Mathematics - 2024-05-28 (page 2 of 3)

Obliv

17:00

@Koro could be sick. Hopefully nothing serious though and will feel better soon. Can be a million reasons though so if you can afford a vet visit just for some bloodwork that would be ideal

Koro

Leslie: these are kittens from my college hostel. I've kind of raised them ever since they could barely even jump.

Xander Henderson

@Obliv Different people write books differently.

Koro

They barely go outside hostel.

And mostly live around me.

Xander Henderson

I am currently working on a book which is essentially notes from teaching a precalculus class over the last several years. It is a beefed-up version of the activities I give students in class, held together with a bit more exposition.

Jakobian

@XanderHenderson and that's great, there's candies for everyone out there

leslie townes

17:02

koro i would second obliv's recommendation to seek vet care if it is possible, but i have no idea if it's possible. (around here vet care is often prohibitively expensive, or at least expensive enough that it involves a really delicate balance of the ideal and reality)

Jakobian

I don't like every style of writing and I imagine different people just naturally have different preferences towards writing

Koro

Behavior change in just 2 hours!! How does that happen? What catalysed it?

leslie townes

koro i have an indoor only cat and she does have mood swings. not unusual for her to shadow me around the house, or for her to vanish for 20 hours of the day and come out just to eat

Koro

@leslietownes perhaps vet care is not available nearby here. Also, everything seems exorbitant in US.

leslie townes

if she weren't eating that would alarm me, but, it's only because she's indoor only that i can monitor her eating and drinking and bathroom habits

with an indoor-outdoor cat, who knows

Koro

17:06

So, perhaps, it could be part of their growth process then.

leslie townes

it could also be that someone else just started feeding them and they only have so much room in their stomach. cats are pretty faithless companions :)

Obliv

i wonder if it's possible to send blood samples to labs yourself so you can DIY vet care

Koro

@leslietownes some people do feed them. But the kittens never refused a slice of meat until today no matter how full they were.

But it may be possible.

I'll see him tomorrow then.

Sine of the Time

@Obliv 💀

Koro

I've also been sick for last few days.

Obliv

17:09

I highly recommend getting them vaccinated from some dangerous viruses at least

like rabies

Koro

(body pain + stomach issues + fever + appetite loss)

Sine of the Time

it also happened to my cat when he eats a lot of meat

Koro

@Obliv sadly, rabies is very common here.

Almost every dog around is rabid.

Soumik Mukherjee

@Koro Take care, I hope you get well soon

Koro

Thanks. It sucks to take so many medicines.

And I don't know why Bengali doctors prescribe so many medicines.

So many...

Obliv

17:12

well, maybe whatever the cat has spread to you or vice versa? Try to see a doctor if you don't eventually feel better

Koro

It's heavy sometimes on health resulting in weakness, dizziness and all.

For fever once, I was prescribed about 8 tablets and some syrup.

Eight!!

It only got worse. I stopped taking the medicines and took tablet of another medicine instead and I was fine next day.

@Obliv yeah, I've seen a doctor today.

giannisl9

@Jakobian I feel I miss some understanding.

Koro

All 500 - 650 mg tablets.

giannisl9

I can see how using $(r, \theta, \phi)$ we can describe all space.

But out of those, the position vector is just $r\hat{r}$ since $\hat{r}$ is a function of $\theta$ and $\phi$.

$\hat{\theta}$ and $\hat{\phi}$ are also functions of $\theta$ and $\phi$.

Koro

17:28

The cat (the mother of the kittens I talked about) has given birth again.

giannisl9

But the position vector is not a linear combination of $\hat{r}$, $\hat{\theta}$ and $\hat{\phi}$.

Koro

Their eyes haven't opened yet.

giannisl9

So what is $r\hat{r} + \theta\hat{\theta} + \phi\hat{\phi}$?

Is that vector in space?

I am thinking that all $\hat{r}$, $hat{\theta}$ , $\hat{\phi}$ are in space.

Obliv

@giannisl9 position vector is specified by all coordinates $r,\theta,\phi$

giannisl9

@Obliv yes, but not all basis vectors?

