« first day (3587 days earlier)      last day (1425 days later) » 
02:00 - 20:0020:00 - 00:00

Yuvraj
Yuvraj
2:45 AM
@robjohn thank you sir,i might gave u so many troubles!
 
user462942
Hi @robjohn
 
robjohn
robjohn
@Yuvraj Do you see how to adapt that answer to $z_1,z_2,z_3$?
@Joanna hello
 
Yuvraj
Yuvraj
3:00 AM
@robjohn yes sir
 
a perspicaciously curious mind
a perspicaciously curious mind
@Joanna was I right?
 
user462942
3:16 AM
@robjohn what do you do for a living, if you don't mind me asking?
 
user462942
@robjohn you've answered some pretty hard questions on Math.SE, I've enjoyed reading them.
 
user462942
@aperspicaciouslycuriousmind yeah ...
 
user462942
I've gone from proof-y mathematics --> proof-y research --> applied math research --> very applied math research, and now I regret landing up in a lab-based lifestyle of research, where I feel the mathematician is devalued while the domain experts such as the physicists get all the glory / get to tell the story / is most senior in rank.
 
user462942
Experiencing this, I now want to move back ... away from a lab-based research culture, and perhaps avoid interdisciplinary work altogether -- and focus on purely mathematical problems.
 
Pig
Pig
what kind of research do you do if i may ask?
 
user462942
3:22 AM
Fluid dynamics at the moment, though I'm hesitant to say much more, since I'd like to remain anonymous @Pig
 
user462942
I'd open to hearing comments and suggestions
 
user462942
I'd love to chat about this
 
user462942
Brb, stepping out for a bit.
 
a perspicaciously curious mind
a perspicaciously curious mind
cya
 
 
1 hour later…
Yuvraj
Yuvraj
4:40 AM
@robjohn converting them as argument of this complex numbers
 
Knight
Knight
5:16 AM
@robjohn Sir I wanted to find the limit of $$\lim_{x\to 0} \left(\frac{3}{2} x^2 +1 \right)^{1/x^2}$$
As we know $$\lim_{t\to 0} (t+1)^{1/t} = e$$
So, I thought as $x\to 0$ the expression $\frac{3}{2} x^2 $ will become very small and $1/x^2$ will become very big, so we would get $e$ as the answer.
 
 
1 hour later…
robjohn
robjohn
6:30 AM
@Joanna I've done many things, from teaching math at UCLA, to designing software at Apple. Currently, I am writing software to teach Logic at UCLA.
@Knight $e^{3/2}$. Take a look at this answer.
 
user462942
@robjohn Nice -- I have a friend who's headed to UCLA this fall for grad school.
 
robjohn
robjohn
@Joanna what is their area?
 
user462942
biostats, I think @robjohn
 
user462942
6:46 AM
@robjohn do you participate around the mathoverflow site?
 
user462942
My research-level questions in math are on the applied side, so I wonder if they're suitable for mathoverflow.
 
user462942
e.g. applying differential geometry tools
 
robjohn
robjohn
@Joanna I have only one post there currently.
 
Pig
Pig
@robjohn are you a lecturer at UCLA? I'm curious about roles in academia that's not professorships
 
user777
user777
7:03 AM
Hi, I have a question: I just came through a "cross product" and I find that it is defined in only 3-space (any 3-space), I wonder Is there a cross product in 4 or n-space?
or do you have any comments about this. Thank you
 
Pig
Pig
@user777 there's exterior product, which can be used in calculating volume.
 
robjohn
robjohn
7:29 AM
@Pig I was an assistant prof in the Math dept.
@user777 the "cross-product" can be extended as a function of $n-1$ vectors in $\mathbb{R}^n$, which gives a vector perpendicular to all $n-1$.
 
Pig
Pig
ah that's cool
writing software to teach logic is really cool though
 
robjohn
robjohn
@user777 $$ \begin{bmatrix} x_1&x_2&x_3&\boldsymbol{i}\\ y_1&y_2&y_3&\boldsymbol{j}\\ z_1&z_2&z_3&\boldsymbol{k}\\ w_1&w_2&w_3&\boldsymbol{l}\\ \end{bmatrix} $$ is what it looks like in $\mathbb{R}^4$
 
Knight
Knight
8:13 AM
@robjohn Thank you sir, that link really helped me a lot.
 
robjohn
robjohn
@Knight great!
 
Knight
Knight
8:34 AM
Which is your favourite joke sir?
 
Yuvraj
Yuvraj
@robjohn hi
i got the result sir
 
robjohn
robjohn
9:18 AM
@Yuvraj I thought you said that here, or was that about something else?
 
Yuvraj
Yuvraj
@robjohn yes sir,but i was referring to your comment,and in my comment i said sir i got the idea so i said yes
sorry for such confusion
 
 
1 hour later…
geocalc33
geocalc33
10:42 AM
can you take the logarithm of a category?
 
