I think my conclusion was not totally correct, there's not just one trivializing open set, but all trivializing open sets must have the same cardinality.

Hello, I'm seeking the topic "proof without words" or something like that. I have some troubles to find it can someone help me. Thanks

does anyone here has a work-sheet or something else on "Stolz–Cesàro theorem", my syllabus doesn't include this but still i want to practice?(i am unable to find on internet)

there's an tag for questions on 'proofs without words': math.stackexchange.com/questions/tagged/proof-without-words

Yes but I remember of a very big list

not sure how useful the questions in there are for that purpose, but that's a starting point

I remember one with the Fourier's transformation it was brillaint

brilliant

sounds like this one, though i'm not sure it is: math.stackexchange.com/q/733754/137524

How yes thank you !

See you all

can someone with a Springer Link account download an article for me?

Wow, that was a great animation @Semiclassical

@ForeverMozart DOI?

what is doi?

is it an axiom/assumption to assert that if a set is closed and associative/commutative under a binary operation, then so will its subset?

@ForeverMozart Every article has a "digital object indentifier"

I have to prove that the subset of a set is a group under the restriction of the binary operation on the set

oh you will get the pdf for me?

doi is 10.1007/BFb0061685

Sure, If I can

@Obliv No, not all subsets of a set with some operation are closed under that operation

@ForeverMozart expirebox.com/download/57490afd12ffecd6f5f2f50c29392272.html

alright well assume the subset is closed under the operation

@ForeverMozart Did it work?

what about commutativity/associativity? are those assumed to hold for the subset then?

@Obliv that does not make sense

YES thank you!

fixed. sorry i'm rushing

@ForeverMozart Cool

those are properties of an operation, not something that can be closed (but they are inherited by subsets which are closed under the operation, as can be directly checked)

alright thanks

@MikeMiller Well, sort of, it's kinda complicated. We have 5th grade to 12th grade in here, most of which classifies as high school, but it's partitioned into 5th-10th and 11th-12th. The first is called "secondary education", the second is called "higher secondary education". To get from the first to second one needs to go through an exam, and one has to do like really well there because everyone does.

What's a good source for reading about deck transformations/ automorphisms of covering maps?

are all groups assumed finite?

ah nvm they aren't

The perks of the hard work is that one gets to choose science subjects or other nonsense subjects as they prefer.

is $\{x^{n} ~|~n\in \mathbb{Z}\}$ a subgroup of $G$ if $G$ is finite? I don't think this subgroup would be closed.

it's called the cyclic subgroup of $G$ generated by $x$ in my book

@Obliv Stop using the word closed like that, it will just cause confusion. Add "under the operation". Anyway, yes, why wouldn't it be?

@Sebgr Yeah, that conclusion isn't correct but you have a proof.

(it is also a subgroup even if $G$ is infinite)

I would prefer that you didn't call things you don't like nonsense.

@MikeMiller what math are you working on today?

I've been sick and not doing much for the past couple days. I'm starting to feel better, so I'll probably do some light reference-chasing related to my project and some light unrelated reading.

@MikeMiller Although it is nice to say that I study nonsense.

gotcha

@KarlKronenfeld Yes, it's fine to call things you like nonsense. ;D

Recalls my "symbol-pushing" answer

You have a point there.

I hate symbol pushing. Symbol pushing is why I didn't like my first Riemannian class.

@tobias I guess a better question that would solve my confusion would be: does there exist non-modular operations that allow for a finite set to be closed under the operation?

@Obliv what is a modular operation?

hi... which mathematical constants that are well known are hard to compute accurately?

uh

@Obliv If you mean taking mod some number then yes, lots (for most sizes of sets)

right now I'm sitting at a campus pc with my laptop. both the pc and my laptop have mathematica open running the same kind of code for different parameters

I think he's asking if there are finite groups other than $\Bbb Z/n$.

@Semiclassical lol

@Semiclassical Alex Clark ran some GAP code for me recently, ending up running it in parallel on like 30 computers for a few days

that'd be nice

Fine; I'll call them sickening. I do not have much respect for history as taught in my school, although I am not rebuking the subject itself if that was the impression you got - that was not my intention.

especially since this one is basically just the same computation run for different parameters

'sickening' is not a much better choice of adjective, but fine, I won't engage on this.

except that my laptop doesn't run it nearly as fast as the lap PCs do

it's a little weird to me that a physics grad student mostly programs

it's the mathematical equivalent of watching ice melt

well, it depends on what kind of physics

if i was an experimentalist, i wouldn't be doing a lot of mathematica programming

and if i were a particle physicist, i would be doing a lot of software stuff

what i'm doing right now is basically just PDE stuff

so what kind of physicist are you?

