Mathematics - 2021-11-26

Dan Donnelly

00:07

Happy TM day

TM = turkey m__d_r

not turing machine :'(

except they probably use turing machines to systematically murder all the turkeys

I mean turing complete languages

copper.hat

00:42

that would be turkey complete

anakhro

rip turkeys

Dan Donnelly

04:12

We are taking care of 12 cats all together now

grandma's cats

Koro

Hi DanDonnelly!

I have one confusion in the definition of adjoint: Suppose $V$ and $W$ are finite dimensional inner product spaces over field $F$. Let $T\in L(V,W)$ then $T^\ast: W\to V$ is defined as follows: For all $v$ in $V$ and for all $w$ in $W$, $\langle Tv, w\rangle=\langle v, T^\ast w\rangle $

leslie townes

nitpicking a little bit, i'd say some work needs to be done before you know that defines anything, but so far so good

Koro

My confusion is: I believe that the definition should be actually $\langle Tv, w\rangle_W=\langle v, T^\ast w\rangle_V $ to indicate that the inner product may be different in $V$ and $W$.

leslie townes

yes. but it's fairly nonstandard to decorate the inner product with notation indicating the space, when it is otherwise unambiguous

for example the + of V and the + of W are different operations (in general) but are written the same. it's an extension of that.

if you had a number of spaces all going at once or were worried that it wasn't clear, you'd decorate just as you suggest.

Koro

Ah, so my understanding is correct. Thanks a lot. :)

But then the problem is: adjoint should be defined with respect to inner products -I mean that if inner product in V changes then so will $T^\ast$. There is no reason to believe (so far with me) that $T^\ast$ will remain the same when the inner products change.

leslie townes

04:29

yes. but (outside of textbook problems, maybe?) it's fairly rare to consider all of these inputs as variable at the same time

this is reflected in the common notation $\langle \cdot, \cdot \rangle$ not explicitly denoting any particular choice of inner product. if you had more than one, you'd decorate this some how and you would need to decorate the * in T* too

Koro

I was thinking about matrix of a linear transformation $T\in L(V,W)$. We make it very clear while writing a matrix of $T$ that the matrix is w.r.t. bases B and C in V and W respectively. But it seems it's not the practice with adjoint.

leslie townes

they do in some textbooks. outside of that setting, if there is some orthonormal basis around, people generally do not make it very clear

again, just as context generally allows one to get away without writing $+_V$ and $+_W$, many people will talk about "the matrix of $T$" without naming the bases, let alone working it into a notation for the matrix.

especially, god help us, physicists.

they invented some of it so we have to indulge them.

or, we do whether we have to or not.

Koro

My confusions about the adjoint definition are clear now. You beat my confusion to death. Thanks a lot , Leslie. :)

Leaky Nun

@Koro rip confusion 2021-2021

leslie townes

it's very good to keep in mind as a matter of the theory of it all, but it melts away in practice. e.g. in many applications in hilbert space, one tends to work with orthonormal bases only, or if one has other bases floating around, one would not write matrices with respect to non-orthonormal bases.

nobody's annoyed by the fact that juxtaposition is used both for scalar multiplication (which can vary depending on the spaces) and composition of linear maps (same), but let's remind ourselves of that too. might need it on the first day of teaching it.

Koro

04:40

@LeakyNun hahaha

leslie townes

if you abstract beyond operators on hilbert space you will need to think about this again. e.g. if your set of 'operators' is just an algebra with an involution * on it, it might not be the one you expect it to be. same with norms, topologies, you name it. although there are various theorems to the effect that when you specify enough of the structure, the rest fills in uniquely.

dc3rd

06:42

I missed the discussion on adjoints...sigh....and here I am working on self-adjoint and normal linear operators....fun but abstract as f**k...

copper.hat

07:54

think of adjoints as transposes, as in $x^T Ay = (A^Tx)^T y$

Adil Mohammed

09:32

@robjohn no i dont think there is any (x,y) where $f(x)\gt f(y)$ isnt satisfied when $x\gt y$...

but my textbook tends to consider the function as neither increasing nor decreasing when f'(x)=0 even for functions like $x^3$ which IMO is always increasing throughout the domain

maths researcher

12:09

does $\mathbb{Q}(\sqrt[3]{2})$ have a subfield of dimension 2 over $\mathbb{Q}$?

