90
$\begingroup$

The Whitney graph isomorphism theorem gives an example of an extraordinary exception: a very general statement holds except for one very specific case.

Another example is the classification theorem for finite simple groups: a very general statement holds except for very few (26) sporadic cases.

I am looking for more of this kind of theorems-with-not-so-many-sporadic-exceptions

(added:) where the exceptions don't come in a row and/or in the beginning - but are scattered truly sporadically.

(A late thanks to Asaf!)

  • 3
    There are many examples coming from Waring's problem: Every sufficiently large number is a sum of at most 7 cubes, for example. But, removing "sufficiently large" requires changing the 7 to a 9.2012-08-24
  • 1
    @Andres: Please allow me to count Waring's problem as *one* further example.2012-08-24
  • 134
    **Theorem:** All natural numbers are larger than $10$, except $0,1,2,3,4,5,6,7,8,9$.2012-08-24
  • 0
    Go ahead. Ellison's and Vaughan-Wooley surveys (both mentioned on the wikipedia page) are excellent.2012-08-24
  • 0
    @Asaf: Your theorem is not of **this kind**. (Forgive me for not having defined precisely what *this kind* actually is - just by examples.)2012-08-24
  • 11
    @Hans: I know it's not. I'm trying to prod the discussion into helping you shape the definition in your head, so you could write it up. If I thought this is something you're looking for, I'd post it as an answer.2012-08-24
  • 0
    @Asaf: If you know what I'm aiming for - could you **please** help me shaping it in my head, so I can write it up?2012-08-24
  • 2
    I just knew that you're not aiming for this. I'm not very good at reading minds in general... I also think that what I said above had to be said.2012-08-24
  • 0
    It would be so great if people were better at reading other people's minds ;-) @Asaf: I did hope you not only guessed what I was *not* aiming for but also in the positive. I was hopeful that my two examples were enough. But alas - they were not. For the moment, I am helpless.2012-08-24
  • 20
    @Asaf We need to add 10 to the list of exceptions, no?2012-08-24
  • 0
    Wouldn't theorems with no exceptions be in this list too? (Such as all integers are divisible by 1)?2012-08-24
  • 1
    Specifically, are we talking about theorems of the form "every $x$ with properties $A$ has property $B$, except for the following explicit list"? Or do you allow "for every $x$ with property $A$, all but finitely many $y$ that satisfy $B$ have property $C$" (where the list of exceptions can depend on $x$).2012-08-24
  • 1
    And what about "every $x$ with $A$ has $B$, except for finitely many" (but we don't necessarily know which, or even how many)?2012-08-24
  • 0
    @Hans: commenting on the answers you upvoted was _completely_ unnecessary. Please don't do this.2012-08-24
  • 5
    Actually all answers to [this question](http://math.stackexchange.com/q/178183/34930) should qualify.2012-08-24
  • 0
    @Qiaochu: I guess you are right. (I found it a not so bad way of bookkeeping: which answers meet the intention of the question. But I understand that you don't like this, and I will not do this anymore.)2012-08-24
  • 2
    @AsafKaragila: Of course your **theorem** is a fallacy. It fails for number **10**.2012-08-24
  • 3
    @ypercube: It was pointed out before. Sadly, MSE do not allow editing comments after some time, and there is no point in deleting and re-posting it again. I am very happy that you were able to find a mistake in my sarcasm, it is *very* important to do that!2012-08-24
  • 1
    @AsafKaragila, Have we not yet learned the perils of sarcasm?2012-08-24
  • 1
    @Steve: Who are you? [Sir Lancelot](http://www.imdb.com/title/tt0071853/quotes?qt=qt0470578)?2012-08-24
  • 2
    Somehow, a lot of answers (even Asaf's marvellous theorem) are of the type “no object is of low complexity, except from a finite number of example”, which is not as satisfying as the examples given by the OP.2012-08-24
  • 2
    Many of the answers are not about sporadic exceptions, but rather the first n cases are exceptions before the theorem actually begins to work.2012-08-24
  • 0
    I don't think my own answer about Heegner numbers is really a good example either. All class numbers are greater than 1 is not a great theorem in the context of this question. All it is saying is that the class number considered as a function f(n) is equal to 1 at these values of n and not equal to 1 at other values of n.2012-08-24
  • 0
    @AsafKaragila, Am not.2012-08-24
  • 0
    Someone *please* make this community wiki!2012-08-26
  • 0
    @SteveD: Uhh it's CW for **two** days now...2012-08-26
  • 0
    Weird, when I answered it wasn't indicated as such...2012-08-26
  • 0
    Interesting question. But, I think you've sort of contradicted yourself. If we take classical logic for granted, and a theorem has any exceptions, then such theorems have instances where they end up false. That implies contradiction, and thus we don't have a theorem in the first place. The statement of these theorems actually hold in *all* cases, e. g. with "all primes are odd except 2" holds in all cases since the cases are 3, 5, 7, 11 ... I'd suggest you write instead something like "theorems which come as easily described by a pattern which admits of a few exceptions."2012-09-18