nominally, condensed matter physics. that's what my advisor does

it's the "physicists" who don't have to program ;D

well, there's two ways for a theorist to get 'experimental data'

one is to rely on actual experimentalists. that takes too long :P

the other is to program stuff and call that your data

i say that a tad sarcastically, but to the extent that one uses numerics both to get simulations and to test analytical approximations that's pretty much true

I am looking at our math syllabus right now. It's actually nicer than I expected.

which I guess means I'm pretty disconnected from physical reality. But hey, at least i can admit that :P

You still know all of it, though.

It's got a chapter on probability & statistics at the very end. That I do not know, and would like to know. But other than that, yeah. Also no ambiguous word problems anymore - which is nice.

i find it a little funny that probability and statistics are always stated together. mostly because i like probability much more than i like statistics

@MikeMiller Robert Israel gave an explicit parametrization for that winding-number problem i linked earlier: math.stackexchange.com/a/1789352/137524

eh

explicit parameterizations don't really get me off

how about the plot? :)

that plot wasn't actually there when i linked that, which makes it even nicer

Can someone help me on a easy question? I don't understand the answer

I got a seminar talk tomorrow that is supposed to last for 45 mins, problem is I've got 6 pages of text for the board and at least 2 if not 3 more are gonna come to it D:

@notorious How literally is one supposed to read the statements?

Why isn't it False?? Why is it can't be determined from information given?

well its not determined, you dont know if wei-pin and neelam play games or not, only that they prefer to read

Have you given many talks before? Do enough and you'll be able to figure out what your page:minute ratio is.

My notes tend to be around 10 minutes per page.

ohhhhhh wow that was so easy okay I get it now. Just because they prefer it doesn't mean they don't play it ok

i had a talk today that was 1 hour for 6 pages, talk before that i had prepared 16 pages for 2 hours and only did 8, so i guess the speed is variable lol

Ideally one should do the talk fully in a room by yourself before giving it, though I have done talks without doing that.

did that in the one I held today^, I did the first half alone and then went to hold it and forgot how the things in the second half worked

What do math majors do? Do they all teach? Is there research and if so what is the research for? Is it science related? (sorry just starting out university but computer science, not math)

for some reason I'm not bothered though, I'm going to do the stuff thats cool in detail and if there isn't time for the rest then there isn't time. This is probably exactly the wrong attitude and the listeners might be annoyed...

@notorious yes, there is plenty of research, and no, it is not all science related (in fact, most of it is quite far from the other sciences)

Like statistics? Or is it like proving things?

@notorious proving things. It is kinda hard to explain if you have not seen much math. But think of something like Fermat's last theorem

I like manifolds, which are higher-dimensional versions of curves and surfaces. I would like to know which 3-manifolds embed in 4-dimensional space. I try to invent "obstructions", usually numbers associated to the 3-manifold so that if the number isn't zero, it doesn't embed. Defining and calculating those obstructions is one part of my research.

Hi there! Does anyone know the difference between the kernel of a linear transformation and the basis of the kernel (again, of a linear transformation)?

@Fredefl it is the same as the difference between any vector space and a basis for that space

yes.

@Fredefl do you know how a basis of a vectorspace and the vectorspace itself are different objects?

Thanks guys! I have reduced the matrix to a rref. Next I have identified the variables of the R3=>R3 room. x1 = 2*x2, x2 = independent/arbitrary, x3=0. Thus I can conclude that the basis of the kernel is [2, 1, 0], right?

@MikeMiller That's a nice precis. Of course, hearing that description, I bet you can guess what would tickle my fancy.

namely, is there a nice physical example of that :p

Depends what you mean by physics.

i'm willing to be pretty generous

something involving field theory, I would imagine.

Even being pretty broad it's a hard sell. The closest thing I can think of is "one can be defined using instanton homology groups", where instanton homology groups are groups that count instantons on Y x R somehow.

in the sense of electromagnetism, say, not QFT.

hmm

We don't have many obstructions yet.

