Back to symplectic geometry. So far, everything I did in my last post only used the fact that the symplectic form was skew symmetric, not that it was closed. Indeed the “closed” property is rather mysterious, (as far as I’m concerned, although in the literature it is called “geometric”), since I don’t know of any really good geometric intuition for the action of exterior derivative on 2-forms. Still, it is a hugely important condition, and key to many of the special properties of symplectic geometry, notably for us, the Darboux theorem. Note that there is no real equivalent condition for Riemannian structures, and therefore it takes us in a whole new direction. I would love to have a better sense of how to “explain” why certain symplectic arguments don’t work in the Riemannian world (eg Darboux theorem), but I haven’t delved deeply enough into the proof to do this, since it’s not coherent to just say “the Riemannian form isn’t closed.”

For an application of closedness, recall the Lie derivative of a tensor with respect to a vector field : it’s just the derivative of along the flow lines of . In the case that is a differential form, one can prove formally, using induction and certain algebraic properties of , that

where of an -form is just contraction of with that form (ie evaluate that form on to get an form). Again, this formula means nothing to me other than some algebra (I wish I could change that!), but it’s called “Cartan’s (magic) formula” and it’s very nice. In particular, suppose is Hamiltonian, and we want to evaluate ; contraction of and gives by definition which is exact and therefore closed; what’s more, because is closed, the other term in Cartan’s formula is 0, and we see that —the symplectic form is invariant under all Hamiltonian flows, which is to say, the flow of is a one parameter group of symplectomorphisms (diffeomorphisms which preserve the symplectic form, the analogy of isometries in Riemannian geometry)! The equivalent concept for Riemannian geometry is a Killing vector field, and generically, these are very hard to find, which one can see because a generic Riemannian manifold has a finite group of isometries (right?). But in the symplectic case we’ve just produced lots of (one-parameter subgroups of!) symplectomorphisms, indeed the space is infinite dimensional, and in this way is much more akin to volume preserving smooth maps (of which it is a subset, since is a volume form). In particular, we’ve shown that for every smooth function we get a family of symplectomorphisms, tangent to . In general a vector field whose flow is a family of symplecticmorphisms is called symplectic—by Cartan’s formula, this is the case if contracted with , it yields a closed 1-form—and we see that Hamiltonian vector fields correspond exactly to the case that this closed form is exact.

Questions to do with Hamiltonian symplectomorphisms (those that come from Hamiltonian vector fields), and related subjects, are at the heart of symplectic geometry. For example, one can ask about their fixed points, which is the Arnold conjecture, and there is the related Weinstein conjecture, just proved by Taubes. For now I won’t go into these, and will let the comparison with Riemannian geometry be the main motivation.

Concerning that comparison, we remarked earlier that on a fixed vector space , both symmetric and skew-symmetric forms, when non-degenerate, can always be put into a standard form, in other words, their only invariant is the dimension of the underlying space. Therefore at any point in our manifold, we know what the form looks like at that point. However, in Riemannian geometry, if we could construct some neighborhood around a point for which, in local coordinates, the metric restricted to every point in that neighborhood was standard, then we would have shown that that neighborhood was isometric to Euclidean space, and therefore flat, for example. But not all Riemannian manifolds are locally flat! Thus even locally, Riemannian geometry is quite complex. The same thing, remarkably, is NOT true for symplectic manifolds. This is Darboux’s theorem.

The proof I will give starts with a lemma called Moser’s trick; there is also a more hands on and dirty, “down-to-earth geometric” proof, I believe. Anyway, suppose is compact, and we have a smooth family of symplectic (closed, non-degenerate) forms , (here is the space of differential 2-forms), with an “exact derivative,” meaning that if we look at for each (which we can define in the usual way, since is an -vector space), then the resulting 2-form can be written as for some smooth family of 1-forms. This is the assumption. Then Moser tells us that this path of 2-forms, which lives only in some associated space of forms, can actually be realized geometrically on via a family of diffeomorphisms ( such that ) (NOT symplectomorphisms, duh, since they change the symplectic form). The proof is quick: each is non degenerate, so it gives us a vector field for each , which is just the contraction (as an anecdote, note here that is actually , so we’re plugging the primitive into its own derivative! This always seemed a little incestual to me. On a more serious note, it comes up all the time: is there some more concise or intuitive way to describe what it’s measuring?). Since is compact, we have a flow of diffeomorphisms generated by . And this is the desired family of maps! Because

for all ! (the last equality follows from the linearity of , and the first is one of these fancy Lie-derivative identities (applied to the flow of a time-dependant vector field) which one simply proves directly). So it’s constant, and it starts at (because is the identity).

