Mathematics - 2016-06-11

Ted Shifrin

01:00

@Jake1234: You might look at section 1.3 of my notes and read a little about the tangent indicatrix and Fenchel's Theorem and Fary-Milnor. Gauss-Bonnet is all about surfaces, not curves.

Mike Miller

Hi @Ted. Did you see the question about degree and divergence?

Ah, great, you had a simple example. I had a picture but was having trouble writing down an honest function realizing it.

user 1618033

Working here, 04:15 am.

The Substitute

03:04

Hello, off topic algebra. When constructing Galois automorphisms on a splitting field $\mathbb{Q}(a,b)$ by defining their image on the generators, how does one show that these maps extend to automorphisms of the Galois group?

Agawa001

05:10

@user1618033 keep getting us updated with ur rss feeds

user 1618033

07:34

@Agawa001 I'm exhausted. These days I submitted the most important papers in my all mathematical activity.

Owatch

08:12

Hey, can someone assist me with transformations again?

It should be quite simple.

I'm trying to find the transformation matrix $A$ from a transformation: $T:R^{2} \rightarrow R^{2}$/

aras

If $L/K$ is a field extension, is it automatically assumed that $L>K$ or is it also possible that $L=K$?

which definition is more widely accepted

Owatch

All I'm given are two sets of vectors and their images.

Apparently I need to break down the images into what happens with the standard vectors. But I'm not sure what I do after that.

aras

@Owatch What is given your specific case?

Owatch

Is there an easy way to write vectors in latex?

It's whats making me avoid trying to write this in chat, since I use ugly ascii formatting.

Anyways, the transformations are:

{1,0} -> {2,5}. {0,1} -> {-1,6}

aras

I would just say use points, like $(1,2,3)$. You can also use \langle and \rangle to get brackets or make a column matrix with \begin{pmatrix}, but this is probably easiest.

OK as for the question, I would say the best way to approach it if you're still learning the concepts is to first write out your matrix $\begin{pmatrix} a&b\\c&d \end{pmatrix}$.

Owatch

08:20

I've tried writing out what happens to $e_{1}$ and $e_{2}$.

Then solving for:

$2e_{1} + 5e_{2}$ and $-e_{1} + 6e_{2}$.

But it got me nowhere.

aras

Then multiply that general matrix by your two given vectors (1,0) and (0,1). You will get some expressions in a,b,c and d. Since you also know what the images are equal to, you have a system of equations. Then you can find the values of a,b,c,d.

Owatch

I don't know what you mean.

I just get.

{a,c} = {2,5}. {b,d} = {-1,6}

What is there to solve?

aras

Yeah you're almost done. So you have a=2, c=5, b=-1, d=6. That's your matrix.

You can double check by multiplying that this gives you your transformation.

Owatch

But I've done no solving..

Probably because input is standard...

aras

Exactly, if the input vectors were more "random" you would have to solve a linear system of equations.

Owatch

08:30

So then what is the method with standard vectors supposed to do?

This one gives me them as input, but many do not. And I'm somehow to decompose it into standard vectors and then use that to find the transformation matrix.

Same method?

aras

The same method works, yeah. You can try it with nonstandard input vectors. The system of linear equations won't be "trivial" anymore but you will get the matrix.

And as you are solving the system of linear equations, you will implicitly solving for the standard input vectors, which is what you were trying to do above.

So for ex if your problem is finding the transformation from (1,1)->(2,5) and (2,3)-> (-1,6) (no idea if this will end up ugly)

Owatch

I'm going to try another with nonstandard input.

aras

Yes do that, I think it will illustrate the process much better.

Owatch

So I've got one for: $T: R^{3} \rightarrow R^{2}$/

With: T({0,-2,0}) = {6,2}. T({-1,2,0}) = {6,0}. T({1,2,3}) = {0,-1}

Not sure how to solve the system of linear equations.

Owatch

08:58

Yeah I do not know.

