In the December 2010 issue of Scientific American, an article "A Geometric Theory of Everything" by A. G. Lisi and J. O. Weatherall states "... what is arguably the most intricate structure known to mathematics, the exceptional Lie group E8." Elsewhere in the article it says "... what is perhaps the most beautiful structure in all of mathematics, the largest simple exceptional Lie group. E8." Are these sensible statements? What are some other candidates for the most intricate structure and for the most beautiful structure in all of mathematics? I think the discussion should be confined to "single objects," and not such general "structures" as modern algebraic geometry.

Question asked by Richard Stanley

Here are the top candidates so far:

1) The absolute Galois group of the rationals

2) The natural numbers (and variations)

4) Homotopy groups of spheres

5) The Mandelbrot set

6) The Littlewood Richardson coefficients (representations of $S_n$ etc.)

7) The class of ordinals

8) The monster vertex algebra

9) Classical Hopf fibration

10) Exotic Lie groups

11) The Cantor set

12) The 24 dimensional packing of unit spheres with kissing number 196560 (related to 8).

13) The simplicial symmetric sphere spectrum

14) F_un (whatever it is)

15) The Grothendiek-Teichmuller tower.

16) Riemann's zeta function

17) Schwartz space of functions

And there are a few more...

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    With great respect for Richard-and without going so far as to call for it's closure,because it is an interesting question-I think if anyone of lesser stature in the mathematical community had posted this question,there would have numerous calls to close it as too general and subjective. – The Mathemagician Dec 12 '10 at 18:16
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    The main output of this question will likely be a reinforcement of the ego of each domain/community. Not that good for the unity of mathematics. – Denis Serre Dec 12 '10 at 18:38
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    IMO, the statements are sensible because they include the words "arguably" and "perhaps". In other words, I think there is little objective content to them. With all due respect to the OP, I think this question is the epitome of "subjective and argumentative", and I have voted to close for that reason. – Pete L. Clark Dec 12 '10 at 19:04
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    I think the question is really good and deserves its place on MO. This said (and this is in no way in reference to the OP, it is just a general rant) I've noticed that often on MO when someone with no points posts a soft question/big list type of question there is always the same bunch of people who rush to close it and to say basically it's lame. When a professor posts a question virtually of the same order, he gets 80 votes up and congratulations on the "amazing question". I've seen many examples of that and it is this same "police" going to every post and deciding what's good, what's bad. – 16278263789 Dec 12 '10 at 19:46
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    Since this has already attracted a vote to reopen (at time of typing) I've opened a thread on meta MO (… ); if you feel the question should be reopened please take the discussion there. Also, please vote this comment up for visibility. – Yemon Choi Dec 12 '10 at 20:08

40 Answers 40

The absolute Galois group of $\mathbb{Q}$. It contains the information of all algebraic extensions of the rationals - and is therefore the most important single object of algebraic number theory. Representations of the absolute Galois group are central to many diophantine questions; see for example the Taniyama-Shimura conjecture (aka modularity theorem) which led to a solution of Fermat's last theorem and states in some form that certain Galois representations associated to elliptic curves come from modular forms.

One of the most intricate set of conjectures is dedicated (partly) to the study of representations of the absolute Galois group of $\mathbb{Q}$: the Langlands program.

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    +1. And if that's not intricate enough you can upgrade to the conjectural motivic Galois group; the absolute Galois group being the quotient corresponding to varieties of dimension 0. – AFK Dec 13 '10 at 23:56
  • How about the equally conjectural Langlands group ? But frankly, these are all "derived" from the monoid $\mathbf N$ by some "constructions". – Chandan Singh Dalawat Dec 14 '10 at 6:15
  • I would like to know what can I read to get an appreciation for this absolute galois group? – muad Dec 19 '10 at 6:32
  • I'm no expert, but I think a good place to start is almost any book on algebraic number theory/class field theory eg. Algebraic Number Theory by Cassels and Fröhlich. It should be noted that such texts mainly consider the abelianizations of absolute Galois groups, which are, while difficult enough, of course much simpler than the full story. – Lennart Meier Dec 19 '10 at 23:39
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    Maud: "Fearless Symmetry" by Ash and Gross is a non-technical introduction to the subject. – Simon Lyons Mar 26 '11 at 13:11

How about the Leech lattice. This is a 24-dimensional packing of unit spheres where each one touches 196560 others. It is the densest 24-dimensional lattice packing (and very likely the densest 24-dimensional sphere packing, although this has not been proved - EDIT: it has been now). It has a remarkable amount of symmetry, and most of the densest sphere packings known in dimensions < 24 are derived from it (and known sphere packings in dimensions > 24 are nowhere near as dense when normalized for the dimension).

