Tricki
a repository of mathematical know-how

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  • tao 9 years 32 weeks ago To calculate a contour integral, use theorems rather than direct calculation

    It may be better to use "contour integral" here rather than "path integral" - I think the latter is more often used for the integrals in quantum field theory than in complex analysis.

    Also, there's something odd with the use of the "Add" tag at the end of the general discussion that I don't see how to easily fix.

  • tao 9 years 32 weeks ago How to use the method of stationary phase to control oscillatory integrals

    I renamed it (and added another example, while I'm here...)

  • Sameed 9 years 32 weeks ago Elementary number theory front page

    That's a nice one. Another example might be a proof of the irrationality of \sqrt{2} or maybe a proof that the diophantine equation x^{4}+y^{4}=z^{2} has no positive integral solutions.

  • gowers 9 years 32 weeks ago Elementary number theory front page

    Definitely agreed. A nice example would be the infinite-descent proof that every positive integer is a sum of four squares.

  • gowers 9 years 32 weeks ago How to use the method of stationary phase to control oscillatory integrals

    Vicky, if you spot more of those it would be great. I've just found one of my own: using the law of trichotomy. I'm about to change it to something more descriptive and imperative.

  • Sameed 9 years 32 weeks ago Elementary number theory front page

    I believe Fermat's 'descente infinie' deserves a mention here. What's your opinion?

  • Vicky 9 years 32 weeks ago How to use the method of stationary phase to control oscillatory integrals

    Is there perhaps a more helpful title for this article? If you don't know what the method of stationary phase is, then it's entirely unclear what this article is about and which problems the method tackles!

  • gowers 9 years 32 weeks ago Using generators and closure properties

    Obviously that proof is not appropriate in this article. But it might be nice to write an account of it, decide what general trick it illustrates, and write a geometry article about that general trick. Then one could link from this article saying that there was an alternative, more conceptual proof.

  • Sameed 9 years 32 weeks ago Using generators and closure properties

    An even better approach might be to take the Riemann sphere point of view.It is then natural that the Mobius transformations are the rotations of this sphere and the preservation of circles follows from the fact that stereographic projections map circles to circles.

  • gowers 9 years 32 weeks ago To show that a group element is non-trivial, show that it has a non-trivial action or image

    One other thing is that in an article I wrote on group presentations I gave two proofs that the group with two generators a and b and the relations a^3=b^3=1 is infinite. One was a direct attack and the other was to find an action on \N with an infinite orbit. It's similar enough that some kind of cross-linking might be good (or alternatively one might just repeat the example here – my view is that it's fine if the same example appears in more than one article). I won't do anything yet, given that the structure of this article is not yet clear.

  • wilton 9 years 32 weeks ago To show that a group element is non-trivial, show that it has a non-trivial action or image

    Yes. BS(1,2) is isomorphic to the semidirect product of Q_2 and Z, where Z acts via multiplication by 2. Which gives yet another way to see do it, by considering actions on pairs of these things.

    I rather like the idea of including all these different approaches. There are interesting stories to be told about which generalize to other, similar, groups. For instance, Tim's nice observation works because BS(1,2) is residually finite. But BS(2,3) isn't, so it won't work there.

  • tao 9 years 32 weeks ago To show that a group element is non-trivial, show that it has a non-trivial action or image

    One can map a into the translation x -> x+1, and b to the dilation x -> x/2, and it's looking like this is a faithful representation. Well, it's still a good example of the general principle, at least...

  • tao 9 years 32 weeks ago To show that a group element is non-trivial, show that it has a non-trivial action or image

    More generally, I guess one can go look for nontrivial matrices a which are conjugate to their square a^2 (not hard to guess such matrices, after thinking a little about what their eigenvalues must be like), and that gives a whole bunch of useful representations. Easy in hindsight :-) [and, annoyingly enough, I actually knew this; this is how one proves the Frobenius lemma that the irreducible representations of SL_2(F_p) are large. I should have known better than to try to attack the word problem directly...]

    I'm not sure how to structure the article now; I feel that my pedestrian way of doing things is instructive, if perhaps only in the negative sense of showing what not to do. Perhaps we can give several different proofs of the same result?

    I didn't know about Britton's lemma, so I learned something today!

  • gowers 9 years 32 weeks ago To show that a group element is non-trivial, show that it has a non-trivial action or image

    Isn't the relation ab=ba^2 satisfied in the dihedral group D_6? In other words, if we add the relations b^2=a^3=1, we get a group where we know that a and b are non-trivial (e.g. because it acts faithfully on a triangle). So there's our homomorphism. Or am I missing something?

  • wilton 9 years 32 weeks ago To show that a group element is non-trivial, show that it has a non-trivial action or image

    Well, there's a theory for dealing with this. When you use it, then things are certainly simpler!

