Mathematics - 2019-02-14 (page 1 of 2)

nCm

00:21

How to find an orthonormal basis for a orthogonal projection operator?

quallenjäger

00:33

Is it true for the inverse triangle inequality that $\mid a-b \mid\leq \mid a \mid - \mid b \mid$

Thorgott

consider $a=0$, $b=1$

topologicalmagician

@TedShifrin y is function of x and x is a function of u

user193319

00:57

If $A \subseteq B \subseteq X$ and we know that $X$ deformation retracts to $A$, can we say that $B$ also deformation retracts $A$?

Let $H$ be the deformation retraction of $X$ onto $A$. I was thinking that $F(b,t) = H(\iota (b),t) = H(b,t)$ would be a deformation retraction of $B$ onto $A$, but there is no reason to think that $H(b,t)$ will stay in $B$...

Secret

01:58

New thoughts about how to think about matrices and tensors:

Linear functions y=ax+b: one input value gives one output value, and such output can be normalised

Matrices and 2nd order tensors Ax=b: one ordered pair of inputs (x,y) gives one output value. The output can be normalised in the sense that the output is the combination of the unit output of the first and the unit output of the second

3rd order tensors: one ordered triples (x,y,z) gives one output value. and the output can likewise be decomposed similarly

Zee

02:38

@LeakyNun Zorn’s Lemma requires a non empty set

Come on,, you should know that

JBis

03:30

$8^{x-2} = 20^x$

how do I do?

I figure a log base 8 of both sides

$x - 2 = x * log_8(20)$

Jonathan Perales

Laurent series: Let's say I have to get the laurent series for (1/(z)(z-1)) in powers of z. Can I multiply 1/z by the series expansion I get from (1/(z-1)) to get my result?

anakhro

03:55

You will want to do partial fractions.

So let $\frac{1}{z(z-1)} = \frac{A}{z} + \frac{B}{z-1}$, and then solve for $A$ and $B$.

Rudi_Birnbaum

05:10

Good morning!

Would anyone like to explain to me what it means that the/a Hesse-Matrix is positive semidefinite ?

As far as I know, you can have imaginary eingenvalues in the Hessematrices I am dealing with, what in a sense means that you can have negative curvature components (interpreting the Hesse Matrix as second derivative of some scalar function $\Bbb R \to R^n$.

Though "they say" its positive semidefinite and give the "whole thing" a ">0" sign.

In some sense. Which is it?

Leaky Nun

05:44

@Rudi_Birnbaum ich glaub, das es heist "Hessian" in Englisch

Rudi_Birnbaum

@LeakyNun stimmt natürlich!

but please forget the question I have found the answer!

Leaky Nun

For smooth functions, the Hessian is symmetric, so it's orthogonally diagonalizable

in particular it has only real eigenvalues

Rudi_Birnbaum

(its only so for convex functions ... which is kind of circular ..)

Leaky Nun

circular?

Rudi_Birnbaum

@LeakyNun I am not sure, in my case that is not true in general.

Leaky Nun

05:47

enlighten me

Rudi_Birnbaum

The general statement is

"The Hessian matrix of a convex function is positive semi-definite."

Leaky Nun

sounds believable

Rudi_Birnbaum

So if the function is not (in all dimensions, I guess) convex, but instead you have a saddle point of some order, than you get imaginary eigenvalues.

This what I knew.

Leaky Nun

are your functions smooth?

Rudi_Birnbaum

Yes

Leaky Nun

05:49

3 mins ago, by Leaky Nun

For smooth functions, the Hessian is symmetric, so it's orthogonally diagonalizable

3 mins ago, by Leaky Nun

in particular it has only real eigenvalues

Rudi_Birnbaum

But now I stumbeled across something where a kind of hessian occures that is always psotive semidefinite

Leaky Nun

In fact you only need C2 to have symmetric Hessian

Rudi_Birnbaum

but then the EV are possibly negative then

Leaky Nun

indeed

Rudi_Birnbaum

if its nor convex

Leaky Nun

05:50

but never imaginary

Rudi_Birnbaum

My cases are QM cases which are not symmetric but hermitian

(self-adjoined)

Leaky Nun

hermitian... you're working with complex functions?