Thorgott

12:36

do you know the tower law?

maths researcher

12:56

yeah

Thorgott

do you know the dimension of $\mathbb{Q}(\sqrt[3]{2})$ over $\mathbb{Q}$

maths researcher

3 right?

Thorgott

yes, now think

maths researcher

13:11

alright, what does the notation $\mathbb{Q}(X^3)$ mean

?

Thorgott

13:31

the rational function field in the variable $X^3$, what's that got to do with anything

maths researcher

14:14

it's another question.

A hint sufficed

robjohn

14:45

@AdilMohammed "increasing" is really about $f$ at two separate points. One can look at "instantaneous increase" which means $f'(x)\gt0$. instantaneous increase at every point of an interval implies increasing, but not vice-versa.

Daniel Adams

15:05

Hi guys I have a pretty simple probability question/reference request if anyone could assist me. Consider the diffusion process $$dX(t)=dW(t), X(0)\sim \rho $$ where $W(t)$ is a 1-dimensional brownian motion. Let $\mu_t$ be its associated distribution at time $t$. Does anyone know if the LDP of $\mu_\epsilon$ has been studied i.e the behaviour of $\mu_\epsilon$ as $\epsilon\to 0$

Koro

what happened to market today?

Kevin Driscoll

@Koro my guess is panic about new SA COVID variant

Koro

:(

Mad Spaces

15:27

Hey. According to the mean value theorem of integrationcalculus... can we say for a a riemann integrable function $ f $ that $\int_w f = \int_w f *1 $ since $g(x) = 1 \forall x $ is a monotonous function thus we conclude $ \int_w f = f(x_o) * vol (w)$ ?

bjorn93

16:02

Does anyone know the name of this identity involving the floor function? I'm fairly certain I saw it named on Wikipedia but can't find it right now.
https://math.stackexchange.com/questions/1894725/proof-about-floor-function-lfloor-x-rfloor-lfloor-x-frac1n-rfloor-c

Vrouvrou

hello, how to prove this $\forall a,b\in \mathbb{R}, a+b<a^2+b^2+2$

bjorn93

* Found it. Wikipedia calls it Hermite's identity. I'm not sure if that's a widespread name. en.wikipedia.org/wiki/Hermite%27s_identity

Leaky Nun

@Vrouvrou move everything to one side and see what circle it determines

Vrouvrou

$a^2+b^2+2-a-b$

soupless

16:21

@TedShifrin Hello! Can I ask about your comment in my recent answer?

monoidaltransform

16:49

every splitting field is always a normal extension?

123

Hi All...

What is the necessary and sufficient condition of rigid body under multiple forces can have pure translation and pure rotation (1. rotation of free body 2. rotation about fixed point within the body 3. rotation when fixed point outside the body)?

Mad Spaces

17:04

This is physics more than mathematics

Generally, movement of rigid bodies (degrees of freedom) is reduced to six degrees.
3 degrees of freedom for center of mass, three for the Angular momentum (a huge reduction from considering the degrees of freedom of each atom)

123

Hello @MadSpaces I am learning mathematical mechanics but still not clear the phenomenon.

Mad Spaces

Your question is not clear to me.

Wht do you mean "sufficient condition" ?

Where is your forces acting?

123

@MadSpaces In simple how can rigid undergo pure translation under the action of multiple forces , each force has different point of application on body.

Mad Spaces

If we do not want the body to rotate, the torque must be zero. (usually taken to the roation axis)

LearningCHelpMeV2

The expression that I got for $Q(x)$ was $x^2-(b+c)x+b^2-bc+c^2$ through long division. Completing the square on $2Q(x)$ gives $2(x-\frac{b+c}{2})^2+\frac{3(b-c)^2}{2}$. The question states that $2Q(x)$ can be rewritten as the sum of three expressions, each of which is a perfect square. But as you can see there is only two perfect squares in my expression. Not sure what they want me to do

Mad Spaces

17:08

If there is a sum force, none zero, and the torque is zero, the body will preform translation.

