Posted by: lewallen | September 2, 2009

Symplectic geometry I

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 {m} in our manifold {M}, {\omega_{m}} should be non-degenerate, so that if {\omega_{m}(X,Y)}=0 for all {Y\in T_{m}M}, then {X} must be 0. Finally, note that because {\omega} is a skew-symmetric 2-form, it is a differential 2-form on {M}, and we require that as a 2-form, {\omega} is closed, i.e., {d\omega = 0}. 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 {\tilde{\omega}: T_{m}M\rightarrow T_{m}^{*}M}. The image of {X} is the functional which sends {Y} to {\omega(X,Y)}. 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 {X_{i}} 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 {T_{m}M\rightarrow T_{m}^{*}M} 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 {{\mathbb R}^{2n}} has a basis of the form {\left\{X_{1},\dots, X_{n}, Y_{1},\dots , Y_{n}\right\}}, with {X_{i}} pairing with {Y_{i}} 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 {\tilde{\omega}} between tangent and cotangent space sends {X_{i}} to the covector (linear functional) which is 1 on {Y_{i}} and 0 on the other basis vectors. As a side note, if we make {{\mathbb R}^{2n}} into a complex vector space by letting {Z_{i}=X_{i}+iY_{i}}, then the standard form {\omega} can be written in terms of the standard dot product as {\omega(A,B)=\langle iA,B\rangle}, indeed, we have {iX_{i}=Y_{i}} (I hope I have my signs ok). Thus the standard complex, Riemannian, and symplectic structures on {{\mathbb R}^{2n}} 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 {S\subset {\mathbb R}^{2n}}, define {S^{\perp}=\{ X: \omega(X,Y)=0 \text{ for all } Y\in S\}}. Then the non-degeneracy of the form insures that, just like for a normed vector space, the dimensions of {S} and {S^{\perp}} add up to {2n}. However they are not necessarily disjoint! Indeed, for example, with the standard form, {\text{Span}(X_{1},\dots, X_{n})^{\perp}=\text{Span}(X_{1},\dots, X_{n})}. Such a subspace is called Lagrangian, meaning the form restricted to it is identically 0, and it is of maximal dimension ({=n}) 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 {{\mathbb R}^{2n}} 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 {S} 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 {M}, so that the flow is well defined on all of {M}. 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 {f} on a smooth manifold {M}, we can associated the differential {df}, which is the one-form whose value at a vector field {X} at a point {m\in M} is the value of the directional derivative {X(f)} at {m}. The nice thing about having the dual map {\tilde{\omega}} at each point is that from each differential 1-form we can produce a vector field, and this applies to {df}, giving a vector field {\tilde{\omega}^{-1}(df):=X_{f}} which depends on {f}. 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 {f} (indeed, it points towards the direction of most increase of {f}, which naturally is perpendicular to the level sets). To see this, one just applies the derivation {\text{Grad}(f)} to {f}, and unwinding the definitions one gets simply the smooth function {\langle \text{Grad}(f)_{m},\text{Grad}(f)_{m}\rangle }, the norm squared of {\text{Grad}(f)_{m}} for each point {m\in M}. However, doing this with {X_{f}} gives the directional derivative {X_{f}(f)} as {\omega(X_{f},X_{f})=0}! Indeed, {X_{f}} is tangent to the level sets of {f} (so it is just {\text{Grad}(f)} rotated by {\pi/2} in some direction, or, alternatively, hit by the multiplication by {i} map, in an appropriate basis and some compatible complex structure). So the flow of any point {m} under {X_{f}} stays within a particular level set, and so the level sets of {f} are invariant under the flow (the standard example is to take the sphere {S^{2}} with its standard volume form as the symplectic form, and take the function {f} to be the height function. Then the level sets are circles of constant height, and the flow of {X_{f}} can be seen to be rotation around the vertical axis). This is perhaps the moment to mention that {X_{f}} is called a Hamiltonian vector field (rather than gradient vector field as in Riemannian geometry), and {f} is called the Hamiltonian (function) (of {X_{f})}. Indeed, the fact that the level sets of {f} 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 {H} 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 {X_{H}}. 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.