Mike Miller
Mike Miller
11:12 AM
Oops I didn't answer you the last time you asked this since it sounded like a good troll question
Sets are kind of categories, just categories with no non-identity morphisms
So you should ask beforehand if you can take logarithms of sets
The issue is the base of the logarithm. Whatever k you choose, some log_k(n) is not an integer, so not represented by a set
 
Thorgott
Thorgott
11:30 AM
If $\mathbf{C}$ is a locally small category, Yoneda gives you an embedding $\mathbf{C}\hookrightarrow\mathbf{Set}^{\mathbf{C}^{\mathrm{op}}}$. If you had a sensible notion of logarithm (say, functor from the metacategory $\mathbf{CAT}$ to itself, which cancels exponentiation) with base $\mathbf{Set}$, this would give a functor $\log_{\mathbf{Set}}\mathbf{C}\rightarrow\mathbf{C}^{\mathrm{op}}$.
So either the natural Yoneda embedding gets ruined by the logarithm or the resulting category isn't interesting.
 
Balarka Sen
Balarka Sen
I logged in to come and see this
 
Thorgott
Thorgott
I do feel bad
 
Alessandro Codenotti
Alessandro Codenotti
@BalarkaSen lol
 
Mike Miller
Mike Miller
11:48 AM
@BalarkaSen It's Geo's fault
I'm just going to start calling him Giovanni
 
Balarka Sen
Balarka Sen
Catchy name
 
geocalc33
geocalc33
I was playing chess a few days ago
 
Balarka Sen
Balarka Sen
12:15 PM
Just won one by wild flagging
Lol
 
robjohn
robjohn
12:31 PM
@geocalc33 black to move?
 
ICCQBE
ICCQBE
Hello
I have a little problem
Is $|\frac{(-1)^n}{n}|$ equal to $\frac{1}{n}$? ($n\in{N^+}$)
 
Sha Vuklia
Sha Vuklia
guys, where does the calculation go wrong here?
apart from the error at the end (since $e^{i\pi}=e^{-i \pi}$)
and $r=1$
I believe the integral shouldn't yield zero
since the principal value of the square root isn't holomorphic on the whole negative axis
so, the integral of the circle should be the negative of the integral "about" the segment of the negative axis
normally I would compute the integral that way
but someone else suggested doing it like this, but then we end up with zero
but I don't see where the argument fail (as it has to, right?)
btw, hw $\sqrt z$ is the principal value of the square root
 
Mike Miller
Mike Miller
12:49 PM
Sorry, what are you integrating and what are you integrating over? The notation is not clear to me.
 
Thorgott
Thorgott
the second exponential term in the top right should be $e^{i\theta}$, no?
you wanna traverse the whole circle
 
Sha Vuklia
Sha Vuklia
So I'm integrating the principal value of the square root of $z$, that is, $e^{i/2\cdot \log(z)}$, where $\log z$ is the principal value of the logarithm, over the cirkel $C(0,1)^+$
but actually, now that I'm thinking of it
maybe it could be zero after all
I'm omitting the point $(-1,0)$ in the integral, but that should be legitimate I believe
or maybe not hm
hm ye, removing isolated points should be fine I think, as they have measure zero
 
Mike Miller
Mike Miller
What does C(0,1)^+ mean
 
Sha Vuklia
Sha Vuklia
the circle about 0 with radius 1
with the orientation going anti-clockwise
 
Mike Miller
Mike Miller
Why is there an r in your calculation
 
Sha Vuklia
Sha Vuklia
12:55 PM
ye that's unnecessary
 
Thorgott
Thorgott
I don't see where the /2 factor is coming from
 
Sha Vuklia
Sha Vuklia
alright, let me redo it
 
Thorgott
Thorgott
it seems you're parametrizing the circle by $e^{i\theta}$ with $-\pi<\theta<\pi$
 
Sha Vuklia
Sha Vuklia
sry, I knew there are all these erros, but I thought they weren't the main issue
 
Mike Miller
Mike Miller
@Thorgott That's the mistake yeah
My objection to the other things is that they make it harder to extract the point
 
Sha Vuklia
Sha Vuklia
12:56 PM
@Thorgott why is that a mistake?:o
because that's indeed what I'm doing, and I thought that was legitimate?
 
Thorgott
Thorgott
that is legitimate, but you're not doing it correctly
 
Sha Vuklia
Sha Vuklia
alright, let me do it as far as I believe is correct
(without the unnecessary mistakes)
brb
 
Thorgott
Thorgott
sure
 
Sha Vuklia
Sha Vuklia
oh oops haha
yea
I think that solves it
man, I'm so unsure of complex analysis sometimes, that I don't believe my errors could be computational, and I'm immediately looking for fundamental mistakes
 
kahen
kahen
Is Z((x)) / 2Z((x)) isomorphic to (Z/2Z)((x))? It seems that it should be similar to how Z[x]/ 2Z[x] is isomorphic to Z/2Z[x], right?
 
Leaky Nun
Leaky Nun
1:11 PM
@kahen usually when we write k((x)), k is a field
so you just mean power series beginning from some integer power?
 
Thorgott
Thorgott
what's ((x)) again? formal Laurent series?
 
kahen
kahen
The definition of formal Laurent series works just fine for commutative rings
 
Thorgott
Thorgott
why doesn't reduction mod 2 on the coefficients + isomorphism theorem work?
 
