage.
Unless you know about wedge product and differential forms and determinants.
No, I wasn't, @Dair.
 
@Ted yeah I've heard of this, a lot of calculus classes "prove" the product rule by saying something like, $d(uv) = (u + du)(v+dv) - uv = udv + vdu + dudv$ and say $dudv = u'v'(dx)^2 = 0$
 
Well, that's crap.
 
Yup
 
I'd rather make a valid argument based on the area of a rectangle.
 
12:24 AM
The method used to derive the gradient in general coordinates was through taylor expansion then comparing terms ,@TedShifrin

I really don't know much about wedges and differential forms, yet
 
Suppressing limits entirely isn't going to help understanding.
 
I really don't know much about wedges
 
@Daminark I think that argument is used in Calculus Made Easy (for reference)
 
It's just the chain rule, @topologicalmagician. But differential forms is the right way to do differential operators (gradient, div, curl, etc.) in different coordinate systems.
 
Question: what's the difference between the hyperreals, the surreals, and "the long line" anyways?
 
12:25 AM
The long line is totally unrelated, @Rithaniel.
The others are ways of incorporating infinitesimal and infinite numbers into the real number system.
 
@TedShifrin where can I learn more about differential forms?
 
More an exercise in model theory than in analysis.
 
the long line is like the real line
but longer
 
Surreal numbers are some kinda weird thing, I don't think they're even technically a set, but somehow are used for Conway's game theory style
 
You could watch my YouTube lectures if you want to learn them in a careful manner, @topologicalmagician ...
 
12:26 AM
Long line is a topological space used to show that theorems in topology need hypotheses
 
LOL
 
@Daminark just like every other topological space ever
 
Well, some spaces actually are important
 
Okay, that gives me some context.
 
I actually never did the long line when I taught point-set, but I did do the first uncountable ordinal ...
 
12:27 AM
Tbh I don't regret the fact that I'm mostly figuring out the actual spaces that necessitate these hypotheses instead of some contrived ones in point-set
I imagine long line is not first countable?
 
@TedShifrin if you don't mind me asking, which lectures should I focus on before watching your lectures on diff forms?
 
most definitely not first countable
just to get the idea with forms, you can start with them, I think, @topologicalmagician. Eventually you'll want things like the inverse function theorem if you want to understand manifolds and the fancy Stokes's Theorem.
 
Yeah so, the first time I ever really ran into that concept was because I had a problem in functional analysis which was to show that if a space is reflexive or has separable dual, then there's a sequence in the unit sphere converging weakly to $0$
 
yeah, Demonark, I love that.
Of course, the unit balls shrink in volume, too, when you get up past 6 or something.
Shrink to 0, I mean.
 
I was talking to the other undergrads in the class and for some time we were like, wait why are these hypotheses there? Didn't we show for all Banach spaces the weak closure of the unit sphere is the unit ball? Also why is this even a problem?
We realized after thinking about it for a while that the key word was "sequence", since first countability becomes a concern
 
12:31 AM
Math before Cantor: This is fine. Math after Cantor: This is not fine.
 
So running into it there kinda made it make sense to me, like okay this is why nets and countability axioms are a thing! If I had seen the long line or something I would've just thought that it was an irrelevant concept that someone came up with because we forgot some axioms of a topological space and someone found a hole, thought he was being real clever by poking it, and poked it
 
@TedShifrin I'm aware that the gradient has an equivalent definition that involves the metric tensor and the contravariant basis, but it doesn't depend on the unit basis vectors, while I was told that it doesn't make sense to use non-unit basis vectors for the gradient
 
Understandably the issue is that it would take too long to talk about weak topology on Banach spaces in point-set. But yeah idk
 
You can do the gradient, in principle, in any coordinate system. It's just the vector field dual (using the metric) to the $1$-form $df$.
Demonark: Most of us don't have the strength for weak topologies.
 
Heh
Well, time to work through a bit of Morse theory if anyone would like to join
 
12:39 AM
How do I prove that the tangent vectors of a surface aregiven by the partial derivative of the position vector with respect to a coordinate?
 
