Mathematics - 2016-04-23 (page 1 of 3)

Semiclassical

00:57

hi chat

well y naught

hi

Jessy Cat

01:24

Insane in the membrane

Insane in the brain

Insane in the membrane

Insane! Got no brain!

Semiclassical

02:24

grad school getting to you, i see? :) @JessyCat

tylerl-uxai

02:57

Hey, can I get help with this? My euler's number formula isn't computing 1 factorial or 0 factorial

I don't think 0! = 1... Seriously. It just doesn't seem to make sense unless you start with 1 and divide by 1

anon

I don't think bachelor means unmarried, either.

no idea what your "euler's number formula" is so how do you expect us to diagnose the problem with it?

oh, e=1/0!+1/1!+1/2!+... I'm guessing.

but that formula doesn't compute factorials, so it doesn't make sense to say your formula isn't computing 1! or 0!

Adeek

03:19

Hi

I did really good in my presentation. Thanks all for your help @MikeMiller @TedShifrin and @BalarkaSen

anon

what was your presentation on?

Mike Miller

04:13

@anon borsuk-ulam

Adeek

sorry @anon back

it is on Borsuk-Ulam theorem in any dimension

I used homology to prove it

specifically, I proved the following result any map $f : S^n \rightarrow S^n$ which is an odd map must have odd degree.

The way you prove it is you consider the two-sheeted covering space $p : S^n \rightarrow RP^n$ and just construct a short exact sequence in intelligent way. Then you get long exact sequence on the homology, then use f to get a induced map $g : P^n \rightarrow P^n$.

Then you will get a morphism between short exact sequence and itself.

So, by naturality you will get a long exact sequence and itself and you can use that to prove that $f$ must have odd degree

@anon

Semiclassical

06:07

i'm sort've confused by the objection in the comments of this question. Is there really a field $\mathbb{K}$ containing $A$ such that the definition $e^A=\sum_{k=0}^\infty A^k/k!$ doesn't make sense? (Not sure that's enough to prove the relevant theorem, but that's a different matter).

Mike Miller

Take any finite field

Can't divide by n! if 2=0, eg

Semiclassical

hrm. i guess i'd only thought about $A^k$ being in the field, not about $k!$.

Mike Miller

$A^k$ is fine, yeah

Semiclassical

yeah.

but there's no reason to assume that real coefficients make sense.

Mike Miller

so as long as you're of characteristic zero (so that n is never zero for an integer n), you're fine for then in particular k! is invertible

Semiclassical

06:15

right.

Mike Miller

and obviously vice versa.

Semiclassical

not going to pretend i have any idea whether characteristic zero is enough for $\det{e^A}=e^{\text{tr }A}$ to hold.

Mike Miller

i think there's an even worse trouble: what is an infinite sum? you really need to restrict to R and C

Semiclassical

well, in the finite field case, that issue is sort've shortcircuited by the other. not much point worrying in the infinite sum if you can't define more than the first $p$ terms.

tylerl-uxai

06:42

Please tell me this is worth -8 downvotes

I believe that almost everyone is confused about what you were trying to ask/do here. Why do you think $0!\ne 1$? I can also claimed that $1+1=3$ if I write an erratic program for computing the sum of two numbers but that doesn't make the formula valid. You cannot draw general facts from an incomplete computer program. By the way, what's wrong with "e" so that you want to replace it with something else so much? — BigbearZzz 10 mins ago

oops

-8

Q: 0 * anything does not equal 1 (This is completely re-written several times for more and more downvotes)

This deserves a few million downvotes for all the edits I made to it. Right now it's -6. Let's see how many more it gets. Second edit: I guess re-writing math was arrogant of me (read @fleablood's comment). Thanks for looking at this as an attempt to be arrogant rather than seek help with someth...

factorial

I had the same question in the middle of class, I went out of my way to write code for it to prove that I was right... I proved I was right (sorry for sounding arrogant)

And people just keep hitting minus on the question since it started with -2 or -3

I thought if you edit your questions, people would respond by not banning you

Sorry, but I hope you know that my account is locked from stack exchange for editing my question to improve the quality

It just deleted the link I sent

So I won't be able to re-do this since my name is my brand

Valery Saharov

08:55

So does this happen to be an open problem?