Note that the assumption that is exact implies that is constant. Indeed, this latter assumption is actually sufficient for the theorem. So this says that any two “isotopic forms” (cohomologous symplectic forms connected by a path of cohomologous symplectic forms) are actually “strongly isotopic,” meaning the path of forms can be realized by a path of diffeomorphisms, as above.

Finally we use Moser’s trick to prove Darboux’s theorem. Let , here is not necessarily compact. Choose a basis in in which is standard (just at . Now we can extend the coordinates in to a local coordinate patch around , and we have two forms: the form , the one we started with, and the form defined to be the standard form at every point (note that by construction, these two forms agree at ). Now for our family of forms interpolating, we just take the line . We want to make sure these forms are non-degenerate (they are obviously closed), but because non-degeneracy is an open condition in the space of 2-forms, we can take our coordinate chart small enough so that we’re ok. Now note that the -derivative of is just independently of ; this form is closed, so locally it is exact (we’re doing everything locally) so let be an anti-derivative. Fixing up by a constant, we can assume it’s 0 at . Now we define to be the vector field which is contracted with . By restricting ourselves (again!) to a small set inside of our chart, we can define a global flow for (so this is slightly different than Moser’s trick, which was global but for which we used compactness of ). So indeed, applying the reasoning from Moser’s trick, interpolates between and , in particular, , which is the desired result. What’s more, because we fixed things up so that , and therefore its dual vector field , was zero at , the form is actually constant at as we apply the diffeomorphisms. Q.E.D.

Next time: ? Almost complex structures, contact geometry, J-holomorphic curves, examples, knot theory, something completely unrelated, nothing?

One of my summer projects was to try to learn symplectic geometry. In this, my first installment of notes, I discuss some introductory notions; hopefully it’s not too rambling. In the continuation, I’ll prove Darboux’s theorem, a fundamental result which says that locally, all symplectic spaces are isomorphic (something which sharply distinguishes symplectic geometry from Riemannian geometry, where there are many local invariants, such as curvature).

EDIT: Here, I take the point of view that the reader is somewhat familiar with Riemannian geometry, and try to build intuition for the structures in symplectic geometry via an analogy with the Riemannian case. This was helpful for me, to some extent, in order to even have a chance of “breaking into” the field, so to speak. However, there are many reasons why this is possibly a misleading vantage point, so do not believe that it’s the whole story. It may be helpful for some. Please see the comments for additional (undoubtedly better, I am an extreme novice) points of view. Some of these would also be quite suitable for an introduction.

In both symplectic and Riemannian geometry, the main object of study is a smooth manifold equipped with a bilinear form on each tangent space, in such a way that the forms vary smoothly as we move between tangent spaces. In the (possibly more familiar) Riemannian case, this form is a symmetric, non-degenerate, positive definite form, turning each tangent space into a normed vector space. In symplectic geometry, we instead require a skew-symmetric bilinear form on each tangent space, again varying smoothly. We still require that at each point in our manifold , should be non-degenerate, so that if =0 for all , then must be 0. Finally, note that because is a skew-symmetric 2-form, it is a differential 2-form on , and we require that as a 2-form, is closed, i.e., . I’ll introduce examples as we go.

Just like in Riemannian geometry, the fact that the symplectic form is non-degenerate establishes a natural bijection between the tangent space and cotangent space at every point. I’ll write this as . The image of is the functional which sends to . In the Riemannian geometry case, we know that, at a given point, we can just choose an orthonormal basis, and then we have a particularly nice “dual basis:” a basis vector is dual to the unique functional whose value on that vector is 1, and whose value on the other basis vectors is 0. This is a simple way to get a handle on the isomorphism from in the Riemannian case. In terms of this special basis, the symmetric form has been reduced to the standard dot product.

