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18:07
@geocalc33 This might surprise you!! Differential geometry for the win.
If I'm trying to prove that the median $\mathbf v$ is where the minimum value of $\sum_{i=1}^{3}\Vert \mathbf x- \mathbf a_i \Vert$ is at (with $\mathbf a_i \in \mathbb R^2$) but I can only prove that $\frac12 f(\mathbf v) \le f(\mathbf x)$ for all $\mathbf x$, what did I prove? :/
It looks like I only proved that there is a function with a smaller value than f(x)... not a big deal.
however it shows that there IS a min...mhh
18:24
@Simone: What do you mean by median $\mathbf v$?
$\frac13 (a_1+a_2+a_3)$
Oh, the centroid — the intersection of the medians. What makes you think that's the minimum point?
intuition
XD
Um ...
Without the compactness argument, can't you actually use calculus to solve for the minimum point?
@TedShifrin I could but it's a nightmare
18:27
Oh?
You should at least get a geometric characterization of the minimum point, even if you cannot literally solve for the point.
Well the way I showed that 1/2 f(v) < f(x) is through triangle inequality and some adding and subtracting the same terms shenanigans inside the absolute values
I'm telling you that the centroid is not the minimum point in most triangles.
oh?
You could do some actual numerical experimentation.
interesting...
I wouldn't know what could possibly be the min then... should I really calculatethe derivative?
18:31
If you've paid attention to the text, you should know the derivative quite quickly.
How dare you XD
@TedShifrin Square roots! Bah!
shrug
Writing things in coordinates is often not useful.
I can find the derivative, but it has many many terms
You're not thinking.
Hint: It's the sum of three functions. If we've read the text, we know what the gradient of each of those functions is.
18:35
ahhhh right
I forgot
thx
@MikeMiller Literally my favorite example (which gets labelled the anteater example in my lectures) :P
It's like with the ellipse example! silly me
Well, the main example precedes that.
yeah the and example
or was it a bee I can't remember rn
@TedShifrin wow that's neat - differential geometry for the win indeed
18:43
I must be missing something, the computation of critical points is nontrivial when you're taking distance and not squared distance.
Who said it was trivial.
LOL
Shush!!!
lmao
I thought I was the only one who wasn't finished.
To quote me: "You should at least get a geometric characterization of the minimum point, even if you cannot literally solve for the point."
I will comment that this problem has a venerable history, going back centuries.
18:47
@TedShifrin really? neat
Yes. I'm not giving away more right now because of the power of google.
dang I must hurry then XD
I can't remember the name of this point, but I won't comment.
This is also a baby version of a classic question in geometric measure theory (akin to the soap bubble problem).
Say I define a compact, smooth manifold $M$ to be orientable if its tangent bundle $TM$ is orientable as a vector bundle (i.e. by consistently specifying an orientation on each tangent space). Is there a direct way of describing how this orientation characterizes the fundamental class of the manifold? (of course, I can step by step prove the equivalency of this definition with the homological definition, but the process is tedious and I'd like to avoid talking about local homology)
18:53
So you're not willing to discuss nowhere-vanishing $n$-forms?
You don't want local homology; so what do you want to determine the fundamental class?
What exactly do you mean by a characterization?
Exactly :)
Well, not exactly, cuz the volume form can't be exact. smacks self
a description of how an orientation determines a generator of the top homology (already knowing that it is Z)
So are we allowed deRham or simplicial?
@Simone it is straightforward using the subgradient.
You have $0 \in \partial f(x)$ at a solution.
18:57
No more hints, @copper.
And that's too fancy for this level, anyhow.
sry.
i guess rademacher is off the table.
She knows what the gradient of the function is.
@TedShifrin luckily IDK what a subgradient is :P
I know what the gradient is alright, and I'm male XD
Actually, @copper's remark is only relevant if the optimizing point happens to occur at a point where the function isn't differentiable.
Oops. Sorry.
Since when is Simone a French male name?
I think I have interacted with Simone before.
18:59
I'm Italian
So, I think it is certainly in his bag of tricks.
Oh, dopey.
(That was to me.)
My apologies, Simone.
My 2nd last boss was an Andrea. Male, from IT.
@TedShifrin haha, no problem Ted :D
Yes, yes. I thought I had remembered some French conversation with Simone, whence my confusion.
19:00
Andrea is a male name in italy, yes
Yes, and I know Simone is a male Italian name. I just had the French association. Scusi.
