Mathematics - 2019-04-30 (page 1 of 2)

user193319

12:02 AM

4

Q: Open Sets in the Wedge Sum and a Homeomorphism

I am presently working through example 1.21 in Hatcher's book on wedge sums of topological spaces. He makes a few claims which I am having trouble verifying. First, let me set-up some notation. Let $\{X_i\}_{i \in I}$ be a collection of topological spaces. Then $\amalg_{i \in I} X_i := \cup_{i ...

Nicholas Roberts

1:30 AM

Hi all, can anyone provide a hint as to solve the following: Suppose $f$ is a complex-valued harmonic function on a domain $D$. That is, both the real and imaginary part of $f$ are harmonic. Show that if $|f|$ is constant on $D$, then $f(z)$ is constant in $D$.

Semiclassical

Seems like the maximum modulus theorem might sort you out?

Nicholas Roberts

Perhaps! We weren't introduced to the max-mod theorem for harmonic functions yet in lecture, so I am not sure if this was the intended method of proof.

I'm trying to brute force it using all of the assumptions but it gets awfully messy and I'm getting nowhere... So I shall try the max-mod theorem.

Ted Shifrin

You don't need anything fancy. This follows from the Cauchy-Riemann equations quite easily.

Hi @Semiclassic

Semiclassical

hi @ted

yeah, true

Ted Shifrin

Did you type the wrong thing? Oh, $f$ is a complex-valued harmonic function. I missed that.

Yeah, this is a bit less obvious.

Semiclassical

1:37 AM

for convenience, it's probably best to use $|f(z)|^2=f(z)\overline{f(z)} =u(z)^2+v(z)^2$

Ted Shifrin

But all we know is that $u$ and $v$ are totally independent harmonic functions.

Nicholas Roberts

@TedShifrin, How can I use CR if f is not assumed to be holomorphic?

Semiclassical

hmm, true

Nicholas Roberts

Yes.

Ted Shifrin

@Nicholas: I misread.

Semiclassical

1:38 AM

that'll also kick my max-mod idea in the teeth

Nicholas Roberts

Yes, so the modulus squared is constant. So should I take derivatives?

I tried this and got nowhere. I'm probably not seeing the right trick

Ted Shifrin

I was thinking that. So $\Delta(u^2) = 2(u\Delta u + \|\nabla u\|^2)$?

Than we get $\|\nabla u\|^2 + \|\nabla v\|^2 = 0$, which kills it.

Semiclassical

nice

Ted Shifrin

Assuming I didn't do something stoooopid.

Nicholas Roberts

Upside-down triangle refers to $u_x + u_y$, yes?

Ted Shifrin

1:42 AM

No, the gradient. $(u_x,u_y)$.

Nicholas Roberts

Gotcha. And your norm is simply dot product with itself? Thank you

Ted Shifrin

Yup. Better check that I didn't do anything stooopid.

Nicholas Roberts

$norm squared**

Ted Shifrin

Norm squared.

LOL.

Nicholas Roberts

Yes, I will lol

Ted Shifrin

1:44 AM

And "than" = "then" :P

I guess you don't care about that.

Nicholas Roberts

Haha, as long as the math is clear, its all good 8)

Ted Shifrin

OK, I'm leaving for dinner. Bye, all.

Nicholas Roberts

see ya, thanks for the help

Carrick

4:48 AM

Hello everyone...I have a naive question...and I.e. when is the sum of distances from two given points to a point on a given line minimum?? I thought it to be initially on the intersection of the given line and perpendicular bisector of the two given points...but it turns out that I was wrong. So, yh any help will be appreciated

anakhro

5:05 AM

@Carrick sounds like the geometric median.

Karl Kronenfeld

5:32 AM

@Carrick Let $A$ and $B$ be the two points, and let $\ell$ be the line. Let $C\in\ell$ be a sum-of-distances minimizer. The line $\ell$ is tangent to an ellipse with foci $A$ and $B$ at the point $C$, so the reflection property of ellipses holds. Thus, the angle $AC$ makes with $\ell$ (that doesn't contain the vertex angle of the triangle $ABC$ at $C$) is the same as the angle $BC$ makes with $\ell$. This characterizes $C$.

There has to be an elementary way to find the triangle $ABC$ with this information, but I do not see how to do it.