Balarka Sen
Balarka Sen
Z((x)) = Z[[x]][1/x]
 
Thorgott
Thorgott
yuck
 
Balarka Sen
Balarka Sen
1:15 PM
For any ideal I of R, R[[x]]/I.R[[x]] = (R/I)[[x]] and R[x]/I.R[x] = (R/I)[x] are easy checks. This gives you the result
 
Thorgott
Thorgott
does what I suggested fail for some reason?
I don't wanna check the computations
 
Leaky Nun
Leaky Nun
in general the powers can be any poset and the allowed subsets are the well-ordered subsets
in this case the poset is Z
 
kahen
kahen
Yeah, it seems to work. I must've been overthinking things
 
Mike Miller
Mike Miller
@LeakyNun take yourself seriously bro
 
Balarka Sen
Balarka Sen
@Thorgott Formal Laurent series are just localization of formal power series at the uniformizer
Geometric description
"It's not a phase, dad, it's a philosophy" - me justifying localizations
 
Thorgott
Thorgott
1:25 PM
right
 
Leaky Nun
Leaky Nun
@BalarkaSen so formal functions on the punctured neighbourhood
 
Thorgott
Thorgott
I can't wait for my commutative algebra lecture to get to localizations
 
geocalc33
geocalc33
2:09 PM
@robjohn yep it’s a stalemate 😂
 
robjohn
robjohn
@geocalc33 yeah. Black to move: stalemate; White to move: checkmate (but only with Rc6).
Oh, Nd3 works, too. I had thought originally that that left the pawn hanging, but the other knight is still there.
 
 
1 hour later…
Balarka Sen
Balarka Sen
3:42 PM
@Thorgott Have you seen a proof of Morse lemma
 
Thorgott
Thorgott
4:00 PM
eh, I saw one a while ago, but didn't get it then
 
Knight
Knight
Can somebody name some influential mathematicians who were not professional mathematicians ?
 
Stupid Questions Inc
Stupid Questions Inc
Greetings! I've had one small brain fart, how come if $E=[0,1]^2$ then how come $\partial (E\setminus \mathbb{Q}^2)=E$? I thought that since $\partial (E\setminus \mathbb{Q}^2)=\overline{E\setminus \mathbb{Q}^2}\setminus (E\setminus \mathbb{Q}^2)^\circ = E\setminus (E\setminus \mathbb{Q}^2)^\circ$, so $(E\setminus \mathbb{Q}^2)^\circ$ should be empty, ie. it doesn't have any accumulation points (which seems obvious enough).
But is there a simpler way to see that $\partial (E\setminus \mathbb{Q}^2)=E$?
 
Thorgott
Thorgott
$\partial(E\setminus\mathbb{Q}^2)=\overline{E\setminus\mathbb{Q}^2}\cap\overline{\mathbb{Q}^2}=E$ seems faster
 
Stupid Questions Inc
Stupid Questions Inc
@Thorgott that's also nice
 
Balarka Sen
Balarka Sen
@Thorgott OK, I can write out a proof for you in a more general context
I will call a smooth map $f : M^m \to N^n$ with $m \geq n$ an "apparent fold" map if there exists an $(n-1)$-dimensional submanifold $\Sigma \subset M$ such that (1) $df$ has rank $n$ on $M \setminus \Sigma$ and $df$ has rank $n-1$ along $\Sigma$ (2) For any point $x \in \Sigma$, $T_x \Sigma \oplus \ker df_x = T_x M$
A Morse function $f : M \to \Bbb R$ is an apparent fold, where $\Sigma$ is the $0$-manifold of isolated singularities of $f$. The function $f : \Bbb R^2 \to \Bbb R^2$, $f(x, y) = (x, y^2)$ is an apparent fold map, where $\Sigma = \{y = 0\}$ is the $x$-axis.
 
Mike Miller
Mike Miller
4:22 PM
Lmao this is gonna be nuts
 
Balarka Sen
Balarka Sen
I have to get dinner but I will come back
 
Mike Miller
Mike Miller
I just know the flow proof
@Thorgott Let 0 be a critical point of a Morse function. By assumption, Q(v) = (Hess_0 f)(v) is a non-degenerate quadratic form. You want to construct a diffeomorphism Phi so that f(Phi(v)) = Q(v)/2.
Sounds hard. There's a common trick that makes it easier: don't find one diffeomorphism Phi. Find a family Phi_t so that f(Phi_t(v)) = (1-t)f(v) + tQ(v)/2.
The reason this is a good idea is that you can then differentiate this equation to find out what (time-dependent) vector field this is the flow along
At which point you check that the flow your time-dependent vector field generates indeed gives you a flow with the desired properties
This ought to be called the Moser trick
He used something like this to show that any two volume forms on a closed manifold with the same volume are related by a diffeomorphism
 
topologicalorientablesurface
topologicalorientablesurface
4:40 PM
hello
I'm trying to show that the boundary of an n-manifold with boundary is an (n-1) manifold without boundary
My attempt so far:
Let M be an n-manifold with boundary. Let $p\in \partial{M}$. So there exists a boundary chat $(U,f)$ where $f: U\rightarrow \mathbb{H}^n$ is a homeomorphism. So, since $\partial{H}^n$ is closed in $\mathbb{H}^n$, $f|_{\partial{H}^n}: f^{-1}(\partial{H}^n) \rightarrow \partial{H}^n$ is a homeomorphism.
How do I show that $U\cap \partial{M}=f^{-1}(\partial{H}^n)$
?
 
feynhat
feynhat
The last coordinate of that chart will be zero. Isn't that the definition of boundary point?
 
