Not much but I can take a look at it when I get back in about half an hour
Did you have a particular questiomn
Or now, I forgot I have a phone.
Oh, I know something about the contact invariant. This is what you use to prove that L-spaces have no taut foliation (by proving they have no symplectic filling with positive $b^+$)
Hmm, I think I've tracked down the error in my LaTeX code to this line: Prove the $q$-Vandermonde identity $\left[\begin{smallmatrix}r+s\\n\end{smallmatrix}\right]_q=\sum\limits_{k=0}^n\left[\begin{smallmatrix}r\\k\\end{smallmatrix}\right]_q\left[\begin{smallmatrix} s\\n-k\end{smallmatrix}\right]_q q^{(r-k)(n-k)}$. Does anyone have a recommendation for displaying $q$-analogues? Was there already a built-in squarebracket-binomial? And why do you suppose it doesn't like that code?
@PVAL: It's reasonably readable without knowing much about Heegaard Floer homology. The L-space result is in "Holomorphic disks and genus bounds" which I read.
@MikeMiller when topology books say that topologies on X that contain A. Does it mean that A is subset of the topology or A is an element of the topology ?
And the argument says "The lines cut out in the picture below is a cylinder, which occurs to the left of (A) [the infinite torus]. Now draw in two more lines enclosing more holes, and consider the region between the two pairs." @PVAL
Note he is cutting out a cylinder from a hole in the mesh.
@PedroTamaroff Well the point should be they both should be the nested union of surfaces $S_g$ where $S_g$ has genus $g$ and exactly one boundary component. There is obviously such a decomposition of the linear picture (infinitely many glued tori starting from the left), so all you have to do is find a decomposition of the other one and argue from class. of compact surfaces if you like.
@PedroTamaroff: Don't worry about his argument. Pick an increasing decomposition like PVAL recommends and you will no doubt have rediscovered his argument.
@PedroTamaroff Aren't those simple closed curves (i.e. circles)? Is he asking to draw 2 more circles or 2 lines that make a circle? I am confused by his wording
@MikeMiller Are you doing some work on low dimensional topology? I know you're also doing some of this guage field theory stuff, so I can't recall if I'm getting your research confused with someone else.
@mikemiller I mention it because I was trying to recall the details of this 'trick' in higher dimensions that allows one to disentangle two tori without them intersecting each other. I think that was the plan.....
The most-cited thing called a trick is the whitney trick, which isn't really about tori. If you're in a 7-or-more dimensional space, any two embeddings of a pair of tori are isotopic, through arguments I wouldn't call tricks.
@MikeMiller Ah okay, anything more similar to what I mentioned where if you somehow have two tori where in 3D one passes through the hole of the other they're 'stuck' but maybe in $>4$ dimensions there's always a way to take them apart?
This is for basically a trivial reason: you can take the knot, and at an overcrossing, use the extra dimension to push it so that it's an undercrossing.
And thus every link is equivalent, because they're all the unlink.
I don't know, I'd have to think about it. The usual tricks I use to unlink things don't work automatically until $\Bbb R^7$. Maybe they're trivial here?
that distinction between tori and knots, and being able to unlink them, seems like the sort of thing that defies any simple intuition (at least, anything coming from geometry)
In $S^3$ there is no distinction between tori and knots. One side of the torus is homeomorphic to a solid torus, hence the torus is the boundary of a tubular neighborhood of a knot.
In higher dimensions there is not even really a relation between them.
And a torus in $S^4$ has essentially no relation to a knot. Remember, the boundary of a submanifold of $S^4$ is three-dimensional, so a tubular neighborhood of a knot won't have boundary a torus. (The boundary of such a neighborhood will actually be $S^2 \times S^1$.)
@MikeMiller There's a little interesting information in the isotopy class of "how" tori are embedded in $S^3$(where "how" doesn't have anything to do with the image of the map).
@PVAL Fair point, but after you mod out by the mapping class group isotopy classes of embedded tori are the same thing as isotopy classes of embedded knots, yes?
I am curious as to which elements of the mapping class group you can't "ambient isotope" away though
@MikeMiller In dimension 3 (for tori at least), I'm almost certain its none of them. In dimension 4, it is highly dependent on the image ( There are exactly 3 different isotopy classes with image the unknotted torus in $S^4$ and iirc there are examples with infinitely many as well).
I think they're types of graphs and 2-complexes. Kronheimer and Mrowka posted a paper today defining instanton homology for them, apparently they think you can prove the 4-color theorem using this tool.