There is an equivalent construction in the symplectic case. The “standard” symplectic form on has a basis of the form , with pairing with to give 1 and pairing with the rest of the basis vectors (including itself) to give 0 (the form is determined by this data, along with the requirement that it be bilinear and skew-symmetric). There is a theorem, just as easy as Gram-Schmidt, that says that every symplectic form looks like the above in an appropriate basis (a corollary is that if a vector space can be equipped with a non-degenerate skew-symmetric form, then it is necessarily even-dimensional). Therefore after a change of basis, the bijection between tangent and cotangent space sends to the covector (linear functional) which is 1 on and 0 on the other basis vectors. As a side note, if we make into a complex vector space by letting , then the standard form can be written in terms of the standard dot product as , indeed, we have (I hope I have my signs ok). Thus the standard complex, Riemannian, and symplectic structures on all determine each other in this way, and make it into a Kähler manifold. More generally, complex or almost-complex structures compatible with symplectic forms are very important in the subject.

Before I move on to symplectic manifolds, note that just on the level of bilinear forms on vector spaces, there are many features of the geometry of skew-symmetric forms which are quite different than that of symmetric forms (to which we are probably more accustomed), and these all have important implications in the non-linear (= general symplectic geometry) case. For example, for , define . Then the non-degeneracy of the form insures that, just like for a normed vector space, the dimensions of and add up to . However they are not necessarily disjoint! Indeed, for example, with the standard form, . Such a subspace is called Lagrangian, meaning the form restricted to it is identically 0, and it is of maximal dimension () with this property.

I want to give some examples before I go further, but unfortunately, I haven’t really found any amazingly clear and enlightening ones. The simplest and most obvious is itself, with the standard form, which we have already discussed. Already in this case there are interesting questions to be asked. As another, note that if is a smooth orientable surface, then every 2-form is closed, as there are no non-zero 3-forms, and any volume form will be non-degenerate. Therefore all surfaces are symplectic manifolds, and any one-dimensional subspace will be Lagrangian. So until we either introduce some more interesting questions, or go to higher dimensions, we are a little stuck (I could also mention here that every cotangent bundle is a symplectic manifold in a natural way, but I’ll probably talk about this next time). The second thing I want to mention, because it bothered me at first, is the question of motivation for symplectic geometry. Certainly, the formalism arose from physics, and simplified many natural physical models. However, I like to think of it as just another “kind” of geometry on a smooth manifold, another fairly simple structure we can put on it—an evil twin of Riemannian geometry, for which we can ask all the same questions, and see what comes out. This point of view then justifies itself as we find quite elegant behavior, as well as applications to other fields (which I probably won’t discuss this time around).

I have to make a quick disclaimer right now. Throughout, whenever I talk about integrating a vector field to a flow, I’m going to assume implicitly (without stating!) that we’re on a compact manifold , so that the flow is well defined on all of . This is very sloppy: might as well call the rest of these notes “Stuff about compact symplectic manifolds.” Except that Darboux’s theorem is true in general, which I’ll indicate.

OK, on to geometry. To every smooth real-valued function on a smooth manifold , we can associated the differential , which is the one-form whose value at a vector field at a point is the value of the directional derivative at . The nice thing about having the dual map at each point is that from each differential 1-form we can produce a vector field, and this applies to , giving a vector field which depends on . Now when we do this construction with a Riemannian metric, we get (by definition) the gradient vector field, and it has the nice property that it is normal to the level sets of (indeed, it points towards the direction of most increase of , which naturally is perpendicular to the level sets). To see this, one just applies the derivation to , and unwinding the definitions one gets simply the smooth function , the norm squared of for each point . However, doing this with gives the directional derivative as ! Indeed, is tangent to the level sets of (so it is just rotated by in some direction, or, alternatively, hit by the multiplication by map, in an appropriate basis and some compatible complex structure). So the flow of any point under stays within a particular level set, and so the level sets of are invariant under the flow (the standard example is to take the sphere with its standard volume form as the symplectic form, and take the function to be the height function. Then the level sets are circles of constant height, and the flow of can be seen to be rotation around the vertical axis). This is perhaps the moment to mention that is called a Hamiltonian vector field (rather than gradient vector field as in Riemannian geometry), and is called the Hamiltonian (function) (of . Indeed, the fact that the level sets of are invariant under the flow is a hint at the physics origins (as is the name Hamiltonian) of the whole subject. Originally, one would take a particular Hamiltonian to be the energy on a particular phase space (which is itself defined as the cotangent bundle to a state space, giving it a natural symplectic structure, as I alluded to previously). Then one declares that the allowable evolution of the universe is to follow the flow of . That level sets are preserved under the flow gives conservation of energy. This formalism is called Hamiltonian mechanics.

