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16:01
@Secret You know, when people want to show the orientation of a 2d thing in 3-space (like a plane or face), one way to do it is to draw a circle with an arrow on it pointing clockwise or counterclockwise
Example
right, so as you said, chirality needs n+1 points in n dimensions
I don't know of a good similar image in 3D
thus it is not a uniquely 4-nary object
As for 3D, you can e.g. pick any chiral molecule in chemistry, and then label the 4 atoms in the tetrahedral frame
Note also that the circle in that image kinda shows what orientations the arrows on the edges should be in
If you take a polyhedron and give every face the same orientation, then every edge will be given two different orientations by the two faces it's adjacent to
and they'll cancel out
yup, and that observation is also an important part in the proof of stokes theorem
the cancelling of edges of opposite orientation
16:06
Ya
(Example for hexagon, not polyhedron, but a polyhedron is basically the same thing just with no edges)
Oh I was gonna suggest a spinning sphere (as like a generalization of the spinning circle) to show orientation in 3D, but if you view it upside-down it's spinning the other way so it doesn't work
I guess you kinda need a sphere where the northern hemisphere is spinning a different way than the southern hemisphere
(and somehow both are counterclockwise)
(This kinda has to do with how the fundamental group of the set of rotations in 3D and higher is $\Bbb Z/2\Bbb Z$)
chemistry is never short of chiral moelcules
Another example of 3D chirality is the helix
Oh you know what that's actually a perfect image for chirality
The one that goes counterclockwise as you go up is "right-handed", right?
AKA you make right turns if you drive down it
I see
As for 4 dimensions, there's another way to introduce chirality which has something to do with cllifford rotations
But I don't remember the details atm
You can think of the helix as
imagine a cylinder, and draw swirly arrows on the top and bottom faces
and let's say they're both counterclockwise
then the helix joins the two arrows almost
Wait a moment
No it doesn't
Or hold on I need to draw this
OK so if you move along the right-handed helix away from the cylinder (in either direction) then you match the counterclockwise swirls
or away from the center of the cylinder, I guess
Exiting the cylinder
(The bottom swirl looks clockwise but that's only 'cause we're seeing it from the wrong side)
16:22
yup
I realize that the cylinder I drew is a metastable illusion actually
The perspective is ambiguous
necker cube
I meant for us to be seeing it from above
I tend to see metastables as upright, probably due to preference
Drawing 3D objects can be very difficult
16:24
Hm
The helix is metastable also
but only one way feels like it "fits" inside the cylinder
(unless you look at the cylinder the other way also)
How about this: It's actually a 2D surface consisting of two ellipses and several lines
or rather, a 2D space.
Ceci n'est pas une pipe
(unless you print it out, roll it up and smoke it)
Was just about to say
Actually, the helix isn't a good generalization of the swirl, since the swirl has an arrow and the helix doesn't
A better 2D version of the helix would be an S
or Ƨ if it's the other orientation
4D chirality in a hopf fibration, assuming I have not made any mistakes
The two ellipses are actually great circles on the 3-sphere that are orthogonal to each other
16:33
Why are rotations in 4D chiral? I just established that this is not true in 3D.
Hm
Oh I see
It's an even-odd thing
A rotation that keeps the x and y axes constant and rotates along the zw plane is not chiral, but most others are
> The chirality of a rotation can be distinguished, even for clifford ones. The only non-chiral rotation are the great-circle ones.
I've just rediscovered the first sentence of the second post of your link apparently
Maybe I should've read it first
Well I guess if I hadn't thought about it on my own first, I probably wouldn't have had any idea what the poster was talking about
Most of them are not professional mathematicians, and some of them invented their own notations. They once tried to publish a paper on convex regular polytopes, but the supervisor of one of the maths guy said while those shapes are new, they are not significant enough to be published into the literature
From a linear algebra perspective, this is the same as saying that an antisymmetric matrix needs to have even dimension to have nonzero determinant
which is true
(View a rotation as a vector field, showing the velocity of each point in space. The map from the point to the velocity is a linear map, and it's antisymmetric because the velocity has to be perpendicular to the position vector)
(and a matrix is antisymmetric iff $x\cdot Ax=0$)
Arright so this was all some pleasant nonsense
Maybe now I'll try to go back to thinking about the group theory puzzle I had
Hm, Möbius strips work too
@Secret There are two ways to make a Möbius strip
You take a strip, twist it, and glue the ends together, yeah?
clockwise twist vs anticlockwise twist?