37 Answers 37

72

Every automorphism of $S_n$ is inner if $n \neq 6$.

P.S. That $S_6$ has an `essentially' unique outer automorphism is quite a non-obvious fact.

56

I've posted this answer to other questions, but it's worth repeating:

The topological manifold $\mathbb{R}^n$ has a unique smooth structure up to diffeomorphism as long as $n \neq 4$. However, $\mathbb{R}^4$ admits uncountably many exotic smooth structures.

Edit: Other thoughts:

  • The h-cobordism theorem holds for $n= 0, 1, 2$ and $n \geq 5$. It is open for $n = 3$ and false (smoothly) for $n=4$.

  • I've heard that many theorems in number theory only work for field characteristic $\neq 2$ (but my number theory background is sadly lacking).

  • $(\mathbb{Z}/n\mathbb{Z})^\times$ is cyclic iff $n = 1, 2, 4, p^k, 2p^k$, where $p$ is an odd prime, $k \geq 1$. (Granted, there aren't a finite number of exceptions here, but I wanted to mention it anyway.)

  • 0
    Well in Galois theory there is a simple test using the discriminant to see if the Galois group of an irreducible cubic is $A_3$ or $S_3$, but sadly it fails in characteristic 2 because $1 = -1$.2012-09-03
  • 0
    copied from wikipedia2014-04-08
  • 0
    @Herbert: Guilty as charged: some, but not all, of what I've written is indeed copied verbatim. Still, I don't really think that affects the quality of the answer. Also, I like your new username :-)2014-04-08
45

Carmichael's Theorem: Every Fibonacci number $F_n$ has a prime factor which does not divide any earlier Fibonacci number, except for $F_1 = F_2 = 1$, $F_6 = 8$ and $F_{12} = 144$.

This is a special case of his more general result: Let $P,Q$ be nonzero integers such that $P^2 > 4Q$, and consider the Lucas sequence $D_1 = 1$; $D_2 = P$; $D_{n+2} = P \cdot D_{n+1} - Q \cdot D_n$. Then all but finitely many $D_n$ have a prime factor which does not divide $D_m$ for any $m < n$; the only possible exceptions are $D_1$, $D_2$, $D_6$ and $D_{12}$.

  • 4
    It's amazing to me that the exceptions always (potentially) happen at the same members of the sequence regardless of the values of $P$ and $Q$! Is there a brief explanation of why this happens? Are there infinitely many $P,Q$ for which e.g. $D_{12}$ has no prime factors that don't divide $D_n, n\leq 11$?2012-08-24
  • 0
    @Steven: There is a fairly readable proof in [Minoru Yabuta, _A simple proof of Carmichael's theorem on primitive divisors_, Fib. Quart., vol.39, pp.439-443](http://www.fq.math.ca/Scanned/39-5/yabuta.pdf). The reason that these terms are the only possible exceptions is that the problem for all Lucas sequences can be reduced to an analysis of the Fibonacci sequence and the Fermat sequence ($M_1 = 1$; $M_2 = 3$; $M_{n+2} = 3 M_{n+1} - 2M_n$). In slightly more detail, to each Lucas sequence $D_n(P,Q)$ we can associate another sequence $E_n(P,Q)$. (cont...)2012-08-25
  • 0
    @Steven: (...inued) A sufficient condition that $D_n(P,Q)$ has a "primitive divisor" is that $E_n(P,Q)$ is "large enough". For each $n>2$ we can also show that either $E_n(1,-1)$ (Fibonacci) or $E_n(3,2)$ (Fermat) is the minimum for all $E_n(P,Q)$. In this way we reduce the problem to those two sequences. It does seem that the Fibonacci sequence is the only one where the 12th term has no primitive divisor.2012-08-25
40

Every simply connected subregion of $\mathbb C$ is conformally equivalent to the unit disk, with the exception of $\mathbb C$ itself.