It's: T({0,-2,0}) = $-2e_{2}$. T({-1,2,0}) = $-e_{1} + 2e_{2}$. And T({1,2,3}) = $e_{1} + 2e_{2} + 3e_{3}$/

But then what..

Owatch

09:16

Good god this is the most frustrating thing ever.

user 1618033

@robjohn there is a question that bothers me these days. I'm not sure I can get more than I got lately.

Agawa001

10:25

@user1618033 important papers in which sens ?

user 1618033

@Agawa001 In the sense to publish something never done before at such a level of simplicity - something very hard - I won't share the details at this point, but later.

Also some stuff that was not known so far - never published before in articles, papers.

Albas

In one of the proofs if $\sum_{k=1}^\infty = \frac{\pi^2}{6}$ we consider an infinite polynomial and try to look at its roots and find some relation to the above sum using sum of roots. That made me think that does the fundamental theorem of algebra also apply to infinite polynomials. By infinite polynomials I mean polynomials of the form $a_0+a_1x^1+a_2x^2+\cdots$. I think so it might (I have tried a few examples) but is there any rigorous way to prove it?

user 1618033

@Agawa001 I don't think that I slept more than 3 hours this night. The point is that I still need energy to finalize some of my stuff which cannot wait.

Albas

My examples include Taylor series of functions such as $\sin,\cos$

Agawa001

@user1618033 are yu planning to quantify some abstract value in irrational field ?

user 1618033

10:32

@Agawa001 Now I plan nothing, let me sleep some more before making plans. :-)

Agawa001

or maybe bound an abstract limit

planning to sleep is the most famous and effective and wonderful plan in this lived world

3

user 1618033

:-))))))

Max

11:08

Hi

does anyone know where i can find the source for this theorem?

Mean Value theorem for vectors (with integral)

postimg.org/image/ag37j6l3v/

Owatch

I've got another question.

In computing the determinant, may I move row vectors to different locations?

Say I've got a second row of a matrix: 4,0,0,0,5

May I move that up to the first row, and compute with it instead?

Actually, I can just start with any row I want.

So never mind that.

Dieter

11:32

hey, i've a question. e.g. i've several weather-stations (200) across the country.
But i want to assign the temperature to every municipalities (508).
thus, i want to calculate the average temperature of an unknown municipality based on the distance and their closest 4 neighbours.
for 2 municipality, this is quite easy -> α * temp_A + (1-α) * temp_B
whereby alpha α = temp_A / (temp_A + temp_B)

But i don't know how you can do something like this, for multiple C D E stations

Albas

@Owatch if you swap(one swap) rows of a matrix and calculate its determinant it will be negative of the determinant of the original matrix you started with

Balarka Sen

@Albas No, it doesn't. Euler's original proof was non-rigorous. You need Weierstrass' factorization theorem.

Tomáš Zato

12:20

Can I apply $log_2$ on every member of equation? For example I have
$$2^{x^2}+2^x = 3$$
Can I wrap every member on logarithm to get:
$$x^2 + x = \log_2{3}$$

Or do I have to wrap sides of equation, to get:
$$\log_2{x^2+x} = log_2{3}$$

robjohn

13:55

@user1618033 which question is that?

@TomášZato You can do neither. You have to apply the logarithm to both sides in entirety. What you suggest would require that $\log(x+y)=\log(x)+\log(y)$, which is false in general.

user 1618033

@robjohn I wonder if I'm able to get something more extraordinary than the results I got so far. I need a considerable amount of imagination for that.

(referring mainly to some special theorems that allow me to do incredible things)

Tomáš Zato

@robjohn Thanks. After posting this I also noticed that what I did had some consequences. But how to solve the equation then? Is there some formula I have missed?

Jake1234

14:35

log(x+y) is not equal to log(x) + log(y).

Tomáš Zato

@Jake1234 Yes I realized already, but how do I solve it then? I really need help :(

aras

14:57

I'm doing galois theory problems and ran across the following question that I don't understand: "Find two independent vectors in R as a vector space over Q." What on earth is this question asking?