Carrick

6:08 AM

@anakhro i don't think its geometric median..or even if it is...its a rather simple case

@KarlKronenfeld thank you for your time, I actually found the answer in my notes and its like this: we take the image of B about the given line(lets call it B'). So, obviously CB=CB' and then we know to minimize AC+CB, we can minimize AC+CB', which happens when A,C & B' are collinear.

Leaky Nun

once we move the function's domain to the unit disc we gain radial symmetry...

so again let $\phi = \phi(r)$, so $\nabla^2 \phi = \frac1r (r \phi')' = 0$ implies $\phi = C \log r + D$, and what do you know, $\log r$ vanishes on the unit circle

Karl Kronenfeld

6:26 AM

@Carrick Ah, great.

Leaky Nun

$\dfrac{z-i}{-iz+1} = \dfrac{x+iy-i}{-i(x+iy)+1} = \dfrac{x+i(y-1)}{1+y-ix} = \dfrac{2x+[x^2+y^2-1]i}{(y+1)^2+x^2}$

so $G(x,y;0,1) = C \log \dfrac{4x^2 + (x^2+y^2-1)^2}{((y+1)^2+x^2)^2}$

$=C \log \dfrac{(x^2+1)^2 + 2y^2(x^2+1) + y^4 - 4y^2}{((y+1)^2+x^2)^2} = C\log \dfrac{(x^2+y^2+2y+1)(x^2+y^2-2y+1)}{((y+1)^2+x^2)^2} = C \log \dfrac{x^2+(y-1)^2}{x^2+(y+1)^2}$

omg @Semiclassical I derived your answer

well sort of

now $G(x,y;\xi,\eta) = G\left(\frac{x-\xi}\eta,\frac y\eta;0,1\right) = C \log \dfrac{(x-\xi)^2 + (y-\eta)^2}{(x-\xi)^2 + (y+\eta)^2}$

yes indeed it matches with your answer 100%

now to test what $C$ is, we use a test function $\varphi$ right

we want $\displaystyle \int_{\Bbb R \times (\Bbb R_{\ge0})} \varphi(\xi,\eta) \nabla^2 G(x,y;\xi,\eta) \ \mathrm d\xi \ \mathrm d\eta = \varphi(x,y)$ formally

using Green's identity $\displaystyle \int_M u \nabla^2 v - v \nabla^2 u = \int_{\partial M} uNv-vNu$

the LHS becomes $\displaystyle \int_{\Bbb H} G \nabla^2 \varphi + \int_{-\infty}^\infty \varphi(x,0) G_y(x,0;\xi,\eta) - \varphi_y(x,0) G(x,0;\xi,\eta) \ \mathrm dx$

since $G$ vanishes on the x-axis, the third term becomes zero

well $G_y = C \dfrac{2(y-\eta)}{(x-\xi)^2 + (y-\eta)^2} - C \dfrac{2(y+\eta)}{(x-\xi)^2 + (y+\eta)^2}$

if we let $\varphi = 1$ then $\nabla^2 \varphi = 0$, so the equation becomes

$\displaystyle \int_{-\infty}^\infty G_y(x,0;\xi,\eta) \ \mathrm d\xi = 1$

$2C \left[ \arctan\left(\dfrac{\xi-x}{-\eta}\right) - \arctan\left(\dfrac{\xi-x}{\eta}\right) \right]_{-\infty}^\infty = 1$

$2C\left[\left(-\dfrac\pi2 - \dfrac\pi2\right) - \left( \dfrac\pi2 + \dfrac\pi2 \right)\right] = 1$

$C = \dfrac{-1}{4\pi}$

so in conclusion, $G(x,y;\xi,\eta) = \dfrac1{4\pi} \log \dfrac{(x-\xi)^2 + (y+\eta)^2}{(x-\xi)^2 + (y-\eta)^2}$

pregunton

7:41 AM

Hello, I have a (maybe stupid) question. Yesterday I started a bounty on this post, but today I can't find it in the tab of featured questions. Is this normal?

Jacksoja

@TobiasKildetoft Hi Tobias

Tobias Kildetoft

@Jacksoja Hi

Jacksoja

@TobiasKildetoft I have a small question about linear algebra

Tobias Kildetoft

sure

Jacksoja

in the book linear algebra done right, how can we view the elemnts of F^n

as functions from {1,2,..,n} to F

Tobias Kildetoft

7:51 AM

same way we view them outside that book

An $n$-tuple of elements from the set $X$ is the same as a function from $\{1,2,\dots, n\}$ to $X$

Jacksoja

I know ^^ my point is , shouldnt the map be from the set of intergerns up to n to F^n ?

