# June 19, 2002 {#week182} It's been a long time, but in the last Week's Finds I was telling you about my adventures this spring in northern California, and I hadn't quite gotten around to telling you about that cool conference on "Nonabelian Hodge Theory" at the MSRI in Berkeley. I'll continue my story about that now... ...but first, a little detour through the Nile valley! Egyptians liked to write fractions as the sum of reciprocals of integers. For example, instead of writing $$\frac56$$ those folks would write something like $$\frac12 + \frac13$$ Nobody is sure why, but one possibility is that they started with a neat notation for $1/n$, and then wanted to extend this to handle other fractions, and couldn't think of anything better. Of course they *could* have written $m/n$ as $$\underbrace{\frac1n+\ldots+\frac1n}_{\mbox{$m$ terms}}$$ but they preferred to use as few terms as possible. This leads to some tricky questions. For example: clearly every fraction of the form $4/n$ can be written using 4 terms --- but can you always make do with just 3? Nobody knows! This is called the Erdos-Strauss conjecture. Alan Swett claims to have shown you only need 3 terms if $n$ is less than or equal to $10^{14}$. For example: $$\frac{4}{8689} = \frac{1}{2175} + \frac{1}{1718250} + \frac{1}{14929874250}.$$ For much more on this, see: 1) David Eppstein, "Egyptian fractions", `http://www.ics.uci.edu/~eppstein/numth/egypt/` 2) Alan Swett, "The Erdos-Strauss conjecture", `http://math.uindy.edu/swett/esc.htm` Egyptian fraction problems have a spooky way of showing up in various unrelated mathematical contexts... which have a spooky way of turning out not to be unrelated after all! For example, suppose we are trying to classify all the Platonic solids. We're looking for ways to tile the surface of a sphere with regular $n$-gons, with $m$ meeting at each vertex. Suppose there is a total of $V$ vertices, $E$ edges, and $F$ faces. Since the Euler characteristic of the sphere is $2$, we have $$V - E + F = 2.$$ Since each face has $n$ edges but 2 faces meet along each edge, we have $$nF = 2E.$$ Since each vertex has $m$ edges meeting it but each edge meets 2 vertices, we also have $$mV = 2E.$$ Putting these equations together we get $$2E\left(\frac1n + \frac1m - \frac12\right) = 2$$ or $$\frac1n + \frac1m = \frac12 + \frac1E.$$ An Egyptian fractions problem! It's obvious that this can only have solutions if $1/n + 1/m > 1/2$. And interestingly, all the solutions of this inequality do indeed correspond to Platonic solids... at least if $n,m > 2$. Here they are: | $(n,m)$ | Platonic solid | | :------ | :------------- | | $(3,3)$ | tetrahedron | | $(3,4)$ | octahedron | | $(4,3)$ | cube | | $(3,5)$ | icosahedron | | $(5,3)$ | dodecahedron | The cases $n = 1,2$ don't give Platonic solids in the usual sense: after all, most people don't like polygons to have just 1 or 2 edges. Neither do the cases $m = 1,2$, since most people don't like polyhedra to have just 1 or 2 faces meeting at a vertex! One can argue about whether these are irrational prejudices. But it's actually good to study *all* unordered pairs of natural numbers with $$\frac1n + \frac1m > \frac12$$ since they correspond to *all* the isomorphism classes of finite subgroups of the rotation group! The Platonic solids have their symmetry groups, which don't change when we switch $n$ and $m$. The solution $(n,1)$ corresponds to the cyclic group $\mathbb{Z}_n$: the symmetries of a regular $n$-gon, where you're not allowed to flip it over. The solution $(n,2)$ corresponds to the dihedral group $D_n$: the symmetries of a regular n-gon where you *are* allowed to flip it over. In some weird sense, maybe we should think of $\mathbb{Z}_n$ and $D_n$ as the symmetry groups of Platonic solids with only 1 or 2 faces. I'll leave you to ponder the Platonic solids with only 1 or 2 vertices. If you get stuck, look up the word "hosohedron"! The story gets better if we also consider solutions of $$\frac1n + \frac1m = \frac12$$ which formally correspond to Platonic solids where the number $E$ of edges is infinite. In fact, these correspond to tilings of the plane by regular polygons: | $(n,m)$ | tilings of the plane by... | | :------ | :------------------------- | | $(3,6)$ | regular triangles | | $(6,3)$ | regular hexagons | | $(4,4)$ | (regular) squares | Similarly, solutions of $$\frac1n + \frac1m < \frac12$$ give tilings of the hyperbolic plane. For example, Escher used $(n,m) = (6,4)$ in some of his prints, like this: $$\includegraphics[scale=0.68]{../images/escher.png}$$ Let me try to arrange all this information in a table, using lines to separate the spherical, planar and hyperbolic regions: $$ \begin{tikzpicture} \matrix[ column sep=.2cm,row sep=.1cm, nodes={text height=0.65cm}, ]{ \node{${}_m\backslash^n$}; & \node{$1$}; & \node{$2$}; & \node{$3$}; & \node{$4$}; & \node{$5$}; & \node{$6$}; & \node{$7$};\\ \node{$1$}; & \node{$\mathbb{Z}_1$}; & \node{$\mathbb{Z}_2$}; & \node{$\mathbb{Z}_3$}; & \node{$\mathbb{Z}_4$}; & \node{$\mathbb{Z}_5$}; & \node{$\mathbb{Z}_6$}; & \node{$\mathbb{Z}_7$};\\ \node{$2$}; & \node{$\mathbb{Z}_2$}; & \node{$\mathrm{D}_2$}; & \node{$\mathrm{D}_3$}; & \node{$\mathrm{D}_4$}; & \node{$\mathrm{D}_5$}; & \node{$\mathrm{D}_6$}; & \node{$\mathrm{D}_7$};\\ \node{$3$}; & \node{$\mathbb{Z}_3$}; & \node{$\mathrm{D}_3$}; & \node{\footnotesize tetrahedron}; & \node{\footnotesize cube}; & \node{\footnotesize dodecahedron}; & \node[align=center]{\footnotesize hexagonal\\\footnotesize tiling};\\ \node{$4$}; & \node{$\mathbb{Z}_4$}; & \node{$\mathrm{D}_4$}; & \node{\footnotesize octahedron}; & \node[align=center]{\footnotesize square\\\footnotesize tiling};\\ \node{$5$}; & \node{$\mathbb{Z}_5$}; & \node{$\mathrm{D}_5$}; & \node{\footnotesize isosahedron};\\ \node{$6$}; & \node{$\mathbb{Z}_6$}; & \node{$\mathrm{D}_6$}; & \node[align=center]{\footnotesize triangular\\\footnotesize tiling}; & & & \node[align=center]{\footnotesize hyperbolic\\\footnotesize tilings};\\ \node{$7$}; & \node{$\mathbb{Z}_7$}; & \node{$\mathrm{D}_7$};\\ }; \draw[thick,dashed] (-5.3,3) to (5.4,3); \draw[thick,dashed] (-4.4,3.7) to (-4.4,-4.3); \draw[thick] (5.4,1) to (2.8,1) to (2.8,0) to (-0.7,0) to (-0.7,-2.15) to (-2.7,-2.15) to (-2.7,-4.2); \draw[thick] (5.4,0.9) to (4.5,0.9) to (4.5,-0.1) to (0.6,-0.1) to (0.6,-1.15) to (-0.6,-1.15) to (-0.6,-3.1) to (-2.6,-3.1) to (-2.6,-4.2); \end{tikzpicture} $$ Now, the same Egyptian fraction problem comes up when studying other problems, too. For example, suppose you are trying to find a basis of $\mathbb{R}^n$ consisting of unit vectors that are all at 90-degree or 120-degree angles from each other. We can describe a problem like this by drawing a bunch of dots, one for each vector, and connecting two dots with an edge when they're supposed to be at a 120-degree angle from each other. If two dots are not connected, they should be at right angles to one another. So, for example, this diagram tells us to find a basis for $\mathbb{R}^3$ consisting of unit vectors all at 120 degree angles from each other: $$ \begin{tikzpicture}[scale=0.75] \draw (0,0) to (2,0) to (1,1.73) to cycle; \node at (0,0) {$\bullet$}; \node at (2,0) {$\bullet$}; \node at (1,1.73) {$\bullet$}; \end{tikzpicture} $$ It's easy to see this is impossible, since three vectors all at 120 degrees from each must lie in a plane --- so they can't be linearly independent. On the other hand, this diagram gives a solvable problem: $$ \begin{tikzpicture}[scale=1.5] \draw[thick] (0,0) to (2,0); \node at (0,0) {$\bullet$}; \node at (1,0) {$\bullet$}; \node at (2,0) {$\bullet$}; \end{tikzpicture} $$ You just pick two unit vectors at right angles to each other and wiggle the third one around until it's at a 120-degree angle to both. It's not hard. So, the question is: which diagrams give solvable problems? This is actually a very fun puzzle: it's very famous, but most books manage to make it seem really boring and "technical", so you should really spend some time thinking about it for yourself. I'll give away the answer, but I won't say how you prove it's true. First, it's easy to see that if a diagram consists of a bunch of separate pieces, and you can solve the problem for each piece, you can solve the problem for the whole diagram. So, it's sufficient to consider the case of connected diagrams. Second, a connected diagram can only give a solvable problem if it's Y-shaped, like this: $$ \begin{tikzpicture}[scale=0.5] \draw[thick] (0,0) to (9,0); \draw[thick] (3,0) to (3,2); \foreach \x in {0,1,2,3,4,5,6,7,8,9} \node at (\x,0) {$\bullet$}; \node at (3,1) {$\bullet$}; \node at (3,2) {$\bullet$}; \end{tikzpicture} $$ Third, a diagram like this gives a solvable problem only if $$\frac1k + \frac1n + \frac1m > 1$$ where $(k,n,m)$ are the numbers labelling the tips of the Y when we number it like this: $$ \begin{tikzpicture}[scale=0.5] \draw[thick] (0,0) to (9,0); \draw[thick] (3,0) to (3,2); \node[fill=white] at (0,0) {$4$}; \node[fill=white] at (1,0) {$3$}; \node[fill=white] at (2,0) {$2$}; \node[fill=white] at (3,0) {$1$}; \node[fill=white] at (4,0) {$2$}; \node[fill=white] at (5,0) {$3$}; \node[fill=white] at (6,0) {$4$}; \node[fill=white] at (7,0) {$5$}; \node[fill=white] at (8,0) {$6$}; \node[fill=white] at (9,0) {$7$}; \node[fill=white] at (3,1) {$2$}; \node[fill=white] at (3,2) {$3$}; \end{tikzpicture} $$ So for example, this particular problem is not solvable because $\frac14 + \frac13 + \frac17 < 1$. Now, it's easy to see what we can only get $\frac1k + \frac1n + \frac1m > 1$ if one of the numbers is $1$ or $2$. If one of the numbers is $1$, our "Y-shaped" diagram is actually just a straight line of dots! We can also describe this straight line by taking one of the numbers to be $2$, like this: $$ \begin{tikzpicture}[scale=0.7] \draw[thick] (0,0) to (6,0); \node[fill=white] at (0,0) {$2$}; \node[fill=white] at (1,0) {$1$}; \node[fill=white] at (2,0) {$2$}; \node[fill=white] at (3,0) {$3$}; \node[fill=white] at (4,0) {$4$}; \node[fill=white] at (5,0) {$5$}; \node[fill=white] at (6,0) {$6$}; \end{tikzpicture} $$ except for the boring case where we have just a single dot. So, let's assume one of the numbers is $2$. By symmetry we can assume this number is $k$. We are thus looking for pairs $(n,m)$ with $$\frac12 + \frac1n + \frac1m > 1$$ or in other words $$\frac1n + \frac1m > \frac12.$$ This is the same problem as before! So the problem we're dealing with now is very much like classifying Platonic solids! Even better, these diagrams I've been drawing are called "Dynkin diagrams", and we can use them to get certain incredibly important finite-dimensional Lie algebras called "simply-laced simple Lie algebras". For a taste of how this works, reread ["Week 65"](#week65) and some previous Weeks. Similarly, we get certain *infinite-dimensional* Lie algebras called "simply-laced affine Lie algebras" when $$\frac1n + \frac1m = \frac12,$$ and "simply-laced hyperbolic Kac-Moody algebras" when $$\frac1n + \frac1m < \frac12.$$ So, our whole big table above translates into a table of Lie algebras! Let me draw it with the standard names of these Lie algebras below their diagrams. Unfortunately, I'll have to make it very small to fit everything in. So, for example, I'll draw the so-called $\mathrm{E}_8$ Dynkin diagram: $$ \begin{tikzpicture} \draw[thick] (0,0) to (6,0); \draw[thick] (2,0) to (2,1); \foreach \x in {0,1,2,3,4,5,6} \node at (\x,0) {$\bullet$}; \node at (2,1) {$\bullet$}; \end{tikzpicture} $$ as this puny miserable thing: $$ \begin{tikzpicture}[scale=0.15] \foreach \x in {0,1,2,3,4,5,6} \node at (\x,0) {$\bullet$}; \node at (2,1) {$\bullet$}; \end{tikzpicture} $$ This is what we get: $$\includegraphics[max width=0.9\linewidth]{../images/week182table.png}$$ This mysterious way that the same Egyptian fraction problem shows up in classifying Platonic solids and simply-laced simple Lie algebras is actually the tip an iceberg sometimes called the "McKay correspondence" --- though important aspects of it go back to the theory of Kleinian singularities. I talked about the McKay correspondence in ["Week 65"](#week65), so that's a good place to dig deeper, but you should really look at some of the references in there, and also these two --- both of which explain the mysterious word "hosohedron": 3) H. S. M. Coxeter, _Generators and relations for discrete groups_, Springer, Berlin, 1984. 