Strange: The accepted and most-upvoted answer to that question fails to make its point. There are other answers (not by me) that IMHO deserve a better ranking.
ok, take a 3-simplex, cut open into two prisms by cutting along a little square. the identifications you found makes the two prisms into two solid torii $S^1 \times D^2$. the little square is boundary of the two torii.
so your space is two copies of $S^1 \times D^2$ glues along the boundary $S^1 \times S^1$ by the map that pastes the meridian to the longitude.
$S^1 \times D^2 \cup_\partial D^2 \times S^1 \simeq \partial(D^2 \times D^2) \simeq \partial(D^4)$ which is just $S^3$, so you're done.
these kind of decompositions of 3-manifolds into solid genus g surfaces glued along their boundary by some diffeomorphism are called Heegaard decompositions.
All my powers are gone when I leave math and attend other non-mathematical parts of the book. :D
Actually this is the hardest part of my book that will give me headache.
user147690
11:51 AM
Haha I can imagine. I changed out of an engineering major so 1) I wouldn't have to work in groups, 2) I wouldn't have to write any report/discussion etc papers
Hello!!! I have found the eigenvalues and the eigenspaces of the following matrix and I wanted to know if I am right. $\begin{bmatrix} 2 & 0 &0 \\ -3 & 3 &2 \\ 4 & -2 &-1 \end{bmatrix}$.
I have found that the eigenvalues are 2 and 1 and the eigenspaces $V(1)=\{\bigl(\begin{smallmatrix} 0\\ 1\\ -1 \end{smallmatrix}\bigr)x | x \in \mathbb{R}\}, V(2)=\{\bigl(\begin{smallmatrix} 1\\ -1\\ 2 \end{smallmatrix}\bigr)x | x \in \mathbb{R}\}$
One thing is a vector space, the other is a vector.
BTW, if you at some point don't have access to wolframalpha or something similar, you can always check such results by plugging into the equation it should satisfy, namely $Ax = \lambda x$.
Since we're only in dimension 3, you can do it in your head.
@Huy We have the linear map $F: \mathbb{R}^3 \to \mathbb{R}^4$ with $F(1,0,0)=(1,-1,0,2), F(1,1,0)=(1,0,1,2), F(0,0,1)=(2,-1,1,4)$ and we want to find the basis of the image of $F$. Do we have to consider the matrix $\begin{bmatrix} 1 & 1 &2 \\ -1 & 0 &-1 \\ 0 & 1 & 1\\ 2 & 2 & 4 \end{bmatrix}$ and apply Gaussian elimination?
A basis consists of linearly independent vectors. In order to find the basis of the image of F do we find the vectors that form these three vectors: (1,-1,0,2), (1,0,1,2), (2,-1,1,4)? @Huy
@robjohn I like the question, but the accepted answer had previously failed to give an actual proof. Yet it had the most votes by far. It has now been repaired. I like many of the other answers, and I find that Jack's answer (which was right from the beginning) deserves to rank higher. Other answers are nice too.
@ccorn timing has a lot to do with when answers get votes, and once an answer has votes, it tends to get more than the answers that don't have votes. Sad, but true.
@robjohn Most importantly, the flaws of that question's richest answer have now been rectified.
So I removed my downvote (which puts me at 3999 score. I had accidentally crossed the 4k barrier yesterday night with a LHF, but other answers got better, so I could delete mine again. Which is a good thing; it would have made my answer list look dull. Have to concentrate on the interesting questions again.)
We have the linear map $F: \mathbb{R}^3 \to \mathbb{R}^4$ with $F(1,0,0)=(1,-1,0,2), F(1,1,0)=(1,0,1,2), F(0,0,1)=(2,-1,1,4)$ and we want to find the basis of the image of $F$. Do we have to consider the matrix $\begin{bmatrix} 1 & 1 &2 \\ -1 & 0 &-1 \\ 0 & 1 & 1\\ 2 & 2 & 4 \end{bmatrix}$ and apply Gaussian elimination as follows?
in a very brief way, you assume that $\mathsf{PA}$ is $\omega$-consistent and construct a formula $F(\bar{n}, \langle \varphi \rangle)$ for a statement $\varphi$ and a natural number $n$ such that $\vdash F(\bar{n}, \langle \varphi \rangle)$ implies $n$ is the Godel code of a proof of $\varphi$. turns out $T(\langle - \rangle) = \exists x T(x, \langle - \rangle)$ is a schema. now tarskis theorem says that $\mathsf{PA}$ with a schema is inconsistent which is a corollary of diagonal lemma
I think it's possible yea. It'd be the same as performing row operations on the transpose of the matrix you constructed from the vectors that were the result of applying $F$ to the given vectors in $\mathbb{R}^3$ (which as far as I can tell, form a basis for $\mathbb{R}^3$).