Tobias Kildetoft

no

because the elements are $n$-tuples of elements from the field, not from the space

Jacksoja

but what is the map explictly ?

Tobias Kildetoft

what map? Each elements corresponds to a map

Jacksoja

because the way i see it, and what i dont understand , is that we do not get an element of a field

we do get a tuple

f : {1,2, ...,n } --> F

we send an n-tuple x in {1,2, ...,n } to what in F ?

Tobias Kildetoft

7:54 AM

$\{1,2,\dots, n\}$ does not contain any tuples

it contains those $n$ numbers

The correspondence sends the $n$-tuple with $x$ at the $i$'th place to the function which sends $i$ to $x$

Jacksoja

hmm can we do small example ?

like in R^4

Tobias Kildetoft

So $(x_1,\dots, x_n)$ corresponds to the function $f(i) = x_i$

Jacksoja

the idea that this is a function is not clear to me

Tobias Kildetoft

So let's take the vector $(1,2,5,6)$

this corresponds to the function $f$ with $f(1) = 1$, $f(2) = 2$, $f(3) = 5$ and $f(4) = 6$.

Jacksoja

okaaaaaaay

I see now thanks !

so this is the same way as we define maps from R--> R

Tobias Kildetoft

7:58 AM

not sure what you mean

Jacksoja

component wise addition and scalar multip

nothing too important but i get what you said

Tobias Kildetoft

ahh, yes, when we make it a vector space

Jacksoja

I did not think of defining the map f like that

now i see it clearer when you did it component at the time

Tobias Kildetoft

anyway, I need to go shop some groceries now.

Jacksoja

okay thanks again !

and good luck !

Sebastian Alexander B Nielsen

8:20 AM

Are these synonyms?

$$(d/dx) * x^2 = 2x$$

$$(x^2)' = 2x$$

Please, anyone?

Jacksoja

8:43 AM

@SebastianAlexanderBNielsen yes they are , just different notation

Sebastian Alexander B Nielsen

I knew it, thanks for confirming

Rithaniel

9:20 AM

Is there a term for a semigroup of order n which contains a sub-semigroup of order n-1?

For example, the multiplicative structure on a ring is a semigroup. If that ring is an integral domain, then the multiplicative structure on that ring excluding zero is also a semigroup, as I understand it.

Tobias Kildetoft

yes, that is correct

Rithaniel

So, I assume there's not exactly a name for this property?

Tobias Kildetoft

I have never heard one, but I have never really studied semigroups explicitly

(I have actually studied multisemigroups more)

Rithaniel

I've never heard of a multisemigroup. (Also, I'm discovering that it's difficult to find terminology if you don't know where to look.)

Tobias Kildetoft

a multisemigroup is a structure $X$ with a map $m: X\times X\to \mathcal{P}(X)$

which is associative once you do the "obvious" extension of $m$ to subsets

Rithaniel

9:31 AM

Hmmmm, that's interesting.

Tobias Kildetoft

They show up when studying certain types of $2$-categories

and they give a nice way to phrase some properties of positively based algebras

Rithaniel

Ah, category theory. I still want to get into that.

So, this map can never be bijective, so it only goes in one direction.

Tobias Kildetoft

which map?

Rithaniel

$m$, from your multisemigroup structure.

Tobias Kildetoft

I suppose it might be possible for it to be bijective if the set is small enough (explicitly $2$ elements), but I have no idea of it could still be associative

Rithaniel

9:37 AM

I'm used to studying objects with operations which have closure. So this sort of idea seems strange at a glance.

Tobias Kildetoft

A way these show up is the following: Let $A$ be an algebra (say over the reals) with a basis $\{b_1,\dots, b_n\}$ such that whenever we write $b_ib_j = \sum_{k=1}^nc_{i,j}^kb_k$ then all $c_{i,j}^k\geq 0$

We now get a multisemigroup structure on $\{b_1,\dots, b_n\}$ by setting $m(b_i,b_j) = \{b_k\mid c_{i,j}^k > 0\}$

And we also get some partial preorders on that set in a way that works for any multisemigroup, by letting $b_i \geq_L b_j$ if $b_i\in m(b_k,b_j)$ for some $k$

(and extending transitively)

this then gives rise to "cells" consisting of elements that are comparable in both directions using this order, and an example of these are the "famous" Kazhdan-Lusztig cells

Rithaniel

Heh, yes, those cells.