Rudi_Birnbaum

yes

QM = quantum mechanics

Leaky Nun

all holomorphic functions are analytic so in particular smooth

Rudi_Birnbaum

so in general we get sometimes "imaginary frequences"

Leaky Nun

05:52

so you do get hermitian hessians

Rudi_Birnbaum

@LeakyNun nonon

Leaky Nun

?

Rudi_Birnbaum

$\Bbb R^n\to \Bbb C$

Leaky Nun

fair enough

Rudi_Birnbaum

not $\Bbb C \to \Bbb C$

Leaky Nun

05:53

btw the saddle is characterised by $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$

you don't need imaginary eigenvalues

Rudi_Birnbaum

so nothing to do with holomorphic and stuff

@LeakyNun true but you can get them

And well we have singularities somewhere but avoid them by controlling the domain, so to say .. Its just numerical stuff in a way

Leaky Nun

but you should get symmetric complex matrices?

Rudi_Birnbaum

What I am takling about is the dependence of the (total) energy of a molecule from the positions of the nuclei

the positions of the nuclei are $\Bbb R^n$

and the energy is $<\Psi|\hat{H}|\Psi>$.

Where $\Psi$ is $\Bbb R^n\to \Bbb C$ and $\hat{H}$ is an hermitian operator.

Leaky Nun

great

Rudi_Birnbaum

@LeakyNun no hermitian,

$x\in\Bbb R^n$ then my Hessian is

$$\frac{\partial^2 <\Psi(x)|\hat{H}|\Psi(x)>}{\partial x^2}$$

So that can have negative eingenvalues, (no imaginary right)

but the "frequencies" that is the square roots of the eigenvaules (sorry) can be imaginary!

Sorry my mistake!

Leaky Nun

06:04

lol

Rudi_Birnbaum

lol

Now I stumbled across the claim that some Hessian is always positive seimidefinite

that confused me.

Until I found out that its not one for $\hat{H}$ but rather for some $\hat{H}_0$ ...

where $\hat{H}=\hat{H}_0 + \hat{H}'$

which I still do not fully understand (I mean for what reason that one is always positive seimidefinite)...

Rudi_Birnbaum

06:21

Wait, its about the diagonal part of $$\frac{\partial^2 <\Psi(x)|\hat{H}|\Psi(x)>}{\partial x^2}$$, the claim is that this has only positive eigenvalues ... hmmm ..

Ok its the diagonal part of the symmetric part (does it make sense?) @Leaky

Leaky Nun

complex symmetric matrices... not on my expertise :P

Rudi_Birnbaum

too sad

Secret

12:58

Dynamic inconsistency

In economics, dynamic inconsistency or time inconsistency is a situation in which a decision-maker's preferences change over time in such a way that a preference can become inconsistent at another point in time. This can be thought of as there being many different "selves" within decision makers, with each "self" representing the decision-maker at a different point in time; the inconsistency occurs when not all preferences are aligned. The term "dynamic inconsistency" is more closely affiliated with game theory, whereas "time inconsistency" is more closely affiliated with behavioral economics....

nature.com/articles/436150a

Rudi_Birnbaum

13:34

@Secret sick

Zerix

14:17

cosxdy=y(sinx-y) dx, 0<x<π/2

Does anybody know how to solve this differential equation

Rudi_Birnbaum

15:38

@ÍgjøgnumMeg grade meine zweite Radeinheit gemacht 01:15(@5-9°C aber sonnig wenns nicht grad im Schatten war ...).

Jacksoja

@LeakyNun Hello

quallenjäger

16:44

If I reparametrize a path $\alpha$ to a path $\gamma$ with constant speed, could it be that $\gamma$ is no more differentiable?

Aurelius

17:02

I have two sets of numbers

one contains less numbers then the other

however this smaller set is made of numbers that are way bigger than the ones contained in the bigger set

How can I define a function that maps these big numbers into the smaller ones?

I needs to be a bijection

122122122122122122122122122122122122122122122

9711511110010411898119101117121100554950

example of big numbers, or even bigger ones

Rithaniel

You can't form a bijection between two sets of differing cardinalities.

Aurelius

@Rithaniel Yes you are right thanks, I haven't explained correctly, I wanted to say that the bigger set resizes accordingly to the smaller. But this just confuses you. Let's just say that the sets have the same amount of elements

Ted Shifrin

17:22

@quallenjäger If the original curve is $C^k$, the reparametrized curve will also be $C^k$. Note that arclength $s(t)$ of a $C^k$ curve parametrized by $t$ is again $C^k$ (because you integrate a $C^{k-1}$ function).