So Milnor's defining the Hessian here by saying you have a bilinear form on $TM_p$ given by, well if $v,w\in TM_p$, you can extend to vector fields $\widetilde{v}$ and $\widetilde{w}$, and define $f_{**}(v,w) = \widetilde{v}_p(\widetilde{w}(f))$. To this I respond: wut
 
how do you define tangent planes?
assuming critical point, Demonark, that's well-defined
 
Yeah, I'll type check this a bit for the formalism since in my mind manifolds are still in $\mathbb{R}^N$
 
Well, we haven't defined tangent planes in this course, but previously we defined them by a(x-x_0)+b(y-y_0)+c(z-z_0)=0 and a,b,c are the components of the gradient of the surface
 
no, you have different definitions ... level surface versus parametrized ...
standard confusion in multi calculus ...
 
12:43 AM
well, we haven't been given a definition for the parameterized tangent plane
 
you just gave me it a second ago
 
ok so for a parameterized surface, how do I prove that the tangent vectors are given by the partial derivative of the position vector with respect to the coordinate
 
that makes no sense
 
LOL
i just realized what I wrote
i meant to say coordinate instead of gradient
 
So first, the fact that we can extend to a vector field. Fix $v\in TM_p$ and look at a chart $(U,\varphi)$, there you have $TU$ is just $U\times TM_p$, diffeomorphic to $\varphi(U)\times \mathbb{R}^n$. Let's say $\varphi(p) = 0$ and that we contain $B_3(0)$, now take a cutoff function $\eta$ that's $1$ in $B_1(0)$ and $0$ outside $B_2(0)$
 
12:47 AM
I'm saying there's nothing to prove if you have no definition
 
And define our vector field on $\varphi(U)$ by $(x,\eta(x)v)$ (abusing notation $v = d\varphi_p(v)$), pull it back, extend to $0$ on the rest of the manifold.
 
All I have in my notes is that the gradient is a scalar multiple of the normal, and that every surface can be represented by two coordinates and so every point on that surface can be represented by the euclidean position vector
 
gradient of what?
 
gradient of the surface f(x,y)=K where K is a constant
 
no, think about what you just said
have you had multi calc? what is this course>
 
12:52 AM
yeah I had multi cal
its called the calculus of vectors
 
oh geez
 
this is physics or math?
 
I dunno ...
 
12:55 AM
whats wrong?
 
I am leaving now ...
 
:(
 
Have other things to do
 
Okay now, a tangent vector is a curve $\gamma:(-\varepsilon,\varepsilon)\to M$ and acts on functions by $\gamma'(0)(f) = (f\circ \gamma)'(0)$. So if you have the vector field $\widetilde{w}$ and a fixed $f:M\to \mathbb{R}$, then $\widetilde{w}(f)$ presumably eats a point $p\in M$ and spits out $\widetilde{w}(p)'(0)(f) \in \mathbb{R}$?
 
alright, but can you tell me whats wrong? lol
please?
 
12:55 AM
@topologicalmagician I think you broke him.
 
LOL
idk if i'm the problem or what I've written in my notes
is
so any feedback would be appreciated
 
@topologicalmagician I think you should eventually find some book on rigorous multi/linear algebra, maybe Ted's book, maybe Pugh's Real Analysis, or something
And have that completely dislodge everything your professor right now is teaching you
 
@Daminark yeah, I'll get Teds book. I'm just feeling insecure now, because I don't know what the problem is lol
 
I think the reason why Ted doesn't wanna continue here is that the way you seem to have been taught this stuff cannot be salvaged
It's not like there's a problem where you just have to carry the two and it all works out
 
Let the insecurity flow through you.
 
1:02 AM
It'd require undoing whatever your professor did and starting from 0. And he probably has enough to do that there's no time for that now
And also on your side, given that you're in the middle of a class you probably don't have all the time in the world to repeat multi from 0 right now
 
I guess i'll start learning multicalculus from 0 then
 
I think the best thing for you to do is just hang tight for a while
And roll with the heuristics. Keep that discomfort you have inside of you
 
@Daminark thanks Daminark, I really really appreciate it
 
But just go with the flow, do well in this class, and then pretend you heard nothing once you finish the final and learn it properly
 
should I discuss the issues im having with the heuristics with the prof?
 
1:06 AM
@topologicalmagician I would say sure, as long as you aren't rude about it, but there are a lot of ways to be rude about it.
so, if you get in trouble... I am not liable for any of your discomfort.
 
That depends a bit on the professor. There's a risk that he'll think he's being as rigorous as he needs to and get annoyed, and even if you don't have a particularly uptight professor, there are ways to do it and ways not to do it
@Dair I dunno fam my tongue doesn't fit comfortably inside my mouth (now you're thinking about that... you're welcome >:) ) and I'm inclined to sue you over this
 
I feel violated
 
@Daminark alright well thanks so much! i'm going now
@TedShifrin@Daminark Have a good day! and sorry for being annoying
 
Alright, good luck! :)
 
gl.
 