123

@MadSpaces In mechanics book they used the terms moment and couple does both terms refer torque?

Mad Spaces

Moment is not equal to Torque. I am not sure what couple mean?

is this some dirty engineer language? we speak human here sir..

123

@MadSpaces Couple mean two equal and opposite forces their line of action of forces do not intersect.

Mad Spaces

This will produce a torque.

Do you know the none relativistic equations for Torque and momentum?

123

@MadSpaces How moment , couple and torque are different?

Mad Spaces

17:11

Okay i just checked the wikipedia page to see what moment means

Because we didnot use this word in my studies.. i mean i do study in german, still this is not common for physicists.

Torque is defined (none relativistic) $ \vec{T}= \vec{F} \times \vec{r_{rotationaxis}} $ x is here the cross product

123

@MadSpaces Yes i learn't relativistic torque and mometum. But here moment and momentum are different . moment M = r x F ; where r is perpendicular distance through the line of action of force F.

Mad Spaces

Couple is basically a pair of forces attacking the body on two different points. consider for example a piece of paper you push one side with your left finger upwards and hte otherside with your right finger downwards

So basically to answer your question.
If the body is not fixed to a point, IE it is free to move, then you will have translation aslong as the vektoriell sum of the forces is none Zero.
The rotation occurs if the Torque (vector) is none zero.

You really need to take the vectors into consideration, not the scalars

However, i would suggest you writing in the physics chat room, this is for math only :)

123

@MadSpaces Does moment , couple and torque are same? Because they have definition but different condition and all responsible for rotation.

Mad Spaces

Write to me in private. or write in physics stack exchange, there are very good guys there that are wiling to help, lets not upset the fine gentlement and ladies of this room. Mathematicians look down upon us peasents of physics

123

@MadSpaces Thank you dear for your help. I am reading mechanics book which is mathematics , that's why i asked question here.

Mad Spaces

17:18

It might be called mathematics.. it has tho nothing to do with mathematics.

123

@MadSpaces How can i write you private?

Mad Spaces

Click my name and so forth, self explanitory

Now lets get back to the maffssssss

If a function F is differentiable, is $f(x)= F'(x)$continious ?

This whole continious and stuff like that i keep forgetting all the lemmas and statements made about them too many

Thorgott

no, this is famously untrue

the go-to example is $F(x)=x^2\sin(1/x)$ with $F(0)=0$

but the situation can be much worse iirc

Mad Spaces

How can i show that for $F'(x) = f $ then f has the darboux attribute?

Darboux attribute i guess some call it " middle value attribute "

Even if f is none continious, i thought only continious functions have the darboux attribute, hence why i asked

monoidaltransform

If $K\subseteq M \subseteq L$ and $M$ is normal over $K$ and $L$ is normal over $K$, is $L$ normal over $M$?

Mad Spaces

17:34

nvm i proved it :D

Kevin Driscoll

@MadSpaces words like 'moment' and 'couple' are indeed engineer language

Ted Shifrin

@MadSpaces Sign error?

Mad Spaces

What do you mean sign Error @TedShifrin

Ted Shifrin

Wrong order in the cross product .

Mad Spaces

You are right. I have not done newtonian mechanics in ages :)

I broke friendship with newton the moment i discovered Lagrange.

Ted Shifrin

17:39

Anyhow, the intermediate value property of the derivative is a very interesting phenomenon.

@monoidaltransform I think one of your hypotheses is not needed.

Thorgott

17:51

it's far from true that only continuous functions have the Darboux property

iirc, there's nowhere continuous functions enjoying the Darboux property

leslie townes

yes

Thorgott

Conway base 13 function

The Conway base 13 function is a function created by British mathematician John H. Conway as a counterexample to the converse of the intermediate value theorem. In other words, it is a function that satisfies a particular intermediate-value property—on any interval (a, b), the function f takes every value between f(a) and f(b)—but is not continuous. == Purpose == The Conway base 13 function was created as part of a "produce" activity: in this case, the challenge was to produce a simple-to-understand function which takes on every real value in every interval, that is, it is an everywhere surjective...

leslie townes

i've posted an example that is simpler in some sense than conway's. it's still weird that derivatives have them.

to me, anyway.

