@Ted Got a question for you about UGA. In the last few years were you aware of any confrontation among students or between students and the administration over issues like sexual assault, due process, etc?
or perhaps also trigger warnings, students objecting to material being presented in class
@Ted Okay, thanks. It was always an issue at Duke given the Lacrosse case and its been a big thing at a lot of places recently. But I never know if the fact that I never hear about it here at Tech is a Tech thing or a Georgia thing or maybe Im just ignorant
Well, you'll never know if I was forced out for inappropriate behavior ... will you? :D
A friend of mine who taught German was. They arrested him cross-dressing and hiring a male prostitute. Never should have been fired for that. But this is (was) Georgia.
Well, @Kevin, I went and kibitzed at today's duplicate bridge game. I'll play Saturday night. Very scary trying to remember all this stuff after 20 years.
@Ted This is what I can't understand about the movement on the left to empower campus administrators to enforce 'civility' and 'decency' among what the students and faculty say and do. These are the same groups that have done this kind of terrible shit in the past like fire someone for their sexual orientation. But in any case if you WERE forced out, that probably raises my opinion of you.
Theres veyr little intersection between the things deans will fire you over and things that I dislike
Walter Lewin . . . goodness. I felt so, so betrayed when I heard -- I was a fan. Goes to show how you shouldn't put people on pedestals or something like that.
@Kevin: I was totally kidding with regard to me. Although some right-wing religious parents have forbidden their kids to take classes from me (and others have because my courses are too demanding and my grades too low), no students have ever complained. Nor should they. I do not condone sexual harassment.
It seems I may need ot retire soon as well. I'm typing up some notes that I wrote out not 2 months ago. And every few lines I have to stop and spend several hours figuring out why what I said back then was true is in fact true
...lol, anyway, to see if temperature has a statistically significant affect on the final parameter (suppose the parameter description is unknown, but I have numbers for it- we could call it "y")
what analysis (or analyses) would you recommend I run on the data set?
actually, for each of the 9 different temperature and time combinations, there are 4 different observations, so I guess there would be 36 points
@StanShunpike Just like in the Riemannian case, you minimize over lengths of (broken) geodesics between two points. But the way to define length is with the pseudometric, and the lack of positive-definiteness means it's entirely possible for a given geodesic to have negative length, and when minimizing to get $-\infty$.
At least, that's roughly true. I know nothing about pseudo-Riemannian manifolds.
I feel like there should be a map $\pi_1(X)\to\Bbb \{\pm1\}$ where a loop is sent to $\pm1$ depending on whether dragging a local orientation around it preserves or reverses it when getting back to the original point. In fact seems like it should be a natural transformation to the constant functor $\Bbb Z/2\Bbb Z$ with identity map. Is this right, and does it have a name?
nice thing about east coast is that doing a wick rotation in time (i.e. analytic continuation to imaginary time) gives the usual euclidean metric in four dimensions
@TedShifrin Nah all you're doing is taking a problem that as-posed looks ill-defined in Minkowski space and then doing the 'equivalent' Euclidian space calculation. Then analtyically continuing hte result and hoping to God that it works
Oh Wick rotation.... how I barely understand why you work
@Stan you need an apostrophe. Ya'll.
its even in the wrong place since its supposed to stand for you all
@Stan Yea I meant discipline generally. Some of my favorites are philosophy, political philosophy, philosophy of science, linguistics, game design, and constitutional law.
@Fargle I don't know anything about what you've done, so maybe I'm speaking totally out of turn and am 100% wrong, but I hope you find that when you start doing research level work you know essentially nothing plus epsilon
@Lembik me neither, i have used it to post problems that i invented and knew the answer to but received some objections math.stackexchange.com/questions/68701/…
The book I am reading commutative algebra from says, at the very beginning, that the similarity between $\Bbb Z$ and $k[X]$ is a central point in commutative algebra. I am not sure I am convinced, as they never talk about a serious analogy other than how the ideals look like.
I should point out that I know the F_1 business, and don't understand all of it.
okay. so I guess the whole point is that $\Bbb Z$ is the ring of functions on $\mathsf{Spec} \Bbb Z$ while $\Bbb F_q[X]$ is the ring of functions on the affine line $\Bbb A_{\Bbb F_q}^1$. Guess I can live with that.
It just gives you a good depiction of how the group looks like. You guess by looking at the subgroup lattices of quotient groups that the first isomorphism theorem holds.
@Balarka also I had this question. They have not proved that a given lattice of subgroup of a group is the only one for it . I mean how you prove that it is the lattice for that group
@Rememberme As I said, it's very basic, you can't expect it to be used to prove huge stuff. But it's useful visualizing things. For example, the fundamental theorem of galois theory can be "seen" by looking at the lattice structure of Galois groups and field extensions.