Maybe this is already implicitly included in the list, as it is closely related to the monster vertex algebra.

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    I would love to 'see' the Leech lattice somehow. – muad Dec 19 '10 at 6:34
  • I took the liberty to update the part about packings. – Jarek Kuben Dec 27 '17 at 17:41
  • @Jarek: thank you. – Peter Shor Dec 29 '17 at 13:34

The (stable or unstable) homotopy groups of spheres are certainly considered intricate and beautiful by topologists.

Here is an interesting (obvious) fact about the stable homotopy groups of spheres that I learned from Vigelik:

In the category of commutative rings (with unit) there is an initial object, $\mathbb{Z}$. This seems to be one reason the integers are important or rather fundamental. But there is something more fundamental! There is a functor from commutative rings to ring spectra, the Eilenberg-MacLane functor. $H\mathbb{Z}$ is no longer initial in this category, the sphere spectrum is! So somehow the stable homotopy groups of spheres are a pretty cool/fundamental ring.

I do not know a lot about the unstable setting, but there is a lot of extra data that it has.

I think that Lennart's point about intricacy and complexity showing up when you start to try to compute the thing sounds like a confusion of what one means by intricate and complex. But it is not, it is not the messiness of the computation that makes it intricate, it is the way of teasing apart the knowledge we do have in meaningful ways that lead me to believe that it is a very intricate object. Especially all the number theory hidden in the chromatic picture, which is part of what Lennart is referring to when he mentions the moduli stack of formal group laws.

Edit: My advisor pointed out another reason that the stable homotopy groups of spheres are cool: $\pi^S_{*}(S^0)=\pi_{*}(B\Sigma_{\infty})$, so the stable homotopy groups of spheres are the homtopy groups of the classifying space of the the category of finite sets and bijections. This is essentially the Barratt-Priddy-Quillen theorem (I am told, I do not know the precise statement). That is pretty cool too! All that information about finite sets sitting there has to be something.

(seems to look fine now, please ignore the bump)

  • Could you provide an argument for non-topologists? – Deane Yang Dec 13 '10 at 20:38
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    No it didn't. Your answer is about a model for the sphere spectrum which is beautiful for the reasons you give. I read your answer as "The category of symmetric spectra is beautiful and intricate." In that interpretation, I think there is no stolen thunder. Maybe I misunderstood your answer though. – Sean Tilson Dec 14 '10 at 16:04
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    @Sean: Alright, I'll let it slide this time. But don't let it happen again!!! – Harry Gindi Dec 14 '10 at 19:34
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    But seriously, the category of symmetric spectra is beautiful and intricate! – Sean Tilson Dec 15 '10 at 0:25
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    Are you sure, you mean $B\Sigma_\infty$? I think, it should be the topological monoid $\coprod B\Sigma_n$ with monoid structure via block sum. $B\Sigma_\infty$ has homotopy groups concentrated in degree 1 (if I'm not mistaken), which is no good here. – Lennart Meier Dec 17 '10 at 22:43

Since one of the questions is Are these sensible statements?, allow me to answer that one with a resounding NO, and be on record against the size-ism inherent in the statements of Scientific American. As the largest simple exceptional Lie group, E8 deserves credit for both intricacy and beauty. but the author seems to imply that the size record makes E8 not only more intricate but also more beautiful than the other simple exceptional Lie groups.

Granted I don't know much about Lie groups, but it really bothers me that an aesthetic judgment can be based on size alone. Last I checked, paintings are not judged on their size.

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    Have you ever seen Guernica in person? It's breathtaking ;) – Andrew D. King Dec 14 '10 at 20:25

The Mandelbrot Set is widely viewed as beautiful and intricate, although I can't give a mathematical definition for those.

Mandelbrot Set image

alt text

The imperfect self-similarities are no accident. Many of the pieces correspond to the behaviors of the critical point $0$ under iteration of $z \to z^2 + c$.To each point in the plane, there is a corresponding Julia set, and the relationship of the point to the Mandelbrot set indicates some of the structure of the Julia set.