Tobias Kildetoft

as I said "famous" (assuming you have studied Kazhdan-Lusztig theory of course)

Rithaniel

This is a little bit out of my depth, I'm feeling. I feel like I'd want to play around with such an algebra to get a sense of how it works.

Tobias Kildetoft

this was not how they were defined originally btw, since the relevant basis was not known to have the positivity property mentioned until fairly recently

There are some nice example: Take a group algebra with the usual basis

Or for that matter (a bit more interesting), a semigroup algebra

Or, for an example of the above KL stuff, take the group algebra for $S_3$ with the basis $1$, $s+1$, $t+1$ where $s$ and $t$ are the transpositions $(12)$ and $(23)$

Sorry, plus the elements $st + s + 1$, $ts + t + 1$ and $sts + st + ts + s + t + 1$

(I think I got those right)

One can also take $S_2$ with basis $1$ and $s+1$ to start smaller

Rithaniel

9:48 AM

I'm only familiar with one operation on transpositions. This seems to be two operations.

Also, mixing a number with a transposition is new.

Tobias Kildetoft

Ahh, I assume you are not familiar with group algebras then

that makes these examples a lot less of a good starting point.

Rithaniel

I believe not. (I was expecting this would just be referring to groups)

Tobias Kildetoft

no, we needed algebras, not just groups above.

Rithaniel

Okay, then what is the definition of an algebra?

Looked up the definition on wikipedia. Now I need to look into bilinearity.

Tobias Kildetoft

bilinearity is just the distributive law on each side

Akiva Weinberger

10:14 AM

@Rithaniel $(x,y)\mapsto xy$ is not a linear map from $\Bbb R^2\to\Bbb R$

but it is bilinear

(If you hold one of the inputs constant the result is linear)

Rithaniel

Ah, okay, that makes it easier to understand.

Secret

10:57 AM

In general, a multilinear map is a tensor

Tobias Kildetoft

@Secret I hate the term "a tensor"

a multilinear map is a linear map from a tensor product with a suitable number of factors.

Silent

11:10 AM

hi @AkivaWeinberger

In light of this can we say that if restriction of homomorphism on $\Bbb Q[x]$ to $\Bbb Z[x]$ does not change the generator of principal ideal, right?

user193319

11:41 AM

I am trying to show that $\tilde{H}_n(X \coprod Y) \cong \tilde{H}_n(X) \oplus \tilde{H}_n(Y)$, where $\tilde{H}_n(\cdot)$ denotes the $n$-th reduced homology. By Mayer-Vietoris, I was able to show that $H_n(X \coprod Y) \cong H_n(X) \oplus H_n(Y)$, so for $n > 0$ I get the desired formula.

But when $n=0$, things get a little more complicated: $H_0(X \coprod Y) \cong H_0(X) \oplus H_0(Y)$ becomes $\tilde{H}_0(X \coprod Y) \oplus \Bbb{Z} \cong \tilde{H}_0(X) \oplus \tilde{H}_0(Y) \oplus \Bbb{Z}^2$

I don't see why $\tilde{H}_0(X \coprod Y) \cong \tilde{H}_0(X) \oplus \tilde{H}_0(Y)$ holds.

Semiclassical

12:01 PM

@LeakyNun lol nice

I should point out, I guess, that my “guesswork” was a little more informed than it may have seemed

If you do classical electrodynamics, then you’ll calculate the electric field and potential of various charge distributions

And some of those cases correspond to the various free Greens functions

For instance, the potential of an infinite line charge, running along the z-axis, is proportional to log(x^2+y^2)

Away from the z-axis, this is a solution to the 3D Laplacian. But the solution isn’t a function of z, so if we restrict that solution to z=0 we get a solution to the 2D Laplacian

So that gives you the free 2D Greens function

The only extra ingredient needed is the so-called method of images

The idea being that, if you take $(\chi, \xi)$ to lie outside the domain

user123

does anyone have an idea how to solve $x^2 dy= (x^2 +xy+y^2)dx $ ? (with y(1)=1 so y=0 not a solution)..

i thought maybe playing with $(x+y)^2$ but it didn't work..