@Rudi: The real Hessian is a (real) symmetric matrix. If you're dealing with complex functions, probably the Hessian is the matrix of partials $\partial^2 f/\partial z_j\partial \bar z_k$? This will be Hermitian.

quallenjäger

@TedShifrin But does the arc-length has to be differentiable?

Rudi_Birnbaum

@TedShifrin Hi Ted! In general we deal with complex stuff, but in both cases the eigenvalues are real, though can be positive. For time independent QM, you can show that you can write everything real.

Ted Shifrin

@Rudi, well, the eigenvalues of a hermitian matrix are of course likewise real :)

Rudi_Birnbaum

Yes as I said.

Ted Shifrin

@quallenjäger: It will be as differentiable as the original curve. I sketched the argument up there.

OK, @Rudi. I didn't read carefully. My main intent was to mention the complex Hessian.

Rudi_Birnbaum

17:28

but i can be more cosy to have real matrices sometimes.

Ted Shifrin

I'm a big fan of $z$ and $\bar z$ derivatives :P

Rudi_Birnbaum

@TedShifrin Well done you mentioned it ;P

Ted Shifrin

Comes from being a complex geometry person :)

LOL, I'll shaddup now. :)

Rudi_Birnbaum

@TedShifrin No way, we are always pleased about any comments from you. Btw HAPPY BIRTHDAY!!!!

quallenjäger

@TedShifrin Ähm sorry I mean does arc-length has to be invertible?

Ted Shifrin

17:30

Oh, thanks, @Rudi. ;)

Rudi_Birnbaum

Is it a good or a bad sign, that you appear here on your birthday?

Ted Shifrin

@quallenjäger: It's invertible as long as the original curve is regular (nowhere-zero velocity). That's the usual requirement for working with parametrized curves. Otherwise, of course, if the curve stops and sits somewhere for a while, you won't get an inverse.

@Rudi, my birthday was two days ago.

But I probably appeared for a while, regardless.

Rudi_Birnbaum

Oh dear - I lost control of time ..

quallenjäger

@TedShifrin And in such case the differentiability could be a problem?

Ted Shifrin

LOL

Rudi_Birnbaum

17:31

then I wish it belatedly!

quallenjäger

Happy Birthday, Ted!

Ted Shifrin

Thanks, folks. @quallenjäger: What is your precise situation? Is $\alpha'(t)=0$ on a closed interval?

quallenjäger

@TedShifrin I am a bit lost, I mean $\alpha'(t)=0$ could also occur if the path changes its direction? it doesn't really has to be a point, where $s(t)=s(t')$ for some $t'>t$?

So what I am currently struggling is, what would be the sufficient condition to ensure that the reparametrized path to be $C^k$

Ted Shifrin

No, arclength keeps accumulating as you move along the curve. It measures total distance traversed.

quallenjäger

But $\alpha'(t)=0$ does not induce that my path stops right?

Ted Shifrin

17:37

But, changing direction (literally) requires stopping. So there could be problems.

Yes, $\alpha'(t)=0$ means that the velocity is $0$, so the particle stops. The question is: For how long does it stop?

quallenjäger

Ok, then I have a problem. :D I would allow a general $C^1$ path, which I will allow stopping and change direction. I would like to know if I can reparametrize such path to a C^1 path with unit speed.

Ted Shifrin

Note that when your particle stops, the trajectory is likely to fail to be a manifold.

quallenjäger

I am working on $\Bbb R^n$.

Ted Shifrin

No, I'm talking about the image path.

Consider $\alpha(t)=(t^2,t^3)$.

That is a $C^\infty$ function, but the image has a cusp.

quallenjäger

Ah yes, I don't care about that

Ted Shifrin

17:41

So the arclength function is smooth, but its inverse is not differentiable at the point corresponding to $t=0$.

This is a good example to understand, I suspect.

quallenjäger

Exactly such things is of my interest, or as I have trying to work on ((t-2)^2,(t-2)^2)

Ted Shifrin

Why bother with the 2's?

quallenjäger

If I can parametrize such path to a $C^1$ path with unit speed

Ted Shifrin

Just look at the path $(t^2,0)$ in the plane.

Jacksoja

Ted in the definition of a limit

X,Y are metric spaces

why do we have E subset of X , why not work directly on X ?