@Obliv $r\hat{r}$ has all coordinates but not all basis vectors.

Obliv

17:34

$r$ specifies the length, $\theta,\phi$ specify the "direction" so you need all of them to completely describe $\mathbb{R}^3$

giannisl9

@Obliv since $\hat{r}$ is a function of $\theta$ and $\phi$

@Obliv I am referring to the basis vectors not their magnitude.

leslie townes

i don't understand what you're asking, but it might help to be more specific. e.g. what is going on in R^3 with the point that we would call (1,1,1) in cartesian coordinates. what are these r hat, theta hat, and phi hat "basis vectors" in this instance

Obliv

I think working in R^2 first might be easier

giannisl9

@leslietownes a point (x, y, z) maps to $(r, \theta, \phi)$

Obliv

$r^2 = x^2+y^2$ gives the usual transformation of $r$ to/from cartesian coordinates with $x=r\cos\theta$, $y = r\sin\theta$

@giannisl9 you have to use the transformation equations in that example

leslie townes

17:38

as a background observation, it's possible to represent a point in R^3 with a triple of numbers without necessarily thinking of any one of those numbers as a coefficient in some linear combination, and that might be going on here, but i dunno

giannis: okay, so in this example r is sqrt(3), theta is pi/4, phi is pi/4? does that feel right? what about "theta hat"

giannisl9

@leslietownes $\hat{r}$, $\hat{\theta}$ and $\hat{\phi}$ are basis vectors for the new system

@leslietownes but they all depend on position

leslie townes

this is where i'm stuck. you're saying that these vectors are a "basis" for a "new system," but i don't know what they are in the instance of this point that is (1,1,1) in cartesian or (sqrt(3), pi/4, pi/4) in spherical

giannisl9

@leslietownes I think something like that is happening but I struggle to understand it

leslie townes

we may also find it helpful to distinguish between R^3 here and the tangent space to R^3 (which unhelpfully is identifiable with R^3, but i'm suggesting that we not do that)

giannisl9

@leslietownes someone posted this link dynref.engr.illinois.edu/rvs.html

@leslietownes it shows what the basis vectors are

leslie townes

17:43

if we look at little increments in the coordinate variables we can get directions in R^3 we can think of as "based at the cartesian point (1,1,1)" that point in the directions of various coordinate changes, but it's mixing types to want to represent (1,1,1) itself in terms of these directions

okay, that's the e_r, e_theta business on this thing. i'd think of them as a basis for the tangent space, not the space itself

giannisl9

@leslietownes I feel I am missing some knowledge but that you get what bothers me. Can you be a bit more analytic, even just to give some intuition which is not strictly correct.

So, if I were to move in some $\hat{r}$ direction that means I have already picked $\theta$ and $\phi$.

$r\hat{r}$ makes sense since all $(r, \theta, \phi)$ are picked

leslie townes

hat(r) appears to be that page's notation for (r, theta, phi)

giannisl9

but 𝑟𝑟̂+𝜃𝜃̂+𝜙𝜙̂

leslie townes

you might be talking about hat(e_r), which (given a particular point (r, theta, phi) gives a unit vector in the direction of increasing r while fixing the angles)

giannisl9

I cannot understand if this is in the space

because

leslie townes

17:49

i don't know what hat(theta) is. that page does not define it. it defines something called hat(e_theta)

giannisl9

$r\hat{r}$ and we are already somewhere and then if we add $\theta\hat{\theta}$ we go somewhere else

leslie townes

hat(e_r), hat(e_theta), and hat(e_phi) are a basis for R^3, but you would want to think of this as a "copy" of R^3 having the point that you've fixed as its origin, and elements of R^3 represent "direction vectors" based at that point

giannisl9

and if we use the basis vectors at that point we actually change all

leslie townes

you wouldn't naturally think of decomposing the point itself in terms of the basis for the tangent space

a type checker in a computer programming language would throw a fit if you did that

you can certainly form all sorts of expressions for elements of R^3 just using the fact that elements of R^3 can be added together, the question is, what do you hope to do by forming r hat(e_r) + theta hat(e_theta) + phi hat(e_phi)