Semiclassical

Then you can add $G(x,y;\chi,\xi)$ to any solution of the 2D Laplacian, and it’ll still be a solution

@user123 the fact that both coefficients are homogeneous of degree 2 suggests a substation like $x=zy$

(You could instead do $y=zx$. Won’t make a particular difference)

user123

12:17 PM

@Semiclassical nice.. thank you!

user123

12:38 PM

@Semiclassical i get that $tan^{-1}(xy) = ln(|x| ) + C$ is a solution, ain't that problematic?

(i used $v = xy$ )

Semiclassical

Maybe, but that approach seems basically flawed. The point of the substitution is to express the ODE in terms of y/x, not xy

Your substitution should have made the ODE worse, not better

user193319

1:23 PM

How does one show that $H_n(X \coprod Y, \{(x,1),(y,2)\}) \cong H_n(X \coprod Y)$, where $x \in X$ and $y \in Y$?

I guess the LHS could also be written as $H_n(X \coprod Y, \{x\} \coprod \{y\})$...not sure if that's helpful.

Perhaps representing the disjoint union as cartesian product might help...

Mathphile

1:47 PM

2

Q: Is the set of elementary functions which do not have elementary integrals bigger than set of elementary functions which have elementary integrals?

It increasingly seems to me that the functions that have elementary integrals are quite rare in comparison to the ones that don't have them. Even raising an elementary function to a different power may result in it not having an elementary integral. Ex. $(\arctan (x))^{1/2}$ Also many seemingly...

calculus integration

Any ideas?

Semiclassical

2:09 PM

@Mathphile reminds me a bit of this old question of mine: math.stackexchange.com/q/1017447/137524

Rudi_Birnbaum

Hi all!

I got some equation: $$\begin{pmatrix}0 & a & b \\\ -a & 0 & c \\\ -b & -c & 0\end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} t_x \\ t_y \\ t_z \end{pmatrix}$$

say $$A x = t$$

Semiclassical

with $A$ being skew-symmetric

Rudi_Birnbaum

Since $\det(A)=0$

I know there is no unique solution. I think its a "line" of points.

What I want is some $A'=V^T A V$ such that $A'$ gets 2x2 (possibly symmetric?)?

Is this possible or am I dreaming?

Semiclassical

the starting point should presumably be to note that $x=(c,b,-a)^\top \implies Ax=0$

Rudi_Birnbaum

I mean something like $$ A'=\begin{pmatrix}0 & 0 & 0 \\\ 0 & q & r \\\ 0 & r & s\end{pmatrix} $$?

@Semiclassical I know that thats e..g. a cross product

of two columns divided by some coefficient

And now V shall rotate A such that x and that get parallel

just want a ... formula ...

Semiclassical

2:17 PM

Let's proceed more abstractly. Let $v$ be the vector such that $Av=0$.

Rudi_Birnbaum

ok

Semiclassical

for simplicity, I'll take $v$ to be a unit vector

then the corresponding projection operator onto that direction is $P=vv^\top$

and the projector onto the orthogonal complement is $I-P=I-vv^\top$

why?

Why to which?

both

2:19 PM

Suppose I write a given vector as the sum of a part which is in the direction of $v$ and one which is perpendicular to $v$

Rudi_Birnbaum

yes

OK

Semiclassical

So $x=av+bu$ where $u^\top v=0$

Rudi_Birnbaum

I understood

Semiclassical

then $Px = a v$ etc

Rudi_Birnbaum

resolution of the identity

Semiclassical

2:20 PM

yeah

Rudi_Birnbaum

(like)

OK!

Semiclassical

it's of a kind with that, yeah. resolution of identity says that the identity can be written as a sum of projection operators

anyways

Rudi_Birnbaum

So I have the two bits from that

ÍgjøgnumMeg

Hoi @Rudi_Birnbaum !

Rudi_Birnbaum

@ÍgjøgnumMeg Hoi (sorry just leaning here)!