@Rememberme huh? I don't understand the question.
Given a group G, there is only one lattice of subgroups for it. Isomorphic groups have isomorphic subgroup structure.
"one lattice of subgroups" doesn't make sense. Given a fixed group G, the inclusion between subgroups don't jump around. That'd be a silly thing to ask.
@DanielFischer Let $\Sigma$ be the sigma algebra of all subsets of $S$. All functions from $S$ to $\mathbb{R}$ are thus measurable. Let $\zeta_{S}$ denote the measurable set $\Sigma$ endowed with the counting measure on it. Consider any function $$f: \zeta_{S} \to \mathbb{R}$$ This can be considered a simple function $$f(x) = \sum\limits_{k=1}^{n}\alpha_{k}1_{A_k}(x)~~~~\text{where }~A_{1},...,A_{n}~~\text{denotes the singletons of}~ S.$$
For any $x \in \zeta_{S}$ we have that $$f(x) = \sum\limits_{k=1}^{n}\alpha_{k}1_{A_{k}}(x) = \alpha_{k}~~~\text{for some }k = \{1,...,n\}$$
@DanielFischer It might not be the most concise or best proof but I just want to know if the idea is correct.
@Moses You should have mentioned that you're considering a finite $S$ at the start. Without that, the "can be considered a simple function" made me go "Uh, no" at first.
@DanielFischer Not sure if you remember but I defined the function $u$ by $u(x,y) := \frac{1}{4}(\| (x+y)^{+}\|^{2} - \|(x-y)^{-} \|^{2} )$.
Could it be simply observed that this function is not linear by taking $\lambda = 0$ and then noting that $\lambda u(x,y) = 0$ but $\lambda u(x,y) \neq 0$ if we take $y \neq 0$.
@Moses You meant $u(\lambda x,v) \neq 0$ presumably. But then we have (I omit the constant factor) $\lVert y^+\rVert^2 - \lVert (-y)^-\rVert^2$, and $(-y)^- = y^+$, so it's $0$.
But $u(x,x) = \lVert x^+\rVert^2$, and $u(2x,x) = \frac{1}{4}(9\lVert x^+ \rVert^2 - \lVert x^-\rVert^2)$, and that's only equal to $2u(x,x)$ if $\lVert x^+\rVert^2 = \lVert x^-\rVert^2$.
For obvious reasons I will try to maintain some anonymity here.
So I defended my thesis. Which was funded through a government grant.
In order to obtain this type of grant, a professor needs to cooperate with several other research organizations and privately-owned industries. Then it is requir...
@Balarka I was thinking of this: A map between two homeomorphic spaces which is bijective but not a homeomorphism. Will that map from the unit square to the unit disk defined as f(x)={\cos(x), sin(x)}
@MikeMiller Let $V(f)$ be an affine variety, $k[x_1, \cdots, x_n]/(f)$ it's ring of functions. Define the map $V(f) \to \mathsf{maxSpec}(k[x_1, \cdots, x_n]/(f))$ by $(x_1, \cdots, x_n) \mapsto m_{(x_1, \cdots, x_n)}$ where $m_{(\cdots, x_i, \cdots)}$ is the maximal ideal consisting of $f$s in the coordinate ring which vanish at $(x_1, \cdots, x_n)$. A-M says a weak form of Nullstellensatz implies this map is a bijection. Is it also continuous, w.r.t. the Zariski topology?
@MikeMiller Got it. The map's actually a homeomorphism. Fix some $g$ in the coordinate ring. $U_g$ be the open sets $\{x \in X : g(x) \neq 0\}$. These are a basis for the Zariski topology on $V(f)$. Pushforwarding $U_g$ along the map gives the maximal ideals of the coordinate ring which doesn't contain $g$. These form the basis for the Zariski topology on the max-Spec.
ok, ok. well, it suffices to consider the basic open sets in the max-Spec of the coordinate ring. They are of the form $V_g = \{m \in X : g \notin m\}$. Every maximal ideal is of the form $m_{(\cdots, x_i, \cdots)}$, so pullback along the map gives you the set of all points in the variety such that $g$ doesn't vanish on there.
Well since I was thinking that you said the constructing idea of quotient groups a d quotient spaces are the same shouldn't there be some thing called kernel of a homeomorphism? I haven't encountered something like that in munkres yet. Is there something like that@BalarkaSen
oh. I see your point now. showing that it induces a bijection on the bases does it fine because your map is a bijection, so pushforward is inverse to pullback. sorry
@Remember Guess I should point this out : an important distinction to be made between preimage of a point along a continuous map between topological spaces and a homomorphism between groups is that the former might have different cardinalities for different choice of points you're taking preimage of, while the latter will always have the same cardinality. That's why we consider the set of all $g$'s such that $f(g)$ is the identity on the codomain group, rather than any other crazy element.