Obliv

@giannisl9 en.wikipedia.org/wiki/Rotation_matrix maybe you're interested in the rotation space for lack of a better word

leslie townes

17:54

it might help to switch to R^2 like someone suggested. using polar coordinates on R^2, the analogues of hat(e_r) and hat(e_theta) at the cartesian point (1,0), which oddly enough is also (1,0) in polar coordinates, would be (1,0) and (0,1)

Obliv

the basis vectors are given by trig functions on the euclidean space, you don't need $\hat{r}$ here

leslie townes

and it coincidentally happens that your original point, whether in polar or cartesian, is identical to just one of those two basis vectors

but i don't know why you would think to use the basis for the tangent space of R^2 (namely, elements of R^2 thought of as directions pointing away from from (1,0)) as something that you want to represent elements of the original space in terms of

Obliv

@giannisl9 I tend to agree with jakobian that this is not a correct statement because $\hat{\theta},\hat{\phi}$ don't seem to have any meaning, how can the norm of an angle give you $1$

leslie townes

hat(theta) and hat(phi) await definition. i suspect that the question relates to what the page linked above calls hat(e)_theta and hat(e)_phi

but i still don't understand the motivation

giannisl9

@leslietownes yes, the page has an e

Obliv

17:58

\hat(e) vectors are the transformations from cartesian to spherical and vice versa

but hat(theta,phi) aren't a thing I don't think

giannisl9

@leslietownes In your example in $R^2$, $\hat{r}$ is (1,0) and $\hat{\phi}$ is (0,1) at the point (1,0)

and the position vector is $\hat{r}$

leslie townes

@giannisl9 yes. if you change from (1,0) to say (2,2) cartesian, now hat(e_r) is a unit version of (2,2) of the form (a,a) and hat(e_theta) has the form (-a,a)

and if you wanted to write (2,2) in terms of e_r and e_theta you would have (2,2) = ||(2,2)|| e_r

giannisl9

@leslietownes yes and my question is what is

leslie townes

a point (x,y) in R^2 would always be a scalar multiple of hat(e_r) in this decomposition and hat(e_theta) would not enter into it

giannisl9

@leslietownes same in 3D

leslie townes

18:03

yes

giannisl9

@leslietownes but what is then $\theta\hat{\theta}$

Obliv

@giannisl9 R^3 has to be specified by 3 basis vectors so in spherical coords this is length, polar, and azimuthal direction. $r\hat{r}$ is a vector in $\mathbb{R}^3$ that specifies the length and direction

leslie townes

[reading hat(theta) as hat(e)_theta] it's just some random vector that you've chosen to write down that points in the direction of hat(e)_theta

giannisl9

its clearly in 2D

leslie townes

i wouldn't expect it to have geometric meaning, because i wouldn't expect to decompose the point that defines all of these vectors in terms of the e_r, e_theta, e_phi basis for R^3

there's a fundamental distinction between a space and its tangent space that gets blurred when everything is just points in some R^n

if only ted were here

Obliv

18:05

@giannisl9 meteo.psu.edu/holocene/public_html/Mann/courses/METEO470SPR17/…

giannisl9

In my book it says

Any vector A can be expressed in terms of $\hat{r}$ $\hat{\theta}$ $\hat{\phi}$

as a linear combination with some coefficients

but I don't understand that vector at all

Obliv

idk if it's called a jacobian, but there are continuous relationships between those vectors expressed in derivatives

look at the link I just linked for the "variations of unit vectors with the coordinates" @giannisl9

giannisl9

surely its not a position vector or is it? are the coefficients used to describe some position in space but that vector is nothing in that space?

leslie townes

giannis: i would think of these vectors as a basis for a new space. a copy of R^3 whose "origin" is the point at which you define these vectors

i would not think of these linear combinations of hat(e_r), hat(e_phi), hat(e_theta) as living in the same R^3 that (r, theta, phi) or (x,y,z) does

giannisl9

@Obliv this is what I am learning and struggle with its meaning

@leslietownes I see but when the book says any vector

does it mean any vector in R^3?