Maybe it is better for everyone to have userXXXXXXXXXX names and be anonymous :D
nessun problema :P
Anyhow, back to the classic math problem.
@TedShifrin Ack! that would be truly confusing.
I know, @robjohn. I can't keep them sorted in my mind at all.
19:04
bbl
To provide a bit more context, I'm gonna give a seminar talk where I will have to use fundamental classes and since the audience is not necessarily familiar with them, I will introduce them as a black box. The audience is not necessarily familiar with the notion of orientation of a manifold, but they're familiar with the notion of an orientation of a vector bundle. Since we only care about smooth manifolds, I think defining orientability as orientability of the tangent bundle is the most appropriate definition. The reason I'd like to avoid local homology is cause I think it's the least intu
@Thorgott So the problem is you need to say what an orientation is and what homology is. Whatever your homology is, this description somehow needs to come from a passage to "local", because an orientation is local information. One thing you might like is that an orientation of a connected orientable manifold is a choice of one of the two isotopy classes of embedding $D^n \to M$. So somehow what you are saying is "I know what the fundamental class of $D^n$ is...
I know that the fundamental class of $M$ is determined by its restriction (via the diagram in homology of pairs (M, emptyset) -> (M, M - {x}) <- (D^n, D^n - {0})) to any embedded $D^n$. So by choosing which generator of $D^n$ is the standard one, and choosing an orientation of $M$, you choose a fundamental class of $M$ via what it looks like on $D^n$.
So are we rejecting the simplicial, which is the most intuitive? The orientation on a tangent space gives you the orientation on the face of the simplex over which the manifold is locally a graph.
This corresponds to the above description because an orientation of the tangent space $T_x M$ is the same as a choice of path component in the space of isomorphisms $\Bbb R^n \to T_x M$.
So gives you a privileged collection of disc embeddings, or privileged ordering of vertices on a simplex in Ted's picture.
Ah, I figured out the name of that point.
Ssshhhh.
19:13
@TedShifrin I can't solve for x using the derivative, but I was thinking:
If the gradient is the vector that points in the direction in which the function icreases the most
then it's opposite points in the direction in which the function decreases th most
Yes, of course.
so the point will be the sum of the three gradients of the functions $\f_i(x)= \Vert x-a_i \Vert$
19:16
dang it
How is the sum of three vectors determining the point?
sorry the sum of the opposites
I still don't understand what you're saying.
(-1)gradient of f
Yes, I understand that. How am I determining the minimum point from your comments?
19:20
Yeah, you don't :/
at best it tells you the direction
From where?
from x
So I said hours ago that you should be able to give a geometric characterization of the minimum point, not solve for it.
I'm trying...
You've lost the trees in the forest. What should the gradient vector be at a critical point?
19:23
the 0 vector
OK. So think about what that vector must be to get $0$.
It's when $\sum_{i=1}^3 \frac {(x- a_i)}{\Vert x-a_i \Vert} =0$
Right. So I'm suggesting you think about how that can occur. What does it mean geometrically?
I said from the beginning you're never going to solve algebraically for $x$. But you can solve geometrically.
Thanks for the input
The simplical translation is probably the easiest one to grasp visually. But how, explicitly, does an orientation of the tangent space determine an orientation of a simplex in general? If it's embedded in R^n, I can orthogonally project the manifold onto the tangent space and this will be a diffeomorphism in a neighborhood of the point, so that works out, but I feel like in general this would require choosing charts (which can be done of course).
Can a covering map have a critical point?
19:37
@TedShifrin When the sum of the three unit vectors cancel out
But I really can't think of a case when three vectors of equal lenght
@Emolga you mean a covering map between manifolds?
@Thorgott Yes, although I need the very concrete case of a holomorphic covering map between two domain. Is it true that its derivative cannot vanish?
because I can't prove this
a smooth covering map is a local diffeomorphism
really?
yes, pretty much by definition
19:42
Thank you
"He be workin'" has a different meaning than "He workin'"
@Thorgott In a chart where the simplex is an affine simplex with one vertex v_0 at x, you have the oriented basis of T_x M given by the displacement vectors {v_0 v_1, v_0 v_2, ..., v_0 v_n}.
This gives a recipe to go back and forth between ordering vertices and orienting bases of tangent spaces.
@Simone This is the exercise, then. How can three unit vectors sum to $0$?
And, @Simone, does this happen at the centroid?
20:04
@TedShifrin when they form an angle of $\frac23 \pi$ between them
Right!