I know that the field extension R/Q is not finite, which makes this even more confusing...

Are they just asking for a set like {1, \sqrt{2}} where they are a basis for a vector space over Q?

Jake1234

@TomášZato I don't know. Do you know if it is supposed to have a closed form answer? if you change the 2^(x^2) to , (2^x)^2, I think I can see it would be solvable, this way I don't know.

Tomáš Zato

@Jake1234 It's a single choice question and you're supposed to select number of solutions in R for the equation

So no, you don't need to get exact numbers

But I'm not sure what else to do...

Calculating intersections with x axis also requires you to solve the qeuation...

arctic tern

@TomášZato asking to solve an equation and asking to count the number of solutions are two different problems.

should have been upfront about what the real problem was to begin with

Jake1234

Have you tried plotting it/looking at the functions behavior, looking at whether it's increasing /decreasing at certain intervals? The idea would be that if you have a function like say (x^2 - 10) (for simplicity), and you want to know where it's 0, you can look that it's lower than 0 for x=0, on (0,+oo) it's strictly increasing, and goes to infinity, so there's one solution there, analogously (but using that it's strictly decreasing) on (-oo, 0)

Tomáš Zato

So you think the task was to guess what the function looks like and how many times it crosses x axis?

arctic tern

15:11

the task is to count the number of solutions, if I understand you correctly. the way to do that is to first look at the graph and let that inspire you to a proof idea.

for instance, it looks like the function 2^(x^2)+2^x is increasing until it hits an absolute minimum and then it is increasing

you can potentially prove that with some calculus (derivatives etc.)

that's a possible starting point

Tomáš Zato

15:26

People passing this test are not supposed to know derivations. It's entry test to college from high school

Czech technical university's test in particular

Jake1234

The idea is, that if a function from [a,b] is continuous, it's lower than 0 at a, higher than than 0 at b, and it's strictly increasing, there will be exactly one point between a and b, where the function is 0.

Danu

15:42

Hi @MikeMiller

Mike Miller

Hi @Danu

Danu

How are you doing?

Mike Miller

I am ok. I was extremely lazy yesterday and expect to be extremely lazy again today.

Danu

Neat-o. Do you generally work on Saturdays? Do you come into uni?

Mike Miller

work just means sitting down with paper and pen somewhere, which can be at home or a coffee shop or the beach

yeah, I try to. but the quarter just ended and I'm just going to kill the day hanging out with a buddy

Danu

15:49

That sounds nice.

Mike Miller

recently I talked to my advisor about a paper I read, and after I finished, he was like "no, no, that's wrong, no". so I clarified and he told me that was even more wrong. :p so now I'm going to read a survey he recommended so I minimize wrongosity next time.

Mike Miller

16:01

what are you up to?

Balarka Sen

Hi everyone.

Mike Miller

m

Balarka Sen

16:16

$L$ be a line bundle, $X$ it's zero section. $L \setminus X$ def. rets. onto the unit $S^0$-bundle over $X$, i.e., a two fold cover. Saying $L \setminus X$ is disconnected means this $S^0$ bundle is the disconnected cover comprising of two disjoint copies $X \times {-1}$ and $X \times {1}$ in $L$. Isn't sending $x$ to $(x, 1)$ in $X \times 1$ a nonvanishing global section of $L$?

I am not sure if that's quite rigorous. I want to do something like that, at the very least.

Danu

@MikeMiller Me? Not much. Computing pullbacks of forms for now.

Balarka Sen

i.e., I think a section of the two fold cover, which exists by assumption, is also a nonvanishing global section of the line bundle.

@Danu Differential forms are good stuff.

Mike Miller

@BalarkaSen Yes, that's correct. I also want to make it conscious that you used a metric here, which is not given on a line bundle (but I allowed).

Without invoking a metric, can you prove that an $L$ w $L \setminus X$ disconnected is orientable?