30

For squarefree positive integer $d$, every imaginary quadratic field $\mathbb{Q}(\sqrt{−d})$ has class number greater than 1 unless $d$ is equal to one of the Heegner numbers: 1, 2, 3, 7, 11, 19, 43, 67, 163.

  • 3
    While this is true, I don't think it is a very good example of what was being asked, because there are analogous explicit results for other class numbers, and it is known that the class number tends to infinity as $-d$ tends to $-\infty$ (so that it is known that every given class number will belong to only finitely many imaginary quadratic fields).2014-01-19
30

Here's a neat result that holds for all finite groups except if the Monster shows up as a quotient:

Let $G$ be a finite group and $p,q$ primes dividing $|G|$. If $G$ contains no element of order $pq$, then either

  1. the Sylow $p$-subgroups or the Sylow $q$-subgroups are abelian, or
  2. $G/O_{\{p,q\}'}(G)$ is the Monster and $\{p,q\}=\{5,13\}$ or $\{7,13\}$.

[source]

25

Mitchell's theorem: A primitive complex reflection group is either the symmetric group $S_n \subseteq GL_{n-1}(\mathbb{C})$ or one of $34$ exceptions.

Definitions: a complex reflection group is a finite subgroup $W$ of $GL_n(\mathbb{C})$ for some $n$, that is generated by reflections. A reflection is an invertible matrix with codimension one fixed space. A complex reflection group $W \subseteq GL_n(\mathbb{C})$ acting on $V=\mathbb{C}^n$ is imprimitive if there is a decomposition $V=V_1 \oplus V_2$ such that for each $w \in W$, and $i=1,2$, $w(V_i) \subseteq V_j$ for some $j$. Otherwise it is primitive.

  • 0
    Was this proven before Todd-Shephard's theorem?2012-08-24
  • 1
    Well, to be fair I believe what Mitchell actually proved (long before Shepard-Todd wrote down the famous list) is that a primitive CRG is either the symmetric group in its reflection rep'n, or lives in dimension at most $8$. But given this theorem, producing the list of exceptions is sort of inevitable (but still interesting). I think Shepard-Todd probably get a bit too much credit for the classification of irreducible CRGs, when it was really a community effort.2012-08-24
23

How about the Big Picard theorem? http://en.wikipedia.org/wiki/Picard_theorem

If a function $f:\mathbb{C}\to \mathbb{C}$ is analytic and has an essential singularity at $z_0\in \mathbb{C}$, then in any open set containing $z_0$, $f(z)$ takes on all possible complex values, with at most one possible exception, infinitely often.

23

The only consecutive integer powers are 8 and 9.

This was conjectured by Eugène Catalan in 1844 and proved by Preda Mihăilescu in 2002.

21

Map Color Theorem: For any surface $\Sigma$ with Euler characteristic $c \leq 0$, with the exception of the Klein bottle, $$\chi(\Sigma) = \frac{1}{2} (7 + \sqrt{49 - 24c}),$$ where $\chi$ is the chromatic number.

19

There are several theorems in topology and geometry that hold for manifolds of all dimensions save 3 and sometimes 4. For example, of Euclidean space except $\mathbb{R}^3$ and sometimes $\mathbb{R}^4$. There are also some conjectures proved for $\mathbb{R}^n$, n ≠ 3 (resp {3,4}), but which are open questions for n = 3 ({3,4}). Wikipedia describes it thus:

[T]he cases $N = 3$ or $4$ have the richest and most difficult geometry and topology. There are, for example, geometric statements whose truth or falsity is known for all N except one or both of 3 and 4. N = 3 was the last case of the Poincaré conjecture to be proved.