is it required to know some axiomatic set theory to learn mathematical logic? (or the other way around?). i am not planning to read it right now, but well... i just realized it's not as non-mathematical as some people say it is. i guess i'd like to read a bit of this and a bit of that at some point of time
i wonder if my motivation behind really reading up the proof of Godel's theorem today after the discussion with you and Karl was to learn homotopy type theory at some point of time. i think i'd be rather happy if it wasn't.
shivers
speaking of, mathematical logic intersects with higher category theory somewhere, doesn't it? pretty weird.
@evinda Do you need an orthonormal basis, or just a basis? If you just need a basis, the columns of the matrix will do. However, since the columns are dependent, you do need to eliminate one.
@Khallil: It's a type of category. You could define a category and all the words involved so that you know what they mean, but then you'd have to justify "Why do I care about a ____ _____ _____ category with ____s?" And that's a much harder question.
The story goes that Gauss was on Riemann’s defense committee for his Habilitation (super doctorate). In those days, the candidate was asked to submit three topics in advance, with the implicit understanding that the committee would ask to hear about the first topic (the actual thesis was on Fourier series and the Riemann integral.) Riemann’s third topic was “On the Hypotheses which lie at the Foundations of Geometry.” Clearly he was hoping that the committee would select from the first two topics, which were on material he
Namely, my standard solution suggest to look up the proof that the Hopf fibration $f: S^3 \to S^2$ is a Riemannian submersion in that book, that's why. I find that story pretty cool though.
I never read them. Gallot leaves a lot to the reader which helped me build intuition. On the other hand, my taste might be skewed, since I'm not a geometer.
Oh, now that you're here I actually have a question, @MikeMiller: I want an isometric immersion of $\mathbb{R}^2 / \mathbb{Z}^2$ into $\mathbb{R}^4$. A candidate would be $$f(x,y) = \frac{1}{2\pi}(e^{2\pi i x},e^{2 \pi i y}).$$ So I have to verify that the pullback of the standard metric on $\mathbb{R}^4$ by $f$ is the standard metric on $\mathbb{R}^2/\mathbb{Z}^2$. I can compute the pulled back values $g_{11} = g_{22} = 1$ and $g_{12} = 0$. Is that already enough?
@MikeMiller: I'm asking because prior to the exercise I didn't compute the $g_{ij}$ for the standard metric on the torus, so I was surprised to find it is actually just $\delta_{ij}$, as in $\mathbb{R}^n$.
@MikeMiller: The exercise which precedes this one is just a general proof that a group acting freely and properly discontinuously on $(M,g)$ by isometries induces a Riemannian manifold $M/\Gamma$ such that the covering map is locally an isometry, so there were no explicit computations needed.
@Huy: Well, when one says the metric on the torus, they mean the metric such that $p: \mathbb R^2 \to T^2$ is a local isometry, so by definition the metric (at least pointwise and under this identification $T_x \mathbb R^2 \cong T_{p(x)} T^2$) is the same.
It was in the beginning of this year, yes. I already had finished enough courses for my BSc and only needed to finish my thesis. Due to lack of self-discipline, I only submitted the thesis in the beginning of this year. :P
@MikeMiller: There is no MSc in mathematical physics at my university. Officially, I'm doing it in pure mathematics. But the MSc for mathematics allows to choose from almost all mathematical/physical courses, so I choose a lot from mathematical physics because I'm very interested in them.
@Khallil: I think that depends on the university you're at.
@MikeMiller: There's also a MSc in applied maths at my uni. The "bad thing" about it is that MSc in applied maths requires at least one seminar and one semester paper, whereas the MSc in pure maths only requires one seminar (and more courses). I prefer courses over writing papers, so that was the reason I chose to go with MSc in pure mathematics.
(not the only one, but one of them and probably the simplest to explain without trying to explain many more regulations at my university)
I could have studied physics and gone into theoretical physics too, but I would still have been obligated to work in laboratory for a semester or two during undergrad studies, and I can assure you there's few things more dangerous than me in such an environment.
That's a tough question. I'm not really satisfied because I could've worked a lot more. I missed loads of lectures in the second term due to pure laziness.
Indeed, I think I've never been satisfied with my results so far.
Except for algebra, where I got a good grade out of nowhere.
Every other course there's something I wish I had done differently. Started studying earlier, studying some subject better, looking at more examples, etc.