I'll just mention you @TobiasKildetoft, @s.harp :)

i suppose the most obvious stumbling block to a physics example is that 4D there tends to mean 3+1 spacetime, i.e. having metric tensor $dt^2-dx^2-dy^2-dz^2$.

which is definitely 4D but is pretty simple on the face of it

i imagine a string theorist could think of some nice brane example, but um

no

dont know what rref matrix is, but if you have that $A(x_1,x_2,x_3)=0$ if and only if $x_1=2 x_2$ and $x_3=0$ then $\{ (2, 1, 0) \}$ would a be a basis of the kernel, but ´ $\{ (1, 1/2, 0) \}$ would also be a basis

okay, done with computing for now

later

remember a basis is a set, in this case since the kernel is one dimensional the basis of the kernel it is a set with one element

Yes, so if that is the basis of the kernel, how do I then deduce that the kernel (not he basis of it) supposedly is [2, 1, 1]? @s.harp

Also, rref is "reduced row echelon form"

@Fredefl the kernel is not $(2, 1, 0)$ but the SPAN of $(2,1,0)$, this means all elements of the form $(2x,1x,0)$ where $x$ is a number from your field (probably the real numbers in your case!)

@Semiclassical This is also only about smooth embeddings, not isometric embeddings, metric tensors appear in what I do but artificially so (the resulting object doesn't depend on the input metric)

Yes, real numbers. I do not need to determine the kernel per-se. I need to find the basis of the kernel, which I have done, however, I need to find a vector that is not included in the range of T, which I have found to be [0, 3, -1] and [1, 3, 0]. That would be the null space, which would be the kernel, right? Or am I just completely wrong @s.harp?

@MikeMiller I don't know anything, but I've heard that many geometrical things have some applications in (maybe higher dimensional) fluid dynamics, where solutions appear that encode this geometry (even though it may be unstable). Ie here youtube.com/watch?v=YCA0VIExVhg they do knots as vortices in real fluids

@Fredefl range and kernel are different things

Yes, that is not what I am stating. :)

But you should know that if you have a map from R^3 to R^3 so that there are two linearly independent vectors that do not lie in the image (which you are saying I think), then the kernel is at least two dimensional

remember formulas like if A: V to W is linear, dim(ker A)+dim(im A) = dim(V)

@BalarkaSen Highly relevant answer. math.stackexchange.com/a/677158/98602

Right - so let's make i a little more hypothetical. So, I have to find a vector that is not included in the range of a linear transformation. The only vector that is not in the range, would be the null-space, right @s.harp?

@Fredefl no, those are completely unrelated things

Alright - I might not really have a good grasp of this. If I have the basis of the range (image) of T, how do I then go about finding a vector, that is not included in said range? @TobiasKildetoft

@Fredefl You could take a vector of a higher dimension, for example

Taking the cross product is a trick that works in R^3. Another consideration is that you have a set of three linearly independent vectors, namely the standard basis, and you know that at least one is not an element of the range. So, if {v1,v2} is a basis for your range, and {e1,e2,e3} is your standard basis, one of the equations av1+bv2=ei is inconsistent.

Great point @Sebgr! The cross product seems more "clean" though - how'd I go about that @KarlKronenfeld?

You might also find a normal basis for the range, and take dot products, instead of considering a bunch of equations.

@KarlKronenfeld The cross product of k-1 vectors in R^k still has mostly the same properties as the original and is defiend in more or less the same way

@MikeMiller Ah, yes. Resembles Cramer's rule in a way.

If I have a finite field of cardinal $2048=2^11$, I take an element $x\in K*$ and $x\ne 1$, the question is what is the degree of $x$ in $F_2$? What does it means ? Because $x$ is not necessarily algebraic in $F_2$.

yes, it is necessarily algebraic

@Fredefl Just look up cross product and find the formula. (This may be over your head, if you are unfamiliar with dot products) It is a way to produce a vector orthogonal to the plane that is the range of your linear map. Saying that v1 is orthogonal to w is the same as saying that the row vector v1 times the column vector w is 0. So, w is orthogonal to v1 and v2 if and only if it is a solution to the equation Mw=0 where M is the 2x3 matrix with rows v1 and v2.

{1, x, ... x^11} are all vectors in this vector space over F_2, and because there are 12 of them there must be some nonzero relation between them

@MikeMiller arr. thanks, I did a silly mistake, thanks!!

Yes, so cross products I am familiar with - however, which two vectors to take the cross products of, however, I am unsure of.

Ah, two vectors that span the range, of course. Imagine the range as a plane, and the two vectors on that plane. The cross product of those two vectors will be orthogonal to the plane.

Thus, I will have to calculate the span of the range in question ({{0,0,1},{3,-6,3},{-1,2,0}}), which should be (0, 0, 1) and (3, -6, 3).

Crossing those two should then give (6, 3, 0).

Miscalculated.

seems fine to me

lol my bad

wolframalpha.com/input/?i=(0,+0,+1)++cross+(3,+-6,+3)

Yeah, I do that sometimes, sorry

I miscalculated

Haha good one! I much appreciate the help Karl!

