@Jasper it's a neverending job, hundreds of pages, and all the stuff in there is interconnected. I mean I have a clear picture of all is in there together with the network I created.

@Waiting Maybe once you do a degree, you can use parts of your book as a thesis and get a PhD straightaway. =D

@Jasper This is crazy, when you begin to see hundreds of pages like you have in front of your eyes a single picture.

@Jasper It's fascinating.

@Waiting Yes, it must be a very beautiful, complicated picture.

@Dodsy I can better understand the opposite

@Jasper Magnificent.

@Jasper I love mathematics, I couldn't live without it.

On the other hand, I do think it's true that having relationships---not necessarily romantic---helps with anxiety

@Waiting Do you like physics too? Some people here do both, like Semi and Sha.

in the sense that it helps get me out of my own head

Counterpoint: Relationships also add anxiety :P

yep.

NV-US's question is also adding anxiety.

to the extent that relationships / interactions with others get me out of my head, it's helpful

@NV-US why don't you ask it in main, considering that you've asked it here like four times already?

One of the mad starrers is back.

to the extent that relationships would represent new obligations and responsibilities, not so much.

Anxiety helps with anxiety for me.

@TedShifrin It's me, lol.

@Ted you may want to see this question

@TedShifrin I think I am seeing the picture and why your hint is relevant.

cc @AlessandroCodenotti

Jasper: Behave yourself.

@Semiclassical It helps keeping your mind busy and not allowing to develop the anxiety related thoughts.

I thought you might, @Balarka.

i only want hints, on main, they give the answer @LeakyNun

@Dodsy had to head out

i will post it then

but, to continue on

Hell no, @Leaky.

@NV-US Maybe you can ask for a hint on the main site as well.

@TedShifrin alright

I'm not giving up yet, @NV-US. I'm a bit surprised I don't know how to do it (if it's true).

it's embedded the tree at the end, sure. what's its velocity?

this should not require any deep analysis.

haha @TedShifrin

In general, projections needn't commute, of course.

there is a hint given after the question, should i tell u? @TedShifrin

@Jasper Yes, it's fine. However, I might like it more if I studied it more.

What's the hint?

@Semiclassical the opposite of what

?

hint : Use the trace function and ask yourself what the trace of a projection is. @TedShifrin

Oh sorry

the velocity is 350 m/s

that's the initial velocity.

The final velocity is 0 m/s

Yep.

So you know the initial velocity and final velocity. You also know the distance it travels.

@NV-US: The trace depends on the rank.

right.

What you don't know yet is the (constant) acceleration and the time elapsed

@TedShifrin that is all that is given

The acceleration is equal to the change in velocity over the change in time.

right.

@Ted: Suppose I look at a ball of radius $r$ at $0$ in $\Bbb C$. There exists germs $g$ at $0$ such that $g$ does not extend to holomorphic functions on all of $\Bbb{B}_r(0)$. I think this is the key observation.

unfortunately, that equation isn't enough since you have two unknowns in there.

Let's see if I can parse this mathematically. If $p^{-1}(\Bbb{B}_r(0))$ can be written as a disjoint union of neighborhoods above in $|\mathscr{O}(\Bbb C)|$, then there exists a neighborhood $U$ around $g \in \mathscr{O}_0$ such that $p(U) = \mathbb{B}_r(0)$. ($U$ is one of the slices)

what i'd suggest doing is looking at your big 4 and seeing if any of the equations contain exactly one unknown.

okay

I'm out to work (some more).

So @NV-US, that tells us that the sum of the dimensions of the spaces we're projecting onto must be $\dim(V)$. Let me think more.

sure

@Balarka: You're forgetting to say each neighborhood is homeo to the fixed one?

i'll post full pic of the question, if i am missing something.

@Ted Yeah, sorry, $p$ is a homeomorphism $U \to \mathbb{B}_r(0)$, in particular.

@Semiclassical so -350 = a*t

okay. but you don't know a or t, so that contains two unknowns.

@Semiclassical how about $v_f^2=v_i^2+2(a)(d)$?

ding!

:o

you know the initial/final velocity, and you know the stopping distance

Oh, I see the point, @NV-US and @Leaky.

@TedShifrin hint?

so the only unknown in that equation is the acceleration a.

hence...

The images of the $E_i$ must intersect only in $0$. So if we show $E_iE_j = -E_jE_i$, those must both be $0$.

please tell

@Semiclassical so then $0=350+2(a)(0.130)$ $\frac{-350}{0.130}=2(a)$=-1346 m/s ?

@NV-US do we have $k=\dim(V)$?

induction?

No, Leaky.

that does not sound right.

The projections can be onto higher-dimensional subspaces ... not just lines.

no @LeakyNun

@TedShifrin Ah, but by definition, since $U$ is a neighborhood of $g$ it consists of the germs of some holomorphic function $f$ on $p(U)$. Since germ of $f$ at $0$ is $g$, and $\mathbb{B}_r(0)$ is the homeomorphic image of $U$ by $p$, that gives that $f$ is a holomorphic extension of $g$ to $\mathbb{B}_r(0)$, contradicting our assumption about $g$.

you're not being careful with your algebra/substitutions

oh I forgot the squares..