Yeah
It's a bit of a puzzle to see that those are actually different
like, you can't deform one into the other (without letting it pass through itself)
Wait a moment
Even if it can pass through itself
(without making sharp creases)
16:49
though +1 half-twist counterclockwise and +5 half-twists counterclockwise are the same
I used to play with that whenever I was wearing a tie
(which is not that often, to be honest)
I guess a belt also works
Not while you're wearing it though
(You need to allow it to pass through itself for this)
hmm, so passing through itself is not quite a mirror operation
Rough illustration?
But really you need a physical belt or something I think
If you only look at the right half of the middle two things you see that +1 full twist = -1 full twists
aka +2 half twists = -2 half twists
(Half twist = 180 degree twist)
Well, the orientation of the middle two are preserved, but I guess there is no way to transform between each other unless the loop is allowed to pass through itself
Meanwhile, just recalled something that needs exactly 4 points to specify:
A dihedral angle is the angle between two intersecting planes. In chemistry it is the angle between planes through two sets of three atoms, having two atoms in common. In solid geometry it is defined as the union of a line and two half-planes that have this line as a common edge. In higher dimension, a dihedral angle represents the angle between two hyperplanes. == Mathematical background == When the two intersecting planes are described in terms of Cartesian coordinates by the two equations a 1 x ...
Not sure if there are other geometric quantities (other than a quadrilateral) that only need exactly 4 points
17:07
This is also kinda interesting
(ignoring the "pass through itself" thing)
+1=-2=-1 ?
The middle is 2 half twists
(not sure + or -)
and the top and bottom are 1 full twist
Try it
I realized I could experiment with the tab on my jacket's zipper
Hey, remember the "turning numbers" introduced in the Inside-Out video?
The winding number of the tangent vectors
yup
If you look at the the turning numbers of the top and bottom thing, viewed as curves in the plane (so ignoring the crossing information about which strand is on top), they do have different turning numbers
yeah, one is going clockwise and the other is anticlockwise
17:11
@Secret It would help if I had colored the other side a different color, maybe
which would only affect the middle frame
Have seen enough of these to deal with them mentally, though more complicated knots might actually need a physical aid
It has to be an even number of half twists in all three frames 'cause they have the same color on the left as on the right
In other news, figured out a good candidate of a 4-nary operator inspired from an arbitrary quadrilateral. It is technically a subset of the permutator map
This looks similar to $A_4/V_4=\Bbb Z_3$ from group theory
or $S_4/V_4=S_3$
I'm not sure exactly what you're doing but I think it's actually exactly that
Definition: Given an algebraic structure with arity at least 4, then a **Quadrilaterlator** $(a,b,c,d)$ is defined to be the homomorphism:
$(a,b,c,d) \mapsto (a,b,d,c)$
$(a,b,c,d) \mapsto (a,c,b,d)$
17:16
1234 is the identity permutation e. 1243 is the permutation (34)
etc
yeah, cannot think of anything more interesting atm
and if we multiply (34) by an element of V_4 (say (12)(34)) we get the permutation (12)
which has the same shape as (34)
Well, what I want to do is to find a Quaternary operation. Baiscally recall that the commutator $[a,b]$ is defined to be $ab-ba$ in ring theory, and the associator $(a,b,c)$ is defined to be $a(bc)-(ab)c$. So I want to find a 4-nary operator that behaves very differently from both the commutator and associator
One observation is that you need at least a pair of elements in order to define the commutator, and you need at least a triplet of elements in order to define the associator
Thus can go go further and find an operation that need at least 4 elements to define it?
