Mathematics - 2021-01-25 (page 2 of 3)

3:00 PM

They turn coffee into theorems.

@BalarkaSen @MikeMiller @Thorgott useful everyday-topology calculation facts? I'm really aiming low here. Examples of what I have in mind:
Useful decompositions for mayer-vietoris, constructions of spaces as quotients of good pairs, "short exact sequences with last group free abelian are split", "if A is a def retract of U then $H_n(X,A) \overset{\simeq}{\rightarrow} H_n(X,U)$", and such things

Unknown x

what if the field associated with the inner product is not mentioned? can assume it is a real inner product space?

Thorgott

3:16 PM

if $X=A\cup B$ with $A,B$ open, connected, non-empty intersection, then $X$ is connected, so $H_0(X)=\mathbb{Z}$, and you can cut off the corresponding M-V-sequence as $...\rightarrow H_1(X)\rightarrow 0$

and, in general, always use reduced homology to avoid stupidity in degree $0$

I'll think of more useful facts later

user2103480

@Thorgott yeh mood

Mary Star

Which criterion do we use to show the convergence of $$\sum_{n=1}^{\infty}\frac{n \log (n+1)}{1+n^3x^2}$$ ?

user2103480

@Thorgott Also, if one of the maps is injective/surjective, I can squeeze a 0 into the sequence

in a LES

tautological but nice for oversight imo

Mary Star

Do we maybe use that $\log (n+1)<n$ and so we get $\frac{n\log (n+1)}{1+n^3x^2}<\frac{n^2}{1+n^3x^2}$ ?

T_01

Somebody has a hint how could I calculate the determinant of $\begin{pmatrix}1+a_i^2 & a_1a_2 & a_1a_3 \\ a_1a_2 & 1 + a_2^2 & a_2a_3 \\ a_1a_3 & a_2a_3 & 1 + a_3^2 \end{pmatrix}$ generalized for $n \in \mathbb{N}$ ? (The pattern continues, the matrix is symmetric: $A_{ii} = 1+a_i^2$ and $A_{ij} = a_ia_j$ for $i \neq j$)
The exercise even states what the answer is ($1+\Sigma_{1 \leq j \leq n} a_i^2$) but my attempts to show that via induction fail and I did not find an other approach yet

maybe there is also a trick earlier in the exercise and I went to far with multiplying things out? The …

A-Level Student

3:28 PM

Does anyone have any advice for me please?

T_01

:56857517 Never saw this formula

Removed?

Mike Miller

Yeah my statement was incorrect

T_01

ah ok

Astyx

You can compute the eigenvalues of $aa^T$

T_01

I actually tried

But for that i need the determinant of a much more scarier matrix

Astyx

3:30 PM

You don't need to explicit them

Mike Miller

Don't think in terms of characteristic polynomials, Astyx is suggesting just writing down the definition of eigenvalue

Also he means $a a^T$

Astyx

I do, I just realised that

T_01

how do the eigenvalues of $aa^T$ help me?

Astyx

What are they?

Mike Miller

This matrix is $I + a a^T$. If you know the eigenvalues of $a a^T$ you know the eigenvalues of $I + a a^T$. And the determinant is the product of the eigenvalues (repeated w multiplicity).

T_01

3:32 PM

Well the product of them would give me the determinant of $aa^T$

Mike Miller

Yes but we want to know what they are, using the definition of eigenvalue

T_01

Sorry I feel a bit off the road :( Your hint is to "guess" the eigenvalues of $aa^T$, right?

Mike Miller

What is an eigenvalue?

T_01

$aa^Tx = cx$

User 873110

Consider the link math.stackexchange.com/questions/1874740/… and assume $f$ is a proper map. Does this imply $\widetilde f$ is also a proper map?

T_01

3:34 PM

then, $c$ is an eigenvalue

of $aa^T$

Mike Miller

OK. Remember that $a^T x$ is a number. So this says $a$, scaled by that number $a^T x$.

T_01

aha

so then I have $c/aTx$ , an eigenvalue of $a$

Mike Miller

What have you assumed?