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    It is a set with a geometrical depiction of great beauty and intricacy, but... how is it a structure? – Jose Brox Dec 14 '10 at 12:58
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    It is a set so that each point in the set and its complement can be marked up with an associated Julia set and the behavior of $0$. I don't know what more is needed to call it a structure. If you want a more algebraic structure on top, then look at, for example, homeomorphisms on subsets of the Mandelbrot set from quasiconformal surgeries. – Douglas Zare Dec 14 '10 at 18:34
  • I have a question about the Mandelbrod set, which is so naive that I don't dare to ask it as an actual stand-alone question: Are the various ovals (connected components of the interior of the Mandelbrod set) perfect circles? If not, do they have smooth boundary? Are they bounded by algebraic curves? – André Henriques May 14 '14 at 5:40
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    @André Henriques: The biggest region is an exact cardiod. The second is an exact circle. At least many of the smaller regions which look circular are not exactly circular: – Douglas Zare May 14 '14 at 5:58

The monster vertex algebra.

It is (to date) the central object in monstrous moonshine, since its character is the $q$-expansion of the modular $J$-function, its automorphism group is the monster simple group, and the graded trace of any element of the monster is the $q$-expansion of a genus zero modular function. The construction of this structure (by Frenkel, Lepowsky, and Meurman) involves ascending a hierarchy of objects that are by themselves quite intricate and beautiful.

  1. One begins with the extended binary Golay code of length 24. Up to symmetries, it is the unique copy of $\mathbb{F}_2^{12}$ in $\mathbb{F}_2^{24}$, for which any five basis vectors are contained in a unique codeword (i.e., it forms a Steiner $(5,8,24)$ system). The codewords are separated by Hamming distance at least 8, so even if 3 bits in a code word are changed the error can be corrected. The automorphism group of the Golay code is the sporadic simple group $M_{24}$ of order 244823040.

  2. Using the Golay code to produce coordinates of generators, one constructs the Leech lattice $\Lambda$, which is a rather densely packed copy of $\mathbb{Z}^{24}$ in $\mathbb{R}^{24}$. One can also make the Leech lattice as a subquotient of the even unimodular lattice $I\!I_{25,1}$, which has its own exceptional properties. Peter Shor mentioned the Leech lattice in another answer, so I'll just note that its automorphism group is a double cover of Conway's sporadic simple group $Co_1$, which has order 4157776806543360000.

  3. For any positive definite even lattice $L$, there is a canonical construction of a vertex operator algebra graded by that lattice, called the lattice vertex algebra $V_L$. I think physicists say that it is the algebra of chiral symmetries of a conformal field theory describing a bosonic string propagating in the torus $L \otimes \mathbb{R}/L$ (but I may have mixed up the words). It has an action of the holomorph of the algebraic torus $L \otimes \mathbb{C}^\times$.

  4. The "-1" automorphism of the Leech lattice induces an automorphism $\theta$ of $V_\Lambda$, and there is a unique irreducible $\theta$-twisted module $V_\Lambda(\theta)$ that inherits an action of the centralizer $2^{1+24}.Co_1$ of $\theta$ in the automorphism group of $V_\Lambda$. The monster vertex algebra is formed by taking the direct sum of fixed points: $V^\natural = (V_\Lambda)^\theta \oplus (V_\Lambda(\theta))^\theta$.

Apparently, the hard part was proving that the monster acts on $V^\natural$ by automorphisms.

There are some additional conjectural reasons for considering it beautiful:

  1. In the same paper where it was constructed, it was conjectured to be the unique vertex operator algebra with central charge 24, character equal to the modular $J$ function, and representation category equivalent to $Vect$. (Naturally, this does not account for higher structure like twisted modules.)

  2. Witten suggested that it is dual to pure 3-dimensional quantum gravity with minimal cosmological constant by AdS/CFT correspondence.

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    This one seems promising, but any chance you could elaborate on why the monster vertex algebra is beautiful? – Deane Yang Dec 13 '10 at 20:09
  • Dear Deane, I am not sure if the wording of the question: "What are some other candidates for the most intricate structure and for the most beautiful structure in all of mathematics?" insist that the same object is intricate and beautiful. – Gil Kalai Dec 14 '10 at 11:08
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    Gil, I didn't notice that! It seems to me that the question is more interesting if you demand both! – Deane Yang Dec 15 '10 at 1:40

Here an answer in form of a question:

does it strike anyone that many of the candidates for "most intricate and/or beautiful structure in mathematics" proposed here find their natural joint home where they meaningfully relate to each other in... string theory?

These are just the most evident. One could go on about how the motivic Galois group also fits in etc., but I don't want to strain it.

What sometimes makes me wonder is that mathematicians have all that appreciation for all these separate intricate and beautiful phenomena that come out of string theory, one by one, including an impressive list of Fields-awarded work, but that there is little appreciation that closely similar (just vastly "bigger") to how the Moonshine conjecture qualified as a problem in mathematics, the question "What is string theory?" may be one of the deepest open problems possibly not in phyisics, but in mathematics.