leslie townes

18:10

yes. any three-tuple of numbers

the book isn't saying "and by the way, it's going to be very helpful to express the points of your original space in terms of this basis"

giannisl9

leslie townes

it's just something that you would be capable of doing, if you wanted to do it

giannisl9

each of those basis vectors are in R^3 and their sum should also be in R^3 but I cannot understand what that sum gives

$A_r\hat{r}$

I understand, it is the position vector.

leslie townes

A_r, A_theta, and A_phi are not, in general, going to have any meaning

they're just the book's notation for "the coefficient that appears when you do this expression"

Obliv

it's because you're thinking in terms of cartesian coordinates, always, intuitively when you think about sums

giannisl9

18:17

The problem is if we add $A_{\theta}\hat{\theta}$ for that $(r, \theta, \phi)$ we get another point where those basis vectors are different

Obliv

like norms are computed the same so you must think of your typical right triangles to specify the position in space but that doesn't work for this

leslie townes

giannis: i wouldn't think of a 3-tuple obtained as part of a basis expansion in terms of the r, theta, phi vectors as compatible with the 3-tuple (r, theta, phi) itself

i wonder if you want to add them together, i guess just because they have the same data type (elements of R^3) and you can add any 3-tuple to any other 3-tuple

but there's no reason to expect that type of addition to have any significance

Obliv

I have a problem with this notation, it looks like a trap

giannisl9

@leslietownes Sorry, can you describe what you are saying a bit more analytically, I think you kind of answer my question.

@Obliv me too

leslie townes

the same way that (1,1,1) is (sqrt(3), pi/4, pi/4) in spherical and (1,0,0) in cartesian, i wouldn't say "okay, so what is (sqrt(3) + 1, pi/4, pi/4)? what does that mean?" the answer is, it means that's what you get when you add those 3-tuples together, but i don't think that addition is something you would naturally do

the representation of an element of R^3 in terms of the r, theta, phi vector basis is best thought of as a representation of "directions" from the point where you computed those vectors

and is best thought of as existing in a different R^3 than the R^3 where the point is (1,1,1) or (sqrt(3), pi/4, pi/4)

i realize that i'm not doing a great job of explaining this, this is getting into exactly the kind of stuff that i don't understand :)

we need a proper geometer

giannisl9

18:24

@leslietownes I think you get exactly what bothers me but yes, I struggle to clearly understand. I thank you very much for your time though.

Obliv

@leslietownes it's like, $\hat{\theta},\hat{\phi}$ are still orthonormal length 1 vectors pointing in polar, azimuthal directions respectively, but it's not intuitive at all because we think in terms of cartesian coordinate systems almost inherently

wow was that the server or me

I think it was the server since giannis just did the same thing haha

poor leslie probably just got 6 alerts all at once

leslie townes

yes :D

Obliv

@giannisl9 not really helpful at all to visualize, but this answer conveys my issue with the notation math.stackexchange.com/a/2193678/234396

giannisl9

@Obliv your issue but not mine maybe? I have no problem with $|\hat{\theta}|=1$

my problem is to understand how $(r, \theta, \phi)$ representing the point in $R^3$ relates with some linear combination of the basis vectors since only one of them seems to be used.

Obliv

I meant not being able to visualize the rotation matrix

1

Q: Basis Vectors in a General Curvilinear Coordinate System

I'm confused as to how does one find out the basis vectors of a curvilinear co-ordinate system. In the context of a general, arbitrary curvilinear co-ordinate system, the textbook I'm reading states that: If Cartesian co-ordinates $x, y, z$ are expressible in terms of the three curvilinear ...

leslie townes

18:38

giannis the high level idea is that you would not, generally, want to use the basis to represent the point that you started with (although you mathematically could). you would instead be likely to choose that basis to represent "tangent vectors" - i.e. directions of motion away from that point

giannisl9

@leslietownes okay, that makes sense

Obliv

@Obliv @giannisl9 take a look at that link and see that your basis vector $\hat{r}$ isn't sufficient to specify your coordinate system in $\mathbb{R}^3$

giannisl9

so $\hat{\theta}$ and $\hat{\phi}$ do not indeed play any part on the position of a point, only their associated coefficients.

