# Examples where it's useful to know that a mathematical object belongs to some family of objects

For an expository piece I'm writing, it would be useful to have good examples of the following phenomenon:

(1) ${\cal X}$ is a parameterized family of somethings. (Varieties, schemes, manifolds, rings, groups, whatever.)

(2) ${\cal X}_0$ is an member of this family.

(3) ${\cal X}_0$ has some interesting property $P$ that, on the surface, appears to have nothing to do with the inclusion of ${\cal X}_0$ in the family ${\cal X}$.

(4) Nevertheless, the only (or best'', or simplest, or most natural) way to prove that ${\cal X}_0$ has property $P$ is to invoke the existence of the family ${\cal X}$.

So, to put this in the form of a question, what are some good examples of this phenomenon? I'm especially (but not exclusively) interested in examples that could be explained to an undergraduate.

(It was hard to find the right tags for this question; feel free to change them.)

• Perhaps giving one example of such a phenomenon would be useful to people trying to answer this question. – Daniel Litt Sep 1 '11 at 1:28
• Possibly related: mathoverflow.net/questions/40005/… – Timothy Chow Sep 1 '11 at 1:35
• @Daniel: Perhaps giving a family of such examples would be even better? – J.C. Ottem Sep 1 '11 at 10:14
• Vassiliev's approach to finite type invariants gives invariants of individual knots (or links) from the topology of the space of all knots. This strategy generalizes to spaces of maps $M\to N$ without complicated singularities (here $M,N$ are smooth manifolds) and it works best when the set of "very singular" maps, which one wishes to discard, has codimension $\geq 2$. A very readable introduction to all this is Vassiliev's ICM 1994 talk. – algori Sep 1 '11 at 12:07
• Perhaps you could cite this post in the expository paper!! – Spice the Bird Sep 2 '11 at 1:45

Here's another example from algebraic geometry.

For every $g \geq 3$, there exists a smooth curve of genus $g$ with no automorphisms.

To prove this, we first prove the seemingly weaker claim that there exists a stable curve of genus $g$ without automorphisms. This is easy: take any curve of genus 2, pick a point $p$ on it. Since the automorphism of this curve is finite the point $p$ has only finitely many conjugates under the action of the automorphism group, so we can choose another point $q$ which is not conjugate to $p$. Glue $p$ and $q$ together to get a stable curve of genus 3 with no automorphism. Gluing any number of pairs of points together on this curve produces automorphism-free curves of any genus.

Then we invoke the existence of the Deligne-Mumford compactification of $\mathcal M_g$. For in a family of curves, the automorphism group is an upper semicontinuous function, so in particular the condition of having no automorphisms is open. But we know that there is a point in $\overline{\mathcal M}_g$ corresponding to a stable curve with no automorphism, so there must in fact be a whole Zariski open set of curves with no automorphism. On the other hand, smoothness is also an open condition and $\overline{\mathcal M}_g$ is irreducible, so we are done.

• I like this a lot. – Steven Landsburg Sep 2 '11 at 6:51
• It's probably worth emphasizing that this is not an odd example, but instead an ubiquitous technique in algebraic geometry to prove properties of a "general" curve (or higher-dimensional variety). – Arend Bayer Sep 2 '11 at 22:36
• Can you write down such a curve? – Martin Brandenburg Sep 16 '11 at 14:40

In a way, you can think of calculus as using exactly this sort of idea. Say you are interested in determining the area of the region consisting of points (x,y) with $a \le x \le b$ and $0 \le y \le f(x)$ . It turns out the best thing to do is to consider the family of regions $S(c)$ consisting of points (x,y) with $a \le x \le c, 0 \le y \le f(x)$, because then you can associate to this family the function A(c) = area of S(c). Then we know A'(c)=f(c), and this is a much easier problem to solve (and then just evaluate at c=b) than, say, trying to calculate the area of a single region directly, say by exhaustion.