There's a unique point with that property. Is it always in the interior?
If we include the frontier points then yes
20:28
right
weird point, I would have bet my house it would have been the centroid, but it's not.
what's the name of this problem?
20:48
more than frontier points... it can actually be one of the vertices.
21:02
Yes. If it's on the frontier, it will be a vertex. When does that happen? This is called the Fermat point of the triangle.
when the two sides are also at $\frac23 pi$
Fermat point of the triangle... thank you Ted
What if the angle is greater at the vertex?
@TedShifrin are we done btw? The gradient being 0 has a solution (when the unit vectors are at 120° from each other), the solution is unique (the rotation matrix is a function so it has one solution).
@TedShifrin oh in that case it is outside
It can't be outside !!
DANG IT!
I suck at thinking about shapes
21:08
You can still make a compactness argument that there must be a minimum in the triangle (including frontier). It's a little trickier because the function isn't differentiable at the vertices (this gets back to what copper was mumbling about).
The Wikipedia page has a more analytic proof in the vector section
I couldn't have gotten there on my own
Viviani published the solution in 1659... wow
21:23
Good morrows to all................ Half day of Stats and half day of geometry on the way.....**rubs hands**
goodnight
Buon natale or noite in Italian?
buon natale means merry christmas XD
noite is portugese I think?
Almost right on time
I had a feeling..... yes noite is portuguese ...I'm trying to guess the correct word off of my spanish
well I know giorno is day.......
21:26
it's close... buona notte :P
hmmmm, I would've never guessed that........those words put together is a last name in Argentina
oh the Italian community in Argentina is huge
indeed
Are you from Argentina?
if so you probably have an italian last name as well :D
r9m
r9m
21:42
@robjohn It looks more like a mean square with a heart ;)
No, not from Argentina. Just speak Spanish and quite worldly.
awesome
well: night night peeps
22:36
@dc3rd espanol noche = italiano notte
natale es la navidad
23:00
I forgot, was there a coordinate-free characterization of when a critical point of a smooth function on a manifold is non-degenerate?
Yes. You can look at the intrinsic derivative of $df$ as a section of $T^*M$.
23:12
So say we have $A$ a noetherian domain of dim. 1 and we're trying to show that $A_M$ having the property of unique factorization for all ideals $M \in Max(A)$ implies $A$ has the property.

So for $I$ a nonzero proper ideal of $A$, we can get a finite set of maximal ideals $\{ M_1, ..., M_s \}$ of $A$ that contain $I$. We consider all the maps $\varphi_i : A \to A_{M_i}$ and use our assumption to say that $\varphi_i(I) = (MA_M)^{a_i}$ for a unique $a_i >0$. This being so, we can write $I \subset \varphi_1^{-1}((M_1 A_{M_1})^{a_1}) \cap \dots \cap \varphi_s^{-1}((M_s A_{M_s})^{a_s})$.
I think this was one of those where writing everything out cleared things up for me
@Thor: So it's just a transversality check. No coordinates needed to define, but how to decide in a specific case.
23:23
So how do I see these definitions are equivalent? Is this a coordinate computation?
What is your definition of nondegeneracy? That's going to have coordinates, I bet.
23:46
Ok, let's see. Pick coordinates $x=(x_1,...,x_n)$ on $M$, then you get coordinates on $T^{\ast}M$ coming from $x_1,...,x_n$ in the base and picking out the coefficients relative to the basis $dx_1,...,dx_n$ in the fibers. In these charts, $df\colon M\rightarrow T^{\ast}M$ looks as follows: take a point $p$ in Euclidean space, map it to $x^{-1}(p)$, take this to $df\vert_{x^{-1}(p)}$, the first $n$ coordinate of this are just $p$ again, but $df\vert_{x^{-1}(p)}=\sum_{i=1}^n\frac{\partial f}{\partial x_i}\Big\vert_{x^{-1}(p)}dx_i\vert_{x^{-1}(p)}$, so the last $n$ coordinates are just $\frac{
Of course, this misunderstanding is also present without coordinates: $df\colon M\rightarrow T^{\ast}M$ is a section, so its differential has to be fiberwise injective
Can a function have two natural boundaries like Re(z)=-1,1? Such that the function is analytic only in the strip?
Injectivity isn't transversality to the zero section.
The graph of a singular linear map is not transverse to the graph of the zero map.
so I guess my misunderstanding is that "intrinsic derivative" doesn't mean "derivative" in the regular sense
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