Obliv

you need the other 2 planes as well

@giannisl9 without $\hat{\theta},\hat{\phi}$ you'd have no position, you'd only have a length.

leslie townes

giannis: right. although if the point were traveling in a path about the origin, its velocity (which is also an element of R^3, although i would think of it as a "different" R^3 than the one that the point lives in) might point in the theta or phi direction

giannisl9

18:42

@Obliv I think that is not true

Obliv

whatever your basis vectors are for space V, you need all of them to span V

leslie townes

e.g. if a point were moving about the surface of a fixed sphere, its velocity would be a linear combination of the theta and phi vectors only and have a 0 component in the "r hat" direction

Obliv

@giannisl9 okay, give me a point in $\mathbb{R}^3$ in spherical coordinates without usage of $\hat{\theta},\hat{\phi}$

giannisl9

@leslietownes I see!

@Obliv okay $r\hat{r}$

Obliv

what is $\hat{r}$

giannisl9

18:45

whatever you want

any vector

Obliv

how does $\hat{r}$ span $\mathbb{R}^3$

giannisl9

it spans R^3 cause you have a different $\hat{r}$ for different $\theta$ $\phi$

any position vector you can represent as some r and $\hat{r}$

Obliv

what are $\theta,\phi$

giannisl9

the usual

angles

or anyway numbers

Obliv

so they're orthogonal or what

giannisl9

18:48

they are numbers

just numbers

Jakobian

@leslietownes time to become the new Ted, Leslie

its your time to shine, don't give up on your dreams

giannisl9

$\hat{r}$ $\hat{\theta}$ and $\hat{\phi}$ are orthonormal for a specific $\theta$ and $\phi$

Obliv

@giannisl9 you need to specify $\hat{r}$ in terms of $\phi,\theta$ since $\hat{r},\hat{\theta},\hat{\phi}$ are the orthonormal set of basis vectors in spherical coordinates for $\mathbb{R}^3$ you can't just have one

leslie townes

maybe in my retirement

Obliv

yes but $\theta$ isn't just a number it's $\theta\hat{\theta}$ a number in a specific direction i.e. in the polar plane

giannisl9

18:51

@Obliv no that's the thing

its just a number

Obliv

what do you mean it's just a number? this is a vector space we're spanning

we'd like a scalar field + 3 orthonormal basis vectors for $\mathbb{R}^3$

giannisl9

@Obliv if you think of $\hat{r}$ in terms of cartesian coordinates I think you ll see why its enough

Obliv

I'll post it again

1

Q: Basis Vectors in a General Curvilinear Coordinate System

I'm confused as to how does one find out the basis vectors of a curvilinear co-ordinate system. In the context of a general, arbitrary curvilinear co-ordinate system, the textbook I'm reading states that: If Cartesian co-ordinates $x, y, z$ are expressible in terms of the three curvilinear ...

vector-analysis coordinate-systems

the green line in the figure represents $\hat{r}$

the other two lines orthogonal to it represent the azimuthal and polar planes

$\hat{r}$ alone just specifies going away from the origin, but you need the other two to specify which direction away from the origin

giannisl9

no, the green line represents some $\hat{r}$

different lines

different $\hat{r}'s$

Obliv

?

giannisl9

18:55

you think of $\hat{r}$ as the distance from the origin

it is not

it is a vector from the origin to the point with length one

Koro

It's a direction.

giannisl9

yes, a direction

one for every point

Obliv

yes but it's specified in terms of the other basis vectors

giannisl9

no

the basis vectors of cartesian

not spherical

this is how I started thinking about this

Obliv

ok but the basis vectors $i,j,k$ in cartesian coordinates are related through $\phi,\theta$

giannisl9

18:59

@Obliv I don't understand but do you see now why $r\hat{r}$ spans the whole R^3?