• I think this is a really great example since it is at an undergraduate level and is an aspect that is hard to get students in calculus to appreciate. I would make this situation even more concrete: in order to find the area under y = sin x over the interval [0,pi], the simplest solution is to consider the more general problem of finding the (signed) area over [0,t] for all t and then we can use calculus, specializing to t = pi at the end. – KConrad Sep 2 '11 at 12:18
• Moreover, the same issue arises in differential calculus. To find the max/min points for a function, we want to see where the tangent lines to its graph have slope 0 (necessary condition). In order to find the places where the slope is 0, we use differential calculus to find the slope of the tangent line everywhere along the graph first. Only then do we set the result equal to 0. Thus we are considering a horizontal tangent line as lying in the family of all tangent lines in order to locate the horizontal tangent lines. – KConrad Sep 2 '11 at 12:39

This is similar to Dan Petersen's answer, but more elementary. A fact I always mention when talking to students about matrix groups is that the diagonalisable matrices (over $\mathbb{C}$) are dense in the space of all matrices.

This means that various things can be checked just on the diagonalisable matrices. For example, to establish the formula $\det(\exp(A)) = e^{\mathrm{tr}(A)}$ for $A$ a complex-valued matrix: observe that both sides define continuous functions $M_n(\mathbb{C}) \to \mathbb{C}$, and they clearly agree on the diagonal matrices, hence they also agree on the diagonalisable matrices (as $\det$ and $\mathrm{tr}$ are conjugation-invariant), so by continuity they agree on all matrices.

Of course, this example contains within it another application on the principle in question: the proof does not work if we insist on staying within matrices over $\mathbb{R}$ (the diagonalisable matrices are no longer dense), so even to prove the result over $\mathbb{R}$ it is easier to generalise to $\mathbb{C}$ first.

• I'm not sure it's easier to do over the complex numbers. The function det(e^{tA}) solves $\frac{df}{dt}= \mbox{tr }A f$ with $f(0)=1$. The solution to the latter equation is unique since $\frac{d}{dt}(f e^{−t \mbox{tr }A} )=0$. Or you can just use the product definition $\mbox{\det } e^A =\lim e^{n \log \det(1+A/n)}$ and again differentiate $\det$ at the identity to conclude. Isn't proving this density observation you've quoted is a little harder than just differentiating $\det$ at the identity? – Phil Isett Sep 6 '11 at 1:08
• Proving the density observation is nearly trivial. It's easy to see that a sufficient condition for diagonalizability is to have $n$ distinct eigenvalues, so the non-diagonalizable matrices are contained in the closed subvariety of $M_n(\mathbb C)$ defined by the vanishing of the discriminant of the characteristic polynomial. And clearly the complement of a hypersurface is dense. – Dan Petersen Sep 8 '11 at 14:16
• Phil: Your proof using differentiation is another perfect example of how a property of a single matrix $A$ is proven by considering a whole family (actually, the one-parameter family $tA$). – darij grinberg Dec 15 '11 at 18:15

There is a common proof strategy in algebraic geometry which works like this. Here is an example:

Problem: Show that the variety of smooth plane cubics which have flexes at $(0:0:1)$, $(0:1:0)$, $(1:0:0)$ and $(1:1:1)$ is smooth.

Now, you may or may not be able to do this by direct computation, but I think you will agree that is nontrivial. So the following slick proof may impress you.

Proof: Let $U$ be the variety of smooth cubics (an open subset of $\mathbb{CP}^9$). Let $X \subset U \times (\mathbb{CP}^2)^4$ be the variety of ordered pairs: (smooth cubic, quadruple of flexes). We are interested in the fiber of $X$ over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. I claim that the easiest way to show that this is smooth is to use the whole family $X$.

Since a smooth cubic has $9$ distinct flexes, $X$ is a $9^4$-fold cover of $U$, and is thus smooth. Let $\pi$ be the projection $X \to (\mathbb{CP}^2)^4$. By Sard's theorem (or the analogous algebraic result), the generic fiber of $\pi$ is smooth.

But the group $PGL_3$ acts compatibly on $X$ and $(\mathbb{CP}^2)^4$, so the set of points $z$ in $(\mathbb{CP}^2)^4$ for which $\pi^{-1}(z)$ is smooth is $PGL_3$-invariant. The $PGL_3$ orbit through $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is open and dense, so any dense $PGL_3$ invariant set must meet $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$. So we deduce that the fiber over $((1:0:0), (0:1:0), (0:0:1), (1:1:1))$ is smooth, as desired.