That'd be great, a computational proof and a gauge theory proof.
The former, I think the hope is to completely avoid computation. The first paper is here. They have two papers up but this is the one that mentions 4color.
Well, not quite. Here's the second paper, which gives other reformulations
Remember that the word 'instanton' was imported to math quite some time ago and now means something that probably somehow relates back to the physics definition but nobody actively thinks about it that way.
Instanton homology is a type of 3-manifold Floer homology that uses the Yang-Mills functional on connections in its definition, which is where the word 'instanton' comes from
Physicists think about them as particles moving on graphs of functions $R \to R$ by gravity going downward. I do not know what they are or what they are trying to represent, but I'm almost sure that's what an instaton is. One of the better graduate talk-givers here expressed a complete exasperated lack of understanding for what they were talking about. I don't think they knew.
I observed them doing this in math seminars. It would be interesting to see how they behave in their natural habitat.
@KevinDriscoll That's fine, in fact that's the point. The relationship between instanton homology of 3-manifolds and physics instantons is that the word 'Yang-Mills' is used in its definition.
@MikeMiller I think a readable proof of the 4-color theorem which used a lot of topological and geometric machinery is probably a bigger deal than a solution to the Riemann hypothesis and if nothing else, Dr. Z's head would probably explode.
I'm not sure I would ever have the cojones to say that my approach might prove the 4-color theorem in a readable format before I had completed the proof
Why do you think that, @PVAL? Just because it's become a cultural target of "theorem whose proof is incredibly unsatisfying"? I think it'd be a big deal but Riemann would probably still win :p
I'm not really sure what open problems there are in the direction of 4-color. It seems like a (beautiful!) dead end, so I'm not sure this would open pathways
@MikeMiller 4-color is equivalent to this snark theorem for graphs (I just read) so maybe there are more connections in that direction. Just a blind guess, though.
In any case the applications would be graph-theoretic. I suspect KM would not pursue that direction that far. The poor graph theorists would have to learn differential geometry
I think at a lot of mainstream Q&A for mathematicians there is some question involving the four-color theorem. I remember Zelmanov (whose name also starts with a z) recently saying he didn't consider what existed to be a proof. It's a question that is immediately accessible to everyone as well (unlike RH). I have to admit that I don't really know why solving RH would be so surprising (I mean that Deligne guy seemed to prove some nice stuff).
I don't think it's that it would be surprising, it's that its proof would actually be very useful for number theory and would no doubt be a huge breakthrough in related research (so other major nearby problems would start to fall too)
It would have to do something very interesting. Straightforward applications of the old techniques didn't work
Maybe it would have to invent a good theory of $\Bbb F_1$ which I think is the current guess. But if not then it would be something completely new and exciting.
@KevinDriscoll: Not everyone believes the simple groups proof. :P Also the Hauptvermutung is almost the canonical example of your question I think.
"If you have two triangulations of the same space, can you subdivide them both to obtain the same triangulation?"
I think everyone believed it because it's obviously true.
Gordon told me that when Milnor constructed 7-manifolds homotopy equivalent to $S^7$ not diffeomorphic to $S^7$, Milnor initially thought that he had found a counterexample to the topological Poincare conjecture. The idea that smooth/pl/top could be different was not really considered plausible.
The 1978 Akbulut-Kirby conjecture was recently proven false by Yasui (In the last 6 months the paper was published on arxiv). I think that was at least somewhat believed to be false recently though.
That paper certainly doesn't show all such spheres are standard. I think someone here is thinking about that problem using these new fangled things called computers.
Right, I forgot that his proof only covered some subcases before you reminded me. You should tell Kronheimer and Mrowka about it, maybe they'll solve the linear algebra problem with instanton homology of something or other.
Do either of you know a mathematician who "only has a hammer" and "sees every problem as a nail?" I was just talking today with a postdoc about a theorist in physics who solves EVERY problem with a path-integral approach
If I knew a mathematician that had a tool that could solve whatever problem was in front of them I'd be a big fan
I find that hard to analogize. I could say that Kronheimer and Mrowka's work is like that because they solve all the problems they see with gauge theory :p
I mean that it's a bit silly, they're gauge theorists, they do gauge theory. The problems they solve will be applications of gauge theory because they do gauge theory, and they will pick these problems because they have done some gauge theory and see that it applies to some problem
Indeed, although I get the sense that 'gauge theory' is a bit broader than 'path-integral approach.' So I guess this sorta depends on how you classify things as 'hammers'
@MikeMiller $X$ be an $n$-dimensional CW-complex. $H^n(X;\Bbb Z) = [X, K(\Bbb Z, n)] = [X, K(\Bbb Z, n)^{(n+1)}]$. Keeping that in mind, we can build a model for $K(\Bbb Z, n)$ which consists of a $0$-cell, an $n$-cell (and no intermediate cells in between) and all higher cells of dimension $\geq n + 2$, because there is no extra relators to take care of in $\Bbb Z$, and to kill maps from $S^{n+1}$, it is sufficient to attach $(n + 2)$-cells.