Another possible triple thing is $ab+bc+ca-cb-ba-ac$
which I guess is $[a,b]+[b,c]+[c,a]$
so never mind
though still I don't know how to generalize that to 4 either
yeah, the trouble is there are almost no examples of string operations that needs at least 4 letters or 4 strings as inputs to function
17:24
Note that if you permute the inputs, it either stays the same or flips sign
This is like $a^2b+b^2c+ac^2-ab^2-bc^2-a^2c$ that can show up when you're solving cubics
which is also $ab(a-b)+bc(b-c)+ca(c-a)$
and also has two values depending on how you permite the inputs
Something with a 2x2 matrix?
hmm...
Determinant or trace maybe?
determinant and trace can function with any number of inputs
True enough.
17:28
Oh yeah, see if you can combine $\begin{vmatrix}a&b\\c&d\end{vmatrix}$, $\begin{vmatrix}a&c\\d&b\end{vmatrix}$ and $\begin{vmatrix}a&d\\b&c\end{vmatrix}$
I am trying to implement and use the Wasserstein distance as a loss function for a model (I know it's very short and naive description). The images are complex valued and the problem is that I get negative values of the distance, do you have any ideas why is that might be happening?
hmmm....
or maybe $ad+bc$, $ab+cd$, and $ac+bd$?
that sounds reasonable, this expression tend to pop up alot in e.g. matrix algebra, fractions and so on
(which is the same as the determinant but with $+$ instead of $-$)
17:32
that's the permanent of a 2x2 matrix
Oh I forgot that was a thing
Hmm, while permanent and determinant and in general immanents can take any inputs, I have seen that expression $ad+bc$ enough number of times that it can be justified to be defined as one object
hmm...
Given an algebraic structure $S$ with arity at least 4, Let $(a,b,c,d)$ be the <to be named>, which is the homomorphism given by:
$(ad+bc) \mapsto (ab+cd)$
$(ad+bc) \mapsto (ac+bd)$
If the algebraic structure is an n-ring where n>4, then $(a,b,c,d)_1 = ad+bc-ab-cd$ and
$(a,b,c,d)_2 = ad+bc-ac-bd$
Thus $S$ is <to be named>ative if $(a,b,c,d)_1=(a,b,c,d)_2=0$
actually better:
$(a,b,c,d) = ad+bc - (ab+cd)$
$(a,b,d,c) = ad+bc - (ac+bd)$
cyclic permutations saves the day
the only ring with this property is the trivial ring
$(1,1,2,2)=-1$
You'd expect $[a,[b,c]]+[b,[c,a]]+[c,[a,b]]$ to be something nice, but it turns out it's too nice
and the Jacobi identity says it's zero
@Secret Have you heard of the permutohedron and associahedron
associahedron yes, permutohedron nope
and wtf is this:
In geometry, the cyclohedron or Bott–Taubes polytope is a certain (n − 1)-dimensional polytope that is useful in studying knot invariants.The configuration space of n distinct points on the circle S1 is an n-dimensional manifold, which can be compactified into a manifold with corners by allowing the points to approach each other. This compactification can be factored as S 1 × W n {\displaystyle S^{1}\times W_{n}} , where Wn is the...
Also since it is getting late, I might continue on this tomorrow. I think one interesting expression to compute will be the following. If it does not vanish, then perhaps we can name $(a,b,c,d)$ the cyculator as it seems to be very sensitive to cyclic permutations
$$(a,b,c,d)+(a,c,d,b)+(a,d,b,c)=?$$
17:56
How many triangles are there in a 5x5 grid joining 3 points of the grid? Is 2136 correct?
So I guess you take all triples of points, subtract out the triples that are colinear, and then divide by 3!=6
Yeah?
Yeah
The only colinear things are gonna be vertical, horizontal, or slopes 1 or -1, yeah?
On an edge, there are 5*4*3=60 things
and then I guess the off-diagonals have 4*3*2=24 and the off-off-diagonals have 3*2*1=6?
and then see how many of each of those things you have
and subtract whatever you get at the end from 25*24*23=13800
and divide by 6?
