nah, dunbother about it. i am , most of the time, the most unhelpful in this chatroom. [see, @anon asked something, and i was at the point of answering something very wrong]
d'you know the proof of Hilbert's finiteness of invariants theorem? asking because i want to hear the general idea of the proof
Don't know if it is the same proof as the one you are reading. The one I know of is basically generalising a construction of the elementary symmetric polynomials.
So, for the proof, you start of by replacing $\Sigma$ with $G$ and construct $n$ polynomials of that form, by replacing $x_1$ with each of the algebraic generators of $A$.
quick non-serious side-question : is it natural for invariant theorists to write group elements acting on things in a wacky way? Eisenbud uses $(g, x) \mapsto g^{-1}(x)$, you are using $(g, x) \mapsto x \cdot g$ (i.e., right-action), etc.
But you can show that $A^G$ is finitely generated as a module over this subalgebra.
So if you take those ESP coefficients together with the module generators, you can get finite set of generators.
They might not be algebraically independent though.
So it might not be as neat as the ESP for $\Sigma_n$.
For the finite generation of $A^G$ as a module, you consider $x_1$ as a root of the polynomials $p(t)$.
And similarly for the rest: $x_j$.
It lets you write $x_1^n$ as a linear combination of lower powers of $x_1$ over the subalgebra, because the coefficients are the generators of the subalgebra.
@BalarkaSen Right um... as @anon has said, the group action was originally defined as linear maps on $V=<e_1,\cdots,e_n>$. The group action on $A$ is defined by extending the definition on the dual space: $x_j(g\cdot e_k) = [x_j \cdot g](e_k)$.
if you interpret a function $X\to Y$ as a subset of $X\times Y$ (its graph) and have $G$ act on the first coordinate of $X\times Y$ hence act on functions, you get the contragredient action. Since $\{(gx,f(x)):x\in X\}$ equals $\{(x,f(g^{-1}x)):x\in X\}$ we have $(g\cdot f)(x)=f(g^{-1}x)$. (This is when people want a left action.)
@ADG What's the underlying set? I think you usually use the $\epsilon-\delta$ definition. Split the $\epsilon$ in two and then use minimum of the two $\delta$-s.
what you said earlier was right, choosing a minimal set of basis for a module is a tough job. The correct way to go about it is to pick a basis for $M/\mathfrak{m}M$ over $A/\mathfrak{m}$, and then pull this back to a set of generators for $M$.
@Ted so you mean to say if sup A+sup B<a+b+2e and a+b<=k that would mean supA+sup B<k=2e for all e>0 so supA+supB-k=0?? Sorry for any unintentional rudeness
@ADG: You're writing too much stuff. It follows from our lemma above if you just substitute. But sometimes it's better to have simpler things to look at.
You have a typo with $k=2e$ where it should be $k+2e$.
@TedShifrin I was in a ergodic number theory seminar a few days ago. I didn't understand much, plus the lecturer said there is a natural interpretation of all this using "horocycle flows on modular curves". So much to learn :'(
@BalarkaSen Uh... I got this from some notes my supervisor gave me and I also looked at a book by Benson (LMS). Let me see if there are any online proofs. The theorem might also be called Hilbert-Noether.
The weird thing is that so much extraterrestrial machinary is used to prove a quite number theoretic statement, @Ted (it's the Oppenheim-Davenport conjecture, which says for every nondegenerate quadratic form $Q$ of $n \geq 3$ variables which isn't a rational multiple of a rational quadratic form, $Q(\Bbb Z^n)$ is dense in $\Bbb R^n$)
well, at least the proof of FLT uses things surrounding algebraic number theory (iirc?). the proof of Oppenheim-Davenport uses ergodic theory -- heaps of analysis out of nowhere..!
There's all sorts of algebraic geometry stuff in FLT. But there are LOTS of connections between number theory and hyperbolic geometry, so I'm not the least surprised.
Twin primes stuff also used analysis. It should not be surprising that some numver theoretic statements take analysis to prove, some take algebraic geometry, ...