Ted Shifrin

17:43

So you have the same phenomenon as in my example. Basically, the speed looks near $t=0$ like some (nonzero) constant times $|t|$.

@Jacksoja: I'm busy at the moment.

Jacksoja

alright have fun

Ted Shifrin

So $s(t)=\int_0^t |u|\,du$.

You can write that down explicitly. There will be an inverse, but it is far less smooth than the original because of vanishing speed.

quallenjäger

I noticed that, I have a cusp in both cases.

Ted Shifrin

Cusp? I don't understand that.

quallenjäger

Well, in this case I am not sure if the inverse is still differentiable and if the differential is continuous

Ted Shifrin

17:48

So, even though I started with a $C^\infty$ curve (doing $(t^2,0)$ or yours), arclength is only $C^1$. Now let's think about reparametrizing. Obviously, you can't do it continuously because the tangent vector to the arclength-parametrized curve would literally have to switch directions (as a unit vector) at the origin.

So there is no differentiable reparametrization by arclength.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

The same thing happens with my cusp example $\alpha(t)=(t^2,t^3)$.

hi, italic @Alessandro

@quallenjäger: The only hope will be if the curve leaves the stopping point with the same tangent direction with which it enters.

Érico Melo Silva

hi chat

quallenjäger

I didn't quite get your second argument, what the problem if the tangent vector of the arclength-parametrized curve has to switch direction?

Alessandro Codenotti

Hi @Érico

Ted Shifrin

17:54

It's like the signum function, @quallenjäger. If $f'(t)=1$ for $t>0$ and $f'(t)=-1$ for $t<0$, the function cannot have a value for $f'(0)$.

Heya @Eric.

Érico Melo Silva

how goes the livin errbody

Ted Shifrin

@quallenjäger: To be specific, Darboux's Theorem tells you that the derivative always has the intermediate value property (even without assuming $C^1$), so there can be no differentiable such function.

Rithaniel

Need some clarification on the definition of a term. In group theory, does "countably generated" means that the group has (a) countably many elements, or that it needs (b) countably many unique generators?

Alessandro Codenotti

Generators

Rithaniel

Also, the living goes well, EMS. (Danke, Alessandro)

Alessandro Codenotti

17:57

(it also implies that the group has countably many elements)

quallenjäger

@TedShifrin I see, thank you. I will work out these

I might have to come back again :D

Rithaniel

So you could never attain an uncountable set using only countably many generators? Interesting.

quallenjäger

I just noticed that I have totally written up some rubbish in my thesis because of that.

Ted Shifrin

Countable unions of countable things, @Rithaniel ?

My apologies, @quallenjäger. It's your fault for asking :P

Alessandro Codenotti

An element is essentially a finite string of generators, and taking finite subsets doesn't increase the cardinality

quallenjäger

17:59

@TedShifrin At least i noticed it earlier

better than pointed out by some expert to embarrass my self.

Alessandro Codenotti

(I'm assuming AC of course)

quallenjäger

@TedShifrin Would it mean, that if the path is at natural parametrization and is $C^1$, there cannot be such cusp?

Ted Shifrin

Natural parametrization = regular, as I said earlier? Nowhere-vanishing derivative?

Rithaniel

Okay, that makes sense. Though, that makes me think of a though I had a while back. Infinite sums are always of the form $\sum_{n\in\mathbb{N}} f(n)$. What if you were to somehow have a sum over an uncountable set?

quallenjäger

In other words, there cannot be such sudden change of the direction of the tangent vector?

Ted Shifrin

18:01

Tell us how to define that, @Rithaniel :)

Rithaniel

Yeah, that's a good point.

Ted Shifrin

If the derivative is nowhere-zero and varies continuously, then it cannot change direction at a point. It follows from basic stuff that the image is actually a $C^1$ $1$-dimensional manifold, @quallenjäger.

@Rithaniel: Note that you're using more than countability there. You're using an ordering, too.

Vrouvrou

hello, any help for this:math.stackexchange.com/questions/3112370/…

Rithaniel

What about all natural number points on the long line (I have no education on the long line, so my terminology there might be way off)

quallenjäger

But if the derviative is nowhere zero but not continuous, it could be some cusp right?

Ted Shifrin

18:07

No, I don't think so, @quallenjäger. Work out that theorem I told you about real-valued functions. Derivatives must have the intermediate value theorem, always.