To be continued! I’ll say a little more about Hamiltonian flows, and then prove Darboux’s theorem.

Most of my “free math time” has been used to study for quals, but today I’ve made myself post to stop Steven from taking over this blog.

One of my favorite elementary algebric combinatorial results is the Matrix Tree Theorem, which states:

In a nondirected graph with vertices labelled , the number of spanning trees is equal to any principal minor of the Laplacian.

This cute result gets the number of trees on vertices () fairly quickly with some matrix manipulation, which I will leave as an exercise to the reader. I know two proofs of this theorem: the first one involves using the Cauchy-Binet formula on the Laplacian , after making the slick observation that , where is the incidence matrix. Another quick solution can be obtained by invoking the lesser-known version of the Matrix Tree Theorem for directed graphs, which is actually a bit simpler to prove:

In a directed graph with vertices labelled , the number of arborescences into vertex (that is, trees rooted at where all the edges point towards ) is equal to the -th minor of the Laplacian.

But this is not all!

I actually recently learned of a third proof (involving Gessel-Viennot, of all things), but I will not mention it here (see the second reference of this post). The reason I mention the directed version and arborescences is to introduce a lesser-known but closely related result to the Matrix Tree Theorem, the forcibly-named BEST Theorem (‘B’ for de Bruijn, ‘S’ for Smith, ‘T’ for Tutte, and ‘E’ for… van Ardenne-Ehrenfest):

In a balanced directed graph (that is, for each vertex the out- and in-degrees equal) with vertices labelled , the number of Eulerian circuits is equal to , where is the outdegree of vertex and $T$ is the number of directed arborescences into any vertex.

This result is neat, for a couple of reasons. One, it shows immediately that the number of directed arborescences into any vertex in a balanced graph is equal, which is totally not obvious. Second, it implies a connection between Eulerian circuits and spanning trees, which is counterintuitive; in fact, when each vertex has in- and out- degree , which is a kind of graph we get quite often (especially when we consider nondirected graphs as directed graphs), we get that the number of Eulerian circuits is exactly the number of arborescences.

Sketch of Proof: pick any edge to start with. Now, from any Eulerian circuit , construct a directed graph as follows: for each vertex (besides ), consider the last step in the circuit away from it towards a vertex . Then add the directed edge to . This is an arborescence into . Note that besides the last exit from (which must be to ), the other exits can be done in any order. This creates a – fold bijection between arborescences into and Eulerian circuits, which is exactly what we need.

Those familiar with knot theory may recall the Alexander polynomial of a link , which one obtains as a minorof a matrix constructed from any diagram of (though it does not depend on the diagram). I’ll not review the construction since it is best done by example, it is a little tedious to type, and I’m too lazy to draw pictures (though there’s a functional Wikipedia page here). Now, it is well known that is well-defined (and called the determinant of ). However, it has a combinatorial meaning. Construct the following graph on the strands of : each time the strand is crossed above by and comes out the other side as , draw directed edges and (I think this even works if $i = j$, though I believe (I don’t know much knot theory and this is pure intuition, so correct me if I’m wrong!!!) you can always draw diagrams to avoid this). It takes a bit of thinking, but convince yourself that this creates a graph with indegree and outdegree at each vertex. Thus, by considering the construction of , we have:

equals any minor of , which in turn equals both the number of arborescences into any vertex of and the number of Eulerian circuits of .

This is basically all I know about knot theory, so I’ll stop here.

References: Enumerative Combinatorics Vol. 2 (Stanley) for the Matrix Tree Theorem and the BEST Theorem, and A Course in Enumeration (Aigner) for the BEST Theorem and Alexander polynomials.

-Y

P.S. The latter reference deserves some mention, because it has a neat presentation. At the end of each chapter, Aigner gives a “highlight” section with a particularly pretty result (the BEST Theorem being one of them), which serves as fun enrichment material. I wish more math books did this (though this is not the first book I know which does this – Alon’s The Probabilistic Method also has similar sections, which were equally delightful). The exposition is quite good, coming from one of the writers of the beautiful Proofs from the BOOK.

In this post, I want to give an example of a Lie group which is not isomorphic to a subgroup of a matrix group. This contrasts with the algebraic picture, where every affine algebraic group can be realized as a subgroup of a matrix group. I’ll just sketch the details.

