@MikeMiller I think I was told why it was true and the references to look at, but I probably wasn't paying enough intention. I thought there was something on Kirby's list which talks about smooth embeddings of the Poincare sphere minus a point into R^4, though I couldn't find it.
@PVAL: Ok. I haven't talked to experts much about the problem, so I do believe it should be nontrivial to embed $\Sigma(2,3,5)$ punctured into $\Bbb R^4$.
Anyways I'm given two Matrices. $A^{-1}$ and $B^{-1}$ I have to compute: $(AB^T)^{-1}$ So can I just do..: $A^{-1}*(B^{-1})^T$ to get the same thing..? Not sure if it would be.
If you write down using (1 and 2-handles) a presentation of the trivial group which is not known to be trivial by Andrew-Curtis moves and has trivial H_2 the boundary should be an example.
A $\Delta$-complex structure on a space X is a collection of maps $\sigma_{\alpha} : \Delta^n \rightarrow X$ with n depending on index $\alpha$ such that
1)The restriction $\sigma_{\alpha}|int(Delta)$ is injective and each point of X is in the image of exactly one such restriction.
2)Each restr...
@MikeMiller I don't know if that theorem helps much. There are quite a few plausible sounding statements close to SPC4. (the conjectures in the "Man and Machine..." paper. I don't know if Freedman is saying anything that wasn't known at that point. Another nice one is the case of the Akbulut-Kirby conjecture for the case of slice knots (this is implied by the Poincare conjecture).
@PVAL: It seems to me to be completely unhelpful for determining if a manifold I hand you has an embedding. But nonetheless I didn't expect something that looks so nice.
@PVAL: That's not the blog post I was thinking of. I'm thinking precisely of what you mentioned before: Write down a Mazur manifold from a potential counterexample to AC, and double it; you have a canonical embedding of a 3-manifold. now prove that 3-manifold doesn't embed in $S^4$. That idea's due to somebody and for some reason I think it was Freedman.
Likely multiple people thought of that at around the same time, but not knowing at all how to obstruct 3-manifold embedding into S^4 with obstructing embedding in ZHB^4 people probably didn't write anything down.
@MikeMiller I think I have gotten the first examples of contractible 4-manifolds that are not Stein, and shown that the proof of what I thought was the first example must be incorrect. Though I have thought the first thing multiple times and had been wrong each time.
@DanielFischer: I know that for $S$ being any surface, if $\chi(S) < 0$, we can equip $S$ with a hyperbolic metric. is there anything we can say about the Euler characteristic given a surface with hyperbolic metric? is it possible to have a hyperbolic metric on surfaces with $\chi(S) \geq 0$?
@Huy For closed surfaces, $\chi(S) \geqslant 0$ means $S$ is the sphere or a torus, doesn't it? These aren't hyperbolic. For open surfaces, I don't know, can there be an open surface with non-negative Euler characteristic?
@DanielFischer If open surface means 2-manifold with boundary, what about torus mius disk? This deformation retracts onto wedge of two circles, which can be given a cell structure with one 0-cell, two 1-cells and zero higher cells. So $\chi$ is $1 - 2 = -1$, isn't it?
(I can't answer @Huy's question because I don't know what a hyperbolic metric is)
I have learnt to differentiate correctly about a month ago, so sorry to disappoint. But $\Bbb T^2 - D^2$ certainly has something hyperbolic about it, as universal cover of it is homotopy equivalent to a hyperbolic graph, whatever that means.
I usually interpret "surface of genus $g$ is hyperbolic" by the vague picture that the universal cover is upper half plane and the translates of the fundamental domain triangulates it in a "hyperbolic" way. Something like that seems to happen here.
@Huy Yeah, as soon as I was out of the door, I realised that the disk has the same Euler characteristic as the plane. So for open surfaces, we can have hyperbolic surfaces with nonnegative Euler characteristic. The universal covering is a local isometry if and only if the metrics are chosen suitably. If you take a metric of constant - nonzero - curvature, it's an isometry if and only if the curvatures of the base space and the covering space are the same, iirc.