• Terrific --- exactly what I had in mind, except perhaps for the easily-explainable to-undergraduates part. Thank you! – Steven Landsburg Sep 1 '11 at 2:48
• (PS---"Exactly what I had in mind" should not be misread to mean that I had this example in mind. This is, instead, exactly the kind of thing I was looking for. – Steven Landsburg Sep 1 '11 at 2:49
• Nice. How does the fact that the fiber of $X \to U$ has constant cardinality $9^4$ imply that it is a covering? – Bruno Martelli Sep 1 '11 at 10:42
• Your right that I am glossing over steps. In more (but not complete detail): $X$ is closed in $U \times \mathbb{CP}^4$ (exercise). Thus the map $X \to U$ is proper. Proper plus finite fibers means a finite map. Finite map with all fibers the same cardinality is etale (in characteristic $0$). Etale and finite is a covering map. – David E Speyer Sep 1 '11 at 11:25

All the examples of Terry Tao's "tensor power trick" can be seen as an example of this. To be concrete, I'll take a proof of the maximum modulus principle which Mike Steele points out in the comments.

Let $\gamma \subset \mathbb{C}$ be a simple closed curve, let $z$ be a point inside $\gamma$ and let $f$ be an analytic function defined on the interior of $\gamma$, extending continuously to $\gamma$. Let $M$ be the maximum value of $f$ on $\gamma$. I claim that $|f(z)| \leq M$.

Proof: Let $L$ be the length of the curve $\gamma$, and let $r$ be the distance from $z$ to the closest point of $\gamma$. Then Cauchy's theorem gives $f(z) = (2 \pi i )^{-1} \int_{\zeta \in \gamma} f(\zeta)d\zeta/(\zeta-z)$ so $|f(z)| \leq LM/(2 \pi r)$. This is similar to the desired bound but, unless $\gamma$ is a circle an $z$ its center, $L$ will be bigger than $2 \pi r$.

To fix the problem, consider the family of functions $f(z)^n$, as $n$ ranges through the positive integers. Then the same proof shows $|f(z)|^n \leq L M^n/(2 \pi r)$. Taking $n$-th roots and sending $n$ to $\infty$, we conclude that $|f(z)| \leq M$.

Showing that a certain number is equal to some infinite sum of numbers using power series. For example, showing that $\ln(2)=1-\frac{1}{2}+\frac{1}{3}-\cdots$ or $e=2+\frac{1}{2}+\frac{1}{3!}+\cdots$.

• I took a course in deformation theory where this was the first (baby) example of understanding a number by looking at it's deformations in some big family. – Spice the Bird Sep 2 '11 at 1:39

This is perhaps too "computational" and not "structural" enough for your purposes, but there are many instances where an integral or a determinant is most simply evaluated by introducing an extra parameter.

(Actually, as I write this, I'm wondering if maybe some of these "computational" examples can be interpreted geometrically if one looks at them in the right way...)

• Actually, this is more "computational" and less "structural" than what I was envisioning, but it's convinced me that my vision was too narrow. Thanks. – Steven Landsburg Sep 1 '11 at 1:43
• Or lots of parameters! Consider, for example, a special case vs. the general case of the Vandermonde determinant. – Qiaochu Yuan Sep 1 '11 at 5:11

I don't know if this qualifies, because like Daniel Litt, I'm not sure I understand the question. Nonetheless...

Frequently in algebraic geometry to show $\chi_0$ has some open niceness property, e.g. being Cohen-Macaulay, we make a family $\chi$ all of whose fibers are isomorphic to $\chi_0$ (perhaps by being translates under a group action), except for one "bad" one, $\chi_\infty$. Then show that $\chi_\infty$ is good, thereby indirectly inferring that $\chi_0$ must be good.

(Niceness properties seems to almost exactly be the ones that are open in families. A counterexample is given by "normal crossings divisor", a deformation of which may not have normal crossings.)

A current trend in number theory is understanding how modular forms (or more generally automorphic representations) can be put into $p$-adic families, as well as their associated galois representations. One application of this is as follows:

Suppose you know that a certain semisimple reducible representation

$\begin{pmatrix} \chi_1 &0 \\ 0& \chi_2 \end{pmatrix}$

fits into a family of representations

$\begin{pmatrix} a(g) &b(g) \\ c(g)& d(g) \end{pmatrix}$

which is generically irreducible (so the reducible representation is somehow a degeneration of the family). Then one can sometimes use the existence of the function $b$ (or $c$) to "deform" the reducible representation into a nontrivial extension:

$\begin{pmatrix} \chi_1 &\ast \\ 0& \chi_2 \end{pmatrix}$.

While this may or may not impress you, this is exactly the technique used by Wiles in his proof of the Iwasawa Main Conjecture for totally real fields.