I believe the natural numbers are the most intricate and beautiful structure in all of mathematics. Particularly insofar as all of the other intricate and beautiful structure we actually work with can be encoded via the natural numbers.

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    +1: how not to agree! (But then I'd be tempted to say: the empty set! We can describe all natural numbers by simple operations on it, via the von Neumann's ordinal construction) – Pietro Majer Dec 12 '10 at 18:45
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    On the other hand, the natural numbers might not be the most intricate object in mathematics.. – J.C. Ottem Dec 12 '10 at 18:52
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    While the empty set may be a profound notion, how is it an intricate object? – Yemon Choi Dec 13 '10 at 5:27
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    @Yemon: Interesting <cough> argument. The issue seems a bit <cough> subjective though... – Pete L. Clark Dec 13 '10 at 8:21
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    To try and put across my objection to this answer with a metaphor: is sand intricate because we can make stained glass? – Yemon Choi Dec 13 '10 at 22:14

One of the most beautiful structures, in my mind, is the classical Hopf fibration, which allows you to visualize the $3$-sphere $S^3$ as a smooth circle bundle over the $2$-sphere. When you view $S^3$ minus a point as $\mathbb R^3$, one can actually draw very nice pictures of this fibration. It's doubly interesting to me because it involves the isomorphism of $S^2$ with $\mathbb C\mathbb P^1$ from complex analysis.

There are actually 4 such Hopf fibrations (spheres which are total spaces of fibre bundles whose base and fibre are also both spheres):

1) $S^1$ is an $S^0$ bundle over $\mathbb R \mathbb P^1 \cong S^1$.

2) $S^3$ is an $S^1$ bundle over $\mathbb C \mathbb P^1 \cong S^2$.

3) $S^7$ is an $S^3$ bundle over $\mathbb H \mathbb P^1 \cong S^4$, the quaternionic projective line.

4) $S^{15}$ is an $S^7$ bundle over $\mathbb O \mathbb P^1 \cong S^8$, the octonionic projective line.

  • In the interest of full disclosure: I copied and pasted some of this from my own answer to another question:… – Spiro Karigiannis Dec 14 '10 at 17:02
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    I agree that these are among the most beautiful structures in geometry and topology, but I'm not sure that they qualify as being "intricate". – Deane Yang Dec 14 '10 at 17:06
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    @Deanne: You're correct, they're not as intricate as the Mandlebrot set or the Cantor set, but it's amazing to me that one can fill up all of $\mathbb R^3$ completely with disjoint circles (and one line), any two of which are non-trivially linked. [But I deliberately called them beautiful only, not intricate.] – Spiro Karigiannis Dec 14 '10 at 19:07
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    I think they also have intricate highly symmetric triangulations. E.g. "Kuhnell's CP^2". They are beautiful and intricate but perhaps not even aiming to be the most beautiful/intricate. (They are quite modest, just 9 vertices.) – Gil Kalai Dec 14 '10 at 21:50
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    Dror Bar-Natan has a beautiful animation of the Hopf fibration: – Daniel Moskovich Dec 16 '10 at 11:52

The Littlewood-Richardson coefficients. (Or, if one wants a single object, as per the rules of the game: the representation ring of $S_n$ or $GL_n({\bf C})$. Or the ring of symmetric functions. Or the cohomology ring of the Grassmannian with the Schubert variety basis. Etc., etc.)

On the one hand, the Littlewood-Richardson coefficients have fairly simple geometric descriptions (using such combinatorial gadgets as Young tableaux, honeycombs, or puzzles), but on the other hand obey a number of deep recursive properties. (See for instance my Notices article with Allen Knutson on one aspect of these coefficients.) Last, but not least, they are connected to an amazing number of areas of mathematics (see e.g. Fulton's survey article).

The Turing degrees are an immensely intricate poset $\mathcal{D}$. Here are some of their remarkable properites:

  1. Every countable poset is embeddable in $\mathcal{D}$.
  2. $\mathcal{D}$ contains minimal degrees. (a non-zero degree $\mathbf{m}$ with no degree between $\mathbf{0}$ and $\mathbf{m}$)
  3. For every non-zero degree $\mathbf{d}$, there is a degree that is incomparible with $\mathbf{d}$.
  4. $\mathcal{D}$ contains an antichain of size $2^{\aleph_0}$.
  5. No infinite strictly increasing chain in $\mathcal{D}$ has a least upper bound.
  6. For every degree $\mathbf{d} \geq \mathbf{0}' $, there is a degree $\mathbf{c} < \mathbf{d}$ such that $\mathbf{c}'=\mathbf{d}$. (Here $\mathbf{c}'$ denotes the set of indices of oracle Turing machines that halt when using $\mathbf{c}$ as an oracle. Note that one must check that this is well-defined on degrees.)
  7. For any two recursively enumerable degrees, there is a recursively enumerable degree strictly between them.
  8. Any finite distributive lattice can be embedded in the recursively enumerable degrees.
  • @Tony : Nit-picking: 3, 6 should be for non-zero degrees. I would also mention the definability of the map $x\mapsto x'$. – Andrés E. Caicedo Dec 21 '10 at 8:11
  • I know I'm over 4 years late, but shouldn't 6. say that it's so for degrees which are $\geq\mathbf{0}'$? Otherwise a minimal degree serves as a counterexample. – Wojowu Jul 3 '15 at 9:22
  • @Wojowu There is no statute of limitations on comments. I edited accordingly. Since this post is community wiki, I think you could have also edited the entry yourself. Thanks. – Tony Huynh Jul 7 '15 at 19:40

Let $g$ and $n$ be positive integers such that $3g-3 + n > 0$ Let $\mathcal{M}_{g,n}$ the moduli stack of genus $g$ nodal curves with $n$ marked points. There are three obvious families of maps

forgetting a point

$$\mathcal{M}_{g,n+1} \rightarrow \mathcal{M}_{g,n}$$

identifying two marked points on two different curves yielding a new one

$$\mathcal{M}_{g_1,n_1+1} \times \mathcal{M}_{g_2,n_2+1} \rightarrow \mathcal{M}_{g_1 + g_2,n_1 + n_2}$$

and identifying two marked points on a single curve yielding a new one with higher genus

$$ \mathcal{M}_{g,n+2} \rightarrow \mathcal{M}_{g+1,n} $$

This system constitutes the so-called Grothendieck-Teichmüller tower. It is indeed intricate and in my opinion, also beautiful. Moreover, it is a conjecture of Grotehndieck that its automorphism group is naturally isomorphic to the absolute Galois group over $\mathbb{Q}$, namely $\mathrm{Gal}(\bar{\mathbb{Q}}|\mathbb{Q})$.

Ok, I'll throw my hat in the ring: I like the classical Cantor set.

Not only does it demonstrate the complexity that relatively simple subsets of the real line have, it illustrates an important property of measures on the real line - namely, that measurability has nothing to do with cardinality of the set (i.e. this is an uncountable set with measure zero!)

It also gives an example of a completely disconnected subset of $\mathbb{R}$ that literally has no components - it contains no open intervals of $\mathbb{R}$ in its power set.

There are many, many more observations one can make about the Cantor set, but I think the obvious ones make my point very nicely. When I teach real analysis, this is an example I think I'll be using a great deal to illustrate properties of the real line.

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    OK. $ $ – Pete L. Clark Dec 13 '10 at 19:48
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    Perhaps it would be more accurate to say that the Cantor set has many components, since its connected components are its points. – S. Carnahan Dec 14 '10 at 10:29
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    I think that the fat Cantor set is the one that gives the surprising result. It is a bounded nowhere-dense set with positive measure. – Harry Gindi Dec 14 '10 at 12:11
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    Andrew, I disagree. Both $\mathbb{Z}$ and $\mathbb{Q}$ are infinite topological spaces that, under the subspace topologies inherited from $\mathbb{R}$, are made out of singleton components. Both spaces are used quite frequently in mathematics. – S. Carnahan Dec 15 '10 at 14:18
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    As it happens, profinite groups like the $p$-adic integers have a remarkable tendency to be homeomorphic to the Cantor set. One might even add this to the entry in support of the cause. – S. Carnahan Dec 18 '10 at 7:05

The class of all ordinals. The class of cardinals is embedded within it (if AC holds) since one identifies a cardinal with the smallest ordinal such that the set of all smaller ordinals has that cardinality. ($\aleph_0$ is the cardinality of the set of all finite ordinals, $\aleph_1$ is the cardinality of the set of all countable ordinals, etc. $\aleph_\omega$ is the cardinality of the set of all ordinals whose cardinality is $\aleph_n$ for some finite $n$. ($\omega$ is the ordinal that gets identified with $\aleph_0$ in the aforementioned identification) $\aleph_{\omega+1}$ is the set of all ordinals of cardinality $\aleph_\omega$, and so on. $\aleph_\omega$ is the smallest cardinal greater than $\aleph_0$ that is known not to be equal to $2^{\aleph_0}$.)