topologicalorientablesurface
topologicalorientablesurface
you mean $f$? f only has the conditions $f(p)\in f(U)\cap \partial{H}^n$ and that $f(U)$ is open in $H^n$
 
feynhat
feynhat
$\partial H^n$ has all the the last coordinates zero..
 
topologicalorientablesurface
topologicalorientablesurface
right, yes.
But still, if $x\in U\cap \partial{M}$, I don't see why $f(x)\in \partial{H}^n$
i feel like this relates to the fact that $\partial{M} \cap intM$ is empty ,but unsure how
 
feynhat
feynhat
What is your definition of $\partial M$? For me its all those $p \in M$ for which you can find a chart $(U, \phi)$, such that the last coordinate of $\phi(x)$ is $0$ (invariance of domain guarantees that if this happens for one chart, it happens for all charts. So this notion of $\partial M$ is well-defined).
So, $x \in U \cap \partial M$ means that the last coordinate of $\phi(x)$ is $0$ or $\phi(x) \in \partial H^n$.
 
Balarka Sen
Balarka Sen
5:00 PM
@LeakyNun Drunkenstein rekt
 
feynhat
feynhat
Hi Balarka. Any news on when ISI-B will reopen?
The only news we get is "We will let you know in 2 weeks".
 
Thorgott
Thorgott
sounds like a really interesting trick (re: Mike)
diffeomorphisms send interior points to interior points and boundary points to boundary points @topological
 
topologicalorientablesurface
topologicalorientablesurface
i'm only dealing with topological manifolds @Thorgott
 
feynhat
feynhat
homeo too
 
topologicalorientablesurface
topologicalorientablesurface
So, because of the invariance of dimension, it follows that since $U$ is a chart, and $x\in U$ then $x$ is sent to $\partial{\mathbb{H}^n}$ under f? since there exists another chart for which this holds?
i'm not sure how it follows from invariance of dimension though, invariance of dimension is that a manifold cannot be both, m and n dimensional where $m\neq n$
what equivalent formulation are you using?
where $x\in U\cap \partial{M}$
 
Thorgott
Thorgott
5:08 PM
an open subset of R^n and an open subset of R^m can only be homeomorphic if m=n
 
Balarka Sen
Balarka Sen
5:33 PM
@feynhat I don't think the semester's happening. They officially said the rest of the classes will happen online, no clue on how the exams are gonna happen
Ideally there will be no exam
@BalarkaSen OK, continuing from here. Let's investigate the structure of such maps. For any point $p \in \Sigma \subset M$, since $f|\Sigma : \Sigma^{n-1} \to N^n$ is an immersion, we can conjure coordinates $y_1, \cdots, y_n$ around $f(p)$ such that $f(\Sigma) = \{y_n = 0\}$ in these coordinates.
 
feynhat
feynhat
>conjure. kvlt intensifies.
 
Balarka Sen
Balarka Sen
Therefore, $x_1 := f \circ y_1, \cdots, x_{n-1} := f \circ y_{n-1}$ provide coordinates on $\Sigma$ near $p$. Using the fact that $T\Sigma \oplus \ker df = TM|_{\Sigma}$, we can extend this to local coordinates $(x_1, \cdots, x_m)$ near $p$ in $M$. In these coordinates, $\Sigma = \{x_n = \cdots = x_m = 0\}$
Thus, in constructed local coordinates, $f : (M, p) \to (N, f(p))$ is of the form $\Bbb R^m \to \Bbb R^n$, $f(x_1, \dots, x_m) = (x_1, \cdots, x_{m-1}, h(x_1, \cdots, x_m))$ for some function $h$ that we shall analyze.
Since $f(\Sigma) = \{y_n = 0\}$, $h$ vanishes along $\Sigma$. Remember $\Sigma$ is the set of points in $M$ on which $df$ has rank $n-1$. Thus, $\partial h/\partial x_m, \cdots, \partial h/\partial x_n$ must also vanish along $\Sigma$.
So $h(x_1, \cdots, x_{m-1}, 0, \cdots, 0) = 0$, $\partial h/\partial x_i (x_1, \cdots, x_{m-1}, 0, \cdots, 0) = 0$ for all $m \leq i \leq n$.
It thus remains to identify what are the functions $h : \Bbb R^m \to \Bbb R$ which satisfy these criteria
 
topologicalorientablesurface
topologicalorientablesurface
5:53 PM
0
Q: Boundary points and manifolds

topologicalorientablesurfaceLet $M$ be an $n$ manifold with boundary. Fix $p\in U$. Let $(U,f)$ be a boundary chart, for which $f(p)\in f(U)\cap \partial{\mathbb{H}^n}$. Then, for any other boundary chart $(V,g)$, such that $p\in V$ we have $g(p)\in \partial{\mathbb{H}^n}$. I've been told that this follows from the invaria...

general-topology manifolds
 
Balarka Sen
Balarka Sen
Here's a standard observation: Let $F : \Bbb R^m \to \Bbb R$ be a smooth function with $F(\mathbf{0}) = 0$. Then we can write $$F(\mathbf{x}) = \int_0^1 \partial_t F(t\mathbf{x}) dt = \int_0^1 \sum_{i = 1}^m \partial_i F(t\mathbf{x}) x_i dt = \sum_{i = 1}^m G_i(\mathbf{x}) x_i$$ where note that $G_i$ are smooth functions such that $G_i(0) = \partial_i F(0)$
Hm, something's off about what I did earlier.
 