So I'm not gonna do all that but 13800/6=2300 so you're certainly in the right ballpark @Curio
@Secret The reason I brought up the permutohedron was 'cause I was trying to visualize why exactly $[a,[b,c]]+[b,[c,a]]+[c,[b,a]]=0$ and each term of that is a permutation
like, $[a,[b,c]]=[a,bc-cb]=abc-acb-bca+cba$
So now I see - the permutohedron for 3 things is a hexagon
and we basically have 0, -1, and +1 marked on some of the rectangle's vertices (twice each)
and then we rotate it 60 and 120 degrees and add it to itself
and then each vertex gets 0 plus -1 plus 1, so 0 in total
18:14
@AkivaWeinberger Thanks
Can you insert a table into chat using latex?
https://math.stackexchange.com/questions/633979/how-many-triangles-can-be-created-from-a-grid-of-certain-dimensions
https://math.stackexchange.com/questions/8544/there-is-a-5-by-5-matrix-of-points-on-a-plane-how-many-triangles-can-be-formed<
https://www.quora.com/How-many-triangles-can-be-formed-out-of-a-5-X-5-grid-of-dots
@AkivaWeinberger so many different answers XD
18:30
Oh you CAN get slope 2
($\pm2^{\pm1}$)
Whoops
@Curio
Comment left as answer: " Michael Hardy's answer has a small flaw. In his math, he is assuming that there is are 4 5-point diagonals (he adds 30 together) where it should be 30 + 10, as there are only 2 5-point diagonals in the problem. The answer is 2148." — user147263 Feb 12 '15 at 16:15
The first answer writes this:
> With slope $1$, we have two lines of length $1$, two of length $2$, two of length $3$, two of length $4$, and just one of length $5$. Hence the number of degenerate triangles is
$$
2\binom 1 3 + 2\binom 2 3 + 2\binom 3 3 + 2\binom 4 3 + 2\binom 5 3 = 30.
$$
And the same with slope $-1$.
but that should be $1\binom53$ at the end
so he ends up with 20 fewer than he should
hence the discrepancy
so it should be 2148
The Quora answer seems to have forgotten about the things of slope 2
@AkivaWeinberger Ok I've understood my mistake: I had put 24 considering the 2x1 rectangles instead of 12
Thank you
18:51
How does one draw the covering $S_1 \vee S_1$ whose fundamental group is isomorphic to $\ker \phi$, where $\phi : F_2 \to \Bbb{Z}_2 \oplus \Bbb{Z}_3$ defined by $a \mapsto (1 + 2 \Bbb{Z}, 3 \Bbb{Z})$ and $b \mapsto (2 \Bbb{Z}, 1 + 3 \Bbb{Z})$?
I think I understand why drawing will have to have 6 vertices (although this is still somewhat hazy). But it isn't clear how to connect the vertices. Is there some procedure to follow?
19:36
I'm looking at an example in Vakil in which he computes the sheaf of relative differentials $\Omega_{\Bbb P^1_k/k}$ and this turns out to be a line bundle. However since $\operatorname{Pic}(\Bbb P^1_k)=\Bbb Z$ this sheaf of differentials must be one of Serre's twisting sheaves, and from a simple computation it turns out that $\Omega_{\Bbb P^1_k/k}=\mathcal O(-2)$.
This is clear so far, now Vakil claims that the tangent bundle having degree $2$ is related to the hairy ball theorem, but I don't see the connection, can someone explain it?
19:58
A section of O(2) has two zeros - so this is telling you a vector field on P1 has two zeros
(Counting multiplicity)
I'm still a bit confused: I know about the correspondence between vector bundles and locally free of constant rank $\mathcal O_X$-modules, but what's the relationship between (global) sections of such a module and sections of the bundles?
Do you know that the global sections of O(n) correspond to degree n homogeneous polynomials?
The relationship is what you expect! The locally free sheaf is precisely remembering the sections of the bundle
@loch I think I had this as an exercise, I'll look it up again
@loch Oh, that makes a lot of sense now then! Thanks
20:27
Quick question, something I never realized. If A is a sheaf of rings, and F is a sheaf of A-modules, what category does F take values in?
Like, each module is a module over a different ring
What do you mean with a sheaf of $A$-modules? Like $F(U)$ is an $A(U)$-module for all open $U$?