Gladly

Just out of curiosity, do you speak danish @KarlKronenfeld?

@Fredefl No :)

Haha fair enough.

The assignment with finding the vector that is not in the range of T comes right after finding the base of the image of T - thus I am wondering, can I use that same base of the image to find the vector? @KarlKronenfeld

- even though crossproduct of the span seems to do the trick

yeah, use any basis.

Thus if the base of the image of T is (0, 3, -1) and (1, 3, 0)? The cross product of those two will do as well?

Sure. Why not?

A vector orthogonal to the vectors in that basis is orthogonal to the image. So it will be orthogonal to any pair of vectors that make up a basis for the image.

So both the cross product of the span and the cross product of the base of the image will work? I feel incredibly stupid right now haha.

Hello

Help me with that easy task

wolframalpha.com/input/…

Yes, spanning sets work, with the caveat that you need to ensure that you use linearly independent vectors, when you perform the cross product.

I love that correction haha!

But, that amounts to saying that you pick out a basis from within your spanning set.

I'll go with the cross product of the bases of the image since I already have that information :D

Sounds good.

Thanks a lot for the help Karl!

I might return in a minute if I struggle with another one of the LinAlg tasks.

OK, I will stay for a little while

The correct solution found with Wolfram Alpha, by the way: wolframalpha.com/input/?i=(1,3,0,1)%2Bc*(0,-1,1,2)%3Da*(2,-1,7,16)%2‌​Bb*(1,0,3,7)+solve+for+a,b

(where c is instead lambda (on wolfram alpha))

It somewhat depends on the specific vectors, since you are looking at the intersection of a line and a plane not necessarily in general position.

Ok, u1+cu2 is a line, and the first thing to check is whether it belongs to the span of v1 and v2.

Seems like that might not be the "preferred solution". The context of the assignment is that we have two (ordered) bases for the same subspace, B={u1, u2} and C={v1, v2}. Furthermore, the change of base matrixes of B<-C and C<-B have been calcluated. I am unsure how that fits in, though.

Ah, so in the first equation you have a vector [1,lambda] relative to basis B.

Your change of basis matrix will calculate the same vector but relative to C, which begets the coeffients a and b in the second equation.

I am having trouble visualising that - do you mind writing that in a bit of math?

Let $M=[\mathbf u_1,\mathbf u_2]$ be the matrix with columns u1 and u2. Then, the first equation states $\mathbf x=M[1,\lambda]^T$.

Consider how to write the second equation in this way.

That\'ll be $M=[\mathbf v_1,\mathbf v_2]$ with $\mathbf x=M[a,b]^T$, right?

Yes. Perhaps use $N$ for that matrix though.

Yes, sure. I am unsure what the raised T means, though.

Transposed, so that it is a column vector.

Now, the change of basis matrices are related to M and N. Can you spell this out?

No, unfortunately :( - I do not see the relativity.

OK, how are change of basis matrices defined for you?

(sometimes the relationship I have in mind is a matter of definition)

That is a 2x2 matrix, with M and N being 2x1 each, thus perhaps in some way combining then will be useful.

Actually M is a 4x2 matrix :)

Its first column is the 4-vector u1 and its second column is u2.

Sure, so I have calculated them - I'll fetch them, one moment.

Oh yes, it is getting late haha...

I mean, what is a change of basis matrix in general?

I don't need to see the particular matrices in this case.

P_B<-C = ([c_1]_B...[c_k]_B)

k being the number of vectors

And C and B being ordered bases of the same subspace

Ok, and the important part, what equations do the vectors [c_i]_B satisfy?

Can you elaborate?

You have not given enough information to guarantee that you have a change of basis matrix, since I could fill it up with arbitrary numbers.

In other words, there is an important part of the definition of a change of basis matrix (in general) that is missing.

Unfortunately, the only usable description I have found of said matrix type has been in the documents provided by my institution in danish - I'll see what I can find in english.

Oh, I will give the defining property then.

I guess I will simply use the matrices M and N from the start. Recall that the columns of M are the elements of the ordered basis B, and similarly for N and C. Then, $MP_{C\leftarrow B}=N$.

Now, you have equations representing the vector $\mathbf x$ using the matrices $M$ and $N$.

The ability to swap $M$ and $N$ makes it possible to solve for $a$ and $b$.

Do you see how?

Hey

I am trying to process it haha, one moment.

yojo

Can someone help me with a task about commutative rings?

I am supposed to be decent with commutative rings.