I have to run, I'm sorry --- maybe Ted or Astyx can help finish the job

T_01

Sorry, I do'nt know what you mean

Okay, thank you

Astyx

Sure I can

T_01

3:35 PM

Thank you :D

Ok maybe I try to calculate $aa^T$ first

Astyx

You don't need to

T_01

Ah okay

Astyx

So we have $a (a^Tx) = cx$ for a vector $x$ and some number $c$, which we can rewrite $(a^Tx)a = cx$ as $a^Tx$ is just a number

What can you deduce from this?

T_01

i can divide by $a^Tx$ here right? somehow this does not make any sense

Astyx

Not always

T_01

3:39 PM

well under the assumption that this thing is not zero, of course

Astyx

Right.

T_01

but then my matrix seems to be a multiply of a vector this cannot be

Astyx

What do you mean?

copper.hat

look at the range.

T_01

$a \in \mathbb{R}^{n \times n}$, so we have $x \in \mathbb{R}^n$

Astyx

3:40 PM

No, $a$ is just a vector

T_01

ahha

Astyx

$a= (a_1, a_2, a_3)$

T_01

Ahh I did not understand that

Now things start to make sense, one moment

Astyx

Hence $I+aa^T$ is your matrix $(1+a_ia_j)_{i,j}$

T_01

Yes now that makes sense....

Okay Im with you now

@Astyx Then we can deduce that they are linearly dependent?

Astyx

3:42 PM

it depends what you mean by "they"

T_01

$a$ and the eigenvectors of my matrix

Astyx

Well, then again, you're making an assumption

T_01

Yeah of course if things are non zero

Astyx

Then yes

T_01

eigenvectors are non zero

But it could still be that $a^Tx = 0$

for example if $n=2$ and $a1 = -a2$

Mike Miller

3:45 PM

@user2103480 Just saw that. I think your useful facts amount to the Eilenberg Steenrod axioms. :)

Sometimes it is useful to collapse cleverly chosen subspaces that make something look simpler

T_01

or just for any suitable eigenvector

Mike Miller

But I'm gone again

Balarka Sen

@user2103480 The snake map in the homology LES is taking boundary in the literal sense

A very useful topology fact

T_01

Well I can deduce that $aa^T$ and $A$ have the same eigenvalues right?

Since $(I+aa^T)x = x+ aa^Tx$

But actually no

I still do not see how to go on from here.. what am I missing?

Astyx

So do you see what the eigenvalues of $aa^T$ are, and what their multiplicity is?

T_01

3:55 PM

Why is this clear? I mean we just have $(a^Tx)a = cx$, so if (a^Tx) would be non zero I would have eigenvalues of the form $c/a^Tx$, why could'nt there be different multiplicities?

No

Astyx

Ok, let's start with the case when it's nonzero

What can you deduce on $x$

T_01

it equals $c^{-1}*(a^Tx)a$

So basically it is a multiple of $a$ which is interesting

Astyx

Right. Now plug $\lambda a$ in $aa^T$

T_01

What do you mean by plug? calculate $(\lambda a)(\lambda a)^T$ ?

Or $aa^T * \lambda a$

Astyx

No, I mean compute $aa^T x$ when $x=\lambda a$

T_01

4:00 PM

Ok im on it

I think I should get the $\lambda$'s then right?

Astyx

And then deduce the eigenvalue

T_01

And then the product of those should be my determinant

correct?

Astyx

Not yet

T_01

Uff ok i write that down first

copper.hat

what is the range space of $aa^T$?

@T_01 if $aa^Tx = \lambda x$ then either you know something about $x$ or something about $\lambda$.

T_01

4:07 PM

Aha so the i'th entry is $a_i \lambda <a, a> = a_i \lambda ||a||^2$

Astyx

You shouldn't have to do it coordinate wise

Remember, $a^Ta$ is just a number

T_01

Yes of course. This is just $\lambda ||a||^2 a$

not?

Astyx

Ok, so what's the eigenvalue?