4

Q: Approximate spectral decomposition

See attempt below I am interested in effective computations in finding approximate spectral decompositions in some suitable format. Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can ...

linear-algebra hilbert-spaces approximation computability

I start thinking of a publication if even SE can't say anything on this matter

ADG

09:30

$$\large\sum_{\displaystyle\sum_{i=1}^mr_i=n}\binom{2n}{r_1,r_2,\ldots,r_m,r_1,r‌_2,\ldots,r_m}=??$$

Tobias Kildetoft

10:11

My son counts "...13, 14, 15, 0". Does this mean he only has 4 bits memory, and if so how can I upgrade him? :)

Ilan Aizelman WS

10:31

Anyone knows how to prove math.stackexchange.com/questions/1755246/…

Highly appreciated :)

Valentin Tihomirov

10:59

What is faster to write down a math expression, a computer SW or pencil and paper?

Tobias Kildetoft

what does SW mean?

Ilan Aizelman WS

Anyone knows how to find a minimum of function using linear algebra? math.stackexchange.com/questions/1755292/…

Owatch

11:40

Hey.

Why does using Wolfram's QR decomposition function just give me back my input?

I asked it to evaluate: QR decomposition {{0, 0, 6}, {1, 0, 7}, {0, 1, 0}}

This is correct I guess, since I did it by hand..

But shouldn't the QR factorization of a matrix yield it's eigenvalues along the diagonal?

$x^{3} + 0 \cdot x^{2} - 7 \cdot x - 6 = 0$ has companion matrix:

[ 0 0 6 ]
[ 1 0 7 ]
[ 0 1 0 ]

Owatch

12:05

I guess dot product isn't commutative after all...

Valentin Tihomirov

12:17

Is there computer software that allows you to jot down formulas faster than you do it with pencil and paper?

Valery Saharov

mathematica

Tobias Kildetoft

@ValerySaharov depends on how many long complicated shorthands you use

Valentin Tihomirov

Do you mean that long complicated shorthands is the key? How come long typing comes out faster?

Owatch

It seems it depends how you write in the companion matrix too.

There's a form with the coefficients on row 1, and another with them along the last column

Tobias Kildetoft

@ValentinTihomirov Because you can make short commands in LaTeX for long complicated things

user147690

12:27

It is still running btw @TobiasKildetoft

Ilan Aizelman WS

@robjohn @TedShifrin mind taking a look? math.stackexchange.com/questions/1755292/…

Ilan Aizelman WS

13:19

@robjohn Do you have any idea why $A^TAx = A^Tb$ won't yield the same solution $x_1 = 2, x_2 = -5$?

robjohn

@IlanAizelmanWS why should $Ax=b$?

Ilan Aizelman WS

@robjohn $Ax=b$ won't have a solution

because: $||Ax-b||^2$ is minimal if it equals to zero?

@robjohn quote: Taking for granted that the minimum value of $f$ is zero, then writing $f$ in the suggested form implies that $ \|Ax-b\|^2 = 0\iff Ax - b = 0 \iff Ax = b$. So, it sufficient to solve the system $Ax = b$.

@robjohn yup, it works

@robjohn If $P$ is the orthogonal matrix of vector $x \in R^N$ on subspace $U$, and $Q$ is the otho. matrix of the same vector on $U^{\bot}$, can I say, without proving that $Ix = Px + Qx$ , thus $I = P + Q$?

robjohn

@IlanAizelmanWS so you just had the wrong vector in the question.

Ilan Aizelman WS

@robjohn Yup.

@robjohn any ideas about the second question?

robjohn

13:47

@IlanAizelmanWS Do you mean that $P$ and $Q$ are orthogonal projections onto $U$ and $U^\perp$?

Ilan Aizelman WS

@robjohn yup

robjohn

For any $x$, you can obviously write $x=x_1+x_2$ where $x_1\in U$ and $x_2\in U^\perp$.

Saal Hardali

What's the simplest example of a non-locally symmetric pseudo riemanian manifold?

robjohn

Then note that $(P+Q)x =(P+Q)(x_1+x_2) =Px_1+Qx_1+Px_2+Qx_2 =x_1+0+0+x_2 =x$

Saal Hardali

I'm having a hard time imagining a geodesic symmetry which isn't an affine transformation.

Ilan Aizelman WS

13:59

@robjohn I see. thank you!

Mike Miller

14:31

@SaalHardali I guess you want the metric to be indefinite?

