What do named “tricks” share?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

There are a number of theorems or lemmas or mathematical ideas that come to be known as eponymous tricks, a term which in this context is in no sense derogatory. Here is a list of 10 such tricks (the last of which I learned at MO):

Edit: List augmented from the comments and answers:

• the Eilenberg–Mazur swindle
• the Parshin trick
• the Atiyah rotation trick
• the Higman trick
• Rosser's trick
• Scott's trick
• the Craig trick
• the Uhlenbeck trick
• the Alexander trick
• Grilliot's trick
• Zarhin's trick [For any abelian variety $A$, $(A \times A^{\vee})^4$ is principally polarizable.]

Further Edit. And although my original interest was in eponymous (=named-after-someone) tricks, several non-eponymous tricks have been mentioned, so I'll gather those here as well:

• the determinant trick
• the kernel trick
• the W-trick

Some of those listed above do not yet have Wikipedia pages (hint, hint—Thierry).

I (JOR) am not seeking to extend this list (although I would be incidentally interested to learn of prominent omissions), but rather I am wondering:

Is there some aspect or trait shared by the mathematical ideas or techniques that, over time, come to be named "tricks"?

I am aware this is a borderline question; feel free to close if it unduly distracts.

-

10
"An idea which can be used only once is a trick. If one can use it more than once it becomes a method." Quoted from books.google.co.uk/… – Andrey Rekalo Dec 4 2010 at 4:45

8
That construction of Whitney ought to have a more dignified name. – Tom Goodwillie Dec 4 2010 at 4:58

4
The Eilenberg Swindle is another good one. – Sean Tilson Dec 4 2010 at 5:36

5
You could add the Eilenberg Swindle to your list. (A Swindle sounds even more disreputable than a Trick.) – Charles Rezk Dec 4 2010 at 5:36

1
Why is the page linking to the Whitney trick linking to a "Global Oneness" site? <br> <br> Why do they even have a page on Whitney embedding on a site about spirituality? I'm very confused. – Simon Rose Dec 4 2010 at 5:42
show 16 more comments

8 Answers

How about the following (which I think applies to some of these tricks but not others): a trick is something whose usefulness is not fully captured by any particular set of hypotheses, so it would limit its usefulness to write it down as a lemma.

-

3
This is my favorite answer. It captures the notion that a trick is one of the essentially human contributions to math, a creative touch that isn't part of some axiomatic or algorithmic approach that something more mechanical might invent. It's generalizable but not categorizable. – Ryan Reich Dec 4 2010 at 22:04

2
I like this answer, but I'm not sure it really distinguishes between tricks and methods. For instance, I don't know of a lemma one could write down that would do justice to the probabilistic method in combinatorics. But actually I'm tempted to say that "Take a random X" feels like a trick that can be applied over and over again, perhaps for this very reason. – gowers Dec 5 2010 at 22:24

3
@Ryan: I must also leap to the defence of the poor old computer. I don't see any reason in principle that a clever program could not invent mathematical tricks. (I think a proof that a program couldn't invent tricks would prove that humans couldn't either.) – gowers Dec 5 2010 at 22:26

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

I'll take a stab at this.

I think that the term "trick" is used to connote a technique that achieves something as if by magic. If I make a cake by combining flour, sugar, and eggs and baking, that is simply a standard technique, but if I make the cake by putting the ingredients into a top hat and waving a wand over it, that is a magic trick. The way that the Weyl unitary trick makes complex groups behave like compact ones seems like a magic trick. (For those of you trying to follow this half baked analogy, the cake is complete reducibility of representations, the oven is integration, and the hat is ... uhhh.... )

-

13
Ha ha, half baked :) – Zev Chonoles Dec 4 2010 at 7:56
I think this is the answer. – timur Dec 4 2010 at 15:29

3
The hat is orthogonality? – Qiaochu Yuan Dec 4 2010 at 16:12

1
The hat is Fourier transform, of course! – Paul Siegel Dec 6 2010 at 12:20

One well-known trick is a way to evaluate the Gaussian integral $G = \int_\mathbb{R} e^{-x^2}dx = \sqrt{\pi}$ by writing $$G^2 = \left(\int_\mathbb{R} e^{-x^2}dx\right)\left(\int_\mathbb{R} e^{-y^2}dy\right) = \int_{\mathbb{R}^2} e^{-(x^2+y^2)}dxdy$$ which when transformed to polar coordinates becomes $$G^2 = 2\pi \int_0^\infty e^{-r^2} r dr = \pi \int_0^\infty e^{-u} du = \pi$$ via the substitution $u=r^2$. It appears this idea is due to Poisson.