Why must the images intersect only trivially...

But the subspaces can't intersect nontrivially if the sum of the projections is the identity.

@TedShifrin induction?

(A germ which does not extend to a holomorphic function on the ball of radius $r$ at $0$ can be found by choosing a convergent power series of radius of convergence less than $r$)

Well, suppose $E_i$ projects onto $V_i$ and $V_i\cap V_j \ne (0)$.

I wasn't thinking induction, but perhaps so, NV-US. What are you trying to show by induction?

@Semiclassical okay now I'm getting huge numbers

You have the idea, Balarka. I'm not reading details.

Let $E_i \vec x = \vec v = E_j \vec y$ with $\vec v \ne 0$.

well, you are talking about a bullet

@Leaky: There's a better approach.

@Semiclassical should I instead use $d=\frac{v_i+v_f}{2}(t)$?

that if $E_{1}+E_{2}+..+E_{K}=I$, $E_{i}E_{j} = 0$ for $i \ne j$.by induction on k

Suppose there's a nonzero vector $x$ with $E_ix = E_jx$.

ugh.

@NV-US: I thought about that at the dentist. For induction to work, you need to know that the sum of two projections is again projection, and for that you need that their composition is trivial, I believe.

Then $E_j E_i x = E_j E_j x = E_j x = E_i x = E_i E_i x = E_i E_j x$

@TedShifrin For sure; a lot of people are cornering you already, so I'll subtract myself from the crowd :P

I think my idea's okay too.

Good example, though. I like this ten times more than what I gave you at first :P

You saw the point of my hint, @Balarka, so I think that suffices :P

I thought you'd find this more interesting.

ty @TedShifrin @LeakyNun :)

@NV-US I haven't seen it lol

@Semiclassical so using the first equation I get a = -471153.85

The sum of the subspaces (the images) must be the whole vector space. Could they overlap?

for arbitrary transformations yes, and I don't know enough properties of projections to say.

Well, one way to see it is the hint @NV-US gave from his book.

EiEj = EjEi for all j and i, now we can operate Ei on the equation sum (E(i))=I and add all the equations, and subtract from this each equation one by one to get EiEj =0 for all i and j

and I get $t=7.43x10^{-8}$

Wait, @NV-US: How did you get $E_iE_j=E_jE_i$?

@LeakyNun just proved it

@NV-US you didn't see the assumption

That was only for a particular $x$.

:40137248

I was trying to get you guys to see if you could have any nonzero vector in the image of both $E_i$ and $E_j$.

There is a proof, now, but I don't think you see it.

No.

:(

something is not right.

Suppose $E_i$ is projection onto the subspace $V_i$. I'm asking if $V_i$ and $V_j$ can intersect in something bigger than $0$.

well the trace can only be 1 or 0

No, @Leaky. The trace is the dimension of the subspace.

Start by proving that, both of you.

@Semiclassical oh yes, so we can use the equation $d=\frac{v_i+v_f}{2}t$, $0.130=\frac{350+0}{2}*t$, $t=\frac{0.130}{175}=7.43x10^{-4}$

@TedShifrin I know like nothing about trace except tr(AB)=tr(BA)

You know trace is well-defined for a linear transformation, so it's invariant under change of basis.

Trace is also linear. Duh.

lol

But if we use $v_f^2=v_i^2+2(a)(d)$ and then sub the a into $v_f=v_i+a(t)$ we get $7.43 x 10^{-8}$

@TedShifrin that's like saying tr(A'BA)=tr(BAA')=tr(B)?

If you like, yes.

proved @TedShifrin the intersection can only be zero

then I've convinced myself of that, lol

@NV-US Then it follows that $E_jE_i=0=E_iE_j$. Why?

tr(E1)+...+tr(Ek)=dimV

You're not out of the woods yet, Leaky.

if any two images have non-trivial intersection then you won't have enough dimensions...

@Semiclassical nevermind! i was doing the mathematics wrong!

Can you make that rigorous?

@TedShifrin I'm trying hard to

:'(

@NV-US: I thought you had a complete proof?

Hey everyone!

i thought too, thinking again, no hints plz

@Dodsy had to step away

rehi daminark

I'm done with hints.

But you've got the right idea: use what you know to obtain the acceleration and then use that to get the stopping time

Hi Demonark.

How's everything going?

(There's actually a shortcut way to do it, but I'll point it out after)

@Semiclassical well instead I used $d=\frac{v_i+v_f}{2}t$

Damn, I missed a meme during the anxiety discussions

This makes me anxious

But I needed the acceleration anyways because the next question is about the force that the tree exerts on the bullet.

@Dodsy that doesn't work

what doesn't work?

Distance equals change in velocity over 2?

Oh, missed the t

@BalarkaSen missing memes is a horrible experience

whoops put a negation instead of a summation.