T_01

$\lambda ||a||^2$, and lambda was an eigenvalue of $aa^T$

So $a$ is an eigenvector

Astyx

You made a small mistake

T_01

4:10 PM

In the calcalation?

Astyx

We have $\lambda ||a||^2 a = (aa^T)\lambda a$

The eigenvalue is not $\lambda ||a||^2$

T_01

Ah yes, again divided by $a^T\lambda$ if that is non zero

Astyx

That doesn't make sense

T_01

Yes

Wait

we just have to divide by lamdba I guess

Astyx

Why can you do this?

T_01

4:13 PM

So the eigenvalue just remains to be $||a||$ Well only if $\lambda$ is non zero...

Astyx

Yes, why is $\lambda$ nonzero?

T_01

We assumed $a^Tx$ to be non zero

Astyx

More importantly we assumed $x$ to be nonzero

T_01

But the $c$ could be zero as an eigenvalue of $aa^T$

copper.hat

what is the range of $aa^T$?????

T_01

4:16 PM

R^n ? i guess

copper.hat

you need to step back a moment and think about that.

T_01

It is an $\mathbb{R}^{nxn}$ matrix or am I wrong here?

copper.hat

yes and yes

think about what the range of $a a^T$ can possible be.

T_01

Do you mean image instead of range?

@Astyx So in that case i remain with only one eigenvalue $||a||$ with multiplicity $n$ ?

Astyx

No

user2103480

4:19 PM

@MikeMiller yeah tbh the eilenberg-steenrod axioms are what I use mostly since I am bad at explicit computations of maps. I'd appreciate heuristics, like balarkas favourite and only tipp ever about the snake map, and training examples

copper.hat

image range whatever you want to call the collection of values $a a^T x$.

user586228

Where do I ask such questions

Astyx

We've shown that the only possible nonzero eigenvalue is $||a||^2$, and that the corresponding eigenspace is $Span(a)$

Alessandro Codenotti

@user586228 I don't know but definitely not here

user586228

ok

deleteing at once

T_01

4:22 PM

Yes I forgot the square there

Astyx

That doesn't mean that 0 isn't an eigenvalue

Which brings us to the second case: on what condition is $x$ an eigenvector with eigenvalue 0?

T_01

When $a^Tx = 0$ I guess?

Astyx

Do you know what that means?

T_01

That x is in the kernel of $a^T$

Astyx

That's one way to look at it

T_01

4:25 PM

wait no a is not a matrix (ok technically yes but no)

Astyx

it is, just not a square one

What dimension is the kernel of $x\mapsto a^Tx$?

T_01

if $x$ is not zero it is one dimensional (maximal)

Astyx

No

Wrong condition and wrong statement

T_01

Wrote it before you edited the question ^^

Astyx

Ok, but it still didn't make sense

T_01

4:28 PM

So this linear map is represented by the matrix where we have the $a^T_i$ on the diagonal

Astyx

No

T_01

its not?

Astyx

This is a map $\Bbb R^n\to \Bbb R$

Not represented by a square matrix

So talking about diagonals makes no sense

T_01

Ah yes we have a number as result..

Okay so the "matrix" is $a^T$ which has one row, so the image can be highest one-dimensional

So the kernel is up to $n-1$ dimensional

Astyx

not "up to"

T_01

4:30 PM

It is, because the image is not trivial

if $a$ is not zero

otherwise the kernel would be $n$ dimensional

Astyx

Right, so at least n-1 dimensional, and n dimensional iff a is zero

Can you wrap up ?

T_01

Well if $a$ is zero the matrix is $0$ and the determinant is trivially also $0$ :D So lets assume $a$ is not zero so the kernel is $n-1$ dimensional, so looking at $a^Tx$ there should be many zeros as eigenvalues in the situation above

Astyx

Be more specific. Which determinant?

T_01

of $aa^T$, but then my original $A$ is just $I$ so this case is not interesting

Astyx

Right

T_01

4:35 PM

So we have the eigenvalues $||a||^2$ with multiplicity $1$ and $0$ with mult. $n-1$

Astyx

What then?