There is no differentiable parametrization of $y=|x|$, no matter how hard you try, @quallenjäger.

With nowhere zero derivative, I mean.

Of course you can slow down, stop, and turn the corner.

quallenjäger

I see your point. The problem that $(t^3,t^2)$ worked is because we can guarantee the continuity of the derivative by shrinking its tangent vector length. But if we keep the length as constant, then the only way to guarantee the continuity of the derivative is that the direction is closed to each other right?

Ted Shifrin

Yes, sure. If you have a continuous nowhere-zero vector-valued function $f$, then $f/|f|$ is continuous.

quallenjäger

Darboux theorem works only for $\Bbb R$?

Ted Shifrin

18:22

I don't know what "intermediate" means otherwise.

But you can certainly modify this in $\Bbb R^2$ by considering the map to the unit circle, which can be (locally) parametrized by a subset of $\Bbb R$.

quallenjäger

I don't understand, would the intermediate theorem mean, that for any two tangent vectors $f(a)=v,f(b)=w$, which can be thought as a function on the unit circle, there must be a point $\psi$ on the arc enclosed by $v,w $, such that $f'(s)=\psi$ for some $s\in[a,b]$

Ted Shifrin

I mean values in the unit circle.

But I think that's what you're thinking.

quallenjäger

But I can have two arcs, which arc are we considering for the unit circle?

Ted Shifrin

No, the converse. If $f$ maps to a particular arc in the unit circle (you have to remove a point to start with), then every point in the arc between $v$ and $w$ must be a value of $f$ somewhere.

quallenjäger

I see, so it is basically in the direction of from $v$ to $w$?

Ted Shifrin

18:30

I don't know what that means without choosing the arc.

But you can do an "either/or" statement using both the arcs.

quallenjäger

Oh I see

Thank you Ted!

I go back to work

Ted Shifrin

You're welcome.

Vrouvrou

18:51

@TedShifrin hello

Balarka Sen

19:37

Hey @Ted

Daminark

Hey @Balarka!

Balarka Sen

Oh look a nerd

Daminark

D:

Alessandro Codenotti

Looks like this is the nerds club then! Hi @Balarka

Daminark

But yeah how's it going?

Also yo @Alessandro!

Alessandro Codenotti

19:39

Hi Dami

Balarka Sen

Yo @Alessandro

It's going fine

Alessandro Codenotti

What kind of math have you been doing lately?

Balarka Sen

Eh, nothing too specific. I have been thinking about probability now and then

Daminark

Nice

Balarka Sen

I had some vague functional analysis questions that I want to ask actually

Alessandro Codenotti

19:42

I'm a few years late but here isn't the Serre proof the one using Bass-Serre theory to say that a group acting freely on a tree is free rather than going through covering spaces theory? @Balarka

Also my functional analysis is rusty, but I should review it for a course I want to take next term so this might be a good occasion to think about some FA

Balarka Sen

@AlessandroCodenotti I guess so. Bass-Serre theory gives the more general Kurosh subgroup theorem

If $G = A * B$ then subgroups are all of the form (free) * (free product of conjugates of subgroups of A) * (free product of conjugates of subgroups of B)

Alessandro Codenotti

Ah cool

In class we only saw the usual subgroups of free groups are free as follow: By Bass-Serre theory a group acting freely on a tree is free, let $H\subseteq G$ with $G$ free, then $H$ acts freely on the Cayley graph of $G$ (wrt a free generating set) so it must be free

Anyway what about functional analysis?

Balarka Sen

Let's see if I can reconstruct the proof of Kurosh subgroup theorem. $K(G, 1) = K(A, 1) * K(B, 1)$, so $\widetilde{K(G, 1)}$ looks like $Cayley(F_2)$ with vertex spaces being $K(A, 1)$ and $K(B, 1)$ alternatively and edges being one-point unions. This is what is known as a "graph of groups"

$G$ acts freely on this by deck transformations, and the quotient is $K(G, 1)$. The cover corresponding to $H \leq G$ is $\widetilde{K(G, 1)}/H$. The free action on the total space of this graph of groups restricts to an $H$-action on $Cayley(F_2)$

The graph of groups corresponding to this is $Cayley(F_2)/H$ with vertex spaces being stabilizers of the $G$-action on $Cayley(F_2)$. But those are conjugate to $A$ or $B$. The edge spaces are one-point union again.