This immediately tells us that $[X, S^n]$ is countable, but we have to incorporate the data about degree into our consideration, otherwise this fact would not be effective. I am not yet sure how to do this.
@DanielFischer Hi. Another question on importance. The Spectral Mapping Theorem is given by: If $a \in \mathcal{A}:= $ unital Banach Algebra and $f \in \text{hol}(a)$.Then $\sigma(f(a)) = f(\sigma(a))$. Is it's importance simply a way of computing $\sigma(f(a))$?
@IWantToRemainAnonymous These days I work more on research. I managed to discover amazing ways of calculating very hard integrals like $$\int_0^{\pi} \arctan^7\left(\frac{\sin (x)}{2 \sqrt{2}}\right)\csc ( x) \, dx$$
@IWantToRemainAnonymous Of course, I also work on book, I just added another question to my book.
@Danu If you aren't interested in the product quality then you may look for Indian editions, they're in general cheaper, usually available @ebay. For instance, Apostol's Calculus V.1 is at around 200 dollars at amazon, while only 17 dollars for the Indian version at ebay.
:23800157 Very remarkable how $1/\varphi^2$ gets related to the chi function
@DanielFischer Something mentioned previously. If we have $\mathcal{A} \subset \mathcal{L}(X)$. Where $u$ it the unit of $\mathcal{A}$. Then since $u$ is linear and idempotent, it is a projection and we can write $u = E_{1} + E_{2}$ and $Y = E_{1}Y \oplus E_{2}Y$ where you stated that $Y$ is the range of $u$.
Then since $Y = u(X)$ it follows that $$Y = E_{1}u(X) \oplus E_{2}u(X)$$ That's fine, but why do you write $E_{k} = E_{k} \circ u$ to get $Y = E_{1}(X) \oplus E_{2}(X)$. I don't see the reasoning behind that sicne we don't necessarily have $u(X) = X$?
@DanielFischer I am referring to the second decomposition with regard to the spectrum $\sigma(a)$.
@Moses Not sure what your question is. Since $u$ is the unit of $\mathcal{A}$, we have $E_k = E_k\circ u$. So we have $E_k(X) = (E_k\circ u)(X) = E_k(u(X)) = E_k(Y)$.
@Chris'ssistheartist So, your arctan question was removed. Did you succeed in simplifying it? Also, do you plan to publish/tell your method of doing such an integral?
@mickep To many disrespectful comments to a work that I consider to be brilliant. Besides that I was also downvoted. No, I don't plan to share anything about calculating such integrals.
@mickep I might publish something, or just add things in my next books.
off topic algebra. By Dummit/Foote (pg 549) Cor 36, in a finite field of characteristic $p$, every element is a $p$th power (use the Frobenius endomorphism). They define a field $K$ of characteristic $p$ or $0$ to be perfect if every element of $K$ is a $p$th power in $K$. By the Corollary, doesn't this imply that every field is perfect (which is false)?
@TheSubstitute Every finite field is perfect, and every field of characteristic $0$ is perfect. Infinite fields of positive characteristic may or may not be perfect.
@TRiG In theory. In practice, not all editors write useful summaries like you. More like "fixed grammer" or "made more pretty". I can to some extent understand people who have stopped caring about edit summaries.
@IWantToRemainAnonymous I'd only be buying a few books: Ones that I'd like to be able to look up stuff in for the coming 10+ years. Printing is therefore not a real option.
Hello!! Could you take a look at the proof of the Theorem 3 and tell me at which point we use Lemma 6? I have read the proof several times but I haven't understood how this Lemma helps us... The Lemma and the Theorem 3 are the following: http://math.stackexchange.com/questions/1382120/ft-has-undecidable-positive-existential-theory-in-the-language-cdot
@Balarka: Actually the point here is that since you know $H^n(-;\Bbb Z) \cong \text{Hom}(-,K(G,n))$ you can use the Yoneda lemma, and in particular the explicit bijection it gives.