The group we’ll use is the universal cover G of . One can realize as a fiber bundle with fiber over via the map which sends a matrix to its first row, so its fundamental group is . Similarly, we can use this map to show that is a fiber bundle with fiber over and hence is simply connected.

Now let be any smooth map. It induces a map on Lie algebras . We can use this to define a map by . It’s straightforward to check that f preserves the Lie bracket. Since is simply connected, f is the derivative of some smooth map .

To show that r cannot be injective, we will show that the composition is not injective, where the second map is the usual inclusion. I claim that this composition is the same map as the composition , where is the covering space map. By construction, these two compositions induce the same differentials on their Lie algebras. Now using the fact that both G and are connected, this implies that the map on the level of Lie groups must be equal as well. But is not injective: we saw above that the fundamental group for is nontrivial, so we are done.

I keep telling myself that I’m a geometer/topologist, but I keep acting otherwise. I’ve been attending lectures by Imre Leader at Cambridge on Hypergraph Games. I attended the first lecture on a whim, having never thought about games or graphs or combinatorics in my life (other than my odd fascination with topological tic tac toe). Five minutes into the lecture, I realized that topological tic tac toe is a hypergraph game! Moreover, I found the lecture thoroughly entertaining, and I’ve been following the course attentively since. So in this post, which is sort of an appendix to my previous posts (1, 2) on topological tic-tac-toe (henceforth called T4 for short), I will discuss the wider context of hypergraph games. Much of the general presentation here follows Leader’s lectures.

Although I’ll briefly indicate how T4 fits into this larger picture, let me give the short answer up front: game-theoretically, it’s a fairly boring and straightforward example. But as a visualization exercise or as a gimmick, I still think it’s groovy.

Throughout this post, all sets will be taken to be finite; this isn’t essential everywhere, but it simplifies certain things. The data of a hypergraph game are a set called the board and a collection of subsets of which are called the winning lines of the game. Two players (P1 and P2) take turns marking elements of ; each element can only be marked once. The game ends when one player has marked all the elements of a winning line; if the board is exhausted with neither player winning, then the game is a draw. A particular state of the game is called a position; precisely, it is a choice of two disjoint subsets of —those marked by P1 and those marked by P2—such that P1 has marked either one or zero more elements than P2 has. Note that this necessarily encodes whose turn is next.

We are most interested in what happens when P1 and P2 both play with perfect strategy. A winning strategy (for P1) is a function from the set of positions on the board to the set of possible moves such that in any play of in which P1 plays according to , P1 will win. A drawing strategy (for P1) is the same, except the guaranteed result is that the game is a draw or P1 wins.

Proposition 1: For any hypergraph game , exactly one of the following holds:

1. P1 has a winning strategy
2. P2 has a winning strategy
3. both P1 and P2 have a drawing drawing strategy

Proof: The proof is by “backtracking.” We say a position is a P1 winning position if whenever the game is played from this point (with perfect strategy), P1 wins; define similarly the notions of a P2 winning position and a drawing position. (We allow such positions to be such that the game is already finished.) Let . Then every position in which all elements are marked is one of these three possibilities: winning for P1 or P2, or drawing.

Inductively, suppose all positions with at least elements marked is either winning for P1 or P2, or drawing. Let be a position with elements marked; without loss of generality, suppose P1 is next to move. There are finitely many possible resulting positions after P1 moves. If any of these are P1-winning, then is P1-winning. If all are P2-winning, then is P2-winning. If none are P1-winning and not all are P2-winning then by the inductive hypothesis at least one must be drawing. Hence P1 will play for a draw and is drawing. Hence, by induction, the initial position in which no elements of are marked is either P1-winning, P2-winning, or drawing; the Proposition follows, as the status of the initial position is the status of the game itself. Q.E.D.

Hence we can refer to a given game as either a P1 win, a P2 win, or a draw. In fact, we can improve on this. You may wish to try and prove this yourself before reading the given proof.

Proposition 2: No hypergraph game is a P2 win. In particular, if a hypergraph game can never end in a draw, then it is a P1 win.

Proof: We use the technique of “strategy stealing.” Let be a winning strategy for P2; we will find a winning strategy for P1, yielding a contradiction. Note that for either player an extra move can never hurt (think this through!). P1 begins by playing anywhere, say . Then it is as if P1 were the second player with an extra element marked. So playing according to is a win for P1; if at any point says P1 should play at , then P1 plays anywhere unmarked, say , and on his next turn resumes play according to . In the future, if calls on P1 to play at , then P1 plays anywhere unmarked and then resumes , and so forth. Hence P1 has a winning strategy. Contradiction! Q.E.D.