$\widetilde{K(G, 1)}/H$ is $K(H, 1)$, so fundamental group of this recovers $H$. Collect the loops, that gives a contribution of a free group. Collect the "$A$ vertices", that gives a contribution of a free product of conjugates of $A \leq G$. Collect the "$B$ vertices", that gives a contribution of a free product of conjugates of $B \leq G$.

That's all

topologicalmagician

20:05

Hello, I have a question.

Are if and only if statements an alternative ways to proving the uniqueness of a solution?

Mike Miller

@BalarkaSen ?

topologicalmagician

For instance, lets say x satisfies P(x) and if P(x) then x does that mean I have shown that x is unique?

Balarka Sen

@BalarkaSen Er, I meant subgroups of conjugates of A, B, etc. The vertex spaces of $Cayley(F_2)/H$ are vertex spaces of $v \in Cayley(F_2)$ left invariant by some subgroup $K \leq \text{Stab}_v(G)$, quotiented by $K$.

Mike Miller

If P(x) then x doesn't make sense. x is something you plug into a proposition, not a proposition in and of itself.

Your question might make more sense if you had a concrete example in mind you're comparing to.

Balarka Sen

But $\text{Stab}_v(G)$ is always a conjugate of $A$ or $B$ in $G$, so $K$ has to be conjugates of subgroups of $A$ or $B$ in $G$

topologicalmagician

20:13

@MikeMiller Consider the following: $x=g^{-1}h$ $\iff$ gx=h$

Alessandro Codenotti

(I feel kinda bad interrupting one of the few math discussions in this chat, but did you see my album recommendation on DC a few days ago @Balarka?)

Mike Miller

I mean, that tells you that there is a unique $x$ that satisfies $gx = h$, but the reason it tells you that is because the LHS says what $x$ is.

Balarka Sen

@MikeMiller @AlessandroCodenotti Sorry, I'll ask func. anal. in a bit. Just trying to clear out my Kurosh picture

@Alessandro Oh I didn't

Mike Miller

There's only one element named $g^{-1}h$.

Balarka Sen

What's the album?

Mike Miller

20:15

@BalarkaSen Ping me when you do, I am going to step out

Alessandro Codenotti

From the Gallery of Sleep by Night Verses, an interesting instrumental mix of postrock, prog metal and post hardcore

Balarka Sen

For sure! Thanks for the interest!

topologicalmagician

aha, so the left side shows the existence of the solution for gx=h.

Balarka Sen

@Alessandro Oh shit I need a new album to listen to. I'll check this out

topologicalmagician

(Going left to right)

Alessandro Codenotti

20:16

It's a bit heavier than my usual recommendations but you might like it

topologicalmagician

and then going right to left shows that its unique?

Balarka Sen

I have been listening to the new Aesop album on repeat

It's so good

topologicalmagician

Is my reasoning correct, @MikeMiller?

Alessandro Codenotti

I only heard a single or two, I still have to listen to the whole album properly

Balarka Sen

But I need to get out of the rut of listening to same albums over and over again

Mike Miller

20:18

No, @topologicalmagician. The left side tells you everything you want to know. (I disagree that "iff" has much to do with existence and uniqueness.)

Here is the setup. You want to solve the equation $gx = h$.

Using your iff, you see that every solution satisfies $x = g^{-1}h$, and in fact that this solves the problem.

Oh, no, I guess you're right.

Left to right says this element solves the problem. Right to left shows that this element is the only one that solves the problem.

However, this is a very special type of iff, so I stand by my first sentence!

Thorgott

Let $F$ be a field. The minimal polynomial of a Jordan Block $J_d(\lambda)$ of length $d$ to an eigenvalue $\lambda\in F$ is $(X-\lambda)^d$. My professor urged to see this by considering $F^d\cong F[X]/((X-\lambda)^d)$. Multiplication with $X$ induces a linear map on the right-hand space whose minimal polynomial is $(X-\lambda)^d$, but I don't see the correspondence to left multiplication with $J_d(\lambda)$ on $F^d$. Can anyone set me on the right path?

Ted Shifrin

Heya a @Balarka

@Thorgott: You need the right basis for $F^d$.

topologicalmagician

Thanks @MikeMiller, am I correct to say that an alternative way to prove uniqueness is to show that $\exists x(P(x) and \forall y (P(y) \implies y=x)?