In mathematics, a sheaf of O-modules or simply an O-module over a ringed space (X, O) is a sheaf F such that, for any open subset U of X, F(U) is an O(U)-module and the restriction maps F(U) →F(V) are compatible with the restriction maps O(U) →O(V): the restriction of fs is the restriction of f times that of s for any f in O(U) and s in F(U). The standard case is when X is a scheme and O its structure sheaf. If O is the constant sheaf Z _ {\displaystyle {\underline {\mathbf...
" a sheaf of O-modules or simply an O-module over a ringed space"
If you want to be really formal a sheaf of $A$-modules $F$ is a sheaf of abelian groups with a map of sheaves of sets $F\times A\to F$ giving the module structures
ok, cool that is the kind of answer I am looking for.
Can we combine the ratio test together with Weierstrass M-test to show that the series corresponding a bounded nonvanishing sequence of functions converges uniformly?
20:45
@AlessandroCodenotti could we also perhaps be perverse and just define the functor $F$ that yields a sheaf of $A$-modules, and then just define the essential image to be a category where we only specify the maps $F(i)$ where $i$ was a morphism in $OP(X)$.
Does that make sense? I feel like I worded that poorly. Basically take the thing you want to be a functor and use it to define the category so that the "functor" is a functor. Or would that route be poorly behaved
This is getting too categorical for me, I don't know
ok no problem
@PrinceM they form an abelian category with enough injectives
hence sheaf cohomology
21:18
@PrinceM a sheaf of $A$-modules is in particular a sheaf of abelian groups which makes it an actual functor (you just have more structure)
@PrinceM you could define a category that has as objects pairs $(R,M)$ where $R$ is a ring and $M$ is a left module and as morphisms pairs $(\varphi,f):(R,M)\to (S;N)$ where $\varphi:R \to S$ is a ring homomorphism and $f:M \to \varphi^*N$ is $R$-linear
if you do this, then for an $A$-module $F$, the assignement $U \mapsto (A(U),F(U))$ is a functor from $OP(X)^{op}$ to that category
(not sure why you would do that, though)
21:35
Can we have a bijection between areas?
Yes
consider upvoting this question:: math.stackexchange.com/questions/2915911/…
or answering it
someone in the comments said: "I don't know what you mean by a bijection between areas. Integrals are numbers – what's a bijection between two numbers? If you're asking for a bijection between two regions in the plane, well, each region contains a continuum of points, so they have the same cardinality, so of course there's a bijection. Maybe you could give an example of a bijection between two areas, so we could see what you mean."
21:53
But now I've learned that there is no bijection between areas
@Érico and anyone else who's interested: Here's Robert Bryant's retiring presidential address from the AMS on Holonomy. Quite beautiful talk — I just watched the whole thing.
5
Hi
How do I prove that differentials are basis vectors for the space of one forms?
hey @TedShifrin, how are you?
@topologicalmagician: I just gave you the reference in my lectures ... middle of lecture 24.
LOL, have you watched it before, Eric?
21:58
thanks! @TedShifrin, I hope youre doing well
I just got a link to it from Ken Ribet's Facebook post.
Trying to get healthy, but otherwise fine, @topologicalmagician.
@TedShifrin that's great to hear
@loch A holomorphic vector field, of course.
i'm waiting till I get your book, in the mean time i'm using a book, and it defines differentials as $dx_i(v)=v_i$
@TedShifrin
Yes, that's where I started as well.
I originally used $\phi_i$ for $dx_i$, then switched notation a lecture later ...
What book are you using?
22:04
calculus by Robert adams
Oh, I taught out of that book one year. It's not quite sophisticated enough, but it's ok.
heya italic @Alessandro
@TedShifrin yup
yeah, its not as rigorous as id like it to be but im managing
That book, if I recall correctly, has an astonishing error in it ... it states something as true for which it gives itself a counterexample elsewhere. If I remember correctly.
22:06
oh, i'm not sure
What is a set of an infintesimal called?
$S=\{dx\}$
I remember the error ... He asserted that a function that has a minimum at the origin along every line through the origin must have a local minimum at the origin (thought of as a multivariable function). This is indeed very false.