Thorgott

@TedShifrin I agree

leslie townes

f(x) = lim_n tan(n! pi x), if it exists, and whatever you want otherwise, has the property of being surjective on every interval.

Ted Shifrin

17:57

@leslietownes How does a function go from increasing to decreasing without having a critical point in between? Ah, it refuses to be differentiable.

leslie townes

toddler functions, we call them.

they simply refuse to behave.

Koro

@leslietownes I remember that and f(x):=0 if the limit doesn't exist. :)

Koro

18:23

I again have one confusion in definition of an adjoint of a linear transformation: Suppose that V and W are finite dimensional vector spaces. Let $T\in L(V,W)$ and $T^\ast$ be a map from W to V defined as $\langle Tv,w\rangle=\langle v, T^\ast\rangle$. Let me re-write it as: $\langle Tv,w\rangle_W=\langle v, T^\ast\rangle_V$. $T^\ast$ is known to be a linear transformation but I don't understand how this could be. Suppose that I want to verify its homogeneity that is for every scalar $c$,

leslie townes

koro, i may have mentioned this before. first verify that the recipe that purports to define T* actually does.

don't bother verifying properties until you know it defines a map. then use the uniqueness property (from that verification) to do everything else.

it would allow you, for example, to deduce from the fact that c T* w happens to be a vector u satisfying <Tv, cw> = <v, u> for all v, that T*(cw) = c T* w.

Koro

yes, we discussed this but then I misunderstood one step: if $\langle v,u\rangle=\langle v,z\rangle$ for all v then $u=z$ and I agree with that if and only if $\langle , \rangle$ on both sides is the same and this is so because I am allowed to write: $\langle v, u-z\rangle=0$ and since this holds for all v, in particular it does for v=u-z and norm of a vector is zero if and only if the vector is zero so u=z.

leslie townes

it's simpler to just establish, and use, the uniqueness. no need to walk back through the proof of uniqueness every time.

Koro

If the inner product is different on both sides, how can it be concluded that $\langle v,u\rangle =\langle v, w\rangle$ for all v implies u=w

For uniqueness, are you suggesting Riesz representation?

leslie townes

whatever tools you have access to.

note that if you did dip into a proof of uniqueness here, it would exploit the equality of <v, c T* w> and <v, T*(cw)> for all v. same inner product there.

but again, just writing T*(cw) already seems to presume that uniqueness is known, when that's most of the exercise. this is a bad habit that is more or less taught by linear algebra books at all levels.

"let [bleh] be given by [some formula that bleh satisfies, without explicit verification that a linear map exists that satisfies this formula]. now prove stuff about [bleh]." there's a step 0 to any exercise phrased like this, which is to verify that you have a well-defined linear map.

this becomes particularly acute in infinite dimensional settings. people will define a map via an algebraic recipe on an orthonormal basis without checking that the result is bounded (and hence extends to vectors which are not finite sums of basis elements, which is necessary for the recipe to define a map on the whole space). sometimes that is harder than doing whatever else someone might want to know about such a map.

Ted Shifrin

18:37

@Koro: You only use the inner product in one space when you're proving uniqueness. Don't get carried away.

Thorgott

@leslietownes just like defining a map by some procedure and proceeding to claim it's obviously continuous

Koro

I would like to believe that and I think the way linearity of $T^\ast$ has been proven text makes me want to believe that inner product is the same. But I'am confused. For uniqueness, I have Riesz representation theorem and I understand that: given a w in W, suppose $f$ takes $v$ in V to $\langle Tv,w\rangle$ where $T\in L(V,W)$ and I think that this linear product is w.r.t. inner product in W. Then $f$ is a linear function on V.