Ilan Aizelman WS

@robjohn Rob, another theoretical question, if $R$ is the reflection matrix on subspace $U$ and $S$ is the reflection matrix on subspace $U^\perp$, then $R+S = 0$ right? and RS = SR = -I right? (RS = SR = -I is kinda trivial because its eigenvectors are 1,-1, but I proved it with $P_u + P_u^\prep$ = I and $P_u * P_u^\prep = 0$, and used $R = 2P_u - I$ and same for $S$..

By saying reflection matrix, I mean $R^2 = I, R^T = R$.

Saal Hardali

@MikeMiller I just posted a question about this: http://math.stackexchange.com/questions/1755463/affine-manifolds-which-are-not-locally-symmetric

What if I want the metric to be definite and not locally symmetric? Are example of this kind so hideuos that my difficulty of finding them in the literature is for my own sake?

Mike Miller

@SaalHardali No, I asked because I don't have a good mental picture of indefinite metrics, so would not be able to help you. In the definite case the vast majority of metrics are not locally symmetric. Let's work with simply connected and complete things. Then (thm) locally symmetric spaces are globally symmetric. So the isometry group of your Riemannian manifold must be quite big, indeed it must act transitively on your manifold.

So just pick a complete, simply connected manifold it cannot act transitively on. I would probably do this by picking some metric on the plane that has positive curvature somewhere and negative curvature somwhere else.

To see why it shouldn't be locally symmetric more visually, there will be some point that has zero curvature, but that some geodesic through this point, on its "left side" will go through points of positive curvature, and on its "right" will go through points of negative curvature.

And clearly there's no way to have an isometry swap these sides.

It's probably also true in the indefinite case that most manifolds are not locally symmetric, but I have spent more or less no time ever thinking about it :)

Balarka Sen

14:48

OK, I am back home and I should start on reading forms.

Mike Miller

That's true.

@iwriteonbananas Hi.

iwriteonbananas

hey @MikeMiller

how goes it?

and hey @BalarkaSen. where have you been?

Mike Miller

it's ok, need to start working soon today

last day until Tues I have time to get any work done

Balarka Sen

I am a bit confused about whether I should send the solutions of the exercises from the previous chapters I have done to Ted before starting on forms. Maybe I should start write the solutions in my free time and start on forms before because I have been delaying it for too long.

Mike Miller

I mean, it's fine, @BalarkaSen, I'll just be dead before you finish calculus.

iwriteonbananas

14:51

what happens after tuesday, @MikeMiller?

Balarka Sen

@iwriteonbananas To a university.

Mike Miller

@iwriteonbananas no, I just mean I'm busy with other stuff tomorrow and monday

so it would be a shame if i got nothing done today too

iwriteonbananas

right

it's a beautiful day here to get work done. nasty, shitty weather and no construction site noises because it's saturday.

Saal Hardali

@MikeMiller Ah! So basically all I need to find is a metric on the plane which has both negative and positive curvature regions?

iwriteonbananas

@BalarkaSen oh, cool. what did you do there? attend some lectures?

Mike Miller

14:53

@SaalHardali Yeah, that should do the trick for you.

Should I write this as an answer to your question, or would you prefer someone who was willing to write something down with more explicit details?

Saal Hardali

@MikeMiller Is there an example of a manifold which somehow intrinsically has this property? That is both negative and positive curvature regions?

Balarka Sen

@iwriteonbananas studied. attended seminars. studied some more.

Saal Hardali

I'm still thinking about how to find something which isn't pathological since my original motivation for this is why should one care for non-symmetric spaces.

All the ones I'm familiar have a canonical locally symmetric structure

iwriteonbananas

@BalarkaSen sweet

Mike Miller

@SaalHardali I would just cook one up by bashing the standard metric on $\Bbb R^2$ with some scalar function so that what you say is true. For convenience, it's nice to work with scalar curvature, where the change-of-curvature formula is $S(e^{2u}g) = e^{-2u}(S(g)+\nabla u)$, or something like this. I always get these sorts of formulas wrong.

So you could more or less just start with the flat metric and pick a function $u$ with interesting $\nabla u$ whose asymptotics are good enough that you don't make the manifold non-complete. This is not so hard.

@SaalHardali No offense, but that's a crazy question to me! There are so few spaces that actually are locally symmetric. Even in the context of, say, 3-manifolds, we have this wonderful geometrization theorem which says that every manifold can be cut into pieces with very nice geometries. But the 3-manifold itself, before cutting, is probably not going to support a very nice geometry!