Stupid Questions Inc
Stupid Questions Inc
@BalarkaSen lol no exam? So everyone's gotta pass every course?
 
Balarka Sen
Balarka Sen
They should pass everyone, right?
I know some big colleges in my country did that
Just scaled midterm scores and put up some grades
@BalarkaSen Ok, here are the corrections. (1) $x_1 := y_1 \circ f, \cdots, x_{n-1} := y_{n-1} \circ f$ (2) $f: \Bbb R^m \to \Bbb R^n$, $f(x_1, \cdots, x_m) = (x_1, \cdots, x_{n-1}, h(x_1, \cdots, x_m))$ (3) $h(x_1, \cdots, x_{n-1}, 0, \cdots, 0) = 0$, $\partial_i h(x_1, \cdots, x_{n-1}, 0, \cdots, 0) = 0$ for all $m \leq i \leq n$. Think that's it.
 
Stupid Questions Inc
Stupid Questions Inc
6:10 PM
@BalarkaSen i wonder how that will work next semesters which apparently will still need to be done online
@BalarkaSen btw do u think they will organize JEE exams this year?
 
Balarka Sen
Balarka Sen
No clue
 
user462942
@robjohn this generalized cross-product you described evokes thoughts of the gram-schmidt algorithm for turning a linearly independent set into an orthonormal basis -- although in the algorithm, it's the inner-product that's used. Is there a relation between the cross-product and inner product spaces? If so, I've never heard of it ...
 
Mike Miller
Mike Miller
I will be a little surprised if the Math GRE remains very important in the US
Only a little
But I imagine it will become clear that those tests are not terribly useful in the admission process
 
robjohn
robjohn
@MikeMiller I aced the Math GRE in 1981. I don't know what they are like these days.
 
Balarka Sen
Balarka Sen
I will be happy if the admission process comes crashing over in India, it's basically junk. I imagine GRE is similar.
 
user462942
6:15 PM
@MikeMiller Yeah, I've heard it directly from the director of PhD admissions in our math dept -- a top-ranked dept. for many years now. They can't understand why they admit "strong" students, based on the math GRE, and then see poor research ability later on.
 
user462942
@MikeMiller the students admitted also pass written qualifying exams with flying colors, I've been told.
 
topologicalorientablesurface
topologicalorientablesurface
guys, does anyone have an idea on this
0
Q: Boundary points and manifolds

topologicalorientablesurfaceLet $M$ be an $n$ manifold with boundary. Fix $p\in U$. Let $(U,f)$ be a boundary chart, for which $f(p)\in f(U)\cap \partial{\mathbb{H}^n}$. Then, for any other boundary chart $(V,g)$, such that $p\in V$ we have $g(p)\in \partial{\mathbb{H}^n}$. I've been told that this follows from the invaria...

general-topology manifolds
 
Balarka Sen
Balarka Sen
Please don't repost, that increases clutter on the chat. If someone wants to help, they will.
 
topologicalorientablesurface
topologicalorientablesurface
okay.. but new people joined the chat? does every one scroll up and see old posts?
 
Balarka Sen
Balarka Sen
I starred your earlier link for visibility
 
topologicalorientablesurface
topologicalorientablesurface
6:22 PM
alright, thanks
 
Ted Shifrin
Ted Shifrin
@Joanna The issue is that nothing really predicts research ability other than experience with it. Acing the GRE and doing well on exams give some indication about background knowledge. I would argue that's a necessary ingredient (although some of my colleagues disagreed), but it is far from sufficient.
 
Leaky Nun
Leaky Nun
@BalarkaSen oh I didn't follow
also we have chess chatroom now
 
Balarka Sen
Balarka Sen
Oh right
 
user462942
@TedShifrin I see
 
Mike Miller
Mike Miller
Guess we'll see if next year's crop has a noticeably worse background.
 
Ted Shifrin
Ted Shifrin
6:24 PM
Howdy, triple A.
@topologicalorientablesurface: Did you look up the statement of invariance of domain? I think the statement makes it pretty clear. (Is a neighborhood of a boundary point in $\Bbb H^n$ open in $\Bbb R^n$?)
 
Balarka Sen
Balarka Sen
Hi @Ted
@LeakyNun He's having a Twitter breakdown
 
Leaky Nun
Leaky Nun
oh wow lemme see
 
Balarka Sen
Balarka Sen
It's hilarious
Tweet xQc > Magnus at him
 
user462942
@TedShifrin What benefits does the title Emeritus come with? Do you still get an office and salary?
 
user462942
I notice our professors emeritus still have offices -- just not the big corner offices anymore.
 
Ted Shifrin
Ted Shifrin
6:30 PM
No salary, no office in general (although some of the emeritus faculty get a small office to share). But this depends on the place, I suppose. Library privileges is the main thing.
 
user462942
I see
 
Captain Bohemian
Captain Bohemian
I have some questions about Sylvester's law of inertia, hoping someone can help me clarify them.
 