Ted Shifrin

Slow down, @Koro. You want to prove that $T^*w$ is uniquely defined by that equation (for all $v$). If $\langle v,\xi\rangle = \langle v,\eta\rangle$ for all $v$, then $\xi=\eta$. This is all happening in $V$.

Koro

By Riesz representation theorem, there exists a q in V such that $f(v)=\langle v, q\rangle$ for all v. Here, I think that inner product is with respect to inner product in V.

leslie townes

thorgott: there was a guy in my field who was notorious for doing this. he would suggest 'simplifications' of proofs via email. he even published some of his simplifications in weird journals. everybody knew not to cite them but they kept coming out.

Ted Shifrin

18:46

I don't need any stinky theorems.

Koro

Ted, I'll come to that. I'm just building up the background.

Ted Shifrin

Why is Riesz relevant? Who's talking about linear functionals?

Thorgott

heh

Koro

Now, that q is uniquely determined and we denote that by $T^\ast (w)$.

That's how definition of adjoint makes sense.

Ted Shifrin

Huh?

What is $f$?

Oh, $f = \langle T^*w,\cdot\rangle$.

Koro

18:49

given a w in W, f takes $v\in V$ to $\langle Tv,w\rangle$. It's better to denote f by $f_w$.

Ted Shifrin

This still won't get us linearity in $w$. I don't see the point.

Since I interrupted, I'll just leave the conversation.

leslie townes

i just googled that guy. he seemed to have stopped publishing his simplifications

Koro

Ted, the definition I am using for adjoint is: For every v and w in V and W resp. $\langle Tv,w\rangle=\langle v, T^\ast w\rangle$

Then Axler says: this definition makes sense because of what I said above in the comments.

@TedShifrin And using this definition: I have difficulty proving this statement that you wrote because the definition seems to suggest that inner product on both sides of the equality may be different. And the difficulty arises because had it been as you suggest then $\langle v,u\rangle=0$ for all u in V for a fixed v would imply v=0 (I know this.). I hope my question is clear now.

leslie townes

koro, when you want to show the properties of T*, you would be computing <v, -> for two different elements of V. e.g. putting T* (a+b) and T* a + T* b in for - and comparing the results. you might travel through the inner product of W and properties of the inner product to equate them, but that's what you end up doing.

i'm not a huge fan of the phrase 'overthinking a problem' but this might be a good example of that.

<v, T* (a+b)>_V = <Tv, a + b>_W = <Tv, a>_W + <Tv, b>_W = <v, T* a>_V + <v, T* b>_V = <v, T* a + T* b>_V. by definition of T*, linearity of < , >_W, definition of T*, and linearity of < , >_V. that holds for all v so T* (a + b) = T* a + T* b by uniqueness. can you prove that T* (c a) = c T* a along similar lines?

Koro

@leslietownes of course yes!! It's clear to me now. All this time, I was overthinking $\langle Tv,w\rangle_W=\langle v, T^\ast\rangle_V$ (why V on one side and W on the other) completely overlooking the fact that you wrote.

Thanks a lot :-)

leslie townes

19:03

that really does look horrible without chatjax. my apologies, everyone.

Koro

You again beat my confusion to death.

Ted Shifrin

smacks leslie with no remorse

Daniel Adams

Hi folks

Im reading onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160200404

By Varadhan

when he defines the weighted Riemannian distance $d$ is a natural geodesic, I'm confused what specifically makes it natural

leslie townes

19:20

i think i ruin it when i apologize. it's one step too far.

Ted Shifrin

Just stop misbehaving in the first place. Munchkin learns too much from you.

leslie townes

daniel, i'm probably not equipped to answer the question anyway, but i note that the paper is paywalled for many.

Ted Shifrin

Yup, paywalled.

leslie townes

it's on sci-hub, if people want to commit willful copyright infringement.

is what i would say had i checked that, which i didn't

Ted Shifrin

Way too much work.

Your sentence makes no sense, @Daniel.

Daniel Adams

19:24

leslie townes

he defines the length of curves in some space using an arc length formula (the metric is some other ingredient in the paper )and a distance between points as an infimum over lengths of paths between those points.

or that

Daniel Adams

leslie townes

he then says d "is the natural geodesic distance," by which i mean i think he just means d is an instance of a general thing you see in riemannian manifolds where length minimizing curves are geodesics, at least locally.