• And how Skinner–Urban prove the Main Conjecture for GL(2). Wiles also proved Fermat by considering the family of all Galois representations congruent to a given one mod $p$. – Rob Harron Sep 2 '11 at 2:18

One example I know of is the Kazdan-Warner identity: if $(M,g)$ is a Riemannian surface, with scalar curvature $R$, and $X$ is a conformal Killing vector field, then $\int_M R \operatorname{div}(X)\ dg = 0$. You would think that this identity could just be proven by clever integration by parts, but as far as I know there is no such "local" proof known of this identity. The only proof I know of is to invoke the uniformisation theorem to show that $(M,g)$ is conformal to a constant curvature manifold (for which the identity is just an easy integration by parts). One then deforms the original manifold conformally to the constant curvature manifold and verifies that $\int_M R \operatorname{div}(X)\ dg$ is constant with respect to this perturbation.

(The same method can be used to prove more basic global identities of this type, such as the Gauss-Bonnet theorem, although for such identities, intrinsic proofs not requiring deformation are certainly available.)

Among other things, the Kazdan-Warner identity can be used to help classify asymptotic shrinking Ricci solitons in two dimensions, which ends up being one of the components of Perelman's proof of the Poincare conjecture (see my notes on this topic).

This is a bit late, but I thought of another example. Looking at the initial coefficients of the power series $$\sqrt{1+x} = 1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3 - \frac{10}{243}x^4 + \frac{22}{729}x^5 + \cdots$$ we may guess that each coefficient has a 3-power denominator. From Taylor's formula, the coefficient $$c_k$$ of $$x^k$$ is $$c_k = \frac{(1/3)(1/3-1)\cdots(1/3-k+1)}{k!}.$$ To prove the fraction $$c_k$$ has a 3-power denominator, we will show its denominator is not divisible by any prime $$p \not= 3$$ by viewing $$1/3$$ as a $$p$$-adic limit of positive integers: $$1/3 = \lim_{r \rightarrow \infty} a_r$$, where $$a_r$$ is the $$r$$th truncation of $$1/3$$ in its $$p$$-adic expansion. Since the polynomial $$x(x-1)\cdots (x-k+1)/k!$$ is continuous for the $$p$$-adic topology, $$c_k$$ is the limit of $$a_r(a_r-1)\cdots(a_r-k+1)/k!$$ as $$r \rightarrow \infty$$, and these terms are all integers, hence $$p$$-adic integers, so the limit is a $$p$$-adic integer, and thus $$c_k$$ has no $$p$$ in its denominator.

For more discussion of this, see my answer at https://math.stackexchange.com/questions/136206/show-that-sqrt1t-lies-in-mathbbz1-2-t/136288#136288.

In a similar spirit, identities for formal power series (like the chain rule for $$f(g(X))$$ when $$g(X)$$ has constant term 0) may be easier to prove for polynomials using induction and then one can appeal to $$X$$-adic continuity to pass to the limit and get the same result for power series. Here we insert a power series into the family consisting of its truncated polynomial approximations along with the power series itself as the limit.

I don't think this fits the specific criteria you've set out, but I believe it is in the right spirit. It involves using a mathematical concept all undergrads learn to give an elegant solution to a very down to earth puzzle that anyone can understand.

Background

The puzzle is about the game of SET. If you know how the game is played, don't bother reading this paragraph or the next. In the game of SET, you have a number of cards, and on each card there's a simple image. The image consists of 1, 2 or 3 identical shapes. These shapes can be one of three shapes: diamonds, squiggles, and ovals (on a given card, all the shapes are the same, but different cards may have different shapes). On a given card, all the shapes are coloured the same, in one of three possible colours: red, green, and purple. And finally, on a given card, all the shapes are shaded in the same manner, in one of three different possible manners: outlined, filled in, or hatched.

The game involves multiple players, and the goal for each player is to collect the greatest number of sets, where a set is a collection of three cards such that for each category (shape, number, colour, shading), all three cards are either the same or all different. So for example if you have three cards, all three of which are made up of 2 squiggles, but are three different colours and shaded in the three different ways, then this is a set. At the beginning of the game, the cards are all laid out, and then players form sets as quickly as the can until all the cards are gone, and the winner is the one who has made the most sets.

The Puzzle

Suppose you're playing Set, and as you're approaching the end of the game, you notice there are only 11 cards remaining. There ought to be a multiple of three remaining, so you figure one card must have gone missing from your deck. Looking only at the 11 remaining cards, can you figure out which card is missing?