Ted Shifrin
Ted Shifrin
Shocking what actually looking up a statement of a theorem will tell you. :D
Howdy, @CaptainBohemian.
 
topologicalorientablesurface
topologicalorientablesurface
This is what I have so far, @TedShifrin
Claim: Let M be a manifold with boundary. $p\in \partial{M}$ implies that $p$ is mapped to $\partial{\mathbb{H}^n}$ under any chart
 
Captain Bohemian
Captain Bohemian
@TedShifrin what do you mean? I am reading the Wikipedia - Sylvester's law of inertia and have some questions about it.
 
Ted Shifrin
Ted Shifrin
6:34 PM
No, I didn't mean you, @CaptainBohemian.
Sorry about the confused adjacency.
 
topologicalorientablesurface
topologicalorientablesurface
attempt:
$p\in \partial{M}$ implies that there exists a chart $(U,f)$ such that $f(U)$ is open in $\mathbb{H}^n$ and $f(p)\in \partial{\mathbb{H}^n} \cap f(U)$. Let $(V,g)$ be any chart st $p\in V$. By invariance of boundary/dimension, $V$ must be a boundary chart. Where $g:V\rightarrow N$ is a homeomorphism, where $N$ is open in $\mathbb{H}^n$. and $g(V)$ is open in $\mathbb{H}^n$ and $g(V)\cap \partial{\mathbb{H}^n}$ is non-empty. Even with your hint above,
I don't see why
$g(p)\in \partial{\mathbb{H}^n}$
 
Balarka Sen
Balarka Sen
@LeakyNun All that training with GM xQc paid off
 
Captain Bohemian
Captain Bohemian
first, I don't know why Sylvester's law of inertia uses the term inertia. I think the theorem doesn't involve something related to inertia in physics. or does that inertia refer to invariance of signature?
 
user462942
@TedShifrin Sometimes, when a professor emails me with good intent and is very helpful and thoughtful with their response, I don't respond to it because of anxiety. I bet it comes off cold and rude. I'm not sure how to counter such anxiety. One thing that I thought of was to simply respond as soon as I read something, but then the terse response I send back doesn't match the efforts of the professor, ya know? I feel bad ...
 
Ted Shifrin
Ted Shifrin
@topologicalorientablesurface: Restrict domains of charts so that they agree on $M$ and consider $g\circ f^{-1}$. If this is a homeomorphism from a neighborhood of a boundary point in $\Bbb H^n$ to a neighborhood of an interior point in $\Bbb H^n$, what does invariance of domain tell you? (Hint: Perhaps you want to consider the inverse of this homeomorphism.)
@Joanna: Write something, let it sit an hour, revise it, and then send it. A sincere thank you is appreciated, and if you have a follow-up question or appreciate something specific, saying so is cool.
@CaptainBohemian: I think it comes from the moment of inertia tensor in physics, yes.
But, yeah, you could say that the signature is inert (i.e., independent of coordinates).
 
user462942
6:42 PM
@TedShifrin Ok thanks
 
Thorgott
Thorgott
Let $\Phi\colon S^2\setminus\{(0,0,1)\}\rightarrow\mathbb{R}^2$ be the stereographic projection. I'm asked to calculate $\Phi^{\ast}(dx\wedge dy)$. We have $\Phi^{\ast}(dx\wedge dy)=\Phi^{\ast}dx\wedge\Phi^{\ast}dy=d\Phi^{\ast}x\wedge d\Phi^{\ast}y=d\Phi^1\wedge d\Phi^2$ ($\Phi^1,\Phi^2$ being the components of $\Phi$). This is too easy to be the answer, but I'm not sure what more is there to be said; after all, this is precisely the frame given by the chart. What am I missing?
 
topologicalorientablesurface
topologicalorientablesurface
@TedShifrin i'm a bit tired, but why does such a domain exist?
 
Thorgott
Thorgott
 
Ted Shifrin
Ted Shifrin
Intersection @topologicalorientablesurface.
 
Balarka Sen
Balarka Sen
You're supposed to write the form in terms of the ambient $x, y, z$ coordinates of $\Bbb R^3$, in which $S^2$ sits as a unit sphere.
 
Alessandro Codenotti
Alessandro Codenotti
6:45 PM
Hi everyone
 
Balarka Sen
Balarka Sen
Hi @Alessandro
 
user462942
hi Alessandro
 
Ted Shifrin
Ted Shifrin
Hi, demonic.
 
Captain Bohemian
Captain Bohemian
@TedShifrin is the moment of inertia tensor has something to do with Sylvester's law of inertia? I haven't used the moment of inertia tensor for long.
 
Ted Shifrin
Ted Shifrin
@Thorgott: You need the actual formula for $\Phi$.
I'm just suggesting that it's a natural symmetric bilinear form, which is what Sylvester's law is about. @CaptainBohemian
Why do we call $f$ so that $\nabla f = \vec F$ a potential function of $\vec F$? This too comes from physics, but it's off by a negative.
 
Captain Bohemian
Captain Bohemian
6:47 PM
@Thorgott thank you. Let me read this thread to see what it says.
@TedShifrin but there are other symmetric bilinear forms in physics, like stress-energy tensor, metric tensor.
 
user462942
@TedShifrin does the Jordan canonical form show up much in higher-level math? In applied math, it never shows up. I miss abstract linear algebra.
 
Ted Shifrin
Ted Shifrin
It absolutely shows up in dynamical systems (ODE), @Joanna. And in representation theory.
 