We have a spreadsheet online available to the students, and the exam was oral. So many people who had their oral exam in e.g. algebra that year would describe their exam: what was asked, how did the professor behave, how did the student feel, etc. The algebra exams had the worst "reviews" by far and actually made me afraid of the exam. But the prof was the nicest person ever to me.
I'm taking Vector Analysis, Analysis 3, Algebra 1, Intro to PDEs and I'll be writing an essay on a topic of my choosing that we don't learn about in the second year, @Huy. (Maybe Spanish too)
@Huy: Many schools stop complex analysis courses at residue analysis. That course covers roughly what the (1 quarter) UCLA undergrad complex analysis course does.
It was an intro to abstract algebra. Basic intro to groups, binary operations, Lagrange's theorem, rings, fields, cosets, symmetric groups and quotient groups. I took Linear Algebra this year too which was to do with vector spaces, linear maps, matrices/eigenvalues.
Finitely generated abelian groups are a standalone chapter at the end of Algebra 1 which is kinda strange. I'm looking at that first because I need to review my Linear Algebra before doing the other chapters.
Supposedly Algebra 2 is one of the hardest modules in the second year. Averages in the past few years have been at about 50%, @Huy.
@Khallil: Your algebra 2 module is somewhat an equivalent of what I had a really good grade in here. I really still don't know why, because I didn't feel like I had a very deep understanding of any of it.
@Khallil: Over here, we basically do groups in Algebra 1 including Sylow theorems, then we do rings and fields in Algebra 2, leading to a bit of Galois theory. After those two courses, both are tested together, orally.
Oral exams sound a lot better to me. I feel like they reveal more of a mathematical understanding than plugging in formulae and rearranging stuff, @Huy.
@Khallil: They are a lot more relaxed and if you don't understand something or didn't learn some particular chapter (lack of time, no understanding at all), you can tell the professor as soon as he asks about it and pretty much all of them so far have been understanding and proceeded with a very different question. Of course, that rejects a chance for a perfect grade, but I think it's very fair because you can still get a good grade by showing that you understood pretty much everything else.
@Khallil: Usually we are just asked to explain proof ideas, proofs, examples, etc, and when a prof realizes a student is really good, he asks something that requires thinking over what has been covered in the course and exercises.
i don't mind oral exams per se. what i struggle a lot with is writing anxiety, which isn't great when your oral exam is predicated on 1) writing a paper, 2) organizing a presentation
How might I prove that if $a$ is an arithmetical function with polynomial growth, then if for some $s \in \mathbb{C}$ the Dirichlet series $D(a, s) = 0$, we have that $a(n) = 0$ always?
I'm always frustrated by first topology courses that obsess over the definitions and basics like compactness, separability hypotheses, etc, but never actually do any geometric topology
so it's good to hear you're learning some of the fun stuff
It was nice! And I know my advisor taught algebraic topology a semester or two after I graduated, so I'm planning to ask him about it; see if I can work through the material
I haven't been able to sit down and read Hatcher, for whatever reason. I tried, but lost steam before long.
Lemma 1.
For any $x$ in the ring $F[t,t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$), $x$ is a power of $t$ if and only if $x$ divides $1$ and $t-1$ divides $x-1$ (the divisibilities are meant, of course, in $F[t, t^{-1}]$).
Lemma 2.
...
so we have two representation of C* algebra as called equivalent if for representation $(\pi_1,\mathbb{H_1})$ and $(\pi_2,\mathbb{H_2})$ there exists a unitary operator such that $u\pi_1(a) = \pi_2(a)u$
I don't understand I thought unitary operator act in the same space for example action of $\pi_1(a)$ the codomain will be in $H_1$ so u will be in $H_2$ once it act on $\pi_1$ now on the left hand side we have $H_2$ and that is acting on u hmm
It means for $U: \mathbb H_1 \to \mathbb H_2$ to be unitary. Unitary maps make perfect sense between different spaces.
In any case, unitary maps are isomorphisms of Hilbert spaces, so two representations can only be equivalent if the codomains are isomorphic as Hilbert spaces.
@Chris'ssistheartist: Wow... I just got my Mathematica re-registered and it runs about twice as fast. That is probably because this laptop has a quad-core running at 2.5 GHz while I think my other machine was a dual-core running at 2.5 GHz.
Lemma 1.
For any $x$ in the ring $F[t,t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$), $x$ is a power of $t$ if and only if $x$ divides $1$ and $t-1$ divides $x-1$ (the divisibilities are meant, of course, in $F[t, t^{-1}]$).
Lemma 2.
...