1:08 AM
@Dair thank you aswell
 
np.
I should probably go too... either get some fresh air or compose more lol.
 
Yeah I should probably head out as well and do some Morse theory
-..---.-.- -.--.----..-
Oh hi loch!
 
2:04 AM
hi @Daminark !
 
How's everything going?
 
 
2 hours later…
4:08 AM
I have a modest proposal. I think people writing about math should never use the word "undefined" predicatively, only attributively.
In other words, don't say "1/0 is undefined", say "1/0 is an undefined expression".
It's very, very common for people to read the sentence "1/0 is undefined" and come away with the impression that there's some particular thing called "undefined", and that 1/0 is that thing.
 
 
2 hours later…
6:36 AM
@MikeMiller Why haven't you considered 11 for example?
 
 
2 hours later…
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k < 2.5e6 that work
So $\mathfrak a^{ec}/\mathfrak a$ is the kernel of $A/\mathfrak a \to B \otimes_A (A/\mathfrak a) = B/\mathfrak a^e$
 
9:28 AM
Thought: the area of the triangle is half that of the area of the hexagon
(which you can see because the remaining three pieces of the hexagon can be rearranged to fit on top of the triangle)
 
i.e. fold the flaps inside
 
Ya
As an easy corrolary, the side length of the triangle is $\sqrt3$ times that of the hexagon
though I guess we knew that anyway
(Divide the hexagon into six triangles - the big triangle has area three in those units so it has sidelength $\sqrt3$)
Like, imagine if the sidelength of the hexagon is 1 inch, and then instead of measuring area in square inches we do it in triangle inches
but we still get that a triangle of sidelength $s$ inches has area $s^2$ triangle inches
I have trouble remembering that the height of an equilateral triangle of sidelength $1$ is $\sqrt3/2$. I have to rederive it from Pythagoras each time
so maybe this alternate derivation will make it easier to remember
I think, in $n$ dimensions, the height of a tetrahedron is $\sqrt{\frac{n+1}{2n}}$
or $\sqrt{1+\frac1n}/\sqrt2$
I wonder what's the easiest way to derive that
@LeakyNun
The problem is, in $n$ dimensions, volume no longer scales by the square, it scales by the $n$th power
 
9:44 AM
[n-height x (n-1)-volume]/n = n-volume
probably isn't helpful
 
I think that means that the volume of an $n$-simplex is $\sqrt{(n+1)2^n}/n!$?
Inductively
 
maybe
 
Wait no the $2^n$'s on the bottom
$\dfrac{\sqrt{n+1}}{n!2^{n/2}}$
 
10:08 AM
Hi chat
@ShaVuklia No, what is it ?
 
10:36 AM
@LeakyNun I'm back
and I think I see a way to derive the area of an $n$-simplex now
Consider the irregular $n+1$-simplex defined by $0\le x_0,x_1,\dots,x_n\le1$
This has volume $1/(n+1)!$
I'll draw a picture for $n=2$
Right so that has area $1/6$ because it has a base of area $1/2$ and a height of $1$ so we get one-third base times height or $\frac13\frac12=\frac16$
Oh I meant $0\le x_0,\dots,x_n$ and $x_0+\dotsb+x_n\le1$, whoops
So notice also that the face opposite the origin is an $n$-simplex, of sidelength $\sqrt2$
so it has $n$-volume $V_n\sqrt{2^n}$ (where $V_n$ is the $n$-volume of a unit $n$-simplex)
So now let's find the $n+1$-volume of our big $n+1$-simplex in a different way, with that face as the base
What's the height?
It's the distance between $0$ and $(\frac1{n+1},\frac1{n+1},\dots,\frac1{n+1})$
which, by Pythagoras, is $\sqrt{\frac1{(n+1)^2}+\dotsb+\frac1{(n+1)^2}} = \sqrt{\frac1{n+1}}$
so finally we get:$$\frac1{(n+1)!}=\frac1{n+1}\cdot \sqrt{\frac1{n+1}}V_n\sqrt{2^n}$$
$$\frac{\sqrt{n+1}}{n!\sqrt{2^n}}=V_n$$
QED
And now as you mentioned, $h_nV_{n-1}/n=V_n$
which means $h_n=nV_n/V_{n-1}$
and I'm not gonna check that but it seems right
 