Using the regular polytopes as an example:

  • There are exactly 3 regular (convex) polytopes, the n-simplexes, -cubes, and -orthoplexes which exist in all dimensions, except 3 and 4, which have several additional Platonic solids and regular polychora without higher analogues (ignoring the trivial cases of dimensions 1 and 2).
  • There are no regular non-convex polytopes except in dimensions 2, 3, and 4.

Further,

In a sense, 3- and 4-dimensional spaces are privileged. This has implications in the philosophy of physics: why did space (apparently) have 3 spatial dimensions rather than 2 or 527?

See also 3-manifold, special phenomena of 4-manifolds, low-dimensional topology.

15

Vinogradov's theorem that all but a finite number of odd numbers are the sum of three primes.

Not sure about the current state of Goldbach's conjecture.

A theorem at about my level of math: All primes greater than two are odd.

  • 2
    There are effective versions of Vinogradov's three-primes theorem: one due to Liu and Wang http://www.ams.org/mathscinet-getitem?mr=1932763 states that every odd integer $> \exp(3100)$ is the sum of three odd primes. Of course it's likely to be true for all odd integers $\ge 9$ (and that is the case if the Generalized Riemann Hypothesis is true), but there's still some distance to go before that can be verified.2012-08-24
  • 0
    I'm not sure the third one really counts. I could say "all primes other than three are not multiples of three." It's just that we don't have a special name for numbers that aren't multiples of 3.2012-08-24
  • 4
    How about this then: All but one or two or three of my statements are valid answers to the OP question.2012-08-25
  • 2
    Just the other day a proof of [Goldbach's weak conjecture](http://en.wikipedia.org/wiki/Goldbach%27s_weak_conjecture) was announced.2013-06-02
14

Polynomial equations of every degree have no general solution in radicals, with the exception of degrees 1, 2, 3 and 4.

13

Brownian motion is transient in every dimension, with the exception of dimensions 1 and 2.

13

Not sure if this fits the scope of the question, it's hardly as advanced as all the other suggestions but I find it quite elegant and surprising:

There exists no positive integer sandwiched between a perfect square and a perfect cube, with the sole exception of $26$ - equivalent to saying that the only solution of $a^2 \pm 2 = b^3$ is $(5, 3)$.

Of course this is just one among many Diophantine equations which admit only one solution, but it has a special significance when viewed from a less abstract point of view.

Another nice one:

For any prime $p$, the product of its primitive roots is congruent to $1$ modulo $p$, except for $p = 3$ for which the product is equal to $2$ (result due to Gauss).

  • 2
    I liked this fact a lot two years ago. I'm 28.2012-08-24
13

The only spheres which admit almost complex structures are $S^2$ and $S^6$.

11

The "sausage catastrophe" for finite sphere packings comes to mind:

For 1..55 spheres a linear "sausage" is the optimal packing, for higher numbers some cluster packing is optimal.

  • 1
    Optimal by what measure?2012-08-24
  • 1
    Smallest volume of a convex "wrapping". Look at http://de.wikipedia.org/wiki/Theorie_der_endlichen_Kugelpackungen (sorry, no english entry).2012-08-24
  • 1
    +1 not for the $\le 55$, but for the $57, 58, 63 \text{ and } 64$ other numbers of spheres where the sausage is still optimal.2012-08-28
  • 1
    Translated from German Wikipedia: "In 1992 Pier Mario Gandini and Jörg Michael Wills could show, that starting with 56 spheres (with exception of 57, 58, 63, 64) the linear order as sausage isn't the best one, but that in this case a cluster packing is better, which is not a sausage packing. Meanwhile it was shown that for the four exceptions the sausage packing isn't optimal either. So the sausage catastrophe must occur at least at 56 spheres."2012-08-28
11

Fermat's Last Theorem: For any positive integer $n$, except $n = 1, 2$, there is no solution of positive integers $(x, y, z)$ to the equation $$x^n + y^n = z^n.$$

  • 2
    It is somewhat interesting as a result on its own (but for this question, because it succeeds for $n \ge 3$, I don't consider it is relevant), but when dividing throughout by $z^n$ you get that all the paths $s^n+c^n=1$ only have rational points $(s,c)$ for $n=1,2$.2012-08-26
10

How about the Ax-Kochen theorem?

