stabilizer of a point $x \in X$ under action of a group $G$ is the collection of $g \in G$'s which fixes $x$. kernel of the action is the collection of $g \in G$'s which fixes everything of $X$ pointwise. in other words, kernel of a group action is intersection of all the stabilizers.
$\varphi$ is the permutation representation of the action. if $G$ acts on $X$, it's the homomorphism $G \to \text{Aut}(X)$ which takes a $g \in G$ to the automorphism $\varphi_g : X \to X$ which sends $x \to gx$.
so it doesn't make sense to say $\varphi$ maps to $G$.
hi I have a question I know that Probability the three points on a circle will be on the same semi-circle is 3/4. what can be said about Probability the three points on a circle will be on the same quadrant (1/4 of circle)?
@Obliv "act as an identity for the permutated elements of X" is a slightly odd wording, but yes. It's the set of elements $g \in G$ such that $gx = x$ for all $x \in X$.
In other words, elements of $G$ with permutation representation being the identity permutation.
and the stabilizer of that homomorphism is just a single element in $G$ that does the same with the permuted elements? So the kernel is indeed the intersection of all stabilizers.
namely, if I have convex conjugate functions $f(x)$ and $f^*(x^*)$, then what's the convex conjugate of $f(-x)$? I'd think it's just $f^*(-x^*)$ but wanted to be sure
I may have missed a couple, do you recommend me to do all? The ones I felt was nontrivial I did from the transversality section, and most of the nontrivial exercises from the mod 2 chapter seem to follows from my self-intersection vs trivial tubular nbhd ideas. I haven't looked at the exercises from the J-B theorem chapter yet, but I just proved the (smooth, ofc) Jordan-Brouwer myself.
Did you read G&P's proof of Borsuk Ulam? I think it's amazing. Plus you get the generalization of the winding number on $\partial W$ counting roots inside $W$.
okay. The Legendre transform $f^*(x^*)$ of a convex function $f(x)$ is obtained by maximizing the expression $x x^* -f(x)$ for all $x$ as a function of $x^*$
So prove this, @Balarka (It's true without the mod 2 stuff once you use oriented intersection #): Suppose $W$ is a compact $n$-manifold with boundary and $f\colon W\to\Bbb R^n$. Suppose $f\ne 0$ on $\partial W$. If $0$ is a regular value of $f$, prove that the number of roots of $f$ mod $2$ is equal to the mod-2 degree of $f/|f|\colon \partial W\to S^{n-1}$.
I'll admit, this is mostly because I have an inconvenient minus sign showing up and I want to get rid of it without screwing up the rest of my argument :p
I encountered a Legendre transform (which came from physics, as I recall) when we were working in singularity theory 20 years ago, but I've forgotten it all. (Tangent/cotangent stuff.)
@TedShifrin This is actually a research question, btw. It's coming up because I'm ending up with an expression like $p\phi'(p)-\phi(p)$ where $\phi(p)$ is itself a Legendre transform i.e. $\phi(p)=\text{max}_H (pH-\Phi(H))$ where $\Phi(H)$ is a convex function
@TedShifrin: basically "whatever I taught them", so calculus (including diff eqs), a bit of linear algebra and vector geometry, prob, stat, complex analysis
Start with something from exams that (a) is important and (b) you're confident they know how to do. Then go from there, depending on the student. You might have a list of easy questions (that are relevant) and a list of harder questions. Given the time frame, you're not going to get through much. You just want to see that the student can handle basics.
@TedShifrin: I basically prepared a list of easy questions, and solved them today. after I solved them I added some "harder part" to them, depending on how fast a student solves it. I'm still worried that it won't be enough for a very good student, or that I won't be able to help, if a student doesn't know anything
@Huy: Ideally, you want very uncomplicated questions to start, or else you'll end up spending the whole time on one question. Keep calculations very simple.
For a good student, have a few more interesting questions ... not necessarily related to the original one or two.
Good to cover a few different topics since your range is so broad.
@TedShifrin In case you didn't see my proof of T1S^2 being SO(3) yesterday: SO(3) acts on S^2, and the action induces an action on T1S^2 by just moving the unit vectors. This is (1) free because you only have to worry about the axis the isometry fixes, i.e., two points, but clearly the circle fiber rotates at those points (2) transitive because for any two pts (x, v) and (y, w), I can take (x, v) to (y, v) by an isometry first - move along a great circle and then rotate to get (y, v) to (y, w).
So fix any pt (p, v) on T1S^2, and multiplication by an element of SO(3) with (p, v) is the desired diffeo with T1S^2.
@TedShifrin: I intend to help them with calculations, i.e. after they wrote down the complicated expression, I'll tell them "okay, so this simplifies to ... and how would you proceed?"
@Huy: Better to avoid something complicated in the first place, but ok.
For a good student, I'd bias toward something hard in their strength. For a poor student, I'd bias toward something easy in their weakness to see if they learned it finally.
Did you send it to my old math address or to gmail, @Pedro? (Not that it should matter.)