Someone could surely hand you an example of a smooth manifold that is never locally symmetric.

Saal Hardali

14:58

Yeah that's what I really need.

To be satisfied.

Building patholiogies in the plane is fine but it's not really satisfying :(

Mike Miller

These aren't pathologies. These are most manifolds.

I understand why it's not satisfying, but, like, write down your favorite surface in 3-space. Is that not supposed to count as a perfectly good manifold you care about?

The standard torus in 3-space isn't locally symmetric. (The above argument works to see why.)

Saal Hardali

Yeah I get your point. Thing is I'm interested in whether this locally symmetric property is something topological.

Mike Miller

OK. I mean, they're completely classified up to isometry, so you could certainly "simplify" the classification to just listing off the smooth manifolds that support a locally symmetric structure.

But let me think if I can cook up an explicit counterexample easily. I'll probably give up if I can't in the next five minutes or something.

Saal Hardali

In essense what will really shut me up is a smooth manifold which has no compatible locally symmetric structure

Even just an affine connection with no torsion will be enough.

Sorry last sentence was'nt clear.

just ignore it.

Mike Miller

Sure.

@Saal: OK, from looking at the classification, I see that every simply connected symmetric space is either Euclidean space, has nonnegative sectional curvature, or nonpositive sectional curvature. So if I can hand you a closed manifold such that no metric has one of those properties, it cannot be locally symmetric, since its universal cover would be symmetric.

Now a space with nonpositive sectional curvature is aspherical. So let's restrict to manifolds with interesting higher homotopy groups.

Saal Hardali

15:08

@MikeMiller Sounds like this might get nasty.

Mike Miller

OK, so here's a way to finish this off. A manifold with nonnegative sectional curvature has nonnegative Ricci curvature, and the Bochner trick proves that it thus has $b_1(M) \leq n$, $n$ the dimension.

So let's take, like, the connected sum of a bunch of copies of $S^2 \times S^1$.

This has neither aspherical universal cover nor $b_1(M) \leq 3$.

So it cannot support any metric whose sectional curvature is either nonnegative or nonpositive.

And hence cannot possibly be locally symmetric.

This isn't a very nasty manifold IMO.

Saal Hardali

I'l have to think about this.

not so fluent in riemanian geometry.

Mike Miller

Do you want me to write this and details/references as an answer?

Saal Hardali

thanks it looks gof!

Mike Miller

I dunno what gof means

Saal Hardali

15:10

Yeah that would be great!

Thanks

gof = good

Mike Miller

:)

Balarka Sen

it means g composition f

user147690

@SaalHardali Is this from another language, or acronym or just slang?

Mike Miller

Typo.

user147690

Seems unlikely, but possible sure

Semiclassical

15:22

hi chat

user147690

Hello

Adeek

hi @Semiclassical

iwriteonbananas

sup, semi

Mike Miller

it's a party in here

Saal Hardali

@AlexClark Was a typo. Moreover it seems pretty likely looking at my keyboard

user147690

15:32

@SaalHardali Fair enough, I thought the dropping of an o, and moving to an f seemed peculiar

Saal Hardali

@MikeMiller I've asked this on the MO chat with no reply. Do you know if there's some charaterization of groups for which $BG$ is a finite CW complex?

I only know one: $Z$.

Are they really rare like they seem?

Balarka Sen

If K(G, 1) has dimension n, H_{n+1} must vanish.

That's a nice condition, in my opinion.

Mike Miller

@SaalHardali To start with, $G$ must be torsion free. But I disagree, you know many more examples. Every surface but $S^2$ and $\Bbb{RP}^2$ is a $K(G,1)$.

Every hyperbolic manifold is a $K(G,1)$.

Saal Hardali

BG isn't the same as K(G,1) though for topological groups

Ah right.

Mike Miller

It is for discrete groups, and you said you only knew one example. :)

Saal Hardali

15:34

Every closed surface

Balarka Sen

Every knot complement is a K(G, 1) too IIRC but I don't know a proof of that.

Mike Miller

That's also true. You can take the complement of a small tubular neighborhood of a knot so that it's a compact manifold (with boundary) and hence a finite CW complex.

Saal Hardali

Correct. I know thousnds but non for compact connected groups.

hmmm

that's sounds good

Mike Miller

Hm, good point on that, let me think if I know a single example of a connected group.

Actually I should finish writing your answer first.