Every homogeneous polynomial of degree $d$ in $n$ variables with $n>d^2$ has a non-trivial zero in $\mathbb{Q}_p$ for all but finitely many $p$... and the finite set of exceptions depends on the degree $d$.

Here are some things we know about the finite exception set for various $d$: http://en.wikipedia.org/wiki/Ax-Kochen_theorem#Exceptional_primes

This is a special case of the more general fact that any first-order sentence (in the language of valued fields) which is true of all but finitely many Laurent series fields $\mathbb{F}_p((t))$ is true of all but finitely many $p$-adic fields $\mathbb{Q}_p$. Model theory gives us many more examples of the principle "true in char $0$" = "true in char $p$ for large enough $p$".

9

Here's a beautiful theorem of Peter J Cameron from the theory of designs:

Theorem. If symmetric $2-(v,k,\lambda)$ design $\mathscr{D}$ extends, then it is one of the following :

  1. $2-(4\lambda+3,\;\; 2\lambda+1,\;\; \lambda )$1
  2. $2-((\lambda+2)(\lambda^2+4\lambda+2), \;\;\lambda^2+3\lambda+1, \;\;\lambda)$
  3. $2-(495,39,3)$

This appeared in the 1973 paper of Prof. P J Cameron [Cam]. When it was stated, the existence of the design of parameters $2-(111, 11, 1)$2 was yet undecided. It has been now proved with an extensive computer search $[10]$ that this design does not exist.

Some (perhaps) Useful References.

[Cam] Cameron P. J., Extending Symmetric Designs, Journal of Combinatorial Theory, Series A Vol. 14, Issue 2 (Mar., 1973), pp. 215-220.

$[10]$ Lam C. W. H., Thiel L. H., Swiercz S., The Non-existence of Finite Projective Plane of Order 10 Can. J. Math., XLI (1989), pp. 1117-1123.

1 Note that these are the parameters of a Hadamard $2$-design.
2Some readers will recognise that this is a projective plane of order 10.

  • 0
    P.S. This fits in here, because, if we ever have extensible, two designs, there are only finitely many that don't fit into an infinite family (but it could happen that , `...` forbid, there are only finitely many).2012-08-24
9

Any number of theorems about quadratic forms are true over fields where the characteristic is not equal to 2. My favorite reference for the subject, where this sense of "nothing works the same in characteristic 2" is made clear as early as the "Notes to the Reader," is Lam's Introduction to the Theory of Quadratic Forms over Fields. For an even more elementary example of this phenomenon, note that the quadratic formula only holds (or even makes sense) for fields of characteristic not equal to 2.

  • 9
    *A mathematician said “Who / Can quote me a theorem that's true? / For the ones that I know / Are simply not so / When the characteristic is two!”*2012-08-24
  • 0
    +1. The above poem is quoted in the "Notes to the Reader" in Lam's book.2012-08-28
9

The classification of simple finite dimensional Jordan algebras has exactly one exceptional case.

That of simple Lie algebras has five exceptions.

(All this over sensible fields, of course)

There are 3 regular polyhedra in every dimension, except in dimesions 2, 3 and 4.

7

No free group is amenable, with the exception of $\mathbb Z$.

  • 0
    No free group is abelian, with the exception of $\mathbb Z$.2012-10-31
7

Every family of graphs that is closed under minors has at least one forbidden minor, with the exception of the family of all graphs.

  • 3
    As stated, this is trivial, given that every graph is a minor of itself. Perhaps you meant the fact that [every family of graphs that is closed under minors can be defined by _a finite set_ of forbidden minors](http://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem)?2012-08-26
6

Every prime number is odd, with the exception of 2.

  • 28
    *All* prime numbers are odd, and $2$ is the oddest of them all.2012-08-24
  • 4
    @GeorgesElencwajg Although [Hadamard-de la Vallée Poussin](http://en.wikipedia.org/wiki/Prime_number_theorem) and [Erdős–Kac](http://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem) taught us that they *even out*, in the end...2012-08-24
  • 16
    No prime number is divisible by $57$, apart from $57$.2012-08-24
  • 4
    @PseudoNeo Funny, I thought the correct statement was that no prime number is divisible by 42, apart from 42.2012-08-24
  • 0
    @did we must live in a different field, 42 is composite over here :)2012-08-25
  • 5
    @Thomas It is well known that in this universe [42](http://en.wikipedia.org/wiki/42_%28number%29#The_Hitchhiker.27s_Guide_to_the_Galaxy) is everything: composite, prime, finite, infinite, and the answer to every question... (By the way, 57 could have elicited the same kind of reaction from you.)2012-08-25
  • 2
    @PseudoNeo: 57 = 3*19 isn't prime.2012-11-30
  • 3
    @Landei But 57 is a [Grothendieck prime](http://en.m.wikipedia.org/wiki/57_(number))!2014-07-25
6

Other than the sphere, all closed surfaces have list chromatic number equal to chromatic number.