I mean, that's the point (to me) of a lot of category theory. There are no assumptions, it tells you what things are true in "full generality" instead of your special case.
Fair enough, but that's still something at the level of book-keeping, though. Instead of knowing a thousand special-cases, you write down a single concrete fully general thing which you can apply whenever and wherever you want. I know that's useful (I have actually found it to be so), but I'd still like to see some concrete application of category theory where it does something other that book-keeping.
Don't get me wrong, I am sure there exists a lot of them, having already "seen" one of them -- Grothendieck's Galois theory.
Well, another example is of course homological algebra. That's a pretty general nonsensical thing which can be used to do nontrivial non-bookkeeping stuff.
I was merely pointing out what kind of applications of category theory I'd like to see, interpreting that message of yours above written in the context of my ignorance towards category theory I had expressed a few months ago. But if it was not, nevermind.
$\int_a^b f(x) dx= \int_a^b f(a+b-x) dx $ holds not for the reason that one can substitute $u=a+b-x$ for the RHS right? Am I right? Its insane, that I found that written somewhere.....because the limits will definitely change in the second integral.....
I know that $\int_a^b f(x) dx= \int_a^b f(a+b-x) dx $ is True
@BalarkaSen Field theory, some geometry (classical geometry of curves and surfaces) and topology (I'm waiting for the course to get a bit more interesting)
How do you feel about Spivaks Intro to differential geomtry vol 1, as an intro (it looks nice to me, but it does not seem to get as many recommendation)? @TedShifrin @MikeMiller
@PaulPlummer: There's a paper on the arXiv that describes the algorithmic solution to the homeomorphism problem of 3-manifolds. I haven't read it yet, but maybe you'd like it.
@TheArtist The statistics with the larger number of samples (the one where the house always wins) would seem to have more reliability than your 10 samples, but feel free to try again :-)
You can always get your picture in the lobby at a number of casinos :-D
@TheArtist Card counters and others who seem to beat the odds often become unwelcome at casinos, and they will reward you by putting your picture up so their employees will recognize you.
@TheArtist whether you cheat or not is not the point. They won't pursue legal action if they can't prove cheating, but they can kick you out if you cost them money.
Of course, they want people to win some money some of the time. That is what keeps people coming back; but if they start noticing a pattern with someone, they will take notice.
@MikeMiller I am writing down my observations on $[X, K(G, n)] = H^n(X; G)$ explicitly and I found some vague arguments in them that I am trying to patch up First, recall that I tried to consider the analogue for cup products, $[X, K(G, n)] \times [X, K(G, m)] \to [X, K(G, m+n)]$ by looking at the map $K(G, n) \times K(G, m) \to K(G, n + m)$ defined by visualizing $Y = K(G, n) \times K(G, m)$ as a torus, gluing $(n+1)$-balls and $(m+1)$-balls along the meridian and longitude of $Y$, respectively
so that $\pi_n$ and $\pi_m$ vanishes, and then added more cells above $(m+n)$-th level so that higher homotopy groups vanishes. Now, this doesn't work : when I glued the $(m+1)$ and $(n+1)$-balls, that may automatically attract nontrivial homotopy groups at $\pi_{n+k}$ and $\pi_{m+k}$ levels. So one also have to attach cells of dimension between $n > m$ and $n+m$. However, that also adds some stuff to $\pi_{m+n}$, so we don't know if $\pi_{m+n} = G$ anymore. :(
I mean, what you just said is wrong already. $K(G,n) \times K(G,m)$ is not the desired codomain. You want $K(G,n) \wedge K(G,m) = (K(G,n) \times K(G,m))/(K(G,n) \vee K(G,m))$. This is $(n+m-1)$-connected and has the desired $\pi_{n+m}$. Now add higher cells. We talked about this before.
gluing those cells essentially glues a cone over a copy of $K(G, n) \vee K(G, m)$ with it. That's precisely $K(G, n) \times K(G, m)$ quotiented out by that wedge, no?
@MikeMiller Um, what is your advice? I don't see a difference between $K(G, n) \wedge K(G, m)$ and my space, but let's agree that it's not my space. I don't know how to compute homotopy groups of smash products. :S
I mean, it's not clear to me why $\pi_{n+1}(K(G, n) \wedge K(G, m))$ is trivial, say. My homotopy theory is weak.
@MikeMiller Whoops! My space is not $K(G, n) \wedge K(G, m)$, you are right. The wedge product has no cells below $m + n$, so it's $(m+n-1)$-connected.