@Balarka: Yes, what I wanted you to say is this. If $e_1$ is a point on the sphere and $e_2,e_3$ is an orthonormal basis for $T_{e_1}S^2$, then $e_1,e_2,e_3$ is an orthonormal basis for $\Bbb R^3$ and hence defines an element of $SO(3)$ (if you orient correctly).
Hi all! I have a silly question. Does anybody know what a "laplace operator in the Aleksandrov sense" is? Maybe it has something to do with convex analysis...
It's only a suggestion, @Huy. I have only done oral examinations officially at the graduate level. But I did stuff with students in office hours all the time, so I'm basing it on that intuition.
Basically, you should know what grade you're giving. If there was a screw-up on the written on something basic, give the kid a chance to show he's learned it (and improve his grade). If the kid is smart, push him a little, but knowing you're going to give high marks anyhow.
@MikeMiller This reminds me that I had a question. Suppose I look at H^1(M; Z/2). That's in bijection with hypersurfaces in M, by Poincare duality. Also, we proved earlier that H^1(M; Z/2) is in bijection with line bundles on M - 'cause for any line bundle on $M$, equip it with a Riemannian metric and take the unit S^0-bundle, i.e., a 2-cover, which are classified by homomorphisms pi_1(M) --> Z/2.
So is there any direct way to see that line bundles on M are in 1-1 correspondence with hypersurfaces in M? Something something take a section and look at it's zero set (also, ping @TedShifrin)
Given a hypersurface, how do you construct the cover? Pick a tubular neighborhood of the hypersurface. Call the neighborhood $N$ and its boundary $B$. Because $B$ double covers $S$, it carries a deck transformation $\iota$. Now glue $(M \setminus \text{int} N)$ along to itself, sending the boundary $B$ to itself by the involution $\iota$.
namely, suppose $f^*(x^*)=\text{max}_x[xx^*-f(x)]$. then $f^*(-x^*)=\text{max}_x[-xx^*-f(x)]$. but that's not different than $\text{max}_x[xx^*-f(-x)]$
Once you figure out how this one-one correspondence works, see if you can prove that every element of $H^1(X;\Bbb Z/2)$ is realized as an intersection number homomorphism, hence that $H^1(X;\Bbb Z/2) = 0$ iff your favorite theorem holds.
@BalarkaSen It's more or less the same thing as the tangent bundle. You just decide that the key idea is frames, not vectors. (From a different perspective, it's even more true that they're the same thing.)
Basically, for surfaces, one tangent vector is everything, @Balarka. But in higher dimensions you need to keep track of how bases change to understand geometry.
Does anyone know any interesting or difficult equivalence relations problems similar to the last one here? (The link directs to a .pdf on UIUC's website.)
@J.M. Alright. I'm not planning on making any similar tridiagonal matrices then. It doesn't seem like Lanczos is what I need. I'm basically finding the eigenvalues of a hessenberg form non-symmetrical matrix. Since I only plan on finding real solutions, I think a single-shift QR will work. It's just I'm not getting good results with the wilkinson shift I am using and so wondered if it was because I hadn't tridiagonalized beforehand.
@BalarkaSen For silly reasons. Notational, really. In general any element of $Ext^n(M, N)$ can be thought of as being represented by a map $\Omega^n(M) \to N$, where $\Omega^1(M)$ is the kernel of a chosen surjection from a projective to $M$ and then one iterates. It kind of reminds me (likely due to notation) of choosing representatives of elements in $\pi_n$. I suspect I will grow out of these thoughts within the next hour.
Could someone explain the following line to me: "For unitary matrices in case the shift λ following the usual rule is null, we take it to be any nonzero number with modulus equal to unity"
modulus is another word for absolute value. so i suppose it doesn't have to be complex-valued (could be \pm 1) but that seems unnecessarily complicated otherwise
though I honestly don't know what their statement means by the 'shift' lambda
Does anyone know what the name of > **Abstract binomial theorem**. Assume that we have:
> 1. A vector space (or even a module) $V$. > 2. A family of "general terms" $(T_{i,j}\in V)_{i,j\in\mathbb N}$. > 2. A linear operatior $F: V\to V$, such that $F(T_{i,j}) = T_{(i+1),j} + T_{i,(j+1)}$.
> Then $F^n(T_{0,0}) = \sum_{i=0}^n \binom{n}{i} T_{i,(n-i)}$.
For polynomials $f,g \in \mathbb R [X]$ of degree $n$ we can say that they are equal of they coincide on $n+1$ points. Is there a generalization to $f,g \in \mathbb R [X_1, \ldots, X_m]$?
Have you ever seen Euler's proof of Fermat's little theorem? It's quite nice
It hinges upon the fact that $(a+1)^p-a^p-1$ is a multiple of $p$ for all $a$ and prime $p$, from the coefficients of the expansion.
Clearly, $1^p-1$ is a multiple of $p$. Adding $2^p-1^p-1$ (a multiple of $p$ by the above), we get $2^p-2$ is a multiple of $p$. Adding $3^p-2^p-1$, we get $3^p-3$ is a multiple of $p$. And we do this until we get to $a$, getting $a^p-a$ is a multiple of $p$, and — since $a$ is not a multiple of $p$ — $a^p-1$ is a multiple of $p$. QED.