Ted Shifrin

@topologicalmagician: If you remember being taught to check for extraneous roots in high school, this is part of the same thing. For example, when you solve $x=\sqrt{x+2}$ you show that if $x=\sqrt{x+2}$, then $x^2=x+2$, so $x=2$ or $x=-1$. However, checking for extraneous roots is asking whether the converse holds. So, yes, solving an equation is giving an iff ...

Mike Miller

I don't really understand why you want to be so formal. I can't imagine a situation where you'd prove some statement like you envision instead of just showing that something exists, and is unique.

Ted Shifrin

20:32

But the uniqueness check is, typically, showing that $P(y) \implies y=x$. :P

topologicalmagician

@TedShifrin thank you, that makes sense. But shouldn't that be sufficient to show uniqueness?

Ted Shifrin

As Mike says, you're being very symbolic about it, so it's better to understand what you're doing. Yes, once you have a solution $x$, to establish that it's the unique solution, of course you typically show that if you have any ("other") solution $y$, then you must have $y=x$. On the other hand, you might have an algebraic argument where every step is an "iff" and then there's no need to make a separate argument.

2

topologicalmagician

@TedShifrin @MikeMiller Thank you, both of you. That was really helpful.

I have another question, if you don't mind me asking.

If I want to learn differential geometry, should I learn measure theory before learning differential geometry?

Ted Shifrin

What do you mean by differential geometry?

Before you do fancy abstract stuff, I would recommend learning (1) differential geometry in $\Bbb R^3$ (curves and surfaces), (2) differentiable manifolds in $\Bbb R^n$ (Guillemin & Pollack).

For the former (e.g., my text that is freely available), you need solid multivariable calculus and some linear algebra. For the latter, you need multivariable analysis (understanding the derivative as a linear map and the inverse function theorem).

topologicalmagician

What book would you recommend for learning multivariable analysis?

Balarka Sen

20:45

Ted Shifrin, "Multivariable Mathematics"

Daminark

~~Princeton Review's Cracking the Mathematics GRE~~

Balarka Sen

Hi @Ted

Ted Shifrin

Depends on your background. I'm fond of my own textbook (for which there are lectures on YouTube). If you've had a serious real analysis course already, you could look at Munkres's Analysis on Manifolds ... Thanks to a @Balarka for plugging me. :P

hi Demonark

Daminark

How's it going Ted?

Ted Shifrin

(who doesn't know multivariable calculus)

so, a @Balarka, what topology/stratification stuff are you doing these days?

topologicalmagician

20:46

Thank you. Do you have any advice for someone who wants to learn diff geo?

Ted Shifrin

As I asked before, what do you think diff geo is ... and what is your background? And why do you think you want to learn it?

I mean, I spent a whole life working on this stuff ....

topologicalmagician

Reimannian Geometry and Symplectic Geometry

Ted Shifrin

That is serious graduate-level stuff. From the questions you've been asking, I think that might be a few years away for you.

You need multivariable analysis, point-set topology, and a lot of sophistication.

I still recommend doing curves and surfaces first to develop computational skill and intuition.

Mike Miller

And by the time you get there, your interests might (likely will) change, like mine did!

I agree with everything else Ted said.

Ted Shifrin

LOL ...

What a rarity :P

topologicalmagician

20:51

@MikeMiller I'm worried about that. haha

Ted Shifrin

Usually, @Erico is the only one agrees with Ted.

Daminark

It's nothing to be worried about, you like what you like

Mary Star

Hello!! Does someone of you have an idea about my question about cubic splines? math.stackexchange.com/questions/3113106/…

topologicalmagician

I

Ted Shifrin

Totally agree with Demonark.

Balarka Sen

20:51

As long as you don't like higher topos theory you're good

Ted Shifrin

<--- knows nothing about splines

(well, maybe the idea)

Alessandro Codenotti

At some point you'll find out you really like set theory, just accept the truth now instead of wasting time on geometry

Daminark

When you're deeper into grad school and have an adviser then you should hope you remain interested in whatever you chose because it becomes hard to switch

Érico Melo Silva

ive been summoned to agree with whatever it was Ted said

Ted Shifrin

kicks italic @Alessandro out of the room

Daminark

20:52

He said "Totally agree with Demonark"

Érico Melo Silva

uh oh

Ted Shifrin

That was a once-in-a-lifetime thing, Demonark.