 
1 hour later…
12:13 PM
ok so, the main reason why Leaky's zero reciprocal algebra collapsed into the trivial object is because of the following:
By setting $\frac{1}{0} = 0$ and hence $\frac{x}{0}=x0=0$ you are setting countably many elements to be the two sided additive identity. Since in any associative algebra the two side identity is unique, it follows that all zero reciprocals are zero. Now, the sum of fractions axiom then further messed this by making the multiplicative identity into an additive identity, thus by extension, every element x is an additive identity. Hence we have x=0 for all x, thus collapsing into the trivial object
Therefore, the only way it can be saved while leaving as many axioms intact are the following:
1. Change the way the distributive law works that recovers the element $\frac{x0}{0y}$ during the summation step
2. Break associativity so that two sided additive identities are not necessary unique
 
@Astyx it's like whatsapp, but better:p but what I was trying to say is; is there a way to keep in touch, apart from the math chat haha (because I'm not often here)
 
12:28 PM
Sure !
So it's an app ?
 
12:55 PM
@ShaVuklia
 
@MatheinBoulomenos Here's a quote of yours: "we want to show Fatou's lemma, so let $g_n$ be a sequence of non-negative measurable functions. Let $f_n=\inf_{k \geq n} g_n$, then $f_n$ is again measurable and non-negative and we have that $f_n \to \liminf g_n:=f$ (monotonously).

Since we have $f_n \leq g_n$ for all $n$, we have $\liminf \int_E f_n\leq\liminf \int_E g_n$.

If $\liminf \int_E f_n = \infty$, then there is nothing to show. Suppose that $\liminf \int_E f_n = M < \infty$, then from monotonicity, we have that $\int_E f_n \leq M$ for all $n$, so by the assumed property, we have $\in
How did you go from $\liminf \int_E f_n = M < \infty$ to $\int_E f_n \leq M$ for all $n \in \Bbb{N}$? I don't see how monotonicity plays a role.
 
1:18 PM
If $\{f_n\}$ is a sequence of Lebesgue integrable functions on $E$ and I know $\lim_{n \to \infty} \int_E f_n$ exists, can I conclude that $\lim_{n \to \infty} \int_A f_n$ exists for any measurable set $A \subseteq E$?
 
@user193319 since $f_n$ is monotonically increasing, $\int_E f_n$ is monotonically increasing, so $\liminf \int_E f_n = \lim \int_E f_n = \sup \int_E f_n$
 
1:39 PM
Hello math people
Is there a standard notation for gluing a manifold to itself along two of its boundaries
There's one for gluing two different manifolds
$M_1 \cup_h M_2$
But for the same manifold I can't find
 
2:10 PM
Why can't $M_1$ and $M_2$ be the same manifold?
 
I mean I suppose they could
bit awkward notation but that's probably well defined
 
(Disclaimer, I don't actually know the answer to your question. But this would be my approach)
 
gluing manifolds to themselves is one of those Things that's pretty hard to find decent stuff for
It seems to be more of a physics thing
and we're not very good at making it rigorously
 
My mind goes to topology and quotient spaces, but I know little enough about manifolds that I don't know if that leads you down an inappropriate path.
 
It is basically that yeah
the hard part is finding not topology stuff
Pretty hard to find theorems on metrics for glued manifolds
 
2:49 PM
Two disjoint copies of the same manifold? Or literally gluing one to itself
 
There is no standard notation. I would write M/sim.
Gluing manifolds together is not really a physics thing.
 
On associators:
Given that an associator $(a,b,c)$ in any algebraic structure is really a map $f: (ab)c \mapsto a(bc)$ and that if some subset of associativity fails to hold, it often has the form $x \mapsto y$ for some $x,y$ in the algebraic structure. I wonder, can we strung all of these together in a certain way so they form some kind of deductive system
By comparison, juxapositioning commutators in some groups can end up performing a transformation, such as the commutators used in rubik cube moves
 
3:10 PM
Well, managed to finish the chat app in time. But it was a lot trickier than I expected, and I am certain the code shows my inexperience with certain things.
 