4) Joris van Hoboken, _Platonic solids, binary polyhedral groups, Kleinian singularities and Lie algebras of type $A$,$D$,$E$_, Master's Thesis, University of Amsterdam, 2002, available at `http://home.student.uva.nl/joris.vanhoboken/scriptiejoris.ps` or `http://math.ucr.edu/home/baez/joris_van_hoboken_platonic.pdf` Okay. Now --- back to that conference at the Mathematical Sciences Research Institute! You can look at transparencies and watch videos of the talks here: 5) MSRI streaming video archive, Spring 2002, `http://www.msri.org/publications/video/index04.html` If you like watching math talks, there's a lot to see here --- not just this one conference, but all the MSRI conferences! For example, right after the nonabelian Hodge theory conference there was one on conformal field theory and supersymmetry, featuring talks by bigshots like Richard Borcherds, Dan Freed, Igor Frenkel, Victor Kac, and Jean-Bernard Zuber --- just to name a few. You can see talks by all these folks. But anyway, let me start by telling you what nonabelian Hodge theory is.... Hmm. I guess I should *start* by telling you what *abelian* Hodge theory is! In its simplest form, Hodge theory talks about how differential forms on a smooth manifold get extra interesting structure when the manifold has extra interesting structure. To warm up, let me remind you about what we can do when our manifold has *no* extra interesting structure. Whenever we have a smooth manifold $M$ there's an "exterior derivative" operator $d$ going from $p$-forms on $M$ to $(p+1)$-forms on M. This is just a generalization of grad, curl, div and all that. In particular it satisfies $$d^2 = 0,$$ so the space of "closed" $p$-forms: $$\{w \mid dw = 0\}$$ contains the space of "exact" $p$-forms: $$\{w \mid \mbox{$w = du$ for some $u$}\}.$$ This makes it fun to look at the vector space of closed $p$-forms modulo exact $p$-forms. This is called the "$p$th de Rham cohomology group of $M$", or $$H^p(M)$$ for short. It only depends on the topology of $M$; its size keeps track of the number of $p$-dimensional holes in $M$. When $M$ is compact, it agrees with the cohomology computed in a bunch of other ways that topologists like. Fine. But now, suppose $M$ has a Riemannian metric on it! Then we can write down a version of the Laplacian for differential forms. A function is a $0$-form, so we're just generalizing the Laplacian you already know and love. Differential forms whose Laplacian is zero are called "harmonic". Every harmonic $p$-form is closed, but if $M$ is compact life is even better: the vector space of harmonic $p$-forms is isomorphic to the $p$th de Rham cohomology of $M$. This is great: it means the de Rham cohomology, which only depends on the *topology* of $M$, can also be thought of as the space of solutions of a *differential equation* on $M$! This gets topologists and analysts talking to each other, and has all sorts of marvelous spinoffs and generalizations. Some people call this stuff "Hodge theory". But Hodge theory goes further when $M$ has more structure --- most notably, when it's a Kaehler manifold! A Kaehler manifold is to the complex plane as a Riemannian manifold is to the real line. More precisely, it's is a manifold whose tangent spaces have been made into *complex* vector spaces and equipped with a *complex* inner product. Of course the real part of the inner product makes it into a Riemannian manifold. That lets us parallel transport vectors, so we demand a compatibility condition: parallel transporting a vector and then multiplying it by $i$ is the same as multiplying it by $i$ and then parallel transporting it! This makes complex analysis work well on Kaehler manifolds. Now, if you've taken complex analysis, you may remember how people use it to find solutions of Laplace's equation... like when they're studying electrostatics, or the flow of fluids with no viscosity or vorticity --- an idealization that von Neumann mockingly called "dry water". On the complex plane we can talk about "holomorphic" functions, which satisfy the Cauchy-Riemann equation: $$ \frac{df}{d\overline{z}} = 0 \qquad \left( \text{note:}\,\, \frac{df}{d\overline{z}} = \frac{df}{dx} + i\frac{df}{dy} \right) $$ and also the complex conjugates of these, called "antiholomorphic" functions, which satisfy $$ \frac{df}{dz} = 0 \qquad \left( \text{note:}\,\, \frac{df}{dz} = \frac{df}{dx} - i\frac{df}{dy} \right) $$ Both holomorphic and antiholomorphic functions are automatically harmonic, so we can find solutions of Laplace's equation this way. But even better, every