T_01

The determinant of $aa^T$ should be the product of the eigenvalues? (So 0)

Astyx

Why is that?

T_01

Wasnt that the case for symmetric matrices? I dont know, thaught I heared that in la

Astyx

Hum, yes that's true for symmetric matrices. But the broader result is that it's true for diagonalisable matrices

T_01

4:42 PM

Well symmetric matrices are diagonalisable

Astyx

Right

What about the determinant we're interested in now?

T_01

It should be zero then

Astyx

In other words, what are the eigenvalues of $I +aa^T$? Is that matrix diagonalisable?

T_01

I mean the determinant of $aa^T$ is zero. What can one say about the determinant of sums of matrices?

Astyx

I'm not asking about the determinant yet, just about the eigenvalues

T_01

4:45 PM

Well $(I+aa^T)x = x + (aa^T)x = x$ equiv. to $(aa^T)x = x$ so they are the same as of $aa^T$

copper.hat

huh?????

T_01

wait what am i even doing

this is latex math xD

I mean:

$(I + aa^T)x = \lambda x \Leftrightarrow x + (aa^T)x = \lambda x \Leftrightarrow (aa^T)x = (\lambda - 1) x$

wait

copper.hat

think

T_01

Yes

$\lambda -1$ has to match the eigenvalues from above so $\lambda \in \{ 1, ||a||^2 + 1 \}$

copper.hat

yes

T_01

4:48 PM

And then the product is the solution..... what the heck

Astyx

a, not x

T_01

yes, thank you

copper.hat

$\det (\lambda I -A) = 0 $ iff $\det ((\lambda -t)I - (A - tI)) = 0$.

T_01

Sorry for your time guys

Astyx

by the way, since you know the spectral theorem (ie that symmetric matrices are diagonalizable) there's a much faster way of doing that

copper.hat

4:50 PM

np. loosely the eigenvalues of $f(A)$ are $f(\lambda)$.

T_01

How would that look like? I feel like I forgot everything about linear algebra in the period of one semester... we also hat jordan forms and such funny things but I really dont remember anything anymore

Matheus Sousa

Hey guys, can anyone help me? I have to count the number of ways to choose five elements from the set {1,2,...,20} without choose consecutive numbers.

$\left( \begin{array}{c} 20 \ 5 \end{array} \right) - \sum_{k=5}^{19} (20-k)\left( \begin{array}{c} 5 \ 5 \end{array} \right)$

this is my solution

T_01

Thank you very much. I will try to bring this clean on paper and hopefully remember this kind of "trick". I mean I even looked at "could i get the eigenvalues of this thing?" And gave up after 5 minutes because I did not see a way of doing this. Sometimes I feel like I should quit mathematics at all

Astyx

Assume $a\ne 0$. We've shown that the only nonzero possible eigenvalue of $aa^T$ is $||a||^2$, with eigenspace $Span(a)$ - hence with multiplicity 1. Spectral theorem tells us that 0 is an eigenvalue with mutliplicity n or n-1.

copper.hat

@T_01 despair is normal when working with mathematics.

Astyx

4:54 PM

Also, $x, y\mapsto x^Ty$ is a scalar product. So the condition $a^Tx=0$ can be seen as an orthogonality condition between $a$ and $x$.

T_01

Well but I just cannot realize how some genius like you guys can look at such things and after 5 seconds say "ah ok thats easy" :D But learning consistently is the key I guess

copper.hat

it is easy when you have seen it hundreds of times.

Astyx

^

T_01

What are you doing if I may ask? (working, studying..?)

Astyx

Student in grad school

Gotta go, seeya

T_01

4:56 PM

Bye. Thank you again

Matheus Sousa

5:06 PM

Hey guys, can anyone help me? I need to count the number of ways to choose five elements from the set {1,2,...,20} without choose consecutive numbers.