$r\hat{r}(\theta, \phi)$ if you like

Obliv

my point is there is no $\theta,\phi$ if not for $\hat{\theta},\hat{\phi}$

i.e., there is no green line without the red and blue

giannisl9

@Obliv hmm that was my problem also

but many ways suggest other

for example

$(x, y, z)$ mapping to $(r, \theta, \phi)$

well defined and no need for $\hat{\theta}$ and $\hat{\phi}$

Obliv

because $\phi,\theta$ specify which direction and how much direction $\hat{r}$ points, and $\hat{\phi},\hat{\theta}$ give $\phi,\theta$ that direction

giannisl9

@Obliv I don't understand

Obliv

@giannisl9 en.wikipedia.org/wiki/… here's the functions you're talking about

note how each component in cartesian coordinates are related to not just one component in spherical basis

giannisl9

19:04

the mapping is defined yes

Obliv

particularly, every x,y,z component has $\rho$ because they contribute to the norm of $\hat{\rho}$ or $\hat{r}$ if u wish

giannisl9

that was the example I gave

but nothing here is the basis vectors

Obliv

you can go in the other direction to derive the basis vectors

Jakobian

@leslietownes would you name your son Ted

Oh wait you have 2 babies right. Both are girls?

Obliv

see here

giannisl9

19:06

Still you are not showing me vectors, only coefficients

Jakobian

It'd be cute if you named your son Ted

leslie townes

jakobian i have one son but he is not named ted :)

giannisl9

you can go from $(x,y,z)$ to $(r,\theta,\phi)$ back and forth

Jakobian

There's still time, maybe there'll be little Ted in the future

giannisl9

but those are coefficients

Obliv

19:08

@giannisl9 u just make x,y,z $i,j,k$ and $\rho$ has length 1

giannisl9

that is not ok

Am I right @leslietownes?

Obliv

$\rho$ comes from $\rho\hat{r}$

idk actually

giannisl9

In cartesian every (x,y,z) corresponds to the position vector $x\hat{i} + y\hat{j} + z\hat{k}$

x, y, z are numberes

and the other are unit vector

for the system orthonormal blah blah

and you can see by this how this spans R^3

now what about $(r, \theta, \phi)$

$\hat{r}$ $\hat{\theta}$ $\hat{\phi}$

are not constant

they are different for every point

Jakobian

I wonder if I ever become mature enough to be able to rise a child

or be given a chance to

Obliv

$$\left[\begin{array}{@{}c@{}}
\rho\cos\theta\sin\phi \\
\rho\sin\theta\sin\phi \\
\rho\cos\phi \\
\end{array}\right]$$ is the basis

giannisl9

19:14

yes, not constant

its not like (1,0,0) (0,1,0) and (0,0,1)

Obliv

yeah and not like $\begin{bmatrix}r\\ \phi\\\theta\end{bmatrix}$ that's why I said I had issues with the notation

giannisl9

also this you would have to divide by ρ

anyway

my point is

Koro

How common is naming a kid "debt"?

giannisl9

for the position

you don't need the other basis vectors

Obliv

I wonder if $\hat{r} = \cos\hat{\theta}\sin\hat{\phi}$ is correct lol

or makes any sense

giannisl9

19:18

what is cos of a vector

no sense to me

$\hat{r}$ is not related to the other

that means orthonormal

Obliv

they are the angle "vectors" that points in the polar/azimuthal direction, cos is just normal definition of the $\frac{adj}{hyp}$

I guess

giannisl9

I don't know but again the point is the are not needed

for position vector

they are directions

at this point

so if you want the velocity there

maybe you get some other vector in terms of them

but that has nothing to do with position

they are not needed to span R^3

they are needed maybe to span a copy of R^3 in that point

so you get all R^3 again?