6

$A_n$ has no nontrivial normal subgroup unless $n=4$.

5

The sphere $S^n$ is simply connected if and only if $n\geq 2$.

  • 2
    And connected if and only if $n\geqslant1$.2012-08-24
5

There exists an Aperiodic Tiling of $\mathbb{R}^d$ for all dimensions excepting $d=1$. For $d=1$ it is easy to show that an aperiodic set of tiles cannot exist.

Also, for $d \geq 3$ there exists an aperiodic single tile, the problem is still open in $d=2$ (A non-connected single aperiodic tile was discovered 2 years ago, but usually we ask for connected tiles).

4

A graph is planar, unless it contains a copy of $K_5$ or $K_{3,3}$.

Every subgroup of $\mathbb{Q}$ is residually finite, except $\mathbb{Q}$ itself.

The only two perfect squares in the sequence $\lbrace\displaystyle\sum_{i=1}^n i^2\rbrace_n$ are $1^2$ and $70^2$.

2

All triangular graphs $T_n$ are determined by their spectrum except for $n=8$.

2

Just about every single theorem in arithmetic that concerns prime numbers fails for $2$.

2

This paper http://www.jstor.org/stable/2323537?origin=crossref&seq=1#page_scan_tab_contents shows that in some sense, $\mathbb R^n$ can only have a cross product defined on it if $n=3$ or $7$.

1

If $\sin x$ is rational then $x/\pi$ is transcendental, except when $\sin x$ is $0$, $\pm\frac12$, or $\pm1$.

1

Given $R>0$ and $a\in \mathbb{C}$. For every $n\in \mathbb{Z}$ except $n=-1$ we have:

\begin{align} \oint_{|z-a|=R} (z-a)^n \mathrm d z=0 \end{align}

1

The modular curves $X_0(N)$ have no exceptional automorphism except for $N=37$, $63$ and $108$.

Recall the modular group $$\Gamma_0(N):=\left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \operatorname{SL}_2(\mathbb{Z}) \ : \ c \equiv 0 \pmod n \right\} $$ which acts in the upper half-plane $$\mathbb{H}:=\{z\in \mathbb{C}\ : \ \operatorname{Im}(z)>0\}$$ via Moebius transformations. The quotient $\Gamma_0(N)\backslash \mathbb{H}$ is analytically isomorphic to an affine algebraic curve over $\mathbb{C}$, with associated non-singular projective curve $X_0(N)$, the classical modular curve. Suppose the genus of this curve is $\ge 2$ (which happens for all $N$ except a finite number of them, including all $N>36$, $N\ne 49$).

The normalizer $N(n)$ of $\Gamma_0(N)$ inside $\operatorname{SL}_2(\mathbb{Z})$ acts naturally as automorphism in $X_0(N)$ and it is natural to ask if they are all automorphism. Ogg proved in 1974 that this is the case if $N$ is square-free and $N\ne 37$, which it has an exceptional automorphism not coming from $N(37)$. In 1988, Kenku and Momose proved that there was non exceptional automorphism except for $N=37$ and may be for $N=63$. Elkies in 1990 showed that there really was an exceptional automorphism for $N=63$.

But in 2011, after more than 20 years, Michael Colin Harrison, when revising the proof by Kenku and Momose found out that he could not discard the case $N=108$ by their arguments. After doing some computations with the help of MAGMA he was able to find an exceptional automorphism for $N=108$, later confirmed by Elkies. The result was only published in the arxiv.

For more information you can consult the preprint by Harris: A new automorphism of $X_0(108)$.

-1

Have a look at the Strong Law of Small Numbers:

When two numbers look equal, it ain't necessarily so.