@BalarkaSen Surfaces have no boundary (in this situation, we're talking about Riemann surfaces - at least I am), an open surface is a non-compact Riemann surface.
Given $f: \Bbb R^m \to \Bbb R^n$, I can realize $Df(a)$ as a map $\Bbb R^m \to L(\Bbb R^m, \Bbb R^n)$, $a \mapsto Df(a)(-)$, mapping vectors to linear maps. $L(\Bbb R^m, \Bbb R^n)$ is the space of all linear maps from $\Bbb R^m$ to $\Bbb R^n$, which can be identified with space of $n \times m$ matrices, which can in turn be naturally identified with $\Bbb R^{nm}$. Under that identification, we can write down the derivative of $Df : \Bbb R^m \to L(\Bbb R^m, \Bbb R^n) \cong \Bbb R^{nm}$.
When $n = 1$, this is the same as the Hessian, so we have the very convenient formula $Hf = D(Df)$. Trivial, but this is a good explanation of why the Hessian should be the second derivative: so far, my only explanation was that it is the best quadratic approximation to $f$ (modulo appropriate conditions), i.e., the second order Taylor series.
Thinking of the derivative as a linear map-valued function is slick.
@Idle001 Stop acting the downvote martyr. You might even consider providing quality answers, but somehow I have a feeling that is not a realistic option
Find a function in $S$ (set of functions from $\mathbb{Z}$ to $\mathbb{Z}$ with the operation of composition) that has a left inverse, but no right inverse, and decide whether the inverse is unique.
Consider $f(n) = 2n$ and $g(n) = \begin{cases} 2n, & \mbox{if } n \mbox{ is even} \\ 4, & \mbox{if } n \mbox{ is odd}, \end{cases} $. Then $g$ is left inverse of $f$, since we have $(g\circ f)(n) = g(n/2) = n$ for all $n \in \mathbb{Z}$ but not right inverse of $f$ since $(f\circ g)(n) = \begin{cases} n, & \mbox{if } n \mbox{ is even} \\ 8, & \mbox{if } n \mbox{ is odd}, \end{cases} $.
If this is correct, how do I decide whether the inverse is unique?
When I see the things that I had missed while running through every book I saw I missed quite a lot. Actually I missed some very beautiful things indeed. Like for example: Yesterday while reviewing combinatorics I saw that how I had missed things like a combinatorical proof of the binomial theorem(A version you can say). It is very cool in fact.@BalarkaSen
So first building some notation. The book denotes $[n]=\{1,2,3,...,n\}$. Now consider the generating function which gives subsets $$\sum_{A\subseteq [n]}\prod_{i\in A}x_i$$
Now for say $n=2$ you will see that that sum equals $x_1x_2+x_1+x_2+1$ where the empty subset whose product is given by 1
Right, I get it. You can set $x_1 = \cdots = x_n$, and count that coeff of the terms whose exponents is $m$ is precisely the $\#$ of card $m$ subsets, which is $\binom{n}{m}$.
@robjohn in my recent answer the oeis formula is just product of two terms $2^(n-1)*(1+2^n)$, while mine is titanic size !!! is there a secret behind how two forms are outputting same results but different shaped ?
What's your option for calculating this integral? No full solution is necessary, it's optional as usual.
Calculate
$$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...
Work since early in the morning, with almost no break, I gave some food to the dogs, did some other things and continued my work. I didn't even take a break to eat a bit of anything. It's 17 here. Getting on top requires sacrifice @Hippalectryon
(a) Find a function in $S$ (set of functions from $\mathbb{Z}$ to $\mathbb{Z}$ with the operation of composition) that has a left inverse, but no right inverse, and (b) decide whether the inverse is unique.