Semiclassical

morning chat

Balarka Sen

@MikeMiller Right, I used a metric for saying "$L \setminus X$ defo rets onto the unit $S^0$ bundle"

Mike Miller

16:26

You need a metric for saying the $S^0$ bundle.

Danu

It's always disturbing how hard it is to translate between physics and mathematics notation.

Balarka Sen

I agree, unit doesn't make sense otherwise.

I realized that.

Bananach

If we choose to define formally the FT of a function by

$$\hat f(\xi) = {1 \over \sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{-2\pi i \xi
x}dx$$

rather than

$$\hat f(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i \xi x}dx$$

then formulas change. for example

$$\widehat{f \star g} (\xi) = \sqrt{2\pi} \hat f(\xi) \hat g(\xi)$$

rather than

$$\widehat{f \star g} (\xi) = \hat f(\xi) \hat g(\xi)$$

my question is do theorems change also? is the fourier transform still an isometry from $L^1(R) \cap L^2(R)$ to $L^2(R)$? or is their some constant that will pop up?

Forgot to say "Hey everybody"

I would appreciate it if someone can answer this for me. apparently I reached my daily limit for questions

Danu

Why don't you try it out?

Mike Miller

@BalarkaSen So it's not hard to convince yourself there is a bijection between line bundles and $S^0$-bundles. How do you classify the latter?

Bananach

16:30

@Danu I tried it with an example and apparently it is no more an isometry. it gave me two contradictory answers and now I'm confused

Balarka Sen

@MikeMiller Maps from $\pi_1$ to $\Bbb Z/2$. Aka, $H^1(X; \Bbb Z/2)$.

Mike Miller

Whence your observation that simply connected spaces don't have line bundles.

Balarka Sen

Thanks! Now I understand why line bundles are classified by $H^1(X; \Bbb Z/2)$ -- I only "knew" that before.

Right.

Mike Miller

Now you might try to prove that closed hypersurfaces in Euclidean space (not every space!) are orientable.

Jake1234

@TomášZato Have you figured out the problem or do you still need help with it?

Balarka Sen

16:43

@MikeMiller Hmm.

user19405892

hi

Show that for any $n \not \equiv 0 \pmod{10}$ there exists a multiple of $n$ not containing the digit $0$ in its decimal expansion.

can someone explain the proof to this question

Tomáš Zato

@Jake1234 Thanks for asking :)
Friend did help me, the solution was through substitution.

user19405892

Balarka Sen

Eh, my natural instinct here is to Alexander dualize, but then it immediately says the surface has to be orientable and does not use any fact about line bundles at all.

Maybe I should translate that into the context of line bundles or something.

Mike Miller

I'm going to ask you to generalize later, and this proof should be differential topological.

Tomáš Zato

16:52

@Jake1234 Specifically:
$$2^{x^2}+2*\frac{1}{2^{x^2}} = 3$$
Then you solve the equation as if $2^{x*x}$ was some variable $y$, then you desubstitute

Balarka Sen

17:40

@MikeMiller Sorry, I was out. I'll think about it.

Jake1234

17:52

@TomášZato How is (2^x) = 2 (1/2^(x^2) ) ?

Random Variable

@DanielFischer If $\int_{0}^{\infty} f(x) e^{-(a+ib)x} \, dx = g(a+ib)$ for $a >0$ and $b \in \mathbb{R}$, and if $\int_{0}^{\infty} f(x) e^{-ibx} \, dx$ exists as an improper Riemann integral, can we conclude that $\int_{0}^{\infty} e^{-ibx} f(x) \, dx= \lim_{a \downarrow 0}g(a+ib)$?

Daniel Fischer

18:09

@RandomVariable I think so. Maybe there are some problems if $f$ is unbounded at too many points, but if the set of points such that $f$ is unbounded on each neighbourhood of the point is discrete (as a subset of $(0,+\infty)$), everything should work out well.

Mike Miller

@Balarka If you don't have any ideas in the next hour, just move on to the next seftion for now.