Alternatively, draw up an example of some 11 cards, and ask the student to solve the same problem. When presented with a specific example, the student may not think that there is a general solution, and so making the problem more concrete actually adds an interesting twist to the problem, it makes it a nice test of analytical thinking.

The Solution

The set of all cards can be regarded as a 4-dimensional vector space over $\mathbb{F}_3$ in an obvious way. Why give it this vector space structure? Because a collection of three cards is a set iff, regarded as vectors in this vector space, their sum is the zero vector. Also, since you're supposed to be able to make sets until you've eliminated all the cards, you know that the sum of all the cards is 0. So:

0 = sum of cards put into sets so far + sum of remaining 11 cards + missing card

Clearly the sum of the cards put into sets is 0, and so:

missing card = -(sum of remaining 11)

Consider the following weak form of Darboux's theorem:

Let $\omega$ be a symplectic form defined on a neighborhood of $0$ in $\mathbb{R}^{2n}$, meaning that $d \omega =0$ and $\omega$ has no kernel as a skew symmetric form on the tangent space at $0$. Then, passing to a possibly smaller neighborhood $U$ of $0$, there are coordinates $x_1, \ldots, x_n$, $y_1, \ldots, y_n$ on $U$ such that $\omega = dx_1 \wedge dy_1 + dx_2 \wedge dy_2 + \cdots + dx_n \wedge dy_n$ on $U$.

Proof: Make a linear change of coordinates so that $\omega(0)$ is $du_1 \wedge dv_1 + du_2 \wedge dv_2 + \cdots + du_n \wedge dv_n$. Define $\omega_0$ to be the differential form $du_1 \wedge dv_1 + du_2 \wedge dv_2 + \cdots + du_n \wedge dv_n$ and set $\omega_t = t \omega + (1-t) \omega_0$. We will use this family of differential forms.

So all of the $\omega_t$ give the same skew symmetric form at $0$. Also, $d \omega$ and $d \omega_0$ are both $0$. Shrinking our neighborhood of the origin to be contractible, by Poincare's theorem, we have $\omega = d \theta$ and $\omega_0 = d \theta_0$ for some one forms $\theta$ and $\theta_0$.

All of the differential forms $\omega_t$ are nondegenerate at $0$. Shrinking our neighborhood, we can arrange that they are nondegenerate everywhere on $U$. Thus, there is a vector field $X_t$ such that $\omega_t(X_t, Y) = \theta(Y) - \theta_0(Y)$ for any vector $Y$.

Flowing along $X_t$ defines an automorphism of a neighborhood of $0$. One can compute that this flow pulls back $\omega_0$ to $\omega_t$. So the pullbacks of the coordiante functions $u_i$ and $v_i$ are our desired $x_i$ and $y_i$.

I like the calculus answers above such as evaluating integrals and infinite series as values of functions satisfying differential equations. The following are much more specialized but seems also to illustrate the point in the question, of deciding something about an individual case by virtue of its membership in a family.

Mumford's proof that a smooth cubic surface has exactly 27 lines is of this type. He proves that cubic surfaces form a family, that the number of lines is finite and constant precisely on the connected family of smooth surfaces. Then he computes the number on one smooth surface.

Hershel Farkas conjectured in 2004 that one can decide whether a 4 dimensional principally polarized abelian variety with an isolated "vanishing theta null" (an isolated double point on its theta divisor at a point of order two) is or is not the Jacobian of a genus 4 curve, by noting whether the rank of the double point is three or four.

One way to check this is as follows:

i) in the universal family of 4 dimensional ppav's, generic Jacobians are characterized by having two ordinary double points on theta.

ii) an isolated double point at a point of order two splits into two odp's under deformation in any reasonably general family if and only if its rank is three.

Thus a ppav with the stated hypotheses is a limit of Jacobians, hence is a Jacobian.

The infinitesimal approach to the Torelli problem is another example. Classically one tried to show the Torelli map t:M(g)-->A(g) (assigning to a curve its Jacobian) is injective by using information in the image point t(C) = J(C) to reconstruct C, either the theta divisor or a birational model. This did not take advantage of the fact that C belongs to a family M(g) and that t is defined on the whole family.

The infinitesimal approach uses the fact that the Torelli map is defined also at all curves near C, and thus uses information in the image of the differential, or in the kernel of the codifferential t*, to reconstruct C.