But if grading is only based on "intricacy", maybe the class of all sets, conventionally denoted "V" because it looks like the letter V (?) might be in first place. Some people have tried to embed all of mathematics within this thing.

Later edit: The "\aleph"s and the "\omega"s are failing to get rendered when I view this thing. Look at the code and you'll see them.

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    I've heard "V" is called that way because of Von Neumann. I haven't seen definite evidence one way or the other. – Andrés E. Caicedo Dec 12 '10 at 18:26
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    This is not even a mathematical object. Next! – Harry Gindi Dec 13 '10 at 22:19
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    @Harry: I suspect that what you meant is either that it's not a set or that it's not one of the things referred to in the first-order language of set theory. Or something like that. And I have to suspect that David Roberts had the same thing in mind. But I think the idea that that is the essence of mathematical-objecthood is debatable. – Michael Hardy Dec 14 '10 at 21:48
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    @Harry: it is without question a mathematical object! Conway’s manifesto for the “Mathematicians’ Liberation Movement” is worth reading in this connection — the liberation being from foundational constraints. It can be read as dismissive of questions of foundations — and I’d strongly disagree with that; I work largely with foundational structures, I think they can be very illuminating, and I care passionately about them. [cont’d] – Peter LeFanu Lumsdaine Dec 16 '10 at 16:29
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    But the point I do agree with, and have never seen a serious argument against, is that if mathematicians are working with something, and working with it mathematically, then it is mathematics — and if it doesn’t quite fit into a particular foundation, this is a problem with the foundation, not the mathematics. If it looks like a mathematical object, smells like a mathematical object, has theorems about it like a mathematical object, then it’s a mathematical object. – Peter LeFanu Lumsdaine Dec 16 '10 at 16:33

This has been forgotten so far:

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    I have just rolled back an edit whose author took it upon himself to expound on universality of the zeta function. While interesting, I see no indication that this is what the OP @JohannesEbert intended – Yemon Choi May 13 '14 at 3:40


It can also be thought as the set of prime numbers. I don't know if it can really be considered "intricate"...

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    as an aside I saw a 'simple question', what does Spec(Z) look like with the etale topology?, and the response given was that answering that question is essentially the point of arithmetic geometry. so perhaps not so simple – Samantha Y Dec 22 '17 at 17:44

In my view it is difficult to come up with an alternative to any of the exotic Lie groups, which are unquestionably quite intricate but are also beautiful because they express the properties of certain geometric spaces using both fundamental algebra (i.e., groups) and geometric structures of their own (i.e., Riemannian geometry). I don't know $E_8$ particularly well, but I still have vivid memories of Robert Bryant's lectures describing the structure of $G_2$.

  • Could you elaborate on $G_2$ then? I don't know much about lie groups, but would love to hear about the geometry. – Sean Tilson Dec 13 '10 at 23:49
  • @Sean: post it as a question, and I will be happy to give a fairly detailed answer. – Spiro Karigiannis Dec 14 '10 at 1:36
  • If Sean doesn't do this soon, I will. My vague recollection is that you look at the 7-dimensional space of imaginary octonions. Since multiplication is not associative, there is a naturally defined 3-form that expresses the non-associativity, and $G_2$ arises as the group that preserves the 3-form. The first person to try to explain the octions to me was Calabi, when I was still an undergraduate. Then Bryant explained it again, right after he showed that $G_2$ can be the holonomy group of a non-symmetric Riemannian metric. – Deane Yang Dec 14 '10 at 1:57
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    Thanks for the encouragement: – Sean Tilson Dec 14 '10 at 5:50

The outer automorphisms of the group $S_6$ of all permutations of a set of six objects.

$6$ is the only number for which $S_n$ has any outer automorphisms.

The group inner automorphisms of $S_6$ is a subgroup of index $2$ in the group of all automorphisms of $S_6$, and --- here's a (probably) unexpected fact: It's one of exactly three subgroups of index $2$, no two of which are isomorphic to each other.