harmonic function is a linear combination of a holomorphic and an antiholomorphic one! All this stuff works much more generally for $p$-forms on Kaehler manifolds. To get going, let's think a bit more about the complex plane. If we have any $1$-form on the complex plane we can write it as a linear combination of $dx$ and $dy$, where $x$ and $y$ are the usual coordinates on the plane. But things get nicer if we work with *complex-valued* differential forms. Then we can form linear combinations like $$dz = dx + idy$$ and $$d\overline{z} = dx - idy$$ and express any $1$-form as a linear combination of *these* in a unique way. We call these the $(1,0)$ and $(0,1)$ parts of our $1$-form. This means that if we have a function $f$, we can take its exterior derivative and chop it into its $(1,0)$ part and $(0,1)$ part: $$df = \partial f+\overline{\partial}f.$$ These guys are called "Dolbeault operators". Anyway, it turns out that $$\overline{\partial}f = 0$$ is just a slick way of writing Cauchy-Riemann equation, which says that $f$ is holomorphic. You should check this for yourself! Similarly, $$\partial f = 0$$ says that $f$ is antiholomorphic. Now let me say how all this stuff generalizes to arbitrary Kaehler manifolds. We can decompose any $p$-form on a Kaehler manifold into its $(i,j)$ parts where $i+j = p$. For example, a $(1,2)$-form in $4$ dimensions might look something like this in complex coordinates: $$f dz_1\wedge d\overline{z}_3\wedge d\overline{z}_2 + g dz_2\wedge d\overline{z}_3\wedge d\overline{z}_4.$$ We have $$d = \partial + \overline{\partial}$$ where $\partial$ maps $(i,j)$-forms to $(i+1,j)$-forms, while $\overline{\partial}$ maps $(i,j)$-forms to $(i,j+1)$-forms. This allows us to take the de Rham cohomology groups of our manifold $M$ and write them as a direct sum of smaller vector spaces, which I'll call $$H^{i,j}(M)$$ for short. So far I don't think I've used anything about the metric on $M$, so all this would work whenever $M$ is a so-called "complex manifold". But if we really have a Kaehler manifold, and it's compact, we can say more: a $p$-form is harmonic if and only if all its $(i,j)$ parts are. This means $H^{i,j}(M)$ is isomorphic to the space of harmonic $(i,j)$-forms. Alternatively, you can describe $H^{i,j}(M)$ just in terms of $\partial$: you just take the $(i,j)$-forms in here: $$\{w \mid \overline{\partial}w = 0\}$$ modulo those in here: $$\{w \mid \mbox{$w=\overline{\partial}u$ for some $u$}\}$$ This is called the "$(i,j)$th Dolbeault cohomology group of $M$". That's Hodge theory in a nutshell. There's even *more* you can do when $M$ is a Kaehler manifold, but I'm getting a little tired, so I'll just let you read about that here: 6) R. O. Wells, _Differential analysis on complex manifolds_, Springer, Berlin, 1980. This is a really *great* book for learning about all sorts of good geometry stuff, starting with differential forms and working on up through Hodge theory, pseudodifferential operators, sheaves and so on. But anyway, I've given you a little taste of Hodge theory. The main thing to remember is that when your manifold is complex, the cohomology becomes "bigraded": instead of just $$H^p(M)$$ you get $$H^{i,j}(M).$$ So now, what's nonabelian Hodge theory? The basic idea is simple: instead of askng what extra structure the *homology groups* get when $M$ is a complex manifold, we ask what extra structure the *homotopy type* of $M$ gets when $M$ is a complex manifold. The homotopy type includes invariants like the homotopy groups, but also more. How are these constrained by the fact that $M$ is complex? Unfortunately, to describe the answer --- even a little teeny part of the answer --- I need to turn up the math level a notch. For starters we can consider the fundamental group $\pi_1(M)$. But this is hard to relate to differential geometry, so we will immediately water it down by picking an algebraic group $G$ and looking at homomorphisms of $\pi_1(M)$ into $G$. These are basically the same thing as flat $G$-bundles over $M$, so it's easier to see how $M$ being a complex manifold affects things. We can even be sneaky and study this for all $G$ at once by forming a group $\Pi_1(M)$ called the "proalgebraic completion" of $\pi_1(M)$. This