This makes the problem of finding equations for C much easier since in general the kernel of t* contains all quadrics containing the canonical model of C, while from J(C) one can only reconstruct such quadrics of rank ≤ 4.

In some sense all proofs by induction have this form. I.e. one proves a non trivial statement by checking it in a trivial case n=1, then bootstrapping up one case at a time by induction. Indeed the algebraic geometry example above of finding a curve of every genus with no automorphisms is a geometric form of induction from the easy case of genus 2. The inductive step is the upper semicontinuity statement.

• Concerning the comment at the end about proofs by induction, the case $n=1$ is not always trivial, but sometimes is in fact where all the meat of the argument lies. Consider the proof that if $A$ is a UFD then $A[X_1,\dots,X_n]$ is a UFD. Only the case $n=1$ is hard. – KConrad Jun 12 '14 at 1:56
• Well, you can start the induction with the trivial $n=0$ case. – Tom Goodwillie Jun 12 '14 at 2:19

In differential geometry, the use of geometric flows seem to fit in this framework. One could see Hamilton-Perelman proof as an instance of this phenomenon, but maybe the lack of canonicity of Ricci Flow with surgery disqualifies it.

However, Hamilton's theorem on 3-manifolds with $Ric>0$ and Brendle-Schoen differentiable sphere theorem would be interesting examples maybe. In these case, Ricci Flow (without surgery) creates a deformation between your initial metric (with $Ric>0$ or strictly quarter pinched) and metric of constant curvature $1$, since only quotients of the sphere bear such metrics, the initial manifold was a quotient of a sphere.

I'm sure there are also interesting examples in harmonic map heat flow (starting with the work of Eels and Sampson) but I don't know them well enough.

There are $12$ elliptic plane cubics with fixed $j$ invariant passing through $8$ points in the plane ($4$ or $6$ if $j = 0, 1728$).

Proof: Use the pencil of cubics argument to show that there are $12$ nodal cubis pasing through $8$ points. Then move the result along the family $\mathcal M_{1,1}$.

A somewhat amusing example in the theory of automorphic forms is that the constant $1$ function is the residue of an Eisenstein series (considered as a meromorphic vector-valued function). So, any $1$, anywhere, can be replaced by $Res_{s=1}E_s$.

As a quick application, take a cusp form $f$ for $GL_2$ over a number field $k$ that generates an irreducible representation $\pi_f$. Identifying $1$ with the residue of an Eisenstein series, the $L^2$-norm of $f$ turns into the residue of a Rankin-Selberg L-function times the norm of the first Fourier-Whittaker coefficient of $f$, $\rho_f(1)$: $$||f||_{L^2}=\int_X |f|^2\ dx=Res_{s=1}\int_X |f|^2E_s\ dx=|\rho_f(1)|^2Res_{s=1}\Lambda(s,\pi_f\otimes\tilde \pi_f)$$ If we want, we can break up the Rankin-Selberg L-function a bit to get something more tangible. $$||f||_{L^2}=|\rho_f(1)|^2L_\infty(1,\pi_f\otimes\tilde \pi_f) L(1,Sym^2\pi_f)Res_{s=1}\zeta_k(s)$$ where $L_\infty$ is a certain product of Gamma functions (whose parameters depend on $\pi_f$), $\zeta_k$ is the Dedekind zeta function of $k$ (whose residue at $s=1$ we know from the class number formula), and $L(1,Sym^2\pi_f)$ is the symmetric-square L-function of $\pi_f$, which is more mysterious (though a certain amount is known about how it changes as $\pi_f$ varies). Typically, we assume either $||f||_{L^2}=1$ or $\rho_f(1)=1$, so the formula turns information about L-functions into information about Fourier-Whittaker coefficients or $L^2$-norms (or vice-versa).

I've written a paper (or two) about collection $\mathcal{R}$ of all pointed topological spaces $Y$ satisfying the property $\mathrm{map}_*(X,Y) \sim *$ (for fixed $X$). The interesting fact is that if $\mathcal{R}$ contains $S^{2n+1}$ for all sufficiently large $n$, then $\mathcal{R}$ contains all finite-dimensional simply-connected CW complexes. My proof works by induction on the cone length of $Y$ with respect to the collection of all wedges of spheres, and the passage from spaces of cone length $n$ to those of cone length $n+1$ requires information about a huge array of spaces with cone length $n$ (or less), and not just the ones in the given cone decomposition.