  • And this is nicely explained in Rotman's book on group theory. – Pedro Tamaroff Jul 31 at 7:36
  • @PedroTamaroff : Does Rotman explain that the group of all automorphisms of $S_6$ has exactly three subgroups of index $2$? $\qquad$ – Michael Hardy Jul 31 at 12:35
  • Hm, I don't think so. Not at least in the edition I have. – Pedro Tamaroff Jul 31 at 19:06
  • @PedroTamaroff : Nonetheless I'll probably look at Rotman. $\qquad$ – Michael Hardy Jul 31 at 19:25

I have voted for ${\mathbb N}$; but let me nevertheless propose an object living in the analytical realm, namely the Schwartz space ${\cal S}$ of infinitely differentiable functions $f:{\mathbb R}\to{\mathbb C}$ that for $|x|\to\infty$ together with their derivatives go to zero faster than any power $1/|x|^n$. The "intricateness" of this space stems from the many operations you can perform in it and from the fact that these operations are intertwined with each other in miraculous ways. $$ $$ Responding to a comment: You have (a) ordinary multiplication and convolution, (b) "multiplication" with arbitrary polynomials $p(x)$ and operations $p(D)$, (c) multiplication with functions of the form $x\mapsto e^{iax}$ and the translation operator $T_a: f(\cdot)\mapsto f(\cdot-a)$ and (d) scaling of the variable $x$ resp. $\xi$. The Fourier transform $\Phi$ interchanges in each of these three cases the respective operations; and at heart of it all is Gauss' normal distribution $x\mapsto {1\over \sqrt{2\pi}}\int e^{-x^2/2} dx$ which stays fixed under $\Phi$. And, last not least, there is a scalar product which is preserved by $\Phi$.

  • I actually discussed this space with a friend a few months ago. I thought geometrically,this is the space of sequences of rotations in the complex plane and thier subsequences such that the rotations have no nonempty intersection with each other-is this correct? – The Mathemagician Dec 14 '10 at 22:03
  • More detail would make a better case. Which transformations? (Fourier transform is one, but which others do you have in mind?) – Yemon Choi Dec 16 '10 at 7:13

The generalized cohomology theory known under the name Topological modular forms:

Here's a picture of the graded ring $\mathit{tmf}\;^*(pt)$:

And here's the spectral sequence used to compute it:
Its $E_2$ page is $Ext_{A(2)}(\mathbb F_2,\mathbb F_2)$, where $A(2)$ is this beast:

Here's another spectral sequence, that computes the closely related graded ring $\mathit{Tmf}\;^*(pt)$:

I heard good things about F_un!

Another one: Chaitin's Omega constant.

Original answer by: none

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    Why did this get such a low rating? If one knew the Omega constant, then one could solve the halting problem. Moreover, it has a simple, natural definition. – Richard Stanley Dec 18 '10 at 1:08
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    It looks like the person who wrote this answer couldn't be bothered to write a sentence or two of explanation or justification. Anyone with suitable expertise and at least 100 points is welcome to add to it. – S. Carnahan Jan 12 '11 at 15:23

Consider the canonical pointed symmetric sequence of simplicial sets $S$ defined such that $S_0=S^0$ and $S_n=S^n$. This is an ordered sequence of simplicial sets with a natural symmetric group action defined by permutation of the suspensions. By abstract nonsense, we can show that this category has a symmetric monoidal closed product. The object $S$ admits the natural structure of a monoid for this tensor product. The category of symmetric spectra becomes the category of modules over this monoid.

What's deep and intricate about this object? Well, I just read a paper by A. Salch that shows that the category of commutative $S$-algebras models the proposed theory of the field with one element!

The category of modules over this monoid is the category of symmetric spectra!

  • If you had come to dinner after the midwest you could have seen the easy proof that the $K(\mathbb{F}_1)=\pi^S(S^0)$. Is this really all it takes to get the category of Symmetric Spectra? Also, I imagine you should like the monoid that Schwede uses to compute homotopy groups of Symmetric Spectra. – Sean Tilson Dec 13 '10 at 23:53
  • @Sean: Yeah. You look at the category of pointed simplicial combinatorial species à la Joyal. This is the functor category $sSet_*^\mathbb{P}$ where $\mathbb{P}$ is the core groupoid of the category of finite sets. The category $\mathbb{P}$ has a natural monoidal product given by the disjoint union. By generalized abstract nonsense (Day convolution), this induces a symmetric monoidal closed tensor product on $sSet_*^\mathbb{P}$ called the tensor product of pointed simplicial species. Then the sphere spectrum is a monoid in this monoidal category... – Harry Gindi Dec 14 '10 at 3:54
  • and the category of all symmetric spectra with its natural tensor product is given by the category of pointed simplicial series with an $S$-action on them. Then the smash product of spectra is then given by the coend formula $M\wedge N:=colim(M\otimes S\otimes N\rightrightarrows M\otimes N)$ tracing out the action. – Harry Gindi Dec 14 '10 at 4:14

I don't know if it's the most beautiful or the most intricate, but I certainly think the random graph $G(n,p)$ deserves consideration, if only for philosophical reasons.