Thorgott
Thorgott
@Balarka what does it mean to write a form in terms of ambient coordinates?
 
Ted Shifrin
Ted Shifrin
@CaptainBohemian: Fair enough. Another argument might be that even the moment of inertia is positive-definite, so signature isn't very interesting.
He means what I said above, @Thorgott.
 
Leaky Nun
Leaky Nun
@BalarkaSen lmao
 
user462942
6:55 PM
@TedShifrin Is representation theory a natural next step for someone who likes the abstract linear algebra and knows it all the way through to the end of a standard proofs-based book, e.g. knowing the Jordan form, SVD / polar decomposition? My prof. suggested that functional analysis was the natural next step.
 
Captain Bohemian
Captain Bohemian
@TedShifrin the second fundamental form is also a symmetric bilinear form.
 
Thorgott
Thorgott
I do have a formula for $\Phi$, I'm not sure how that relates
 
Ted Shifrin
Ted Shifrin
@CaptainBohemian: We're not here to list all possible symmetric bilinear forms.
Probably "inert" in the sense that it doesn't change (not matter how you change coordinates) is the right thing. What did Thorgott's link say Sylvester said?
@Thorgott: That's how you compute pullbacks. You differentiate the (coordinates of the) function.
 
Balarka Sen
Balarka Sen
Think of $\Phi$ as a function from $\Bbb R^3 \setminus \{x_1 = x_2 = 0, x_3 \geq 0\}$ to $\Bbb R^2$
 
Thorgott
Thorgott
the link says Sylvester had the nature of it being invariant in mind
ah, so you're saying to differentiate it in ambient space
and then restrict it back?
 
Balarka Sen
Balarka Sen
6:59 PM
Yeah
 
user462942
@TedShifrin do you do geometric representation theory?
 
user462942
^ not sure if that is a thing
 
Ted Shifrin
Ted Shifrin
No, I don't do representation theory at all, but it certainly shows up in basic ways in the mathematics I have taught and used (e.g., vector bundles and principal bundles).
 
user462942
I see
 
The Substitute
The Substitute
7:14 PM
Hi, are the non-crossing partitions on n elements isomorphic to all structures that correspond to Catalan numbers?
 
user462942
Time for some vegan chili - bye everyone. Thanks @TedShifrin :)
 
Ted Shifrin
Ted Shifrin
Bubye.
 
Balarka Sen
Balarka Sen
@BalarkaSen Let me continue from here. Apply this to the function $h : \Bbb R^m \to \Bbb R$ which satisfies $h(x_1, \cdots, x_{n-1}, 0, \cdots, 0) = 0$ and $\partial_i h(x_1, \cdots, x_{n-1}, 0, \cdots, 0) = 0$ for all $m \leq i \leq n$. Let $p = (a_1, \cdots, a_{n-1}, 0, \cdots, 0)$. Clearly, $\partial_i h(p) \equiv 0$ is the zero function for $i < m$, since $h$ is zero along $\{x_m = \cdots = x_n = 0\}$.
 
Ted Shifrin
Ted Shifrin
You're talking to yourself, a Balarka?
I assume your first $m$ needs to be an $n$?
 
Balarka Sen
Balarka Sen
Nah, $m \geq n$.
 
Ted Shifrin
Ted Shifrin
7:16 PM
Then your $m\le i\le n$ is wrong. Confuzling.
And your last thingy, too.
 
Thorgott
Thorgott
(this started by him talking to me; I'll read it after finishing all my homework)
 
Balarka Sen
Balarka Sen
$h$ vanishes along $\Bbb R^n \times 0 \subset \Bbb R^m$, and the normal partials vanish at a point. That's what I mean.
I did mean $n \leq i \leq m$
 
Ted Shifrin
Ted Shifrin
I'll ignore it all.
 
Balarka Sen
Balarka Sen
That's fine, Ted
 
Ted Shifrin
Ted Shifrin
Agh. Someone trying to do a masters in math just emailed me, asking me to explain how you compute the arclength of the usual parametrization of a circle. This isn't promising.
 
Balarka Sen
Balarka Sen
7:18 PM
You get $h(\mathbf{x}) = \sum_{i, j \geq n} h_{ij}(\mathbf{x}) x_i x_j$ for some functions $h_{ij}$, by applying the lemma twice.
Where $(h_{ij}(0))$ is the Hessian of $h$ at $0$.
Or rather, it's the $(m-n+1) \times (m-n+1)$ block in the Hessian of $h$ which is nonzero.
We do not know a-priori that this is a nondegenerate symmetric matrix, but if we do know this, we can change the coordinates $(x_n, \cdots, x_m)$ so that this nondegenerate symmetric bilinear form gets diagonalized (this is pointwise LDL), and $h(\mathbf{x}) = x_m^2 \pm \cdots \pm x_n^2$
(@Mike This is basically what a Morse-Bott function is, right? $h : \Bbb R^m \to \Bbb R$ with critical value $0$, $h^{-1}(0) = \Bbb R^n \times \{0\}$ and the normal Hessian $h$ is nondegenerate along $h^{-1}(0)$?)
 