OK anyways I did that something similarly. now i'm finding sup and inf of 2<x^2<3 for x in Q but i don't think there is one. sqrt2 and sqrt3 are irr so we can find another and contracdiction
If you wrote proofs in linear algebra and some in algebra, then proofs in analysis are not so bad. But the sentences in analysis are more complicated than the sentences in algebra.
@BalarkaSen, the official description says this: "Some topics covered include the Sylow theorems, solvable and simple groups, Galois theory, finite fields, Noetherian rings and modules."
ill do my homework of real analysis later and study for inorganic quiz and maybe do physics practical filework of diffraction and interference of light and maybe then sleep i have a class tom at 9 AM @Rememberme life here is tough
@Rememberme A short exact sequence is a chain of homomorphisms $A \stackrel{f}{\to} B \stackrel{g}{\to} C$ such that $f$ is injective, $g$ is surjective, and $\ker g = \text{im} f$
the point is that certain things get very easy once you write down a short exact sequence. for example, to write down all the information give in the short exact sequence mentioned above in words, you have to say : $f : B \to C$ is a map, which has kernel $A$, so that it induces the isomorphism $B/A \cong C$ by the 1st isom theorem.
that's too much to write down.
also, one of the beautiful facts about short exact sequences is the splitting lemma. i mean, you can write down the statement of the splitting lemma in words if you want but that'd be too much writing.
to illustrate, to write "$f : A \to B$ is surjective" in paper you need 17 characters, but to write "$0 \to A \stackrel{f}{\to} B$" you just need 6 characters.
@evinda gut, danke, obwohl ich zurück nach Italien umziehen musste (wegen burokratischer Probleme konnte ich nicht diesen Jahr in Deutschland studieren :( )
One of the integrals I came across these days (during my studies) is $$\int _0^1\int _0^1\text{Li}_2(x y) \text{Li}_2((1-y) x) \ dx \ dy$$
that can be turned into a series, or can be approached by using the integration by parts, but these
ways do not look like as a promising way to go, or I migh...
@TedShifrin got any good problems to teach me about adjoints? Was reviewing my quantum mechanics today and think I need to understand adjoint and self-adjoint operators better
Does someone have a PDF with a proof of the normality of Champernowne's constant? The original article on the journal of london mathematical society's website needs to be payed
@DanielFischer This is something I came across about 2 years ago, but I've never been able to understand: For what sort of functions can $f(a+z)$ be expanded in the form $f(a+z) = \sum_{k=0}^{\infty} c_{n} e^{-kz}$? In other words, for what sort of functions could you find another function with a Maclaurin series expansion such that $f(a+z) = g(e^{-z})$? One author claimed this was possible for $(a+z)^{-n}$.
@RandomVariable Since $z\mapsto e^{-z}$ has period $2\pi i$, it is necessary that $f$ has period $2\pi i$. And if $f$ is holomorphic and $2\pi i$-periodic on a strip $s < \operatorname{Re} z\rvert < t$, it induces a holomorphic function $g$ on the annulus $e^{-t} < \lvert w\rvert < e^{-s}$ via $w = e^{-z}$.
Now when we look at the desired expansion, $g$ has zero principal part, so $g$ extends to a holomorphic function on the disk $\lvert w\rvert < e^{-s}$, and that means that $f$ has an analytic continuation to the half-plane $s < \operatorname{Re} z$, and such that $\lim\limits_{\operatorname{Re} z \to +\infty} f(z)$ exists.
So, since $(a+z)^{-n}$ is not $2\pi i$-periodic, that doesn't work.
So I'm looking at this website which states:
One of the questions an instrutor [sic] dreads most from a mathematically unsophisticated audience is, "What exactly is degrees of freedom?" It's not that there's no answer. The mathematical answer is a single phrase, "The rank of a quadratic form....
I need to "distort" a sigmoid function, here is its wolfram alpha code: y(x) = (1/(1 + e^(-8(x - 0)))) from -1 to 1 .....I need the same "kind" of function except want it to start "ramping up" further along the x-axis. That is, it should be close to zero until almost 0.6 on the axis..and then ramp up rapidly. Anyone know how to manipulate the above function to do that? Or can give me code for a function that does what I need?