In a 2005 note in the American Mathematical MONTHLY, R. Dawson has observed that this is a trick that only works once; there are no other integrals that can be evaluated by this method. Specifically:

Theorem. Any Riemann-integrable function $f$ on $\mathbb{R}$, such that $f(x)f(y) = g(\sqrt{x^2+y^2})$ for some $g$, is of the form $f(x)=ke^{ax^2}$.

See: Dawson, Robert J. MacG. On a "singular" integration technique of Poisson. American Mathematical Monthly 112 (2005), 270-272.

So if a technique is a trick that works twice, this one is definitely still a trick.

-

1
Note that this trick has something in common with the Rabinowitsch, Cayley and Eilenberg tricks and probably some others on the list: in order to solve a $k$-dimensional problem, you go into more than $k$ dimensions. This also seems to be a distinguishing feature of things called tricks. – darij grinberg Dec 5 2010 at 22:04

2
I think Dawson was anticipated by James Clerk Maxwell. A result called Maxwell's theorem says that if $X_1, \dots, X_n$ are independent real-valued random variables and their joint density is spherically symmetric, then all of them are normally distributed, i.e. the probability density of each of them is a Gaussian function. – Michael Hardy Dec 5 2010 at 22:17

1
Constantine Georgakis, "A Note on the Gaussian Integral", Mathematics Magazine, February, 1994, page 47 This paper gives what its author considers "a better alternative to the usual method of reduction to polar coordinates" for evaluating this integral. See en.wikipedia.org/wiki/Gaussian_integral . – Michael Hardy Dec 5 2010 at 22:19
While the specific form $f(x) f(y) = g(\sqrt{x^2+y^2})$ applies only to Gaussians, there are further uses of this kind of transformation: in one direction, to the volumes of Euclidean spheres in higher dimension (imagine you already know $\Gamma(1/2)$ but not the area of a circle); in another direction, to the usual proof of $B(x,y) = \Gamma(x) \Gamma(y) / \Gamma(x+y)$; combining these, to Dirichlet integrals; and by analogue, to the relation between Gauss sums and Jacobi sums — and probably others that don't come to mind right now. – Noam D. Elkies Jul 9 2011 at 6:11
@Michael Hardy: that "better alternative" in the paper by Georgakis in fact is due to Laplace. See york.ac.uk/depts/maths/histstat/…. – KConrad Mar 26 2012 at 3:15

http://en.wikipedia.org/wiki/Rosser%27s_trick

"A technique is a trick that works twice"

Note that Grothendieck never published his proof of the Grothendieck-Riemann-Roch theorem because he felt that the proof depended on an "astuce" (trick) rather than flowing naturally.

-

5
It is interesting that the accepted English translation of "astuce" is "trick", whereas that of the adjectival form "astucieux/euse" is "clever" or "astute". This seems to give the concept of a trick a better connotation in French than in English. (I am tempted to load up Lord of the Rings with the French subtitles on to see whether Gollum accuses Frodo and Sam of being "astucieux".) – Pete L. Clark Dec 4 2010 at 16:32

2
@Pete: you're absolutely right. For instance, to translate the expression "dirty trick" into French, you would not use the word "astuce" because "astuce" has a uniformly positive connotation of praise (yes, even in that Gollum context). – Thierry Zell Dec 4 2010 at 16:53

Scott's Trick is called a "trick" because it is not actually necessary for the completion of the proof in which it is involved; however, without the trick the proof is massively more tedious. Although the other tricks may not have a widely-agreed-upon-reason for being a trick, I suspect that they may be called such for similar reasons.