Saal Hardali

:)

Every Hyperbolic manifold is a K(G,1) for its fundamental group I assume right?

Ah sure. I'm tired.

Ignore last comment

Mike Miller

15:41

OK, I wrote the answer.

I also expanded on the torus comment a little bit, because even though that's not really what you were looking for, I think "Every child's third or fourth manifold isn't locally symmetric" is a good reason to care about things that aren't locally symmetric :)

Saal Hardali

@MikeMiller Best. Answer. Ever.

thanks.

Mike Miller

Sure, glad to. I really like a lot of the questions you post.

Saal Hardali

Thanks.

That's too flattering

Mike Miller

@SaalHardali If $X$ is a simply connected, finite dimensional CW complex that's not contractible, the loop space $\Omega X$ is never finite-dimensional. This follows from a spectral sequence calculation on the pathspace fibration $\Omega X \to \mathcal PX \to X$ - if $H^*(\Omega X)$ was zero after some point, then some element of the second page ($H^p(X;H^q(\Omega X))$ for the highest $p$ and $q$ for which this is nonzero) would survive the spectral sequence after it stabilizes

Because we chose $p$ and $q$ as large as possible, no differentials go to that, and no differentials go from it

That would give us a nonzero element of $H^*(\mathcal PX)$ which is nonsense.

Now apply this to $X = BG$, where $G$ is a connected compact nontrivial Lie group. Because $G$ is connected, $BG$ is simply connected. If $BG$ was finite-dimensional, then $\Omega BG = G$ would fail to be, by the above.

Balarka Sen

It's so darn hot in here.

Mike Miller

15:55

@BalarkaSen Move to California, then.

Balarka Sen

Cruel world. Now that I am back home I can't do math peacefully without turning on the air conditioner.

@MikeMiller How's the weather there?

Mike Miller

It's around 70. It's a beautiful day outside.

Balarka Sen

I envy you.

Saal Hardali

@MikeMiller Wonderful!

Thank you.

Mike Miller

Sure thing. Also a fun problem.

Balarka Sen

15:57

it's about twice the temperature here then.

Saal Hardali

Great problem. Seems pretty important for representation theory of compact groups to know that all of them must have infinite classifying spaces.

Mike Miller

@SaalHardali Lie group didn't matter above, since the only time I used that was to have the formula $\Omega BG = G$. Probably you can get away with something like "finite-dim H-space" or "grouplike monoid"

Saal Hardali

Yeah figured that,

only compact matters here.

Mike Miller

I don't think even that matters, just non-contractible & finite-dim'l

It's just that for Lie groups everything deformation retracts onto a compact subgroup, so we may as well just work with those :)

Saal Hardali

Ah right compact is stronger then finite.

That's really good to know. I feel relieved.

Mike Miller

16:01

I think it's mostly important that $H^*(BG)$ is a nice, big ring to get characteristic classes from.

Here's a question, though. In every example I know it's finitely generated.

Is it finitely generated for all Lie groups $G$?

Saal Hardali

Need to think about that.

That's lie algebra cohomology problem

Mike Miller

I don't know the answer, but my money is that it's true...

Saal Hardali

Which I don't know

Mike Miller

How do we relate it to Lie algebra cohomology? I know the relation for $H^*(G)$ but I've long forgotten any relationship to $H^*(BG)$.

Saal Hardali

I'll need to think about this to make this precise

anyway I need to run.

Mike Miller

16:03

have a good day!

Saal Hardali

thx

Mike Miller

@BalarkaSen AC on or off, just get to work. ;)

Balarka Sen

Working right now.

Mike Miller

StackExchange: asking the important questions

29

Q: Why does the terminator have genitalia?

At the start of The Terminator we see him fully naked and fully equipped. He doesn't need them, so why does he have them?

terminator

Balarka Sen

i have seen it on the hot questions list. not gonna click on it.

Semiclassical

16:20

that's almost impressive in its awfulness

Mike Miller

Hi @Ted. Saal asked some interesting questions in here earlier you might like.

Valery Saharov

Can someone check my solution please?

Ted Shifrin

Good night, @MikeM. You feeling better?

Valery Saharov

4

Q: Approximate spectral decomposition

See attempt below I am interested in effective computations in finding approximate spectral decompositions in some suitable format. Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can ...

linear-algebra hilbert-spaces approximation computability

Mike Miller

Yeah.