So: knowing that $Y$ has a finite cone decomposition leads to the apparently unrelated conclusion that $\mathrm{map}_*(X, Y)\sim *$; and the proof requires the collection approach.

• What (sort of objects) are you talking about? – Laurent Moret-Bailly Dec 16 '11 at 7:51
• pointed topological spaces -- I clarified above. – Jeff Strom Dec 17 '11 at 2:05

One of the proofs of the riemann mapping theorem!

The mapping is constructed by considering a family of mappings with a certain property. Then we prove that the maximal member is the mapping we are looking for. It almost feels like cheating.

The proof is tedious but certainly accessible to undergraduates.

Some problems in graph theory can be solved most elegantly by appealing to matroid theory; good luck in trying to explain matroid theory to undergraduates though.

When proving that every natural number is a sum of four squares, one puts the number of possibilities (counted in the right way!) $r_4(n)$ into a series $\sum r_4(n)q^n$ which turns out to be a modular form (a "familiy of coefficients") of some level and can therefore be written as a sum of suitable Eisenstein series identifying the $r_4(n)$ with simpler terms.

When proving Iwasawa's theorem on class numbers in $\mathbf{Z}_p$-extensions, one considers the whole field tower at once.

To prove that the Hilbert series (the generating function of the sequence of dimensions of homogeneous components) of a finitely generated commutative graded algebra is a rational function, the easiest way is to prove more general statement (Hilbert-Serre theorem) stating that the Hilbert series of a graded $k[x_1,\ldots,x_n]$-module is a rational function.

• I think this is one of a long list of cases when proving a result about algebras is harder than proving the same property of modules over these algebras. It seems to tell us that algebras are just modules that accidentally happen to be over themselves... – darij grinberg Dec 15 '11 at 18:12
• True. Though in this case, you rather consider $A/I$ as an $A$-module, of course. – Vladimir Dotsenko Dec 16 '11 at 7:05

When proving the weak Mordell-Weil theorem for an abelian variety $A$ over a number field $K$:

One way is to prove that one has a monomorphism $A(K)/n \hookrightarrow H^1(K,A[n])$, and one can prove that the image of $A(K)/n$ in the cohomology group consists of classes unramified outside a finite set, and finish off using the finiteness of the class number and the finite generation of the unit group.

A better way is to remember the unramifiedness beforehand (so you don't have to analyse the image in $H^1(K,A[n])$): Use that $A$ can be spread out to an abelian scheme $\mathcal{A}/\mathfrak{O}_{K,S}$ over an open subset of the ring of integers of $K$. By further enlarging the finite set of places $S$, one can assume that $n: \mathcal{A} \to \mathcal{A}$ is surjective on the étale site of $\mathfrak{O}_{K,S}$ with kernel $\mathcal{A}[n]$. Now take the long exact cohomology sequence associated to $0 \to\mathcal{A}[n] \to \mathcal{A} \to \mathcal{A} \to 0$ and use the Néron mapping property $\mathcal{A}(\mathfrak{O}_{K,S}) = A(K)$ and again the finiteness of the class number and the finite generation of the unit group.

Basu's theorem in statistics. You have a parametrized family of probability distributions. Call the parameter $\theta$. The probability distribution of a random variable $X$ depends on $\theta$.

A thing like $X-\theta$ is a random variable but is not a "statistic", since its value---not just its probability distribution---depends on $\theta$. Think $\theta$ as some unobservable quantity like the average height of 21-year-old males and $X$ as the observable $n$-tuple of heights of 21-year-old males in a random sample. The average of the heights in the sample would be an observable random variable, i.e. it depends on the pair $(X,\theta)$ only through $X$.

Now a statistic $g(X)$ (i.e. an observable random variable) whose probability distribution does not depend on $\theta$ is called an ancillary statistic.

A statistic $g(X)$ that "admits no unbiased estimator of zero" is called a complete statistic. The phrase "admits no unbiased estimator of zero" means there is no function (not depending on $\theta$) $h$ such that the expected value of $h(g(X))$ is $0$ regardless of the value of $\theta$.

A statistic $g(X)$ is called a sufficient statistic if the conditional probability distribution of $X$ given $g(X)$ does not depend on $\theta$. (Intuitively, $g(X)$ contains all information available in $X$ that is relevant to drawing inferences about the value of $\theta$.