I don't know

Obliv

you definitely need length, and 2 orthogonal directions but if $\hat{r}$ contains all of that information then obviously it spans R^3 but if you want to split up your basis into 3 "vectors" you get redundancy

since you're breaking apart the idea in cartesian coordinates of having 3 orthogonal vectors with length & direction

into a vector $\hat{r}$ with direction and magnitude, with $\hat{\theta},\hat{\phi}$ being pointless

Idk tbh, but I gtg now lol sorry. hope you figure it out and ping me if you do

giannisl9

they are relevant if you want for that basis to represent something that is not in the direction of $\hat{r}$

so maybe velocity

I think I get it more by explaining it :ㅖ

:P

they are directions to use to span a copy of R^3

R^3 again having chosen a direction already

aaaaha

imagine being at a point

position vector $r\hat{r}$

if you want all R^3 again having chosen $\hat{r}$

you need $\hat{\theta}$ and $\hat{\phi}$

if you just want to go to your point

just use $\hat{r}$

@Obliv ping

leslie townes

@giannisl9 👍

Sine of the Time

19:31

first time leslie pings

giannisl9

@leslietownes yes, I think now I kind of understand what you were saying! Thanks ^^

Pizza

i have a question

i was reading this

Consider the vector space $M(\mathbb{R})_2$, and let:
$V = \left\langle \begin{pmatrix} 1 & 0 \\ 1 & -2 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} \right\rangle$
$W = \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in M_2(\mathbb{R})_2 : 2a+b=c-2d=0 \right\}$
be two subspaces of $M(\mathbb{R})_2$. Determine $V+W$.

whats the difference beetween the notaion $M(\mathbb{R})_2$ and $M_2(\mathbb{R})_2$

leslie townes

19:47

one has an extra 2 in it? the notation is not as significant as what the notation has been defined to mean

in my thesis i think i used M_n(X) to denote the nxn matrices with entries from the set X

but i spelled that out in writing, and those options you listed would also work

the second one makes me wonder what the second 2 is doing

i taught out of a book that used M_{axb}(X) for the a x b matrices with entries from X

Pizza

but shouldn't it be written the same way?

leslie townes

maybe? i don't know. where is this coming from?

in the above excerpt, they seem intended to convey the same meaning

i wouldn't assume that everybody who writes math is paying careful attention to everything they write

Pizza

it's an exercise that's in my slides

@leslietownes think so

leslie townes

yeah, that seems a safe assumption

Pizza

yes, now I will try to do the exercise

Shaun

20:02

in In the search of a question, 18 secs ago, by Shaun

I'm looking for the relationship between the subgroup rank of a subgroup and that of its ambient group. This is likely to have been covered before. I can't find it though

Pizza

@leslietownes as you said the correct notation is $M_2(\Bbb R)$

leslie townes

well, the "correct" notation is any one that has been defined so that you don't have to guess about what it means

:)

i do like M_n(X)

Jakobian

20:25

@SineoftheTime not the first time, but definitely a rare occurrence

Pizza

can someone take a look and see if its correct

i think yes

zeta space

20:46

I'm interested in defining a notion of deformed spheres like a homotopy (fixed point homotopy) that is

$h_t:S^2 \to S^2$ for $t\in[0,1]$ (here's what i have so far)

correction: fixed ENDPOINT homotopy

FEP

homotopy

Here's why it's relevant to you: at $t=0,1$ we have a Croissant surface. At $t=1/2$ we have a unit sphere

and no it's not a pinched torus

Shaun

21:16

Re: the above, I asked this:

0

Q: Let $\operatorname{sr}(K)$ be the subgroup rank of $K$. When is $\operatorname{sr}(H)\le\operatorname{sr}(G)$ for $H\le G$?

The Question: When is $\operatorname{sr}(H)\le\operatorname{sr}(G)$ for (not necessarily abelian) groups $H\le G$ (in general$^{\dagger}$)? Here $\operatorname{sr}(G)$ is the subgroup rank of a $G$. The Details: Define: $$\operatorname{rank}(G)=\min\{|X|\mid X\subseteq G, \langle X\rangle =G\}...

group-theory combinatorial-group-theory

Obliv

21:46

@leslietownes i was sitting in a math class and the professor mentioned that u can find the textbook on "some russian websites" lol

do u lecture right now?

Xander Henderson

22:13

My summer calc class has ten enrolled! Yay! It's gonna run!

2

Obliv

22:31

Nice!

Dannyu NDos

Does the notation $R[x,y]$ refer to the polynomial ring such that $xy \neq yx$?

Even if I want the commutative version, I can take $R[x,y] / \langle xy - yx \rangle$, right?

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