Note that the strategy-stealing proof of Proposition 2 is completely non-constructive! One theme in the theory of hypergraph games is the search for explicit winning strategies to particular games. This is very hard! Many (most?) of the known winning strategies for small games are by case analysis. If you constructed winning strategies to the various flavors of T4 similarly to the way that I did, then you were essentially doing clever case analysis, in which you used observation and/or symmetry to cut down the number of cases considered. This quickly becomes unwieldy for larger boards.

While there seems to be a dearth of general techniques for constructing winning strategies, there is one which helps for constructing drawing strategies—the technique of a pairing strategy. Let be a hypergraph game. Suppose one can choose exactly two elements from each winning line such that all elements are distinct; we call this a splitting of the board. Then P2 has an easy drawing strategy: whenever P1 plays by marking one member of a given pair, P2 responds by marking the other member. Since by Proposition 2 a P2-winning game is impossible, it follows that any hypergraph game which admits such a splitting is a draw. Exercises 1-3 below give some practice with splitting strategies.

We conclude with what may be a surprising fact to some.

Warning: Based on our intuition and what we have seen in the various flavors of T4, one might conjecture the following: if a hypergraph game is a P1 win and is a collection of subsets of which contains as a subset, then the hypergraph game is a P1 win as well. This is false, and the phenomenon is known as non-monotonicity. Exercises 4 and 5 below are a do-it-yourself example.

Exercises:

1. Use case analysis to show that the games of 3×3 tic-tac-toe and 4×4 tic-tac-toe are draws.

2. Construct pairing strategies for 5×5 and for 6×6 tic-tac-toe, showing both are draws.

3. Given a pairing strategy for x tic-tac-toe, construct one for x tic-tac-toe. Conclude that for , the game of x tic-tac-toe is a draw. Why wouldn’t this work for x tic-tac-toe?

4. Let be the set of vertices of a binary tree of depth ; that is, and the bottom row has leaves. Let consist of the subsets of which are direct paths to leaves along the tree; there is one of these for each leaf. (By direct, we mean each step descends one level.) Show that the hypergraph game is a P1 win in moves.

5. Let be the game of the previous exercise on a binary tree of depth . Add one winning line to the set such that the resulting game is a draw, proving that non-monotonicity can occur.

Inspired by Tim Gowers’s illuminating informal discussions of mathematical topics, and in particular this one on vector spaces, I am starting my own “Shoulda Series.” That is, a series of notes on things about which I might have said: “Someone should have told me this a long time ago!” In particular, I don’t want to use this series to “give away” the sorts of things you should really figure out for yourself.

Question: Why do we avoid choosing bases for vector spaces whenever possible? In particular, why is an isomorphism defined independent of bases “better” than one which uses bases?

One of thousands of possible answers: Since I’m a geometer at heart, I think the following is a great reason. In short, the moral is: basis-free isomorphisms generalize to vector bundles and basis-dependent ones usually do not. This is because “choosing a basis” of a vector space is the point-wise analogue of the “choosing a local trivialization” of a vector bundle. To do anything globally, one generally needs to use several different trivializations; a bona fide isomorphism of vector bundles requires a family of vector space isomorphisms compatible with the transition functions between the various trivializations. (For those who know what this means, the precise check is that the Cech 1-cocycles corresponding to the bundles in question differ by a Cech 1-coboundary, where the coefficient sheaf is the constant sheaf associated to .)

For a wealth of examples, consider the following. Any two finite-dimensional vector spaces (over the same field) are isomorphic, but not naturally so. But there are usually several different isomorphism classes of (real or complex) vector bundles of a given dimensionon on a given space.

A more specific example will also bring out a slight improvement of our moral above. Let be a real vector space and be a real vector bundle. Recall that the isomorphism between and its dual is not natural. Hence is not always isomorphic to its dual . (Here, is the trivial vector bundle .) But if is given an inner product , the isomorphism becomes natural: the map is simply . So if a vector bundle admits an inner product (in the sense of vector bundles), then we should expect an isomorphism . Indeed this is the case, and in fact this is a common phenomenon: using a partition of unity, you can check that every real vector bundle on a paracompact base space admits an inner product.