    The group you're considering here is the Baumslag–Solitar group BS(1,2). The standard way to prove that a is non-trivial is to observe that the group is an HNN extension of \Z and appeal to Britton's Lemma. You're basically reproving Britton's Lemma in this case.

    Another (morally equivalent) point of view is to say that the object on which your group acts is the universal cover of a space constructed by gluing one end of a cylinder to itself via a 2-to-1 covering map. The fact that the circle in the cylinder lifts to a line in the universal cover shows a has infinite order.

    I'd be tempted to do an example like the (2,3,7)-triangle group, and construct the action on the hyperbolic plane.

    Perhaps we can segue this article into a discussion of Britton's Lemma, graphs of groups etc. Hmmm.

  • tao 9 years 32 weeks ago Complete the square

    Hmm, I thought this was a common term, but I just googled for it and the first mathematical occurrence of it was... one of my own web pages. But "shifting the contour" is a reasonably common term. It's not _quite_ the same as complex substitution (it's based on Cauchy's theorem for contour integrals, rather than the change of variables formula), but it of course combines very well with such substitutions. But it also can be used independently of complex substitution, e.g. to evaluate integrals such as \int_{-\infty}^{\infty} \frac{e^{ix}}{x^2+1} by shifting the contour to a big semicircle etc.

    I link to this nonexistent article in a couple other places, so I think I'll start a stub on it, and perhaps return to it later.

  • Sameed 9 years 32 weeks ago Look at small cases

    Maybe you are right. There are loads of problems that can be solved by doing specific cases only. Lucas's identity in elementary number theory might be a good example.(Adding the terms in the Pascal's triangle diagonally yields the Fibonacci numbers)

  • gowers 9 years 32 weeks ago Look at small cases

    I know I suggested putting this example here, but now I look at it carefully, I wonder whether it really is a prime example of Gel'fond's advice. It seems to me that the basic proof (even if you dress it up with matrices) is this: expand out both sides and observe that you get the same answer. And it's not clear to me that doing that for n=3 is any easier than doing it for the general case.

    A truly convincing example of this trick is one where the only realistic way of doing the general case is to do small cases and spot a non-obvious pattern.

  • Sameed 9 years 32 weeks ago Look at small cases

    I have added the example mentioned above. It might still require some editing though. You may state some prerequisites for the above example but I personally think that it's fairly straightforward and that elementary linear algebra would suffice.

  • b.jacelon 9 years 32 weeks ago Discretization followed by compactness arguments

    The above example is arguably an instance of a more general discretization strategy: that of replacing compact spaces by finite complexes. For example, every compact manifold has the homotopy type of (and hence the same cohomology as) a finite CW complex. Perhaps more in the spirit of this article, every compact subset X of \mathbb{R}^n can be written as the intersection of a decreasing sequence (X_j) of compact sets, each of which is a finite simplicial complex. (Just take successive triangulations and keep the bits that intersect X.) One can hope to use simple arguments (such as induction on the number of simplices) to deduce some property about the X_j, and then use a limiting argument to deduce the property for X. For example, the C*-algebra C(X) of continuous functions on X is the direct limit of the algebras C(X_j). If one is, say, interested in proving something about the K-theory of X, then the relevant continuity property is that K_*(\lim C(X_j)) = \lim K_*(C(X_j)) (and a useful observation might be that K_*(C(X_j)) is finitely generated).

  • gowers 9 years 32 weeks ago Complete the square

    I've started an article about making complex substitutions in integrals. My guess is that this is what you mean by contour shifting, but I have not put this link in yet because I am not quite sure. If it is, and if this is the standard name for the technique, then I'll add a remark to that effect to the article.

  • tao 9 years 33 weeks ago Complete the square

    I put up some quick examples to reflect them. Feel free to edit, of course.

  • Anonymous (not verified) 9 years 33 weeks ago Complete the square

    I've seen this used multiple times, in a context similar to the usual proof of Cauchy-Schwarz inequality, where completing the square of a real-valued expression shows that some grouping of terms is non-negative.

  • Sameed 9 years 33 weeks ago Look at small cases

    Absolutely.There should be a string of articles.Hopefully, many more articles would come up with an explanation.Tricky is going to be a great ride.

  • gowers 9 years 33 weeks ago Look at small cases

    I marked it as a stub because it had neither a "General discussion" section nor any examples. It is certainly interesting and valuable advice, but doesn't quite count as "mathematically interesting information" in the intended sense, because it doesn't yet contain any actual mathematics. But if, say, you were to add the example that is explained on the page you suggest, then it would instantly lose its stub status.

    Just to repeat, it is possible for a Tricki article to contain something interesting, but still to count as a stub. So you should not take the stub designation as in any way suggesting that Gel'fond's suggestion is uninteresting. But I would see the quotation as the beginning of what should ultimately be a quite long article with several examples: a principle this important deserves nothing less.