Balarka Sen

I have one, I just don't know how to make it work.

$M$ be a hypersurface inside $\Bbb R^n$. Take it's normal bundle. If it's trivial, I'm done because I can take the outward normal field as a section.

Suppose it's not, then $NM \setminus M$ is connected, and inside it lives a connected two fold cover of $M$ disjoint from $M$ (the unit S^0-bundle).

So take a point $x$ on $M$. Then on that unit two fold cover, because it's path connected, I can take a path going from $(x, -1)$ and $(x, 1)$ (connecting the ends of the two unit normals) inside the cover. That will be disjoint from $M$. This shouldn't happen.

I am trying to figure out why. Give me a few more minutes.

Mike Miller

Remember that in all of this you don't just live in the normal bundle, you live in $\Bbb R^n$.

Balarka Sen

Ah, I got it, I think.

So I have a path going from $(x, -1)$ to $(x, 1)$ not hitting the surface. I can homotope by straightline homotopy to the path which hits the straightline going from $(x, -1)$ to $(x, 1)$ - that hits $M$ once. But homotopy shouldn't change the intersection number mod 2!

Mike Miller

I would have preferred you just closed it off to get a loop that intwrsects once but is null-homologous, but that's my taste. :)

Balarka Sen

18:25

Right, that's a better parsing.

The $\Bbb R^n$ hint was helpful.

Random Variable

@DanielFischer Can you think of a particular example where the conclusion might not hold?

Mike Miller

Prove that a closed hypersurface in a smooth manifold $M$ (not necessarily compact) determines a homomorphism $\pi_1(M) \to \Bbb Z/2$. Conclude both Jordan-Brauwer and that hypersurfacees are orientable in any manifold with $H^1(M;\Bbb Z/2)=0$.

This is an iff. Jot the reverse implication down for another day.

Bananach

Hi everyone

Mike Miller

if you also have your thing about self-intersextion and normal bundles written down, ask me about it another time

Bananach

Are real continuous functions of compact support uniformly continuous over R?

Tomáš Zato

18:29

@Jake1234 that part was $2^{(1-x^2)}$ so you put the - from the exponent in order to get plus x^2

Mike Miller

@Bananach Yes. You should try to prove it.

Bananach

@MikeMiller many thanks!

Balarka Sen

@MikeMiller I don't have the habit of writing things. I'll note it down.

Jake1234

@TomášZato are you saying (2^x) = (2^(1-x^2) ) ?

Balarka Sen

@MikeMiller Suppose I have a hypersurface $S$ in $M$, then triangulate it to get a homology class in $H_{n-1}(M; \Bbb Z/2)$, dualize to get a class in $H^1(M; \Bbb Z/2)$ hence a homomorphism $\pi_1(M) \to \Bbb Z/2$ so that bit is obvious. I am trying to figure out the other bit.

Jordan Brouwer means $S$ disconnects $M$, right?

Mike Miller

18:42

yeah

Balarka Sen

Take a path going from one side of $S$ to the other, without hitting $S$. Close it off like you did to get a loop hitting $S$ at one point. So I have two things intersecting at one point, one of them representing 0 in homology. This dualizes to "take cup product of something in $H^{n-1}$ with 0 in $H^1$ and you get the generator in $H^n$".. nonsense.

I am a bit afraid because I didn't use your $\pi_1 \to \Bbb Z/2$ description. Is this fine?

(by one of them representing 0 in homology I mean the dual of $[S]$, not the loop)

Also, side question. Every hypersurface $S$ in $M$ gives rise to a homomorphism $\pi_1(M) \to \Bbb Z/2$. Is there a direct way to see how hypersurfaces in $M$ are in one to one correspondence with line bundles over $M$?

Take a section of a line bundle and look at it's zero set (i.e., intersection of it with th zero section), maybe? I vaguely remember this from algebraic geometry, but I am not sure.