is a proalgebraic group --- an inverse limit of algebraic groups --- which contains $\pi_1(M)$ and has the property that any homomorphism from $\pi_1(M)$ into an algebraic group $G$ extends uniquely to a proalgebraic group homomorphism from $\Pi_1(M)$ to $G$. It's nice to ask what extra structure $\Pi_1(M)$ gets when $M$ is a complex manifold, because this question has a nice answer. To get ready for how nice the answer is, first go back to plain old abelian Hodge theory. Note that making the cohomology of $M$ bigraded gives an obvious way for the algebraic group $\mathbb{C}^\times$, the nonzero complex numbers, to act on the cohomology. The reason is that for each integer there's a representation of $\mathbb{C}^\times$ where the number $z$ acts as multiplication by $z^n$, so gradings are just another way of talking about $\mathbb{C}^\times$ actions. Since the cohomology of $M$ is automatically graded, putting *another* grading on it amounts to letting $\mathbb{C}^\times$ act on it. So in plain old Hodge theory, the answer to "What extra structure does the cohomology of $M$ get when $M$ is complex?" is: > "It gets an action of $\mathbb{C}^\times$!" And it turns out that in nonabelian Hodge theory, the answer to "What extra structure does $\Pi_1(M)$ get when $M$ is complex?" is: > "It gets an action of $\mathbb{C}^\times$!" This is incredibly cool, but the story goes a lot further. The fundamental group is just the beginning; you can do something similar for the higher homotopy groups --- but it's a lot more subtle. In fact, you can do something similar directly to the homotopy type of $M$! When $M$ is a compact complex manifold, there's a homotopy type called the "schematization of $M$" whose fundamental group is $\Pi_1(M)$ --- and there's an action of $\mathbb{C}^\times$ on this homotopy type! By the way, when $M$ is a compact Kaehler manifold the action of $\mathbb{C}^\times$ on its cohomology extends to a natural action of $\mathrm{SL}(2,\mathbb{C})$, as explained in Wells' book. I wonder if $\mathrm{SL}(2,\mathbb{C})$ acts on the schematization of $M$? I learned about most of this fancy stuff from an incredibly lucid talk by Bertrand Toen. Unfortunately there seems to be no video of his talk, since he gave it down the hill at U. C. Berkeley instead of at the MSRI --- and the handwritten notes at the MSRI website are rather illegible. So you want to learn more about this, you should probably start with this quick summary of abelian Hodge theory: 7) Tony Pantev, "Review of abelian Hodge theory", `http://www.msri.org/publications/ln/msri/2002/introstacks/pantev/1/index.html` and then take the deep plunge into this paper: 8) Ludmil Katzarkov, Tony Pantev and Bertrand Toen, "Schematic homotopy types and non-abelian Hodge theory I: The Hodge decomposition", available at [`math.AG/0107129`](https://arxiv.org/abs/math.AG/0107129). There are a lot of model categories and $n$-categories lurking in the background of this subject, as well as ideas that originated in physics, like "Higgs bundles". For the brave reader I recommend these papers: 9) Bertrand Toen, "Toward a Galoisian interpretation of homotopy theory", available as [`math.AT/0007157`](https://arxiv.org/abs/math.AT/0007157). This answers the question: "the fundamental group is to covering spaces as the whole homotopy type is to... what?" The fact that it's in French probably makes it easier to understand. 10) Bertrand Toen and Gabriele Vezzosi, "Algebraic geometry over model categories (a general approach to derived algebraic geometry)", available as [`math.AG/0110109`](https://arxiv.org/abs/math.AG/0110109). This is only for badass mathematicians who find algebraic geometry and homotopy theory insufficiently mindblowing when taken separately. Ever wondered what an affine scheme would be like if you replaced the ground field by an $E_\infty$ ring spectrum? Then this is for you. (I thank David Eppstein for pointing out the work of Alan Swett.) ------------------------------------------------------------------------ > *Geometry may sometimes appear to take the lead over analysis, but in fact precedes it only as a servant goes before his master to clear the path and light him on his way.* > > --- James Sylvester