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    What philosophical reasons? – Gil Kalai Dec 14 '10 at 21:45
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    I would say the Erdős–Renyi/Rado graph rather than this, if you're looking for something from graph theory. This occurs as the Fraïssé limit of the category of finite graphs and embeddings – David Roberts Dec 14 '10 at 23:57
  • David: isn't that graph also known as "the" random graph on a countably infinite vertex set? – Yemon Choi Dec 15 '10 at 0:16
  • Gil: For the reason that it can easily demonstrate the existence of a graph with a beautiful and intricate structure without explicitly exhibiting it. Although you could argue that we may as well say that a coin flip is is an intricate and beautiful structure. Also, it should maybe be disqualified by your "single object" specification. – Andrew D. King Dec 15 '10 at 3:27

Maybe this wouldn't be my first choice, but I still think it's worth being on the list: Gödel's constructible universe $L$.

I would argue that it is intricate because it can serve as a model for "all of mathematics" (i.e., ZFC), furthermore answering many combinatorial questions left open by ZFC alone. Even though most(?) set theorists will probably argue that it is not "the" true model giving the right answer to these questions, it is still undoubtedly a rich and complex structure, moreover one in which the axiom of choice and the continuum hypothesis are not only true but "explained".

But it is also beautiful because of its connections with higher computability theory (e.g., the sets of integers constructed at the level $\omega_1^{\mathrm{CK}}$ of the constructible hierarchy, where $\omega_1^{\mathrm{CK}}$ is the smallest nonrecursive ordinal, are exactly the hyperarithmetical sets, i.e., the (lightface) $\Delta^1_1$ sets of the analytic hierarchy), and, in a related manner, because of Jensen's results on the "fine structure" of $L$. In a very intuitive way, I'd say that $L$ consists of sets that are ultimately "computable" (iterating the Turing jump as far as it can be), a perfectly regular construction that prohibits any randomness.

So even if set theorists are unhappy with $L$ because it forbids really large cardinals, and even if they try to construct something better (the core model), I argue that Gödel's original $L$ is still something immensely intricate and beautiful.

The Surreal numbers as constructed by Conway.

They contain a copy of many objects already on the list ($\mathbb{N},O_n$) and so surpass them in complexity, and they are recursively defined from the empty set using $O_n$ length recursions which is incredibly beautiful.

I would also propose the absolute Galois group of the field of fractions of the Grothendieck ring of the ordinals, however I have very little understanding of this object (nor does anyone else to my knowledge).

  • ${\bf R}^3$ contains a copy of many objects already on the list – does that mean it surpasses them in complexity? – Gerry Myerson Aug 1 at 13:01
  • @GerryMyerson If we allow definitions of all order over $\mathbb{R}^3$ then I'd say yes; if we stick to first order definitions it may be simpler, since RCF is complete but PA isn't. – Alec Rhea Aug 1 at 17:06

How about $\mathbb{R}^n$? I hope people don't consider this example too simplistic. After all, the structure of $\mathbb{R}^n$ gives rise to all of the theory of topological and differentiable manifolds. Specific important highlights include the theory of "algebraic" equations (inverse and implicit function theorems) and the local theory of differential equations (jets, forms, integral submanifolds).

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    You should try to repeat Lisi's publicity stunt with $E_8$ replaced by $\mathbb{R}^n$ :-) – Urs Schreiber May 13 '14 at 22:04

The rotational symmetries within Young's lattice, discovered at the beginning of the 21st century by Ruedi Suter. I think the bilateral symmetry has been known many decades, but rotational symmetries were not, and they seem quite surprising.

The Wikipedia link below clearly explains the idea. The first link is Suter's paper.

This is a less-known example. Consider the semigroup $$ A_0 = \big\{ \big[\begin{smallmatrix} 0&0\\0&0 \end{smallmatrix}\big], \big[\begin{smallmatrix} 1&0\\0&0 \end{smallmatrix}\big], \big[\begin{smallmatrix} 0&1\\0&0 \end{smallmatrix}\big], \big[\begin{smallmatrix} 0&1\\0&1 \end{smallmatrix}\big] \big\} $$ under usual matrix multiplication. The variety $\mathrm{var}A_0$ generated by $A_0$ is finitely universal in the sense that the lattice of subvarieties of $\mathrm{var}A_0$ embeds all finite lattices. The semigroup $A_0$ is a minimal example since the variety generated by any semigroup of order three or less is not finitely universal.

Brownian walk (or Wiener process, to conform to "single object" stipulation).

protected by François G. Dorais May 13 '14 at 5:18

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