Daniel Fischer

19:01

@RandomVariable No. Depending on what you allow as an "improper Riemann integral", things may get complicated when you have unboundedness at too many points, but I think even then it should work. Any $f$ one is likely to be actually interested in should be unproblematic to handle with integration by parts.

Balarka Sen

Hi @TedShifrin

Ted Shifrin

Hi @Balarka, g'night @MikeM. Time for me to give you one of my favorite questions (that was given to me in grad school): Use intersection theory to prove that a compact hypersurface in a simply connected manifold is orientable.

Semiclassical

hi @ted

Ted Shifrin

Hi @Semiclassic

Balarka Sen

@TedShifrin Did something like that above. Take normal bundle of the hypersurface $X$ inside $M$. Assume the unit $S^0$ bundle (i.e., two fold cover) inside the normal bundle is connected - then I can take a path inside that cover going from $(x, 1)$ to $(x, -1)$ over any pt $x \in X$ not hitting $X$.

This is homotopic to the path going from $(x, 1)$ to $(x, -1)$ hitting $X$ once. But homotopy doesn't change intersection mod 2.

Contradiction. So the normal bundle is trivial, hence there exists a outward unit normal flow on $X$ - making it orientable.

Ted Shifrin

19:08

OK. Cool. Except you really need closed paths to apply simple connectivity, don't you?

Balarka Sen

(for the unit S^0 thing I have assumed existence of a metric on the normal bundle)

@TedShifrin Well, if I have two paths with same endpts in a simply connected space, they are path-homotopy equivalent.

Ted Shifrin

OK.

Semiclassical

this question is a bit scattershot: math.stackexchange.com/q/1822421/137524

Balarka Sen

I fixed the retract proof yesterday, by the way.

Ted Shifrin

Questions like that give me a headache and I ignore them, @Semiclassic.

How? @Balarka

Semiclassical

19:11

a sensible policy

Balarka Sen

$r : M \to M$ be the smooth retract onto $A$. I proved that $Dr$ is of constant rank near $A \subset M$. So all that was required to do is to figure out when a map $f : M \to N$ of constant rank has image a manifold.

Ted Shifrin

OK, that's one approach.

Balarka Sen

So assume $f : M \to N$ has constant rank everywhere. By constant rank theorem, for any point $x$, there exists charts around that and $f(x)$ where $f$ is locally $(x_1, \cdots, x_m) \mapsto (x_1, \cdots, x_k, 0, \cdot, 0)$.

If for any point $f(p) \in f(M)$, there is a chart $U$ around $p$ in $M$ such that $f(U) \cap f(M)$ is a neighborhood of $p$ in (the subspace topology of) $f(M)$, then we'd have that neighborhood of $p$ in $f(M)$ cut out by $x_{k+1} = \cdots = x_m = 0$, hence $f(M)$ would be locally level set thus a submanifold.

This excludes the figure eight example, where image of a chart in the circle is never a neighborhood of the bad point.

Ted Shifrin

Oh, have you proved the rank theorem?

Balarka Sen

Yeah.

Ted Shifrin

19:17

It's not in my book or in G&P. I added it as a graduate exercise the last time I taught the course.

Balarka Sen

I stumbled upon that when I was reading about the implicit function theorem.

It's not in your book, yep.

Ted Shifrin

Right. The proof is a slight generalization thereof.

Mike Miller

@Ted I just gave him that ;)

Balarka Sen

(just to finish what I was saying earlier, the retract $r$ satisfies this in a neighborhood of $A$ - 'cause retracts are open maps. It was a fairly straightforward condition, I was being dumb).

Ted Shifrin

Not I, @MikeM.

Hmm, how do we prove a retraction is open?

Tomáš Zato

19:23

@Jake1234
No no no. The original equation was actually more complicated and I wrote it incorrectly. It was in fact:
$$2^{(x^2)} + 2^{(1-x^2)} =3$$
Sorry for the confusion

Balarka Sen

I think Mike was referring to closed hypersurfaces in simply connected manifolds being orientable.

@TedShifrin Well, there is a left inverse so it's in fact a quotient map.

