My dissertation is about simulating conservative physical systems on a computer; but it is based in symplectic geometry, which exists only in even dimensions.
Put it this way: Riemannian (and as a special case, "ordinary") geometry is based around the inner product <a,b>, which gives you both angles and projection (and 2D shadows from 3D objects, for example) and lengths. You learn a lot about this (not in the "curve" differential-geometric sense, but most of the intuition carries over) in college linear algebra: you learn to think about entire vector subspaces having orthogonal complements as defined by the inner product. On the other hand: length (even if curved) in that direction is the same as length in this direction, i.e. the inner product is a symmetric bilinear map.
Symplectic geometry is built on a skew-symmetric bilinear map (the symplectic form) so that w(u,v)=-w(v,u). In 2D, the symplectic form is equivalent to the determinant: note that this is an oriented or signed area rather than an absolute-value one. And then in 3D, there is no symplectic form (it's rather fun to prove this) such that there is no vector (apart from the origin) that collapses all the other.
Anyway, the challenging thing about this, writing my dissertation, is that unlike with linear algebra where you can gradually extend intuition from the real line to the plane to 3D space and then think "ok, I think I can grok this in 500 dimensions" (and go do multivariate statistics, for example), in symplectic geometry there is either the trivial case (and again, this is the determinant and not particularly "new" to you) or the 4D case next.
And then you can't draw pictures. YOU CAN'T DRAW PICTURES. This is such an intuition-fucker. Books that need the symplectic form right away and can't waste time on the geometry merely note that the symplectic form at higher dimension is the sum of projections of 2D determinants. Ok, wait why?
But here's the fun thing (that maybe I've spoiled by talking about inner products first): while you're proving that there can't be a symplectic form in odd-dimensional spaces, you stumble upon a parallelism between symplectic complements, i.e. the sets
symp(W) = { u | w(u,v)=0 for all v in W}
and orthogonal complements, i.e. the sets
orth(W) = { u | <u,v>=0 for all v in W}
Namely that they're kernel sets of the bilinear maps <.,.> and w(.,.) (and because w(u,v)=0 implies w(v,u)=0, they're both left- and right-kernels). Moreover any vector z can be uniquely expresseed as
u+v where u is in W and v is in orth(W)
or alternately
p+q where p is in W and v is in symp(W)
So Riemannian geometry and symplectic geometry are like ways to split a vector space in complementary parts. Now, because there's Riemannian/euclidean geometry on a line and on the plane, you can build a geometry where there are planes orthogonal to lines, i.e. a 3D geometry. And maybe you have a timeline across which 3D spaces are strung together, ie. 4D space. But this 4D space of yours isn't symplectic. So it's realy hard to see.
So if you look for "symplectic geometry" on YouTube you're bound to find Dusa McDuff's lecture where she starts by writing in big bold letters in the blackboard:
4 = 3 + 1
4 = 2 + 2
... and that's a way to "see" four dimensions: to see entirely different geometries built on it.
Ok, think of two vectors in "three dimensions" (but not really). Draw X,Y,Z axes. So these pseudovectors will have a determinant which you can project on XY space and YZ space. Now, because our vectors have 4 dimensions and not 3, these projections will not be the same! Then your four-dimensional symplectic form (in some applications an "exterior product") is the sum of those two projections.
So you're pseudovisualizing a 4D object in 3D space as you might pseudovisualize a 3D object in 2D space but with a different concept of projection/perspective/etc. because we're not in the inner product geometry, we're in the symplectic geometry.
> Moreover any vector z can be uniquely expresseed as p+q where p is in W and v is in symp(W)
Of course, crucially, this is only true if `W` is non-degenerate; as you'll know, one of the important things about symplectic spaces is the existence of half-dimensional, totally isotropic subspaces. (In fact, there's nothing particularly symplectic per se about this issue; it is not the (conjugate-)symmetry of the usual inner product so much as its positive definiteness that means that no such extra non-degeneracy condition must be imposed. Indeed, the study of non-positive definite but symmetric inner products, as in the geometry of relativistic spacetime, carries its own challenges to intuition.)
Put it this way: Riemannian (and as a special case, "ordinary") geometry is based around the inner product <a,b>, which gives you both angles and projection (and 2D shadows from 3D objects, for example) and lengths. You learn a lot about this (not in the "curve" differential-geometric sense, but most of the intuition carries over) in college linear algebra: you learn to think about entire vector subspaces having orthogonal complements as defined by the inner product. On the other hand: length (even if curved) in that direction is the same as length in this direction, i.e. the inner product is a symmetric bilinear map.
Symplectic geometry is built on a skew-symmetric bilinear map (the symplectic form) so that w(u,v)=-w(v,u). In 2D, the symplectic form is equivalent to the determinant: note that this is an oriented or signed area rather than an absolute-value one. And then in 3D, there is no symplectic form (it's rather fun to prove this) such that there is no vector (apart from the origin) that collapses all the other.
Anyway, the challenging thing about this, writing my dissertation, is that unlike with linear algebra where you can gradually extend intuition from the real line to the plane to 3D space and then think "ok, I think I can grok this in 500 dimensions" (and go do multivariate statistics, for example), in symplectic geometry there is either the trivial case (and again, this is the determinant and not particularly "new" to you) or the 4D case next.
And then you can't draw pictures. YOU CAN'T DRAW PICTURES. This is such an intuition-fucker. Books that need the symplectic form right away and can't waste time on the geometry merely note that the symplectic form at higher dimension is the sum of projections of 2D determinants. Ok, wait why?
But here's the fun thing (that maybe I've spoiled by talking about inner products first): while you're proving that there can't be a symplectic form in odd-dimensional spaces, you stumble upon a parallelism between symplectic complements, i.e. the sets
symp(W) = { u | w(u,v)=0 for all v in W}
and orthogonal complements, i.e. the sets
orth(W) = { u | <u,v>=0 for all v in W}
Namely that they're kernel sets of the bilinear maps <.,.> and w(.,.) (and because w(u,v)=0 implies w(v,u)=0, they're both left- and right-kernels). Moreover any vector z can be uniquely expresseed as
u+v where u is in W and v is in orth(W)
or alternately
p+q where p is in W and v is in symp(W)
So Riemannian geometry and symplectic geometry are like ways to split a vector space in complementary parts. Now, because there's Riemannian/euclidean geometry on a line and on the plane, you can build a geometry where there are planes orthogonal to lines, i.e. a 3D geometry. And maybe you have a timeline across which 3D spaces are strung together, ie. 4D space. But this 4D space of yours isn't symplectic. So it's realy hard to see.
So if you look for "symplectic geometry" on YouTube you're bound to find Dusa McDuff's lecture where she starts by writing in big bold letters in the blackboard:
4 = 3 + 1 4 = 2 + 2
... and that's a way to "see" four dimensions: to see entirely different geometries built on it.