Alessandro Codenotti

I guess that was deserved @Ted :P

Ted Shifrin

I don't think owners can kick out other owners. Maybe I should try.

topologicalmagician

I'm pretty interested in measure theory, is there any connection between measure theory and diff geo?

Alessandro Codenotti

20:53

Or maybe I should try

Ted Shifrin

It still shows up as an option, @Alessandro. Who's gonna try?

topologicalmagician

@AlessandroCodenotti who doesn't love set theory?

Balarka Sen

something something geometric measure theory

Érico Melo Silva

geometric measure theory has a big part of it devoted to DG and MT

Alessandro Codenotti

There's plenty of connections between measure theory and set theory, just saying

Daminark

20:54

I think some types of geometric measure theory

Ted Shifrin

@topologicalmagician: If you get into very analytic parts of geometry, like geometric measure theory or geometric partial differential equations.

Daminark

Lmao, 3 points for GMT and 1 point for set theory

Mike Miller

@topologicalmagician Just learn what you like now. Don't worry about connections. You'll find them if they're there. :)

Alessandro Codenotti

@topologicalmagician We should throw a coin

Mike Miller

This is too far out to think about very hard.

Daminark

20:54

@MikeMiller a budding differential geometer must think about connections!

Ted Shifrin

@Alessandro: You gonna try?

Érico Melo Silva

hah

Alessandro Codenotti

Nah, I was just joking, but I think it should work

Mike Miller

@Daminark yeah I was deciding whether or not to make such a joke

Ted Shifrin

smacks Demonark for reverting to his old bad humor

Mike Miller

20:55

And decided that only real nerds would say that

Daminark

I got got D:

Érico Melo Silva

u got him good bud

Ted Shifrin

OK, Alessandro, I'll do it to you. :P

Oh, it said it would notify mods. I certainly don't want that.

Alessandro Codenotti

If you want to try :P first time it's only for a minute right?

topologicalmagician

NVM i'm gonna do number theory

Daminark

20:56

Oh now we're talking

Ted Shifrin

It's OK if they notify room owners, but I don't want outside mods summoned.

Alessandro Codenotti

Oh, nevermind then, I thought only the third kick would notify mods

Érico Melo Silva

oh no he's gone over to the dark side

Ted Shifrin

I dunno the rules, @Alessandro.

Daminark

>:D

Mike Miller

20:56

He might care, but I don't.

Daminark

I got notified actually

Balarka Sen

OK, finally free from other bull

Mike Miller

I think only room owners should be notified.

Ted Shifrin

Oh ... Mike did it.

Yeah, I got notified.

topologicalmagician

@TedShifrin, I found your youtube series. I feel like binge watching them.

Ted Shifrin

20:57

Some is very elementary, but some is quite advanced, @topologicalmagician.

Alessandro Codenotti

I'm back!

Missed me?

Ted Shifrin

Oh darn, @Alessandro is back.

Balarka Sen

@Alessandro So, you read my proof of Kurosh subgroup theorem?

Alessandro Codenotti

I have an exam in two days so I didn't, I'm busy studying

Balarka Sen

Ah ok

Ted Shifrin

20:58

Evidently very busy studying.

Mike Miller

I sure as hell didn't read it

Anonymous

@MikeMiller And mods who are active in the room at the time of kicking.

Alessandro Codenotti

I will on Saturday though, it looks interesting and now that we went through Bass-Serre theory I might as well see the applications

Ted Shifrin

Cover your eyes, @Blue.

Mike Miller

@Blue So you should leave! :D

Alessandro Codenotti

20:58

@TedShifrin Maybe I should ask to be kicked more often

Ted Shifrin

You hardly need to ask, @Alessandro.

Mike Miller

Damn, I can't kick him. He's too powerful.

topologicalmagician

@TedShifrin do you have advice on how to improve ones mathematical sophistication?

Mike Miller

@topologicalmagician You are thinking too hard!!

Anonymous

@TedShifrin It's okay. I can handle it. :P

Mike Miller

20:59

Stop thinking about very long-term goals. Just keep doing math.

Ted Shifrin

Lots of practice, @topologicalmagician.

Mike Miller

It's a matter of time and practice.

probably bad advice to tell someone to not think while doing math

Transcript for

$Mathematics$ Mathematics