I forget which is left or right. I mean $r \circ i = 1$.

Ted Shifrin

Oh, sorry, @MikeM. "Just gave him that" is ambiguous. :)

That's a right inverse.

Balarka Sen

If I write it as $A \stackrel{i}{\rightarrow} M \stackrel{r}{\rightarrow} A$ it becomes left, thus the confusion :P

Ted Shifrin

Well, that explains why some algebraists write their functions on the right. Herstein, for example. To that I have to say only UGH.

Balarka Sen

:P

Andrew Thompson

19:28

@TedShifrin They did that in our reptheory course. Nightmare.

Balarka Sen

I and prof did an observation together that algebraists do a lot of things backwards. Proving the Noether normalization and then proving Nullstellensatz with it, for example.

Writing functions on the right adds to the list, then.

@MikeMiller I am not sure if you saw these messages, so pinging you with it just in case.

Mike Miller

I'm busy right now, sorry. You can tell me about them later if you like.

Balarka Sen

Oh, sure. Thanks.

I should either do my schoolwork or get more stuff done from G-P then.

Ted Shifrin

G'night, @Balarka.

Balarka Sen

I am not sleeping!

Mike Miller

19:35

@Ted: DId you say stuff above? I missed it, sorry

Ted Shifrin

I only apologized for confusion about what you were referring to with "just gave him that."

Mike Miller

Oh, sorry for the lack of clarity myself

Semiclassical

Whether you are sleeping and whether you should be sleeping are two different things :p @BalarkaSen

arctic tern

@TomášZato that can be solved. it's u+2/u=3, where u=2^(x^2).

Semiclassical

with one root immediately obvious, in fact. (and the other is easy to find at that point)

Random Variable

19:39

@DanielFischer So basically it holds because $\int_{0}^{\infty} f(x) e^{-(a+ib)x} \, dx $ converges uniformly for $ a \in [0, \infty)$?

Jake1234

@TomášZato Oh right, no problem.

Balarka Sen

@Semiclassical fair enough

I covered thermal expansion yesterday. expansion coefficients, etc.

user 1618033

Solve the elementary equation

$$2^{\sin(x)}+2^{\cos(x)}=3$$

(without pen and paper - almost missed to mention it)

Agawa001

/me hides

Danu

o/

Balarka Sen

19:51

\o

Daniel Fischer

@RandomVariable Yes.

user 1618033

@Agawa001 :-) It's a very nice equation, one needs no pen and paper for.

(just to notice something very simple)

Agawa001

i m sorry i am cheating now

i dont have a good brain pointer as my pen is

user 1618033

@Agawa001 OK OK, cheat then (but not too much) ;)

Agawa001

i found an equation of the form f(t)=f'(log(t)) which implicates omegawright

i ll post it in 10 mins after regathering my thoughts wait

user 1618033

20:05

Great

In the meantime, let's take a look at that

$$2^x \ge x^2+1, \ \forall x\in[0,1]$$ where we get equality for $x=0$ and $x=1$.

Q.E.D.

Balarka Sen

20:19

Hi @Anubhav.K

Anubhav.K

heyy @BalarkaSen

Balarka Sen

How are you doing?

Anubhav.K

I've a flight to catch in the morning but don't feel sleepy at all

Balarka Sen

Yikes.

Anubhav.K

so try to doing math as well as reading live commentary of football

what about you?

Balarka Sen

20:30

Learning smooth manifolds, mostly. Getting schoolwork done too.

Anubhav.K

Good,which books are you following?

for a long time I've not post any good answer in MSE

:P

Balarka Sen

@Anubhav.K Guillemin-Pollack. Occasionally Hirsch, but not much.

Anubhav.K

Ohh that's great

:)

Balarka Sen

How's your study of algebraic geometry coming along?

Anubhav.K

Last 1 week I've not get much time for studying, becasue I went to my home

And there it is very difficult for me to study in presence of my big family :P

Balarka Sen

20:35

Ah.