@Semiclassical pls conztinue

ÍgjøgnumMeg

2:21 PM

:D

Rudi_Birnbaum

@Semiclassical from here

Semiclassical

right. let me label that second projector as $P_\perp= I-vv^\top$

Rudi_Birnbaum

ok

I now what to do

3. binomial

?

yeah?

(I-v)(I+v)

Semiclassical

well, following my nose, I'd note that by construction I've got the resolution of identity $I=P+P_\perp$. So I can write $Ax=AIx = A(P+P_\perp)x$

But $AP=0$ by virtue of $Av=0$

Rudi_Birnbaum

yes

Semiclassical

2:24 PM

so $Ax = AP_\perp x$

trying to remember what to do next. Should be obvious

Rudi_Birnbaum

we need a transpose I guess

Semiclassical

well, we do know at this point that $P_\perp x$ lies in the orthogonal complement of $v$

Rudi_Birnbaum

yes

Semiclassical

But that doesn't seem to clarify matters as much as I'd like.

I guess, ideally, we're looking for a rotation which maps $v\to e_1$

Rudi_Birnbaum

yes something like that

Semiclassical

2:32 PM

well. following my nose again: We should have vectors $v_0,v_+,v_-$ such that $Av_0=0,Av_+ = \lambda v_+,Av_- = - \lambda v_-$

Rudi_Birnbaum

yes makes sense

well no

$A v_0 = v_0$

Semiclassical

No. That'd mean $A$ has 1 as eigenvalue

Rudi_Birnbaum

right ...

well cant we just diagonalize A ... by guessing?

I mean "diagonalize"

Semiclassical

I'm more used to diagonalizing symmetric matrices

I think there's some subtleties for skew-symmetric

Rudi_Birnbaum

I try to diagoanlize anything that is not up in the trees upon counting to $n \times n$

Semiclassical

2:38 PM

for instance: $v_+^\top A v_+ = v_+^\top A^\top v_+ =- v_+^\top A v_+ = \lambda v_+^\top v_+$

Rudi_Birnbaum

@ÍgjøgnumMeg: whats the hight of a $pizza$?

Semiclassical

which looks all sorts of bad, since the first and third terms imply that that expression should equal zero

hmm, take a look at this: en.wikipedia.org/wiki/Skew-symmetric_matrix#Cross_product

Rudi_Birnbaum

@ÍgjøgnumMeg $a$ since the whole is $\pi z^2 a$ ...

Semiclassical

so one should have $Ax = t\implies \vec{t} = (-a,b,-c)\times \vec{x}$

Rudi_Birnbaum

@Semiclassical Oh wait ...

ÍgjøgnumMeg

2:42 PM

@Rudi lol nice

@Rudi ich hab ein Stipendium bekommen!

Semiclassical

Note that $\vec{t}$ must be perpendicular to $(-a,b,-c)$

and therefore should lie in the plane through the origin with normal $(-a,b,-c)$

Rudi_Birnbaum

@Semiclassical yes!

Very nice. So the equation corrsponds to an inversion of the corss product ...

But $V$ is still open ...

@ÍgjøgnumMeg Geil Mann! Glückwunsch!!

Semiclassical

Well, I'd say $x\mapsto Ax$ corresponds to the cross product with $(-a,b,-c)$ (acting from the left)

Rudi_Birnbaum

@ÍgjøgnumMeg Wo was wie?

Semiclassical

the example $(-a,b,-c)=(1,0,0)$ may clarify things

oh, blah, I didn't transcribe that right

ÍgjøgnumMeg

2:50 PM

@Rudi danke :) Vom deutschen akademischen Austauschdienst

Rudi_Birnbaum

@ÍgjøgnumMeg Das ist gut!

Semiclassical

$$\vec{a}=(a_1,a_2,a_3)\implies A=\begin{pmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{pmatrix}$$

ÍgjøgnumMeg

Ja voll gut!

Semiclassical

So that's $\vec{a}=(-c,b,-a)$

Anyways. Taking the example of $\vec{a}=(1,0,0)\implies (a,b,c)=(0,0,-1)$

and therefore $A=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0\end{pmatrix}$

In which case it's of the form you expected

$x\mapsto Ax = (0,-z,y)^\top$ has the following effect: We project $x=(x,y,z)^\top$ to the $yz$-plane, and then rotate it by 90 degrees around the z-axis

Not sure what else one can say there

Rudi_Birnbaum

Well some explicit form for $V$ such that $VAV^T$ has that form would be cool.

