@DanielFischer: so if I look at the action via Möbius transformations SL2, for $gz := \frac{az+b}{cz+d}$ I have $g^{-1}z = \frac{dz-b}{-cz+d}$ and $(gz)' = \frac{1}{(cz+d)^2}$, so $(g^{-1}z)' = \frac{1}{(-cz+a)^2}$?
Hey @TedShifrin. I took the subject GRE today. Confidently answered 51 of 66 questions, and didn't guess on any of the remaining problems. Not sure if that's good or bad, but I felt decent. Probably could've done a bit better though.
OK. Well, the first question is on gradient. In exercise 3, you have asked to draw the steepest path up towards the mountain top. This can be just done by starting at a point in the bottom level set, and draw a bunch of gradients as you move through each previous gradient infinitesimally. This is your desired path.
@MikeM: It just came across my FB feed that the head of the Benghazi committee has been "caught using private email server." F***ing hypocrites, all of 'em.
BTW, you can avoid integration by considering $h(t)=t^{-k}f(tx)$ as suggested in the problem in the first place. Then you just show the function is constant because the derivative is $0$.
Hello @TedShifrin Could you take a look at math.stackexchange.com/questions/1493072/… I am confused about how we get only $B=Pb$ and not $B= \pm Pb$. Could you explain it to me?
OK, I don't seem to have any other question right at this moment. By the way, derivative of dot product of function just follows from writing them down componentwise, use the algebraic formula for dot product to write it as sum and product of function from $\Bbb R^n \to \Bbb R$, and use sum and product formulas, right?
By the way, @TedShifrin: bananas was complaining about PDE's he'd have to do in his functional analysis and diffgeo classes this semester. I linked him to the lecture video where you talk about applications of PDE. I am sure he will be motivated by the minimal surface equation :)
@Balarka: I guarantee that when you get through the inverse function theorem stuff, you'll love transversality and differential topology, not even mentioning differential forms.
So, @user159870, what you put in your post is right to a point. We have $T=Pt$ and $N=Pn$. Instead of using cross-product, what happens if we use $N'=-\tau B$?
Oh, @Balarka, you meant just that one exercise you asked about?
@TedShifrin There's an 11 y.o. diffgeo guy here too, as I recently discovered. We're corresponding, and I think he may tell me a rough sketch of the proof of Fary-Milnor.
And then I'm done differentiating. After finishing off the exercises I left out, going to skip chapter 4 mostly, but I think you also have something to say on smooth manifolds in the last section of chapter 4, right?
Yes, it seems that $T=Pt$, $N=Pn$, and $B=(\det P)Pb$. If we do a reflection, a right-handed basis turns into a left-handed basis, so this has to happen.
I don't know a formula for $Pv\times Pw$, but I can figure it out by dotting with a third vector and then using the property of determinants.
@TedShifrin do you know the meaning of an arrow that looks like $)\longrightarrow$ but without the gap at the base of an arrow. It isn't a hooked arrow and is without any doubt a bracket shape and size, used on a diagram in the context of defining quotient modules? If you do, please share.
@AlecTeal If you share an picture, we might be able to clarify.
@AlecTeal Rings with $1$ are automatically additively commutative. =)
Proof $(1+1)(a+b) = (1+1)a+(1+1)b=a+a+b+b$ is also equal to $(a+b)(1+1)=(a+b)1+(a+b)1=a+b+a+b$, cancelling $a$ to the left and $b$ to the right gives $a+b=b+a$. TA-DA!