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Responses

  1. jesus at least make it 2 parts

  2. I think you underestimate the importance of the physical motivation. The guiding example of a symplectic manifold is the cotangent bundle of a manifold. The motivation for symplectic geometry and, more specifically, dynamical systems on a symplectic manifold comes from Hamiltonian mechanics.

    The modern subject focuses on compact symplectic manifolds, but I think students should first learn about the origins of the subject before restricting their attention to the compact case. For example, check out Arnold’s book, Mathematical methods of classical mechanics.

  3. Thanks for a great post.

    I really like the comparison to the Riemannian case. I tried to learn symplectic geometry once in the past. But, with just definitions, it is hard to visualize and get some intuition.

    I am wondering what is an analogue for Lagrangian submanifold. It seems to be a mix of transverse and orthogonal submanifold.

    (By the way, I’m one of Yan’s classmate at MIT. Right now, I’m trying to learn some Seiberg-Witten theory.)

  4. I think it is a mistake to try to view symplectic geometry as somehow analogous to Riemannian geometry. The two are quite different and in fact somehow complementary to each other (which is why you can have both simultaneously, for example, in a Kahler manifold).

    A Lagrangian submanifold is characterized by the fact that a tubular neighborhood of it is symplectically isomorphic to a tubular neighborhood of the zero section of the cotangent bundle of the Lagrangian submanifold.

  5. […] geometry II 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 […]

  6. So, on the one hand, there is no Darboux theorem for Riemannian geometry, because Riemannian manifolds have intrinsic curvature. I expect — I am not an expert — that a better parallel for Riemannian geometry might be a Poisson structure, where the Darboux theorem only holds in part, and the local structures are not entirely classified as far as I know.

    On the other hand, there are very nice unifications of Riemannian and Symplectic geometry. Deanne Yang has already mentioned Kahler structures, which are Riemannian, Symplectic, and Almost Complex (you get any one of these for free given the right compatibility condition of the third).

    Another, in my mind better, unification is in supersymmetry. The most naive definition of Riemannian and/or Symplectic structure, as you gave, is of a smooth symmetric/antisymmetric bilinear form, i.e. a smoothly varying map T_xM^{\otimes 2} \to \RR for each x\in M. Well, supersymmetry switches the words “symmetric” and “antisymmetric”. So a Symplectic (normal) vector space is precisely a purely-odd Riemannian vector space, and a Riemannian (normal) vector space is precisely a purely-odd Symplectic vector space.

    Oh, I should define “supervector space”. As a monoidal category, the category is supervector spaces is precisely the category of representations of the group Z/2 of order 2. Any such space V splits as a direct sum of eigenspaces V_+ \oplus V_-, where the generator of Z/2 acts trivially on V_+ and as -1 on V_-. But as a braided category SVect is different from Z/2-rep. A _braiding_ in a monoidal category is a (nice) map V\otimes W \to W\otimes V. In SVect, we have V = V_+ \oplus V_-, W = W_+ \oplus W_-, so V\otimes W = (V_+ \otimes W_+) \oplus (V_+ \otimes W_-) \oplus (V_- \otimes W_+) \oplus (V_- \otimes W_-), and the braiding is the usual flip map v\otimes w \mapsto w\otimes v on each of the first three direct summands, but the map v\otimes w \mapsto -w\otimes v on the last summand.

    To do calculus/geometry, we take the algebra of smooth functions on V = V_+ \oplus V_- to be the algebra C^\infty(V_+) \otimes \bigwedge V_-, where \bigwedge V_- is the exterior algebra of V_-, or perhaps of its dual, depending on your convention. Put another way, take a basis of the dual space V^*, and take the supercommutative algebra freely generated by this basis. “Supercommutative” is to emphasize that we’re demanding that m(a\otimes b) = m(flip(a\otimes b)), where “flip” is the braiding I defined above and m is the multiplication map. Anyway, a supermanifold is then something that’s locally a supervector space, in that it is a sheaf so that small enough pieces look like the algebra of smooth functions on a supervector space.


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