Semiclassical

2:57 PM

Agreed.

Rudi_Birnbaum

given your $A$.

Semiclassical

if you glance down to the section on spectral theory in the wiki link

Rudi_Birnbaum

Then I know how to transform my space such that I get a problem with a unique solution.

Semiclassical

you are looking for an orthogonal transformation to get it of the desired form

one thing to note: By the nature of this mapping, you shouldn't expect to get real eigenvectors

Rudi_Birnbaum

a special unitary...

yes since antisymmetric matrices have imaginary ones ...

Semiclassical

2:59 PM

your transformation is (basically) to do a 90 degree rotation of your vectors

Rudi_Birnbaum

really 90?

Semiclassical

yeah. $(-c,b,-a)\times \vec{x}$ is perpendicular to both $(-c,b,-a)$ and $\vec{x}$

if you restrict $\vec{x}$ to the $(-c,b,-a)$ plane, that's just 90 degree rotation

(since then $\sin \theta=1$)

Rudi_Birnbaum

but we start with $e_\alpha$ ...

with the unit vectors in the original system

and we must rotate them onto $(c,-b,a)$

Semiclassical

The upshot, I think, being that you look for perpendicular vectors $v_\pm$ such that $Av_+ = \lambda v_-$ and $Av_- = -\lambda v_+$

Rudi_Birnbaum

@ÍgjøgnumMeg und wo gehts jetz hin??

ÍgjøgnumMeg

3:04 PM

Heidelberg!

Rudi_Birnbaum

@ÍgjøgnumMeg Super!! Wann??

@ÍgjøgnumMeg Ich glaub da gibts auch ganz fesche Hasen ;-)

ÍgjøgnumMeg

@Rudi :D Im Oktober faengt das Sommersemester an

err

Winter*

Semiclassical

To be a bit more precise about this line of thinking: Note that, if such vectors exist, then $A^2 v_\pm = -\lambda^2 v_\pm $

Rudi_Birnbaum

yes

Semiclassical

But $A^2 = \left(
\begin{array}{ccc}
-a^2-b^2 & -b c & a c \\
-b c & -a^2-c^2 & -a b \\
a c & -a b & -b^2-c^2 \\
\end{array}
\right)$

which has determinant 0 and trace $-2a^2-2b^2-2c^2$

mathematica moreover says that the eigenvalues are just $0,-a^2-b^2-c^2,-a^2-b^2-c^2$

Rudi_Birnbaum

3:09 PM

degenerate

Semiclassical

and that's pretty easy to believe. Note that $A^2+(a^2+b^2+c^2)I=\begin{pmatrix} c^2 & -b c & ac \\ -bc & b^2 & -ab \\ ac & -ab & a^2\end{pmatrix}$

Rudi_Birnbaum

sure

Semiclassical

and the eigenvectors there are pretty nice

the eigenvector for 0 is still just $(-c,b,a)$. the orthogonal complement is spanned by $(a,0,c)$ and $(b,c,0)$

Not totally happy with that yet tho

blah. eigenvector for 0 is $(-c,b,-a)$ and the orthogonal complement is spanned by $(a,0,-c),(b,-c,0)$

Rudi_Birnbaum

3:26 PM

am away for 1 h.

nbro

3:55 PM

Is there like an operator in mathematics that swaps the dimensions of a tensor?

For example, say I have the tensor $a \in \mathbb{R}^{n \times m \times h \times k}$. I would like to make $a$ like e.g. this: $a \in \mathbb{R}^{n \times k \times m \times h}$.

I think that, in theory, these are equivalent

However, I would like to know if there is a formalism (an operator) that does this

Secret

3

Q: Defining tensor transpose without representing them as matrices

In the comments of this post there was a discussion about why I hesitate to use the conventional tensor notation. There I briefly mentioned that I find it illogical and inconsistent. One of my main issues is the transpose of a tensor and what it entails. As far as I have understood tensors, in t...

abstract-algebra vector-spaces tensors

Rudi_Birnbaum

4:10 PM

@Semiclassical isn't that what we want:

I suppose the "$H$" in $$A=U\Sigma U^H$$ is a typo?

That means we have $\Sigma$ (since we have the eigenvalue).

And now we 'just' need to solve it for the $U$ ..

or does it mean conjugate?

nbro

@Secret Isn't what I asked related to tensor transpose?