A class of continuous maps between topological spaces which has the same cardinalities of it's preimages, independent of the choice of the point, is a covering map.
OK, time for me to start on multivariable calculus. Loads to do :(
@Huy I have a question (if you are free then pls do give me some kind of hint): Can you give an example of a bijective continuous function between two homeomorphic spaces (both of the spaces are distinct) such that the function is not a homeomorphism
I hope this doesn't turn out to be a silly question.
There are lots of nice examples of continuous bijections $X\to Y$ between topological spaces that are not homeomorphisms. But in the examples I know, either $X$ and $Y$ are not homeomorphic to one another, or they are (homeomorphic) disconnec...
That's a pointless distinction. Why would you ever care about distinct but homeomorphic spaces? In any case, if you want, then pick another space that's distinct but homeomorphic to the spaces linked and compose with that homeomorphism.
@MikeMiller: Is there a simple argument to show $S^7$ has a global frame even though it is not a Lie group, or does one have to construct one explicitly?
@MikeMiller: And also, is it possible to show that all other $n$ except $n=1$ and $n=3$ are not parallelizible, just using differential geometry? A friend of mine showed me a proof using algebraic topology, but I don't know much about that.
@Huy There's various elementary proofs for the $n=2k$ case, but I don't know one for the general case. jstor.org/stable/2320860?seq=1#page_scan_tab_contents I have heard nice things about this paper. ("Analytic Proofs of the "Hairy Ball Theorem" and the Brouwer Fixed Point Theorem " J. Milnor)
@MikeMiller: Another question that I think should be simple to answer: I have shown that if two isometries $\phi, \psi: (M,g) \to (N,h)$ and their differentials coincide at some $p \in M$, then they are the same. How can I use this to find an upper bound for the dimension of $\operatorname{Isom}(M,g)$?
@PVAL: What is actually the definition of being parallelizable? I'm reading from different sources, some define it to mean "trivial tangent bundle", some something like "global $n$ frame" which are equivalent, and some write "it is parallelizable if and only if the tangent bundle is trivial" - which to me implies this is equivalent to the definition. But what actually is the definition?
@Huy These are equivalent, and used somewhat interchangeably. Given a global $n$-frame it should be pretty easy to construct a vector bundle isomorphism between $TM$ and $M\times \Bbb R^n$. For instance if you have $n$ linearly independent vector fields $\{v_i|}$ The map $(x, \sum a_iv_i(x) \mapsto (x,\suma_ie_i)$ gives your isomorphism.
The thing that is less obvious is that say a vector bundle is topologically trivial and smoothly trivial are equivalent. They are, but at this point I'd recommend just thinking about the smooth case (in other words all our vector fields are smooth).
Man, I knew it. I'm just now reading that it's true that every parallelizable manifold is orientable.
At the exam after showing many Lie groups were orientable I was asked if I thought all Lie groups were, and I thought it could be the case because we get an $n$-frame by left-translation, but then I was too afraid to say something wrong because we hadn't done it in class before, so I didn't say it =_=
Hmm. I once answered a question there, and now comes a question of (subjectively) higher quality and even more predisposition to an answer like the one to the older question. Still I feel I have to flag the new one as duplicate. Sigh
@ccorn I am having some students tommorow participate in a quiz of some sorts. I had an interesting problem for them, and was wondering if there existed some generalization of the problem. However I am not sure how to phrase it.
@N3buchadnezzar My english may be better than the one in japanese data sheets, but not much, I'm afraid. That said, my impression is that english lends itself much better to short, straight-to-the-point expositions.
@ccorn I am giving them the problem that they have a mug (8dl) of beer and two glasses of 5dl and 3dl, the problem is to split the beer into two glasses, each holding 4dl. I was thinking about generalizations of different even sizes mugs (m), and primes p and q representing the glases. I had a tough time trying to figure out which choices of m, p and q makes it possible to divide the liquid in a finite amount of pours so to say.
@JoeStavitsky upload.wikimedia.org/wikipedia/commons/8/84/… Looks like this. You need $(B+A) \cdot H / 2$ Now notice that $B+A = \text{side}$ and $H = \text{side} \cdot \sin \theta = \text{side} \cdot \sin \frac{\pi}{3}$.
@ccorn Poor students might do the cumbersome math, to split the tiny amount of beer fairly ;)