In mathematics, and in particular universal algebra, the concept of n-ary group (also called n-group or multiary group) is a generalization of the concept of group to a set G with an n-ary operation instead of a binary operation. By an n-ary operation is meant any set map f: Gn → G from the n-th Cartesian power of G to G. The axioms for an n-ary group are defined in such a way that they reduce to those of a group in the case n = 2. The earliest work on these structures was done in 1904 by Kasner and in 1928 by Dörnte; the first systematic account of (what were then called) polyadic groups...
So, the generalisation of an associator is an n-associator, which are homomorphisms relating (abc....)gh, a(bc...)gh, ..., abc(...gh)
 
@AkivaWeinberger to itself
like a cylinder glued to a torus
this kind of shenanigans
@MikeMiller It is when you do wormhole related business :p
Physicists do it pretty fast and loose, though
 
Then that's just a quotient
 
yeah
 
$S^1\times I/(\langle x,0\rangle\sim\langle x,1\rangle)$
 
3:20 PM
just wanted to know if there was a more compact notation
bit cumbersome otherwise
 
$S^1\times I/{\sim}$ and then define $\sim$ somewhwre
Yeah I dunno
 
One notation I've seen which is sort of related is $SK(M)$
 
But I wonder, what is a n-ary generalisation of the commutator, a "permutator"? : $abcdef... \mapsto \pi(abcde...) $
 
not quite that but close enough to reuse it maybe
 
What's SK(M)
 
3:22 PM
Apparently SK is the initials in German for ~ cut and paste
 
Still, I think we can do something more interesting than commutators and associators when going to 4-nary algebraic structures
I am not sure though on what that could be. Perhaps it has something to do with abcd, a(bc)d, (ab)(cd)
I guess, it really boils down to the possible classes of string operations available to be operated on a string:
The commutator captures transposition of strings
The associator captures the order of grouping of strings
So there should be an inherenty 4-nary operation on a string, that requires at least 4 letters to carry out
 
Thought: take a tetrahedron, and mark any two opposite edges
The rotational symmetries of the result will not depend on which pair of edges you choose
but the reflectional symmetries will
 
There are examples of quaternery operators in Processing, such as the function rectangle (a,b,c,d) which draws a rectangle
but that does not seemed to exploit anything inherently 4-nary about strings
 
3:38 PM
@Slereah I don't deny that you use the language.
M/~ is perfectly fine as long as you explain what the equivalence relation is, which you will.
 
Aight
 
Also
If you visit any face of the tetrahedron and go around its edges counterclockwise, you will always visit the three "edge/opposite edge" pairs in the same order
and if you visit any vertex of the tetrahedron and go around the edges incident to that vertex counterclockwise, you visit the three pairs in the opposite order as the faces
Hm. What about cubes? You can group the edges by what axis they're parallel to, and then you can go around a vertex, or you can go around one of those hexagons that cuts the cube in half
Hm. It depends on what vertex, now
Oh and you can't specify "clockwise" or "counterclockwise" for the hexagons because it depends on how you're holding the cube
 
Chirality is a property of asymmetry important in several branches of science. The word chirality is derived from the Greek χειρ (kheir), "hand," a familiar chiral object. An object or a system is chiral if it is distinguishable from its mirror image; that is, it cannot be superposed onto it. Conversely, a mirror image of an achiral object, such as a sphere, cannot be distinguished from the object. A chiral object and its mirror image are called enantiomorphs (Greek, "opposite forms") or, when referring to molecules, enantiomers. A non-chiral object is called achiral (sometimes also amphichiral...
Ah...
Chirality need at least 4 points to be specified, that is an inherently 4-nary quality
But... what does the "chirality" of a string means...
 
n+1 points in n dimensions
Knots can have chirality
In fact, most knots are chiral
like the trefoil
The figure eight knot is not chiral, though, somewhat surprisingly
 
well, it has an axis of rotation thus you can rotate its mirror image
Hmm... it sounds like interesting 4-nary algebraic structures may be closely related to knot theory, should study that more some day...
 
Aren't those the same ?
 
yeah, it has 2-fold rotational symmetry
 
Those are the same (even if you require that the left side of the string - in this case the iPhone charger part - stays on the left, and the right side of the string - in this case the USB part - stays on the right)
@Astyx No - look at the crossing on the top-right
In the first image the horizontal strand goes over; in the second image it goes under
 
Yeah but if you turn it 180 degrees
 
Ah yeah that's what I was just saying about keeping the left bit on the left and the right bit on the right
In other words, keeping the orientation as well
 
3:55 PM
Huh, ok
 
Tie it, try it
(third rhyme)
And this one is chiral
and I think that the smallest achiral (equal to mirror image) knot with an odd number of crossings has 15 crossings
 
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