Daniel Adams

Sorry I didnt clip the full picture

leslie townes

daniel and i have got to stop echoing one another. i'll go first.

Daniel Adams

19:26

when he says "the natural geodesic distance"

with respect to what

I guess this is a silly question...

The paper is concerned with this heat equation ^

leslie townes

daniel if i'm not missing something, from his hypotheses on the a_ij (not in that image), your a_ijs define a riemannian metric on the space (i.e. a way of computing lengths of tangent vectors) and this is the usual recipe for turning that into a metric on the space (i.e. a way of computing distances between points).

i'm looking mostly at assumption (A) on the bottom of 659.

Daniel Adams

So I guess he means natural metric for studying this problem.

leslie townes

whatever he's saying, it seems to be expository in nature and not a statement that one would need to prove or be able to prove.

Ted Shifrin

Yeah, this is the standard definition of a metric space structure on a Riemannian manifold. You can read about it in any standard differential geometry book. As something accessible, I recommend Boothby.

What do you mean by "with respect to what?" You're asking about the word "natural"? It's because it is naturally associated to the Riemannian metric structure, which gives you lengths of paths.

robjohn

@DanielAdams it looks as if there are two $\dot\omega(\tau)$s inside the square root. Is that correct?

Daniel Adams

19:35

@robjohn yes

Ted Shifrin

@robjohn Sure. The metric is a (symmetric) bilinear form.

It's horrid notation, if you ask me, but ...

robjohn

It is interesting that they don't bring it out as $|\dot\omega(\tau)|$

Ted Shifrin

They're being explicit about the dependence on $a$.

Just like we would write $\sqrt{\sum g_{ij}\dot x^i\dot x^j}$

robjohn

Okay; weird, but okay.

Daniel Adams

@TedShifrin @leslietownes right thanks for the reference, I guess I was just confused since he starts with the heat equation pictured above, on flat $\mathbb{R}^k$, and now your talking about manifolds, and a metric induced from a manifold

leslie townes

19:37

i note that it's quite possible for me to introduce some unchatjaxxed < and > here, but i'm resisting the impulse.

Ted Shifrin

He's treating $\Bbb R^k$ as a Riemannian manifold, i.e., using non-standard metric structure.

gets ready to smack leslie

robjohn

Is $\omega$ a vector?

Ted Shifrin

Yeah, vector valued function.

Daniel Adams

yes

robjohn

Ah, okay, so normally, one would have a transpose on one of them. I get it.

sorry for derailing the thread.

Daniel Adams

19:40

@TedShifrin sure hes using a non-standard metric structure, based on the equation he is solving, I just wondered why this specific one is the right choice.

Ted Shifrin

It all depends on the definition of $a$, of course. You can solve the heat equation on any Riemannian manifold. The Laplacian depends on the Riemannian metric.

Daniel Adams

any ideas how to mimic this if $a$ was not invertible (of course the equation would be a bit silly in that case ) ?

Ted Shifrin

I have no idea why they're writing $a^{-1}$. As I indicated above, we ordinarily write the metric as $g$. One can try to do Riemannian geometry without positive definiteness (e.g., relativity). I don't know what happens with degeneracy (non-invertibility), though.

Daniel Adams

sure that wasnt really a fair question

Thanks btw guys!

Kevin Driscoll

20:09

When I studied DeRham cohomology, the coefficients came from the Reals. I got the impression from some comments from the instructor that having a (co)homology theory with real coefficients is less desirable than one where the coefficients are only integers. I know a bit more now about singular and simplicial homology theories where its pretty clear that the "obvious" coefficient are in Z. Is there a short, consensus reason why one would prefer theories with Z coefficients to those with R?

Ted Shifrin

Depends on the situation. Torsion carries a lot of topological information that will be invisible using DeRham, for example.

It's more information to know $H^n(X,\Bbb Z)\cong \Bbb Z/2$ than to know $H^n(X,\Bbb R) = 0$.