However, the situation is different with complex vector bundles. Let be a complex vector bundle. While every complex vector bundle paracompact base space admits a Hermitian inner product, this product induces an isomorphism between the dual bundle and the complex conjugate bundle ; neither of these need be isomorphic to itself.

For example, consider the tangent bundle to the Riemann sphere and its dual, . Considered as real 2-plane bundles, these are isomorphic since is paracompact. But they are non-isomorphic as complex line bundles. For readers familiar with characteristic classes, there is an easy way to see this algebraically. The Chern classes of a complex vector bundle take values in the integral cohomology of the base space. They obey the relation , so if for any odd , then is not isomorphic to . The Stiefel-Whitney classes of a real vector bundle obey the same law. But since they take values in cohomology, and the relation is an equality!

Aside: Unoriented phenomena (e.g. Stiefel-Whitney classes) tend to use relations, while oriented phenomena tend to use relations. Complex vector spaces have a canonical orientation on their underlying real vector bundles, so complex phenomena (e.g. Chern classes) fall under oriented phenomena. Examples are mod 2 versus oriented intersections (see Guillemin & Pollack, Differential Topology) and the oriented versus unoriented versions of Poincaré duality (see Hatcher, Algebraic Topology, freely available online). While I’m on the subject of references, two great references for vector bundles and characteristic classes are Milnor & Stasheff’s Characteristic Classes and Hatcher’s Vector Bundles and K-Theory (the latter is unfinished and freely available online).

We conclude with our improved moral statement.

Improved moral: A natural isomorphism of vector spaces generalizes to vector bundles. An isomorphism of vector spaces making use of a structure which “globalizes” well (e.g. inner products, when the base space is paracompact) will also generalize to vector bundles. Other isomorphisms often will not.

In my last post we saw how to play and win Torus-TTT, the game of tic tac toe on a torus. Recall that we constructed the game by identifying two pairs of sides of the usual tic tac toe board and defining the winning patterns to be patterns which are the usual tic tac toe winning patterns, viewed via a different identification of the torus with the square. Then our “Easy Criterion” stated that this was equivalent to taking all usual winning patterns and translating them mod 3 in both directions the nine possible ways. In other words, we let act on the usual tic tac toe board by translations mod 3, and defined the Torus-TTT winning patterns to be -translates of the usual winning patterns.

More generally suppose is some 3×3 tic tac toe board with one or two pairs of sides identified, each identification either parallel or anti-parallel. Then the winning patterns of tic tac toe on are defined to be those which can be made into a standard tic tac toe (-TTT) winning position by translating in directions where sides are identified, along with the appropriate reflection in the anti-parallel case (more on this reflection below). Our general plan of attack will be to construct a group of symmetries acting on the -TTT board by transformations which preserve the set of winning patterns.

To be clear (and again, I apologize for the lack of pictures in this post) consider -TTT, tic tac toe on the Möbius band . This surface is formed by identifying the two vertically oriented sides in an anti-parallel fashion and leaving the horizontally oriented sides unidentified. Then winning patterns are those obtained by left or right translations of -TTT winning patterns; here, a translation moves the contents of to and that of to , but when wrapping around there is a reflection: the contents of is sent to . So translation by one or two squares in either direction has order six—that is, performing a translation six times is the identity but no fewer number of iterations yields the identity. So we have an action of on -TTT which leaves winning patterns invariant.

Note that the action of on -TTT has two orbits: the first is the central row and the second is the union of the two other rows. Observe that an opening move is only well-defined up to which orbit it is placed in, so in effect, there are really only two possible opening moves in -TTT! (The solution to exercise 2 of my previous post is then clear: acts transitively on Torus-TTT, so all opening moves are equivalent.) However, once an opening move is chosen, the symmetry is broken—second moves within the same orbit will generally be inequivalent, since their position relative to the first move is part of the data of the game.

Equipped with this action, it isn’t hard to check that -TTT has 16 winning patterns, and that the first player will always win (assuming, as always, perfect strategy). If the opening play is to the six-element orbit, the first player can force a win in three moves.

Of course the use of the groups and isn’t exploiting the maximum amount of symmetry; for instance, rotation by about the center of the board is a symmetry of both these games, but is contained in neither of these groups. But the groups and do capture the full orbit structure, since it is clear that central-strip and outer-strip moves in -TTT really are inequivalent.