My solution: $\left( \begin{array}{c} 20 \ 5 \end{array} \right) - \sum_{k=5}^{19} (20-k)\left( \begin{array}{c} 5 \ 5 \end{array} \right)$

Is this correct?

copper.hat

@MatheusSousa it is not clear what you are asking. are you selecting $x_1,...,x_5$ and you just need $x_{i+1} \neq x_i$ or are you asking if the sorted selection has no adjacent numbers?

is 1 3 2 4 6 a valid selection?

Matheus Sousa

I'm asking a sorted selection. Exactly this. 1 2 3 4 6 is a valid selection

copper.hat

now i am confused.

Thorgott

@Balarka @Mike do all embeddings of a manifold into Euclidean space become isotopic in some larger Euclidean space (perhaps even in $\mathbb{R}^{\infty}$)? isotopic meaning homotopic through embeddings here. interpret embedding in the topological or smooth category, depending on which you prefer.

Balarka Sen

the answer is yes

Thorgott

5:14 PM

nice

is it a transversality argument?

Mike Miller

Do you know that embeddings are dense so long as codomain has dimension >2dim(M)?

user2103480

tfw when you write an email to your prof asking if they can confirm your alternative solution to a problem and help with justifying the steps inbetween, and get back a slightly annoyed "yeah sure you could do it that way but thats not the point of the exercise"

Mike Miller

That's a transversality argument, very standard

Matheus Sousa

@copper.hat is like, if you choose 1 2 3 4 5, this choose isn't valid because all numbers are consecutive. But, if you choose, 1 2 7 8 15, is valid because not all numbers are consecutive

Mike Miller

Once you know that you should be able to convince yourself that the relative fact is true: if S is a submanifold of M, same assumption that dim N > 2 dim M, then [embeddings equal to a fixed embedding on S] is dense in [maps equal to that fixed embedding on S]

user2103480

5:20 PM

professor-chan I did not imply your solution was bad owo

Balarka Sen

@user2103480 lol

Mike Miller

Now apply this to M x I to conclude and S = M x {0,1} to conclude that any homotopic embeddings M -> N with disjoint image are isotopic so long as dim N > 2dim M + 2

Conclude that any two embeddings of M into R^n with n > 2dim M + 2 are isotopic

These aren't the best possible dimensions, this tells you knots are all isotopic in R^5 but that's already true in R^4. But they work

copper.hat

@MatheusSousa surely the answer is $\binom{20}{5}-16$? There are only $16$ consecutive sequences.

user2103480

@BalarkaSen how's this youtube.com/watch?v=Dcmo3gZ38E8

Balarka Sen

2dim M + 1 is the best possible dimension in general

strong Whitney that's all

@user2103480 listening

Sorry, I mean if dim M > 1 then 2dim M + 1 is best possible

Let me rephrase lol, n >= 2dim M + 2 (not a strict inequality) you get by strong Whitney, n >= 2dim M + 1 for dim M > 1 is also true.

Mike Miller

5:28 PM

I am concerned though

Does strong Whitney work relatively?

I guess it ought to, by using relative handle decomps

Balarka Sen

yes

Mike Miller

Probably you ought to assume dim M > 4

Hmmm

Balarka Sen

everything is smooth man

Mike Miller

No

dim M > 3

@user2103480 Yeah but that's all I'll tell you too since I taught him that

user2103480

@MikeMiller smh once again topology will wreck me

Balarka Sen

5:30 PM

lol

what was I asking you that day about Emb(M, R^infty) Mike

I forgot

Emb(M, R^infty)/Diff(M) is a model for BDiff(M) which classifies M-bundles

I was asking how to embed M-bundles in trivial plane bundles maybe

What an awful question

@Thorgott Find an appropriate model for $\text{Emb}(M, \Bbb R^\infty)$ and using what Mike told you prove that it is contractible.

The right question for you

Matheus Sousa

@copper.hat no, the consecutive sequences are $\sum_{k=5}^{19} (20-k)\left( \begin{array}{c} 5 \ 5 \end{array} \right)$

you mean that this is wrong?

copper.hat

@MatheusSousa you lost me. surely the consecutive sequences (after sorting) are $1,2,3,4,5$, $2,3,4,5,6$, $\cdots$ , $16,17,18,19,20$?