**REALLY** basic linear algebra problem here. It has been years since I've done Gauss elimination. I am trying to calculate the rank of $$M = \begin{pmatrix} 1-\frac{1}{n} & -1/n & \cdot & -1/n \\ -1/n & 1-\frac{1}{n} & \cdot & -1/n \\ \cdot & \cdot & \cdot & \cdot \\ -1/n & -1/n & -1/n & 1-\frac{1}{n} \end{pmatrix}$$
Someone told me on SE that a standard way to do this is to do Gauss elimination
@DanielFischer I honestly don't remember, but I know if I were to multiply $nM$ by $-1$, I would get a diagonal matrix with eigenvalue $(1-n)$, multiplicity $n$... I think. grabs book
@DanielFischer Well, actually, wouldn't that just be $1$ with multiplicity $n$? (complete guess)
@DanielFischer I worked with the $2 \times 2$ case. So for a matrix of all $1$s in the $2\times 2$ case, I get $\lambda = 0, 2$ each with multiplicity $1$. But the $\lambda = 0$ is trivial, if I recall, and is not included as an eigenvalue
@DanielFischer If my algebra is right, $\lambda = 1, 1-\sqrt{3},1+\sqrt{3}$. So I would guess that for a $n \times n$ matrix entirely of $1$s, there are three eigenvalues: $1$, $1+\sqrt{n}$, and $1 - \sqrt{n}$
This is a silly question, which would possibly have a silly answer : Consider the affine space $\Bbb A^n_3$. Take two linked circle in your space. This is a variety, and the corresponding coordinate ring is $k[x, y, z]/((y^2 + z^2 - 1)(x^2 + (y-1)^2 - 1))$. Can we say something about linking of the two subvarities -- the two circles -- from the coordinate ring?
@DanielFischer I might have badly misinterpreted what he was talking about. He might be simply assuming that $f(a+it)$ can be expanded in a Fourier series of the form $f(a+it) = \sum_{k=0}^{\infty} c_{k} e^{-ikt}$ that is valid for all $t \in \mathbb{R}$.
Actually, I think not -- the sheaf of rings shouldn't detect anything about linking. The coordinate ring (global section of the ring) might just be coordinate ring of two disjoint things.
@Clarinetist Diagonalization is often paramount when considering eigenvalues because if $A$ is diagonalizable there is a similarity transform such that $A = U^{-1} D U$ where $D$ is diagonal and has the eigenvalues of $A$ as its diagonal elements
OK, so a bit more nontrivial question : what should be the correct algebraic way to detect linking of the two subvarities? Is there even a way to do it?
From the coordinate ring of the subvariety? No. I don't see any reason in general that you should have a good notion of linking.
Nor really why you would want one.
I think it's folly to jump into structure sheaves without first being comfortable with the coordinate rings themselves. That doesn't take much time. But that's just my opinion.
I dunno. You can extract a lot of information about the variety itself from looking at it sheaf of rings (e.g., as I got to know a few days ago, you can tell about "smoothness" of a variety from regularity of the stalk of the standard sheaf at that point). So a natural question would be if you can tell something about the subvarities from it too.
@Huy the new release of euclid the game is ready for download
btw, anyone else here interested in testing out a mathematical iOS app ?
> Euclid: The Game gamifies the 2300 years old book "The Elements" written by the ancient Greek mathematician Euclid in Alexandria.
> Euclid's Elements has been referred to as the most successful and influential textbook ever written. The first level of the game is exactly the first theorem of this ancient book. Throughout the levels you unlock constructions, once you prove you are able to make them.
> The web game (http://euclidthegame.com) is played by 500.000 users in in 213 different countries, we hope that as much people will enjoy the much improved iOS game (to be released september…
Little info about the game above. We are now actively searching for beta testers. You can see screenshots of the app in the facebook page: facebook.com/euclidthegame
The source code is open source (MIT licensed), and will be available from github.com/euclidthegame in a couple of days.
@MikeMiller Ah, well, if you (or anyone else) is interested, you can send your email adress (associated with your appstore account) to beta@euclidthegame.com. And I will make sure you can download it.