-

1
Same for the Rabinowitsch trick. Also as pointed in another answer, the Rabinowitsch trick works only once (although similar ideas must be used, albeit in less famous circumstances). – Thierry Zell Dec 4 2010 at 16:10
Well, it’s only unnecessary if you’re assuming AC (which was not, if I understand right, quite as entrenchedly automatic among set theorists back in the ’60s as it is now). It’s necessary if you want to talk about cardinalities — and more generally, construct quotients of classes by equivalence relations — in ZF and many other set theories. – Peter LeFanu Lumsdaine Dec 6 2010 at 5:31

I've long known the adage that a "trick" works only once whereas a "method" works in multiple instances, or maybe is expected to work in yet unanticipated future instances.

But there's another POV: a trick is something whose efficacy cannot be anticipated, but only by hindsight is seen to work. All methods I've seen of finding $\int \sec x \ dx$ are "tricks". I've always leaned toward viewing unanticipatability as the essence of trickhood.

But I also like Qiaochu Yuan's answer.

-

2
There is an algorithm to find the integral of any rational function of trigonometric functions: use a rational parameterization (e.g. using tan x/2), then use partial fractions (or a residue computation, etc.). I don't see how this is a "trick" in any sense. It is a direct corollary of the fact that the circle is birational to the projective line. – Qiaochu Yuan Dec 4 2010 at 17:37

1
There's the argument often used in calculus texts: $$\sec x = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \frac{du}{u}$$ where $u$ is the last denominator above. That seems like a "trick". I've seen other methods but I don't remember what they are right now. One of them seemed to take some of the mystery out of it, but it was still a trick. But I can't show a residue computation to a first-year calculus class. The "Weierstrass trick" (was it Weierstrass?) of $\tan(x/2)$, etc., does seem a bit more like a "method". – Michael Hardy Dec 4 2010 at 17:50

2
Yes, I agree that that argument is trick-like. The rational parameterization of the circle is not. It is a natural and beautiful geometric argument (just taking the slope of a line between two points) and I think it could easily be presented to a first-year calculus class. – Qiaochu Yuan Dec 4 2010 at 18:05

4
I had a student come up with a different way to do this, on the fly, on the final exam. Write $$\sec(x) = {1\over \sqrt{1 - \sin^2(x)}},$$ then substitute $u = \sin(x)$ and use partial fractions! – Jeff Strom Dec 5 2010 at 2:31

2
I usually integrate this function differently (using something trick-like, but natural one): $$\frac{dx}{\cos(x)} = \frac{\cos(x)dx}{{\cos^2(x)}} = \frac{d(\sin(x)}{1 - {\sin^2(x)}}$$ – Ostap Chervak Jun 14 2011 at 15:00
show 5 more comments

Would regarding a scalar as the trace of a $1\times1$ matrix be considered a "trick"?

Here's an occasion where that's useful:

http://en.wikipedia.org/wiki/Estimation_of_covariance_matrices#Maximum-likelihood_estimation_for_the_multivariate_normal_distribution

-

While tricks have names because they wind up being associated with with some particular mathematician, tricks are tricks because something important goes on "behind the curtain."

For instance, to prove $$(a_1 b_1 + \cdots + a_n b_n)^2 \leq ({a_1}^2 + \cdots + {a_n}^2) ({b_1}^2 + \cdots + {b_n}^2),$$ write \begin{align*} A &= ({a_1}^2 + \cdots + {a_n}^2)\\ B &= (a_1 b_1 + \cdots + a_n b_n)\\ C &= ({b_1}^2 + \cdots + {b_n}^2), \end{align*} then we must show $$B^2 \leq AC.$$ Equality clearly holds when $A = 0$. Otherwise, since $\mathbb{R}$ has no negative squares, for all $x \in \mathbb{R}$, $$0 \leq (a_1 x - b_1)^2 + \cdots + (a_n x - b_n)^2.$$ Expanding the squares, $$0 \leq Ax^2 - 2Bx + C.$$

The quadratic expression vanishes whenever $$x = \frac{B}{A} \pm \sqrt{\left(\frac{B}{A}\right)^2 - \frac{C}{A}}.$$

If $x = \dfrac{B}{A}$, then $$0 \leq A\left(\frac{B}{A}\right)^2 - 2 B\left(\frac{B}{A}\right) + C = \frac{B^2}{A} - 2 \frac{B^2}{A} + C = - \frac{B^2}{A} + C,$$ thus $$B^2 \leq AC.$$

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 12, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9324443936347961, "perplexity_flag": "middle"}