We have now solved tic tac toe on , the torus , the cylinder (since Cylinder-TTT is equivalent to Torus-TTT by exercise 4 of the previous post), and the Möbius strip . There are two cases left to consider. In the following exercises, highlight for hints and answers.

1. Consider -TTT, tic tac toe on the Klein bottle . This is a square with both pairs of sides identified; one identification is parallel and the other is anti-parallel. Who wins? How many winning patterns are there? (Draw them!) Hint 1: Although you may not want to write down the group acting on the Klein bottle explicitly, note that the symmetries do act transitively. Hint 2: The standard winning patterns are three across, three vertical, and diagonals. How do the translations on the Klein bottle board affect the number of symbols in a given column for a given winning pattern? Answer: The first player always wins, with perfect strategy. There are 30 winning patterns.

2. The real projective plane is the last remaining possibility: both pairs identified, both anti-parallel. Solve this one as well: find the winning patterns and figure out who wins. Hint: Keep playing with it! Answer: Any three-square subset of the board is a winning pattern. Thus there are 9 choose 3 = 84 winning patterns, and the first player trivially wins in 3 with any strategy.

3. Try all six patterns with 4×4 boards. To start, are Torus-TTT and Cylinder-TTT still equivalent?

(This post is inspired by one of my favorite books. Also, there really should be pictures in this post; I apologize for their absence.)

I’m going to describe a bit of “recreational topology.”

Hopefully you already know the rules of tic tac toe (“noughts and crosses” for the Brits). And if you’ve played it a few times, you may have figured out that if both players play with perfect strategy, the game will always end in a draw. Now imagine playing on a board which behaves like a Pacman game—that is, the right (respectively, top) edge of the board is identified with the left (respectively, bottom) edge. This is, of course, equivalent to taking a square tic tac toe board and shaping it into a (2-dimensional) torus; we’ll call this game Torus-TTT.

Let’s make this more precise. Recall that one construction of the torus is as a square with opposite sides identified with the same orientation. Then given a tic tac toe board, we imagine it wrapped onto the torus via this construction. Of course we can view the nine resulting boxes on the torus as a square with any of the nine as the upper-left box. Thus we define a winning pattern in Torus-TTT to be any pattern on the usual 3×3 board which, when sent to the torus and then put back onto a 3×3 board with some possibly different choice of upper-left box, is a winning pattern in ordinary tic tac toe (which from now on we’ll call -TTT).

Before reading on, grab a friend and play Torus-TTT a few times. Try and picture each position as on the torus itself, embedded in .

In practice, it’ll be useful to have a handy way of looking at a 3×3 pattern and telling if it’s a winning pattern in Torus-TTT. Luckily, this is easy! Label the nine boxes of a 3×3 board as , where the box is the one in the -th row and -th column. (Take mod 3.) Then the transformation which shifts every symbol one box to the right (mod 3) is clearly equivalent to sending the board to the torus, then taking the box which was formerly as the new . That is, it preserves winning patterns! More generally, the transformations of the form “send the contents of to ” (for fixed mod 3) realize all possible changes-of-origin. Call this transformation . Then we have verified the following:

Easy Criterion: A Torus-TTT pattern is a winning pattern if and only if applying some to it yields a winning pattern in -TTT.

At this point, there are several obvious exercises. A few are listed below. Cheaters can highlight the text following an exercise to see its solution.

1. Enumerate all winning patterns in Torus-TTT. Answer: the usual rows, columns, and diagonals, as well as the following: {(0,0),(1,2),(2,1)} and its image under rotation by 90, 180, and 270 degrees.

2. What is the best opening move? Answer: Anywhere—they’re all equivalent!

3. If both players play with perfect strategies, who wins? And in how many moves? Answer: First player, in 4 moves.

If you’ve completed those three (simple) exercises, then you’ve “solved” Torus-TTT. Of course, different identifications of the edges of the 3×3 board yield different games. There are six possible games: , the torus, real projective space (), the Klein bottle, the cylinder, and the Möbius band. (We don’t allow identifications of adjacent edges.) In the sequel post, we’ll take a look at and solve the remaining cases. Until then, try the following:

4. Prove that Torus-TTT and Cylinder-TTT are the same game.

5. Interpret the solution to Exercise 2 above in terms of a suitable finite group acting on .