(a) Consider $f(n) = 2n$ and $g(n) = \begin{cases} 2n, & \mbox{if } n \mbox{ is even} \\ 4, & \mbox{if } n \mbox{ is odd}, \end{cases} $. Then $g$ is left inverse of $f$, since we have $(g\circ f)(n) = g(n/2) = n$ for all $n \in \mathbb{Z}$ but not right inverse of $f$ since $(f\circ g)(n) = \begin{cases} n, & \mbox{if } n \mbox{ is even} \\ 8, & \mbox{if } n \mbox{ is odd}, \end{cases} $.
(b) Suppose there's another inverse $h$. Then $h(n/2) = n$, therefore $h(n/2) = g(n/2)$ hence $g = h$
Could someone please verify whether the above is correct?
@robjohn@I'manartist I require your help if you guess are ready to give: I am trying to prove that $$\frac{\Gamma(m)\Gamma(1-m)}=\frac{\pi}{\sin(m\pi)}$$ We know that $\frac{{\Gamma(x)}{\Gamma(y)}}{\Gamma(x+y)}$ = $B(x+y)$ where $B(x,y)$ denotes the beta function. Now if I substitute $x=m,y=1-m$ I get $B(m,1-m)$ which equals $\int_{0}^{1}t^{m-1}(1-t)^{-m}{dt}=\int_{0}^{1}\frac{t^m}{t(1-t)^m}{dt}$. Now the methods I have seen after this all involve contour integration something that I dont know yet. So Is there any way of doing this problem by using the beta function and not contour integration
That should be $\Gamma (m) \Gamma (1-m)= \frac{\pi}{\sin(m\pi)}$
@iwriteonbananas There are two. Multiplication of ring-valued functions on chains, and intersection of transverse submanifolds. Which one are you talking about?
Well, we defined intersection numbers of transverse submanifolds of complementary dimensions and then proved some theorem relating that intersection number to some cup product
Actually we didn't prove it yet, just stated it because the time ran out
Gees, you just made me remember I skipped so much from 3.3. Also, there's a lot of things on homology I haven't thought about. Not comfortable with algebraic computations involving cup products either.
@TedShifrin I tried to read the proof that if a function $f : \Bbb R^n \to \Bbb R$ is twice diffable, i.e., $Df : \Bbb R^n \to L(\Bbb R^n, \Bbb R) \simeq \Bbb R^n$ is diffable as a function, then mixed partials of $f$ are equal.
Dieudonne has a twisted mind.
I couldn't read a thing at first glance. I'll try later :P
@MikeMiller So, let $f:M^{2n}\to N^{2n}$ with $b_n(M)<b_n(N)$. Then we can choose some $a\in H_n(N)$ not in the image of $H_n(f)$. Let $\alpha$ be the Hom-dual of $a$ so that $f^*(\alpha)=0$. Now we choose an element $\beta \in H^n(N)$ such that $\alpha\cup \beta$ generates $H^{2n}(N)$, which exists by a corollary of P.D.
From $0=f^*(\alpha)\cup f^*(\beta) = f^*(\alpha\cup \beta)$ it follows that $f$ has degree 0
@TedShifrin Oh, alright. I was just thinking that if I am to know some calculus, I should know a bit more. These are fun stuff, after all... but if you think it'd be unnecessary waste of time, sure, I will do as you suggest. :)
@Balarka: You can generalize the result that $C^1$ implies differentiable, too, but it's a technical point and you should be learning the essentials to get where we're headed.
A $\Delta$-complex structure on a space X is a collection of maps $\sigma_{\alpha} : \Delta^n \rightarrow X$ with n depending on index $\alpha$ such that
1)The restriction $\sigma_{\alpha}|int(Delta)$ is injective and each point of X is in the image of exactly one such restriction.
2)Each restr...
What's your option for calculating this integral? No full solution is necessary, it's optional as usual.
Calculate
$$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...