Anubhav.K

I think from tomorrow I'll again start it with full enthusiasm

So did you finish chapter 2 in G&P?

Balarka Sen

Halfway through.

Anubhav.K

Good

Did you read some Morse theory?

Balarka Sen

Only what's in G-P chapter 1.

Anubhav.K

There are a few problems in chapter 2 as well (I think if I remember correctly)

Balarka Sen

20:39

Haven't encountered them yet.

Anubhav.K

I think in that transversallity section

Balarka Sen

I have done a couple problems from that section. I will see if I get a Morse theory problem.

Anubhav.K

Hmmm...And I think in that section there is a typo in one problem

Balarka Sen

Ted has a list of typos in G-P, I think. I'll check it out.

Anubhav.K

2.3.7

Balarka Sen

20:43

Thanks.

Anubhav.K

I may be wrong with this number

just check it

Balarka Sen

It says if $X$ is a submanifold of R^n, almost every linear subspace of a fixed dimension intersects $X$ transversely. That seems right to me?

Anubhav.K

Yes it is not true ...we need some restriction on the codimension.

let me recall about my counter example

Balarka Sen

Hmm.

I think there should be ctrexamples if $X$ is noncompact.

Anubhav.K

suppose $X$ conatins 0. Then any vector subspace intersects it. If its dimension is less than the codim of $X$
it cannot be transversal.

Balarka Sen

20:48

Ah, right, fair enough.

Mike Miller

Change what you said to affine subspace.

Balarka Sen

I can just take $X$ to be a line passing through $0$ in $\Bbb R^3$. Then a lot of planes intersects it.

@MikeMiller That's what I was thinking.

Anubhav.K

Guys I'm going off, other wise I'll miss my flight

I'll join you people tomorrow

good night

Balarka Sen

22:07

Quiet chat.

user 1618033

22:22

Working very hard here, hence quiet.

(perhaps the same for the rest - but not sure)

Balarka Sen

22:33

Hi @TedShifrin

Ted Shifrin

@Balarka: Here's an exercise for you. I usually presented Morse functions differently when I taught the course. Define $\text{grad}\,f$ as a section of $TX$ (by taking the vector field dual to $df$). Define a function to be Morse if $\text{grad}\,f$ is transverse to the zero section. Prove that agrees with the G&P definition.

Balarka Sen

Hmm.

Ted Shifrin

This has the advantage that you can then apply all the genericity/stability results you're learning that relate to transversality.

And the problem mentioned above is most definitely on my typo list.

Balarka Sen

@TedShifrin Yeah.

Hrm, I have never worked with tangent spaces of the tangent bundle, having a hard time seeing it.

Ted Shifrin

Well, first you should understand that at a point $(x,0)$ of the zero-section, $T_{(x,0)}(TX) = T_xX\oplus T_xX$ quite canonically.

So work it all out assuming $x=0\in\Bbb R^k$.

Oh oh ... Mike is here to spy.

Mike Miller

22:47

@Ted He likes to do this thing wher he says he has a hard time seeing things so he doesn't have to think about them.

Balarka Sen

>:(

Ted Shifrin

You seem to think I'm going to give him a private 10-hour lecture. I'm trying to get him on a reasonable track.

Balarka Sen

Nah, it's fine. I should think about it.

Ted Shifrin

Somewhere in some later problem sets there should be a few exercises using this definition of Morse, too, @Balarka.

Balarka Sen

Noted. Thanks.

Ted Shifrin

22:50

This is one of a number of things I chose to lecture differently from G&P.

Balarka Sen

23:05

Sorry, went to fetch something to eat. Let me think about it.

Balarka Sen

23:17

@TedShifrin Yeah, meh, this is obvious. I should have figured that out.

Ted Shifrin

Most of math is obvious once you've figured it out.

6

When you learn about oriented intersection number, then you can figure out how that relates to Morse.

Balarka Sen

Hmm, alright.

Transcript for

$Mathematics$ Mathematics