Nice okay, I think the tasks isn't even that hard

p is a prime and a,b€R (R is a commutative ring) and I wanna show that (a+b)^p = a^p + b^p

Not enough information, e.g. consider that when R is the integers.

I imagine the characteristic of R is supposed to be p, right?

Seems like something is wrong here @KarlKronenfeld, wolframalpha.com/input/…*%7B%7B-3,1%7D,%7B7,-2%7D%7D%3D%7B%7B2,1%7D,%7B-1,0%7D,%7B7,3%7D,%7B16,7%7D%7‌​D.

Oh that URL got completely butchered

Hmm

goo.gl/YfjTDG

yes I think so @KarlKronenfeld

Fred: you can write [ words ] ( url ), without the spaces, to post links that have punctuation.

are you german?

@Fredefl That wasn't your P_{C<-B}

I took the inverse of the 2x2 matrix and it worked.

I posted my B<-C matrix, though

Exactly

We could always have a difference in notation too.

@Fredefl With the other change of basis matrix

Can anyone give me a hint? I'm a bit lost in this exercise: "Show that any two-sheeted covering has a non-trivial automorphism".

Yes, I noticed that too - with the other one i works.

So, there are matrices P and Q such that MP=N and NQ=M. @Fredefl

I'm not sure where to start, should I re-read about group actions on covering maps or map lifting theorems?

Or maybe it's just about working with the definition of a covering map and playing with the trivialization

@jordan178 The binomial theorem works just fine for arbitrary commutative rings, with the binomial coefficients being scalars in $\mathbb Z$.

And same two P and Q, one of which'll be [1,lambda] & [a,b] just like the base shift matrices?

Could you rephrase that? The P and Q are the change of basis matrices (we seem to have different notations)

Oh alright - how do we then go about tying in lambda, a & b?

You want the equations for $\mathbb x$ to be relative to the same basis.

- and I am very sorry for being horrible at thinking for myself at this point.

ok thank you

I do not quite comprehend what is meant by that

Recall that the basis B is captured in the matrix M, as in the equation $\mathbb x=M[1 \lambda]^T$

Sebgr: Have you tried just doing what you want to do locally (swap the sheets) and showing this is well-defined?

@Fredefl Similarly for N and C. Having the equations be relative to the same basis means using either M for both or N for both.

Do you get the overarching idea here?

actually when you expand (a + b)^n in a commutative ring, the coefficients aren't really in Z. They are sums in the ring, using that ring's addition, of finitely many copies of 1 in that ring. @KarlKronenfeld

@jordan178 That corresponds to the usual scalar multiplication by $\mathbb Z$. :)

(When you have identity)

It is simply defined as, kx=x+..+x (k times). This also works for abelian groups, of course.

ah ok I get it now

Oh, you have to interpret a sum of x (-k times) to mean the corresponding sum of -x's.

I don't seem to, now. Quite frankly, I do find this assignment to be way above my current understanding of linear algebra - really struggling understanding how to deduce a and b.

Think of a and b as being the coordinates for your vector $\mathbb x$, which may as well be a point in the plane spanned by v1 and v2.

You already have a pair of coordinates for $\mathbb x$, namely 1 and lambda. Unfortunately, it is relative to a different system of coordinates, based on u1 and u2.

@EricStucky The best I can think of is using a permutation on the index set $F$ of the local trivialization. But I know that this shouldn't work for $n\geq 3$ (because it's actually the second part of the exercise, to find a counter-example).

I would believe that for $n\geq 3$ because there's no canonical choice of how to swap the sheets, and so it's hard to 'be consistent'. But for $n=2$ this problem doesn't exist.

@Fredefl So, you have to use your change of basis matrix to change coordinate systems.

@EricStucky So, why would it work on $n=2$? It's the only notion I think of for "swapping sheets". I've been thinking about working with the uniqueness of liftings, but that requires extra assumptions over the total space.

@Fredefl The equation $\mathbb x=M[1 \lambda]^T$ "says that" in the coordinate system $M$, $\mathbb x$ has coordinates $[1,\lambda]$.

@EricStucky I just read this. Ok, I'll try to work in this direction

cool :)

lmk either way

I have fiddled around a lot with it, and I still cannot seem to wrap my head around it @KarlKronenfeld - I am afraid I'll have to give up on it in a moment.

@Fredefl Ok, sorry I could not help with this one.

No worries Karl, I much appreciate your help, it was great! If you happen to have some strange, long, all-inclusive equation for it that works - I'll happily take that and try to wrap my head around it tomorrow.