Mike Miller

@BalarkaSen Good Q

For R^inf with the standard topology, a compact subset lies in R^n, yes?

Thorgott

Actually, "relative Whitney" is all we need, right? Two embeddings with disjoint images (which of course can always be achieved) give an embedding of $M\times\{0,1\}$ and if we can extend that to an embedding of $M\times I$, it's automatically an isotopy (because each fiber of $M\times I\rightarrow M$ is embedded).

Mike Miller

No, I don't buy that, the embedding of M x I doesn't need to be time-preserving.

This buys you an isotopy so long as you bump dimension of codomain up.

Thorgott

5:39 PM

@MikeMiller if by standard you mean weak, yes

Mike Miller

To apply Whitney to M x I

Weak topology w/r/t projections p_n. I guess this is obvious: you cover by p_n^{-1}(R - 0).

Thorgott

wdym "time-preserving"

Mike Miller

Preserving second factor

Oh I misunderstood

You're assuming codomain is already > 2dim M + 2

Fine

Balarka Sen

Ideally the model for Emb(M, R^infty) should be a simplicial set

Do away with topology

Thorgott

I'm assuming whatever I need to assume for relative Whitney

just remarking that that's all the input one needs (I think)

Mike Miller

5:41 PM

@BalarkaSen Fuck off

Balarka Sen

lol

copper.hat

he said politely

Mike Miller

@Thorgott That's correct because I just outlined for you the proof of relative Whitney

copper.hat

from a billy Connolly sketch

Thorgott

the denseness result seems a priori stronger, but perhaps it isn't

Mike Miller

5:44 PM

That's how it's proved

Balarka Sen

its a good exercise for you my man

Mike Miller

Any proof of weak whitney proves that

Whether or not they say so

Matheus Sousa

@copper.hat yes, I'm sure about that

Balarka Sen

well the usual trick is boiling dimensions down

by projecting to subspaces

i never liked it very much

Thorgott

yeah, that's the one I know

Mike Miller

5:44 PM

Since you're using transversality technology and you know that you can modify something to be transverse by arbitrarily small wiggles, you conclude that you only need arbitrarily small wiggles to get where you want to go

Yes but that proof still gives the density result

Balarka Sen

this is why i didnt want to explain thorgott anything tbh

Mike Miller

Since the projecting is a transversality trick

copper.hat

@MatheusSousa So there are $\binom{20}{5}$ ways of choosing $5$ numbers and $16$ of these ways are 'illegal'.

Balarka Sen

just treat his questions as yes or no

Mike Miller

LOL

I see

Force him to think about it

user2103480

5:45 PM

topologist gatekeeping

topologistsplaining

Balarka Sen

TOPOSplaining

Mike Miller

@user2103480 I gatekept Balarka so he must pay it forward

user2103480

Gotta keep the number of topologists low. The market's tight

Thorgott

right, I guess that's what it is

Matheus Sousa

@copper.hat ah ok, I understood now. Thanks

Mike Miller

5:51 PM

@user2103480 personally I would like to increase the number of nonprofessional topologists

but decrease the number of professionals

explain

n

o

Hello there.

Random walk on random measures on spaces of random manifolds

The future of topology

Do you see it?

Beautiful

Bright

Wow

Gonna kill myself real quick

user2103480

5:55 PM

space of random manifolds?

you mean just space of manifolds or random variables with values in manifolds

Thorgott

the latter of course

Balarka Sen

this guy

Thorgott

calculate the expected rank of the random homology thus obtained

Balarka Sen

you literally described persistent homology

Mike Miller

Yeah that seems fine

A random thick manifold is a point cloud and a radius

Balarka Sen

5:57 PM

thick manifold lol

Mike Miller

Also known as open set

Balarka Sen

brilliant idea

user2103480

so uh random variables whose values are measures on a space of random variables. and those measures you add together to a random walk. But aren't spaces of random variables pretty large if one doesn't pay attention? Are there common examples of measures on spaces of RV?