Valery Saharov

16:22

if you know linear algebra and complex analysis of course

Mike Miller

There has been lots of interesting activity on MSE lately...

Semiclassical

depends on what you find interesting

Ted Shifrin

Oh, btw, Mike, Alex M. responded to my comment.

Morning, @Semiclassic.

Semiclassical

morning @ted

Mike Miller

@TedShifrin Yesterday there was a question about why we don't use the $\partial$-cohomology groups instead of the $\bar \partial$ ones. There's the obvious answer (holomorphic things are more interesting than antioholomorphic things, and the groups are isomorphic anyway), but Andrew Hwang pointed out that there's no such thing as an antioholomorphic bundle, and holomorphic bundles only have $\partial \bar$ cohomology groups

Which was pretty interesting... I hadn't caught that before.

Ted Shifrin

16:25

Andy's a sharp guy. I told you I've known him since 1990.

You could put an antiholomorphic bundle on an antiholomorphic manifold, I suppose. Ugh.

But it's not totally true that we don't consider $\partial$ cohomology. It definitely shows up.

Mike Miller

I guess... but his point was that antiholomorphic automorphisms aren't a pseudogroup

Ted Shifrin

Hmm, maybe I take back my last statement.

Balarka Sen

Hi @Ted. Back in home. Studying calc.

Semiclassical

that reminds me of a physics thing (albeit a very mathematical point) and i wonder if it's related. might have to do some digging

Ted Shifrin

Welcome home, @Balarka. Are you asking for a gold star already?

Balarka Sen

16:29

What's the gold star for? :)

Ted Shifrin

Balarka: You should take a day break from math and just go do nothing.

Balarka Sen

I'll do it after I get calculus done. Been postponing that for way too long.

Mike Miller

He's just tired of me picking on him.

Semiclassical

bleh, my brain isn't working right now

Balarka Sen

@Ted: I am getting some revision done from chapter 7 done. So, in that exercise, we prove that there are no C^1 space filling curves [0, 1] --> [0, 1]^2 (let U be some open set containing [0, 1], cover [0, 1] by small intervals contained in U by Lebesgue no lemma or by hand, use that C^1 maps are locally lipschitz to get a bound on the diameter of the images of those intervals and bound up the volume of the image to show it's always volume 0). Is it also true if I replace C^1 by just diffable?

Adeek

16:33

hi @TedShifrin @MikeMiller @BalarkaSen

finally this week is my last week of school undergrad

Ted Shifrin

Congratulations, Karim.

Adeek

thank you @TedShifrin

Ted Shifrin

I actually don't know the answer to that, @Balarka. If you have a differentiable function, it cannot be nowhere $C^1$ (i.e., the derivative must be continuous on a set of ? category), but it may well still be space-filling.

Semiclassical

but the linkage i had in mind was with coherent states in quantum mechanics

Mambo

@Adeek Have you accepted the offer at Alberta ?

Adeek

16:37

yeah

Mike Miller

Congrats Karim

Balarka Sen

@Adeek Congrats.

Adeek

thanks @MikeMiller @BalarkaSen

Mambo

congo @Adeek

Mike Miller

@TedShifrin I think I'm out of questions to think about so have to work now ...

Adeek

16:37

and @Mambo

Ted Shifrin

@MikeM: Can there be differentiable space-filling curves?

Mike Miller

"When I think I'm out, they pull me back in..."

Balarka Sen

@TedShifrin Hmm, should I know this, or is it not so easy to see a diff function cannot be nowhere C^1?

haha

Ted Shifrin

@Balarka: It's a beautiful application of Baire Category.

Mike Miller

@TedShifrin Aren't differentiable maps Lipschitz?

Balarka Sen

16:39

Ah, something I don't use very often.

Ted Shifrin

Not without continuity of the derivative, @Mike. So far as I know.

Balarka Sen

@MikeMiller locally? need not be. C^1 maps are locally lipschitz by mean value theorem.

Mike Miller

Clearly not. Good point.

Ted Shifrin

@MikeM: I'd never thought about the composition of antiholomorphic before. The chain rule doesn't cooperate. Interesting.

Mike Miller

OK, I'm not going to think more about this.

Ted Shifrin

16:41

@Balarka: It's closely related to the question I asked you years ago about the pointwise limit of a sequence of continuous functions.

Mike Miller

Your question is answered here on MO.

Ted Shifrin

Quitter @MikeM.

