@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
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
(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 $\Bbb Z/2\Bbb Z$)
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?
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
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
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
Definition: Given an algebraic structure with arity at least 4, then a **Quadrilaterlator** $(a,b,c,d)$ is defined to be the homomorphism: $(a,b,c,d) \mapsto (a,b,d,c)$ $(a,b,c,d) \mapsto (a,c,b,d)$
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?
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?
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:
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
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
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
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
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?
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?
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?
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
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?
@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
@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
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."
@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.
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.
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.
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 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?
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 ...
@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).
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).
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.