Thorgott
Thorgott
Okay, let's see. Let $S\subset M$ be a submanifold of a smooth manifold, let $\iota\colon S\rightarrow M$ be the inclusion. Let $\omega$ be a $k$-form on $M$. The natural way to think of the tangent spaces of $S$ as subspaces of the tangent spaces of $M$ is via $d\iota_p\colon T_pS\hookrightarrow T_pM$ for each $p\in S$. So the natural way to get a $k$-form on $S$ from $\omega$ is via mapping each $p\in S$ to the map $w_p\circ d\iota_p\colon T_pS\rightarrow\mathbb{R}$.
This is actually just the pullback $\iota^{\ast}\omega$. So we have a map $\iota^{\ast}\colon\Omega^k(M)\rightarrow\Omega^k…
 
Balarka Sen
Balarka Sen
If this condition on the Hessian on $h$ is satisfied, we call the almost fold map $f : M \to N$ simply a fold map. We thus obtain that if $p \in \Sigma$, there exists coordinates around $p$ and $f(p)$ such that $f(x_1, \cdots, x_m) = (x_1, \cdots, x_{n-1}, x_n^2 + \cdots + x_{n+r}^2 - x_{n+r+1}^2 - \cdots - x_m^2)$
If $n = 1$, this is Morse lemma.
 
Mike Miller
Mike Miller
7:34 PM
@BalarkaSen You're getting constancy of normal Hessian in a chart from non-degeneracy? That's true but requires proof
Basically that one can not just take a quadratic form into standard form but that one can do so smoothly
 
Balarka Sen
Balarka Sen
I don't do the flow proof. I wrote down a generalization of the standard proof
Using this lemma
 
Mike Miller
Mike Miller
I am not looking right now, but I would be surprised if the Morse lemma proof gave actual constancy of normal Hessian without more work.
It should give you a chart where the function is normally quadratic but the normal quadratic guy varies
 
Balarka Sen
Balarka Sen
I think this works
 
Mike Miller
Mike Miller
Fine I'll look later to see where the error is
;)
 
Balarka Sen
Balarka Sen
You won't find it
 
Pig
Pig
7:40 PM
@Thorgott it's surjective, by a bump function type of argument (extends by zero outside a small tubular neighborhood of $S$)
 
topologicalorientablesurface
topologicalorientablesurface
How would I prove the following: Let $U\subseteq\mathbb{R}^n$ be open and $S\subseteq \mathbb{R}^n$ arbitrary. If $f: U\rightarrow S$ is a homeomorphism then $S$ is open in $\mathbb{R}^n$
?
 
Balarka Sen
Balarka Sen
Well, you need to extend on the tubular neighborhood first, for which you pullback by the natural projection $NS \to S$
I wasn't sure if Thorgott knew the tubular neighborhood theorem
 
Thorgott
Thorgott
I don't, Mike was gonna tell me about it once I have a bit more time
I feel like this should be true regardless. One should be able to write down a "local preimage" in charts and glue those together with a partition of unity. I'm trying to see whether that works formally atm
 
Mike Miller
Mike Miller
Nah I was going to tell you about the equivariant tubular neighborhood theorem
You have all the tools and cleverness to prove the regular tubular neighborhood theorem yourself
You should do that actually
 
Pig
Pig
hmm actually he's probably right
if I only take the special charts so that $S \to M$ looks like $(x_1,\cdots,x_m) \to (x_1,\cdots,x_m,0,\cdots,0)$
then the "same" differential form should already extends
is that true?
 
robjohn
robjohn
7:49 PM
Just watched the Falcon 9 launch. The first stage was back and landed less than 12 minutes after launch!
 
Thorgott
Thorgott
Let $\omega\in\Omega^k(S)$. For each $p\in S$, choose an adapted chart $(U_p,x^1,...,x^n)$ of $M$ around $p$. Then $\omega=\sum_Ia_Idx^I$ on $U\cap S$, where $I$ ranges over ascending multi-indices from $1$ to $m=\dim S$. Then $\omega_p=\sum_Ia_Idx^I$ is a smooth $k$-form on $U$ already.
Let $\{\rho_p\}$ be a partition of unity subordinate to the atlas $\{U_p\}$. Put $\omega=\sum_p\rho_p\omega_p$. Then $\iota^{\ast}(\sum_p\rho_p\omega_p)=\sum_p\rho_p\iota^{\ast}\omega_p=\sum_p\rho_a\omega=\omega$.
this may be too sloppy
 
a perspicaciously curious mind
a perspicaciously curious mind
@robjohn the US finally made it back into space.
 
feynhat
feynhat
wait, did you write p as the index for the partition of unity?
 
Thorgott
Thorgott
yes
 
feynhat
feynhat
oh right. that makes sense.
 
Thorgott
Thorgott
7:55 PM
I should write a restriction in that last line, I think
right, it's $\iota^{\ast}(\rho_p\omega_p)=\iota^{\ast}\rho_p\iota^{\ast}\omega_p=\rho_p\vert_S\omega_p$
but they still sum to $1$
so I believe this works
 
robjohn
robjohn
@aperspicaciouslycuriousmind I love watching space exploration, and it's been way too long since we've had a launch.
 
a perspicaciously curious mind
a perspicaciously curious mind
yup, me too
just reruns on YouTube :(
 
02:00 - 20:0020:00 - 00:00

« first day (3587 days earlier)      last day (1425 days later) »