Oh, cool. It's actually Balarka's question.

Mike Miller

@Ted You know I don't functions without some regularity.

Balarka Sen

@MikeMiller nice, thanks.

Ted Shifrin

Me neither :P

Oh, an extremely recent MO post, I see.

Mike Miller

16:44

Mark McClure's answer is interesting

Lipschitz, Holder, continuous... these are all fine, but it's unhealthy to think too hard about differentiable functions

Balarka Sen

I can believe a space filling curve can be chosen to be smooth outside a cantor set.

Ted Shifrin

But it is worth understanding that a differentiable function must be $C^1$ at plenty of points.

Mike Miller

@BalarkaSen That's not exciting at all, yeah

Balarka Sen

@TedShifrin I'll note that down and try to prove it later on.

Thanks, by the way, interesting fact.

Ted Shifrin

@Balarka: Start here.

Mike Miller

16:47

eh...

Balarka Sen

Hmm, I don't remember the question. Let me have a look at my bookmarks.

Mike Miller

@Ted: Did you see the locally symmetric question I liked?

Ted Shifrin

Very simple. Prove that the limit function must be continuous on a set of ??? category.

No, @MikeM.

Mike Miller

2

Q: Affine manifolds which are *not* locally symmetric

Let $M$ be a manifold equipped with an affine connection $\nabla$. By standard theorems of differential geometry we have convex normal coordinate balls around every point in which geodesics are axis. For every normal ball $N_p \subset M$ We have an obvious involution given by the reflection $s_p:...

differential-geometry riemannian-geometry tensors connections

Someone clever can probably extend what I say to any affine connection

Ted Shifrin

Saal seems to be mis-using the term "affine manifold."

That's different from a manifold with an affine connection on its tangent bundle.

Balarka Sen

16:49

@TedShifrin Ah, so I need to figure out, given a sequence $f_n$ of $C^0$ functions on $[0, 1]$, how much $C^0$ the limit is?

Ted Shifrin

Right, @Balarka.

Balarka Sen

*pointwise limit

OK, I'll do that.

Ted Shifrin

(Assuming that the sequence does converge pointwise, of course.)

Balarka Sen

Interesting question

Yep, sorry about that.

Mike Miller

@Ted: Yes, but it was clear what was meant, so eh

Mambo

16:51

@TedShifrin Hello

Ted Shifrin

Well, still that should be clarified. Bill Goldman spent a lot of his career on affine structures on manifolds.

hi @Mambo

Mike Miller

Sounds like you're the man for the job :)

Ted Shifrin

I'll see if I can find something to link it to, @MikeM.

Mambo

I have a question related to Lie Algebra $M_n(\Bbb{R})$

Balarka Sen

So, I need to find a very discontinuous function which is a ptwise limit of C^0 functions on [0, 1].

Mike Miller

16:53

meh

Balarka Sen

Let's start with something. Say, the popcorn function.

robjohn

@IlanAizelmanWS Since $R^2=S^2=I$ and $RS=-I$, we have $(R+S)^2=R^2+S^2+2RS=I+I-2I$. So it appears that at least its square is $0$.

Mambo

@TedShifrin In $M_n(\Bbb{R})$, the Lie Bracket is given by $ [X,Y] = XY - YX $.

Ted Shifrin

Hi @robjohn.

Balarka Sen

Eh, maybe something simpler. $1$ if rational, $0$ if irrational. Just because popcorn is too hard for me

robjohn

16:55

@TedShifrin Hey there.... it's a beautiful day in LA

Mike Miller

I'll come back when you're doing real things with real functions.

Ted Shifrin

@MikeM, I added a comment.

@robjohn: Here too. I'm trying to decide whether to walk to the park or drive to the ocean and walk a bit.

robjohn

@BalarkaSen The kernels bother me.

Balarka Sen

I think Mike is frustrated by such horrible functions.

Mambo

@TedShifrin But $M_n(\Bbb{R})$ can be thought as set of all left invariant vector fields of $GL_n(\Bbb{R}) also, right?

robjohn

16:56

@TedShifrin I drove by the ocean on the way home yesterday. Today we are going to an art fair.

Ted Shifrin

Sounds lovely, @robjohn.

Sure, @Mambo. Just use matrix exponential.

robjohn

@TedShifrin Tomorrow is the Renaissance Faire.

Balarka Sen

Sure, that seems doable.

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$Mathematics$ Mathematics