Basu's theorem says every complete sufficient statistic is independent of every ancillary statistic.

I have seen at least one instance in which the only convenient way to prove two particular random variables are independent is to embed their probability distribution into a parametrized family of probability distributions in such a way that one of them becomes a complete sufficient statistic and the other becomes an ancillary statistic.

I'm suddenly realizing that I don't entirely remember the details of that example. I think it was something quite similar to Lemma 3 on page 7 of this: http://www.stat.duke.edu/~sd83/Research/das-dey-mult-gamma.pdf Or Example 1 on page 1712 of this: http://www.jstor.org/stable/pdfplus/2239877.pdf?acceptTC=true

I have to like and suggest ideas connected with the Seifert-van Kampen theorem for the fundamental group and its extensions to groupoids and higher groupoids. An anomaly in traditional approaches, centred on the fundamental group, was that this theorem did not compute the fundamental group of the circle, which is THE basic example in algebraic topology. The reason is the the circle cannot be represented as the union of two path connected sets with path connected intersection. The solution was to use many base points rather than just one, and so to work in the context of groupoids; this dates from 1967 and a fairly recent account is in the book Topology and Groupoids.

This extension led to ideas of using higher groupoids in homotopy theory, and so to define higher homotopy groupoids, with higher order Seifert-van Kampen theorems. By 1984 this led to the idea of a nonabelian tensor product of groups which act on each other, see the bibliography. As an example, if $M,N$ are normal subgroups of the group $P$, then the commutator map $[\;,\;]: M \times N \to P$ is a biderivation and so factors through a universal biderivation, a morphism $\kappa: M \otimes N \to P$. For example if $M=N=P$ then Ker $\kappa$ is isomorphic to $\pi_3SK(P,1)$. Thus taking groupoids seriously in algebraic topology has led to new algebraic ideas.

Consider a graph $G$ where each vertex has degree at least two. Prove that this graph has a cycle.

One could prove this by arguing that we can start at a point, and just follow edges, until we meet an already visited vertex, but I think the following argument is slicker:

$G$ belongs to the complement of the set of forests, since every tree has a leaf, but $G$ does not. Forests are the only graphs that do not have cycles. Hence, $G$ has a cycle.

• hmm, how do you prove the claim that forests are the only graphs that do not have cycles? (also, I assume you mean finite graphs?) – bananastack Jun 12 '14 at 14:30

I have two related sorts of example to suggest, probably exhibiting my categorical bias vs. the analytic/geometric-topology weight of the preceding examples.

### Galois-theoretic

Let $P\in K[x]$ be an irreducible (separable) polynomial of degree $n$ over a field $K$, and consider the extension field $L = F[x]/(P)$. It is fruitful (i.e., it's the whole categorical view of Galois theory) to view $L$ as one of serveral isomorphic extensions, with possibly non-trivial automorphism group relative to $F$. In one sense, everything you might wish to know about $L$ and $K$ is contained in your ability to understand $K$ and $P$, but it's evidently fruitful to look at commutative diagrams $L//F\to L'//F$

### Topological-algebraic

So you want to understand some algebraic gadget? A vector space/lie algebra/group? Then study bundles of such things!

For example, it can be a bit stifling to think of a Lie algebra of a Lie group as the tangent space at the identity. Sometimes it's better to think of it as either space of half-invariant vector fields; and then the Lie bracket is given by the Lie bracket! (ha-hah!) But there is plenty of room to study other things, exploting the bundle of isomorphic Lie algebras.

Other times it's helpful to study the universal bundle $EG\to BG$ of some discrete group $G$. These realise the group both as a particular fiber and as the fundamental group of $BG$ at your favourite basepoint, and so you learn things about the group (e.g. related homology functors) by studying the whole collection of groups $\pi_1(BG,b)$ and torsors $EG_b$ for $b\in BG$.

• ...Some discrete group $G$ (else the fundamental group is the set of path components of $G$) – David Roberts Sep 2 '11 at 3:34
• And $EG_b$ is a $G$-torsor, not a group. – David Roberts Sep 2 '11 at 3:34
• Well, yes and yes, but... So, we work in the category of pointed spaces, and then one fiber at least IS a group because it has the basepoint of $E$ in it... but the rest of them, yes, torsors... strictly, a $\pi_1(BG,b)$-torsor, which is ismorphic to $G$ in conjugate ways. – some guy on the street Sep 2 '11 at 17:13