I'm not expecting a rigorous construction here or something, just like to think about what notions are necessary to do this

Balarka Sen

galton watson measures

its a measure on space of galton watson trees

which are tree-valued rvs

user2103480

Sounds cool. A lot of cool point processes and the like come up in biology

Balarka Sen

6:01 PM

lol ok

i think its a meme

user2103480

What do you mean

Balarka Sen

nothing

Thorgott

I think we had an exercise on Galton-Watson something in my probability course

something about, uh, populations dying or something

too applied

user2103480

6:21 PM

But the value of the random variable is a tree isn't it?

Spaces of trees are perfectly fine topological spaces

Spaces of random variables are horrible since normally we don't really put any structure on the probability space

Balarka Sen

@user2103480 Yes

user2103480

I think it can almost always chosen to be a countable product of spaces but if you don't make any identifications with other kinds of objects, this sounds unnatural

Thorgott

small structure = weak

big structure = strong

puny probabilist

dc3rd

@robjohn been attempting to use the diagram you linked to. It has given me some new ways to conceptualize the image, but I'm still stuck. I'm going to give it another 10 mins. I've been working on this problem for far too long and it is buringin me because I know it is something fundamental that will come back to haunt me.

copper.hat

I wonder if someone could do me a favour. I am stuck on a homework problem in the first edition of Munkres, "Topology, a first course" and think (hope) there was an errata (which I can't find) or an update in a later edition. I don't want to buy the book just to check and was hoping someone could look for me please.
The problem is Exercise 7, Section 8.3:
Let $p : E \to B$ be a covering map. Assume that $B$ is connected and locally connected. Show that if $C$ is a component of $E$, then $p\mid_:C \to B$ is a covering map.

user2103480

6:35 PM

"probability spaces don't exists" - probabilist

copper.hat

Hah, just found a 2nd edition copy online and that particular exercise seems to have been removed.

Thorgott

@copper.hat I believe the problem is correct as stated

copper.hat

without the path parts? :-(

i mean is the result true without the path parts?

i'm curious why the exercise was removed in the 2nd Ed.

Thorgott

yes, I'm fairly certain it is true

robjohn

@dc3rd what is the definition of the cross product you're working with? I usually think of it as $a\times u$ is the projection of $a$ onto the plane perpendicular to $u$ and rotating counter-clockwise around $u$ (so that we stay in the plane perpendicular to $u$) then multiplying by the length of $u$

copper.hat

6:48 PM

@Thorgott thanks! i was hoping it was not my inability :-)

Thorgott

I don't know why it was removed either, it's a good exercise

robjohn

@user2103480 almost surely

copper.hat

@Thorgott i found a similar problem here at.yorku.ca/b/ask-a-topologist/2004/1079.htm but it has path in the statement.

dc3rd

@robjohn, I'm not using that form of the definition (at least I don't think so). The one being used is the determinant of three vectors with the first being the basis vectors:

$x \times y = (x_{2}y_{3} - x_{3}y_{2})\mathbf{e_{1}} + .... + (x_{1}y_{2} - x_{2}y_{1})\mathbf{e_{3}}$

Along with knowing that $||\mathbf{x} \times \mathbf{y}|| = $ area of parallelogram

robjohn

@dc3rd Ah, so the coordinate-wise computation.

dc3rd

6:54 PM

@robjohn Also that the cross product is orthogonal to the vectors being used for it

robjohn

@dc3rd Aha. so if it is perpendicular to both vectors, $a\times u$ is definitely in the plane perpendicular to $u$. It is not too hard to see that it is perpendicular to the projection of $a$ onto that plane.

Thorgott

yeah, but connectivity will suffice

dc3rd

@robjohn I'm drawing out the statement you said again to see if it clicks

robjohn

The only thing you need to verify is that the length is the length of the projection of $a$ onto that plane times the length of $u$.

This is taking you away from Ted's approach, and this is a problem from Ted's book, but I do find this definition, or characterization, easier to use from a geometric point of view.

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