Your future just got a bit darker

Yeah, that too

I need to learn way more about linear algebra

I'm completely clueless

@Daminark 50 keks darker

Avg velocity = distance over time = average of initial and final velocities

:O

Are the subgroups of $\Bbb R/\Bbb Z$ easily classified?

This only works because the accel is constant, to be clear

@Semiclassical so now I'm trying to find the force which the tree excerts onto the bullet, but this is equal to the force that the bullet exerts onto the tree. So it's just ma?

@Leaky: Where do the images of $E_iE_j$ and $E_jE_i$ live?

@LeakyNun Yes, every subgroup is dense (so Prufer-like) or finite cyclic.

@BalarkaSen dense?

Ya

Are all dense subgroups the group itself?

Noooo

No.

$\Bbb Q/\Bbb Z$

@Leaky probably a good idea, I remember being told by one of the guys doing algebraic topology talks in the REU that the only things we can really do are linear algebra and kinda combinatorics, so in other parts of math we're trying to turn problems into that

@Semiclassical okay but I get a negative $F_{net}$ so do I switch it to a positive?

Or the p-torsion points

Well, what does negative acceleration mean?

if the bullet exerts a negative net force onto the tree then the tree exerts a positive net force onto the tree?

Good exercise for Leaky: If $\theta$ is an irrational multiple of $\pi$, what can we say about the subgroup generated by $e^{i\theta}$ in the unit circle?

it means that it slowed down at that rate.

@BalarkaSen oh

depending on your conventions.

I still haven't drawn a fbd.

Demonark: Linear algebra is under-appreciated and probably under-studied.

Agreed

Uh, by p-torsion points I meant the p^k-torsion points for all k.

Sure. Equivalently it's accelerating in the -x direction

look, I've devoted my summer to abstract algebra

But geometers and applied mathematicians really need it (as do representation theorists).

Linear algebra is very nice.

@BalarkaSen that's also the torsion subgroup, right?

LOL ... Balarka. Damning with faint praise?

Right, if we set right and up to positive then it is acceleration to the left at 471153.85 m/s

@TedShifrin In some sense that is the entire point of representation theory. We feel like we have a chance to understand vector spaces

so the force of the tree on the bullet points away from the tree

Sure, @Tobias :)

@Alessandro Yup

right.

Oh, we almost needed you for someone the other day, @Tobias, but luckily he showed me enough of his book that I figured it out.

Interesting question. If $V^G$ is at least 2-dimensional, show that $V$ cannot have a cyclic vector.

@TedShifrin a grad student I talked with today tried to explain me representation theory in 5 minutes, what I got is mostly that linear algebra is easier than group theory so why not try to reduce the latter to the former?

So by N3L the force of the bullet is equal and opposite i.e. +x direction as it should be

@TedShifrin Hah. No, really, I have understood why linear algebra is crucial as I progressed (major credit goes to you for forcing me to learn multivariable calculus, of course :P).

@AlessandroCodenotti Yep, that is pretty much it

@TedShifrin What is a cyclic vector?

@Semiclassical so then I'm getting 848.07 N. Or 848.1 N.

Though I will say that I appreciated linear algebra more by picking it up/using it in other contexts than in linear algebra itself.

kk. I don't have a good sense of what the numbers 'should be' unfortunately

@Tobias: As I recall, a vector so that its $G$-orbit generates?

@Semiclassical well I converted grams to kgs

I ended up liking it because it has a nice theory

@TedShifrin i am not able to prove that $V_{i}$and $V_{j}$ are distinct, just so i am understanding right, $E_{i} : V_{i} -> V{i}$.

and multiplied it by the positive version of the acceleration.

@TedShifrin Ahh, I would just call that a generator

Right

otherwise I'd have to change all of my calculations, right?

I thought of cyclic vector in the module sense, Tobias.

the d would be negative.

@NV-US: No, $E_i$ projects all of $V$ onto $V_i$.

I didn't see this lol

@TedShifrin Hmm, not sure I recall the vectors themselves being called cyclic in that context either. But it has been a long time since I did commutative algebra

ok, thinking again

Well, keep in mind that N3L tells you that F_21 = -F_12

@Tobias: Like for Jordan form ... a vector that generates the invariant subspace.

and $V_{i}$ is?

@NV-US: It's the image of the $i$th projection map :)

So if F_12 = ma then F_21 = -ma

Right.

@TedShifrin I do think that it's cyclic and it's countable.

should I say in my answers which NL I am uisng?

or is that overkill.

(well cyclic does imply countable)

but I hesitated for a long moment

So if a<0 then this cancels the minus sign. No need to put the minus sign in by hand

@TedShifrin sorry, i am very dull :(

and it's also dense @TedShifrin

Can't hurt to include it

Well, sure, it's cyclic. @Leaky. But can you describe it topologically, say?

Okay that's key

Aha ... yes, it's dense. Can you prove it?

I hesitated for a long moment because there does exist elements between the identity and the generator

@TedShifrin I saw a proof using pigeonhole