Secret

well if you are swapping the dimensions of a tensor, it is a transpose in some form

user193319

Is it true that $\text{supp } (f+g) \subseteq \text{supp }(f) \cup \text{supp }(g)$?

nbro

I just found this paper: Tensor Transpose and Its Properties

It defines a transpose of a tensor using a permutation

But I am still trying to understand if it generalises my example or it is something completely unrelated

Semiclassical

4:26 PM

@Rudi_Birnbaum probably H means hermitian conjugate, though I can’t see why it’s at all necessary here

But yes, that’s what we’re looking for

Rudi_Birnbaum

well skew symmetric matrices are in general only diagonizable over $\Bbb C$.

take:

$$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}$$

Semiclassical

yeah, you wouldn't expect a matrix that does 90-degree rotations to have three real eigenvalues

Rudi_Birnbaum

right

nbro

@Secret Check this: Multidimensional Matrix Mathematics: Multidimensional Matrix Transpose, Symmetry, Antisymmetry, Determinant, and Inverse, Part 4 of 6.

Secret

4:43 PM

that paper looks pretty junk and that "journal"

And the author seemed to be dealing with multidimensional arrays, which are not necessary tensors

nbro

What is the difference between multi-dimensional arrays and tensors?

From a theoretical point of view

Also, why do you say it looks like junk?

Secret

tensors need to be linear maps from a tensor product to some field. Multidimensional arrays need not to obey the rules of linearity and other vector space rules

nbro

What would be an example of a multi-dim array that is not a tensor?

Secret

The paper looks junk because the journal's publisher iaeng is listed in the beall'st list as a predatory journal, meaning that the paper is not peer reviewed

those that are used in programming, there isn't necessary a multiplication structure, nor notions of contra or covariance defined on them

nbro

In my case, I am actually dealing with "tensors" that e.g. you can find in frameworks like TensorFlow

They call them tensors. So, you say that they might not actually be tensors, in a mathematical sense?

topologicalmagician

4:51 PM

Is G\H always a coset for a subgroup H of a finite group G?

user193319

No, G\H will be the collection of cosets.

@topologicalmagician G\H itself isn't a coset.

nbro

I don't get this website: beallslist.weebly.com. In which sense would a journal be "predatory"?

Maybe predatory isn't the most appropriate adjective

topologicalmagician

why are the cosets of the alternating group in the symmetric group A_n and Sym(n)\A_n?

Secret

a predatory journal is one who try to lure young researchers to go to basically nonexistent conferences by paying a sum of money to join said conference. Other predatory journals will publish any paper as long researchers pay money. It is a business and there is no peer review, meaning any junk can end up in there

user193319

@topologicalmagician I don't understand your question.

Secret

4:55 PM

Also the data structure used in tensorflow are indeed only multidimensional arrays, they are not tensors/multilinear maps in the mathematical sense

topologicalmagician

@user193319 The alternating group has 2 different cosets in the symmetric group

because $|S_n : A_n |$ $=$ 2

one of them is $A_n$

what is the other one?

Mary Star

Hello!!

I want to show that $$\bar y\land (x\lor z)\land (\bar x \lor \bar y) = \bar y\land (x\lor z)$$

I have done the following: \begin{align*}\bar y\land (x\lor z)\land (\bar x \lor \bar y) &=\bar y\land \left [(x\lor z)\land (\bar x \lor \bar y)\right ]\\ & =\bar y\land \left [\left ((x\lor z)\land \bar x\right )\lor \left ((x\lor z)\land \bar y\right )\right ] \\ & = \bar y\land \left [\left (\left (x\land \bar x\right )\lor \left (z\land \bar x\right )\right )\lor \left (\left ( x\land \bar y\right )\lor \left (z\land \bar y\right )\right )\right ] \\ & = \bar y\land \left [\left (0\lor \l…

Sebastian Alexander B Nielsen

Given a deck of card, are the two outcomes "drawing a heart" and "drawing a king" independent of each other?
Is it correct to assume there is independence if you have to return the card to the deck after drawing it, and that there is independence if you don't return the card?

nbro

@Secret I see. Thanks for this info! I wonder how can people be lured in this way. Has anyone ever gone to a non-existing conference? You must be pretty stupid to do such a thing