Thorgott

20:32

and the information is stricly less

ah wait, I'm not sure if that claim is entirely true

Ted Shifrin

"more" ≠ "equal"

Thorgott

I was gonna say cohomology with $\mathbb{Z}$ coefficients determines that with $\mathbb{R}$ coefficients

but I think I need a finiteness hypothesis somewhere for that to be true

but I'm too lazy to activate my brain

Ted Shifrin

Well, universal coefficient theorem will involve other cohomology, of course.

Thorgott

I would need to reconstruct the right universal coefficient theorem for this

Daniel Adams

Maybe another question I could ask is ( relating to the metric $d$ pictured above ) when are two points close w.r.t to this metric

Thorgott

20:38

this is the version of the UCT people never talk about

Ted Shifrin

@Daniel I know nothing about the $a$ appearing in your formula, but in Riemannian geometry one has what's called a normal neighborhood of any point given by following geodesics emanating from the point. For a sufficiently small radius, there's a uniqueness result that any two points in there are joined by a unique (shortest) geodesic.

Daniel Adams

I thought there may be some relation to two points $x,y$ being close in this metric, and the solution of the above heat equation at $x$ given it started (dirac mass at $y$), but I guess that doesnt make sense since the metric has no relation to time.

Ted Shifrin

The confusion is how the heat equation is giving rise to motion of the manifold.

I guess in $\Bbb R^k$ we're supposed to have $p$ as the position vector of a point? A LOT of work has been done on the heat equation on manifolds and using it to prove various sophisticated topological theorems. See the work of Peter Gilkey, for example.

I do not remember what I once knew about this stuff many years ago.

Dan Donnelly

20:54

Lost in time, like tears in rain

Ted Shifrin

22:09

Boy, isn't markvs a pip!! He must be just a wonderful professor.

Dan Donnelly

Yes, you got this

^_^

leslie townes

wow, yeah

Thorgott

lol what a loser

and of course resident asshole david stork joins the party as well

leslie townes

oh, i've seen that guy's work before.

not david, the other guy. i ran across his webpage once a long time ago and found something of the same caliber of unnecessary offensiveness on it.

lucky, lucky students.

Ted Shifrin

Yeah, David is a known a**hole, but the other guy wasn't known to me. And, after my comment, of course he had to escalate. I wish I could look him up on RateMyProfessor.

I suppose some people might find some of my strident comments on this site equally repugnant.

leslie townes

22:21

mm, he doesn't have a profile there. i'm not sure people use that site anymore.

Ted Shifrin

I haven't kept up in my dotage.

leslie townes

he has a recording of himself offering sarcastic encouragement to students, as a replacement for them not hearing his encouragement in real life.

Ted Shifrin

Wait ... you know this person's actual identity?

leslie townes

the false encouragement page would be funny if he had a different job.

Ted Shifrin

Figures — a Russian and an algebraist.

leslie townes

22:24

i was going to say, very much fits a stereotype.

Ted Shifrin

He is on RateMyProfessor — 5 responses. 0% would take again. Big surprise.

Centennial Professor. Well, that accounts for his miniscule ego.

leslie townes

also senior enough to not have to care about any of this, yeah.

i'm sure alumni pour zillions of donations into the coffers of his university because of his efforts.

Thorgott

I had a great laugh listening to that

Ted Shifrin

I am fully aware that we have such awesome personalities among the ranks of our professors. But why the h*** is he on this site?

leslie townes

boredom?

these little skirmishes are part of a long campaign in the fight against dotage?

Ted Shifrin

22:28

On that note, I'm going for my walk. BBIAB.

leslie townes

if one looks at it as pure trolling, it's kind of funny. but it's what you'd expect from a teenager.

enjoy your walk.

robjohn

@TedShifrin I was about to delete markvs' comment as it has several flags against it. I hate to do so as it makes your reply moot.

Thorgott

yes, a teenager with a fragile ego

Ted Shifrin

23:01

@robjohn I think my comments were measured and very apt.

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$Mathematics$ Mathematics