@Ultradark set of infinitesimals :P
$S=\{dx_1,dx_2,...\}$
$dx_1+dx_2+...=c$
@TedShifrin oh, thanks for telling me
22:11
$\sum S=c$
Anyhow, I guess you don't want to watch my lectures, but they're quite self-contained. The only warning is that #24 (the one I referred to) had a video glitch the first time we taped, and so there's a complete version (done a year later) at the very end of the list. @topologicalmagician.
@TedShifrin i'm watching your lectures right at this moment
@TedShifrin is it true that for every morse function, if you start at any point on the manifold and follow the gradient flow, you end up at a critical point?
Oh, OK ... Sorry :P
on a compact manifold
22:13
I was about to ask about compactness?
sorry :P
Assuming you don't start at a critical point?
if you start at a critical point then you always stay, right
Yes, then ...
@TedShifrin at Ted, after i'm done with your lectures on diff forms, which ones do I need to start with the ones about pullback?
22:16
$S=\{h_1,h_2,...\}$
Pullback is in there, @topologicalmagician, right after exterior derivative ... You can always jump in wherever you want, but you risk not understanding things, so it makes more sense to be somewhat linear ...
can a finite value be assigned to a divergent integral
22:33
@Ultradark "assigned" in what sense?
@TedShifrin is there a popular definition for "singular foliation"? Or is it another one of those names that have a billion different interpretations?
For dimension 1, I have the definition of an atlas of pairs $(U_i,X_i)$ where $U_i$ cover $M$, $X_i$ is a vector field on $U_i$, and for each pair of indices $(i,j)$, there is a non-vanishing function $f_{ij}$ on $U_i\cap U_j$ such that $X_i = f_{ij}X_j$.
@anakhro: I don't know an official definition. I think of it as a foliation that messes up on a set of measure 0. Such things show up naturally in algebraic geometry, where you have a mapping whose generic fibers are all smooth and diffeomorphic but there are special fibers that are singular.
For example, you can take the curves $y^2=x(x-1)(x-t)$. Projecting to $t$-space, the fibers over $t=0$ and $t=1$ are nodal cubics, and the rest are all smooth cubics (diffeomorphic).
22:53
I don't know whether to be extremely formal about the characteristic foliation or not.
The standard definition is due, I think independently, to Stefan and Sussman. I would not be surprised if various authors weakened or strengthened the definition as appropriate to their needs. It's given by a locally finite submodule of smooth vector fields (from which you can extract the distribution).
So it's basically like a vector bundle whose fibers drop/change rank along a subvariety ...
I guess my total space, as I've defined my example, won't quite be a manifold.
Yes, that's exactly the idea. You want to talk about vector fields because it would be ridiculous to allow any old decomposition into manifolds.
@anakhro $A=\{\int_0^{n_1}f(x)dx,\int_0^{n_2}f(x)dx,...,\int_0^{1}f(x)dx\}.$

$ B=\{\int_2^{k_1}g(x)dx,\int_2^{k_2}g(x)dx,...,\int_2^{\infty}g(x)dx\}. $

$A\mapsto B,$ the first element in $A$ is assigned to the first element in $B$ and so on.

$\int_0^1f(x)dx$ converges.

$\int_2^\infty g(x)dx$ diverges.
so the last term in $A$ assigned to the last term in $B$ assigns a finite value to an infinite value
@MikeM: Well, I was assuming a local product structure, still, as usual, off the "bad points"
23:06
right
even this notion allows for some nasty objects, since it includes any (unparameterized) dynamical system: eg, the span of a vector field vanishing along a cantor set
to get a good example of why this local demand is reasonable you'd want to think about higher-dimensional stuff than i'm willing to bother with right now
let's say we have a sequence of numbers where the next term is the sum of the two previous terms all the way up to a 100. so 1,3,4,7,11,18,29. and we are looking for multiples of 5. We know that there is a sequence that repeats when we take the remainder of these numbers after dividing by 5. 1,3,4,2 , 1,3,4,2, etc... are there any rules that say, if say if I encounter this pattern so many times that there are no multiples of say, 5 in this sequence.
I know this seems like a stupid question, but is there ever going to be instances where a rule like this could be broken where I might see a repeating pattern for some time and then not at all.
23:22
@Ultradark sorry I have no idea what you are desiring. :(
23:47
It's okay @anakhro
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