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I would like to understand what a Symplectic Structure is, and its implications in Classical Mechanics (Phase Space), but in pre-grade terms (If that could be possible). I have not taken any Differential Geometry course, so it has been a bit hard for me to understand it (I do not know what a 1-form, 2-form, neither Exterior Algebra is).

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    $\begingroup$ Did you do any Lie group theory? $\endgroup$ Commented Jul 8 at 3:21
  • $\begingroup$ I'd like to, but I have not (yet). $\endgroup$ Commented Jul 8 at 3:23
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    $\begingroup$ General tip: Check the right margin for related posts. $\endgroup$
    – Qmechanic
    Commented Jul 8 at 4:57

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At the most rough level possible, a symplectic structure (geometrically) is an even-dimensional manifold together with a preferred choice of two-dimensional planes which, taken together, span the space.

To be more concrete, for the $2n$ dimensional symplectic space, give each two-dimensional plane coordinates $x^i,p_i$, where $i = 1, \cdots, n$. Given such coordinates, we can imagine an infinitesimal displacement in either of these directions $dx^i,dp_i$. An infinitesimal displacement is a mathematical object called a 1-form, and you can think of it as something which takes a direction as input and outputs a number. In more technical terms, it is a map from a vector (the direction of the displacement) that outputs a number (the amount of displacement in that particular direction), so it is a dual vector.

But to specify a plane, we need to have two directions as input, not just one. The fancy math way to talk about the plane "spanned" by these two coordinates is via the wedge product of their differentials, written as $dx^i \wedge dp_i$ (no sum over $i$ yet). This is a map which takes TWO directions as input and outputs a number. This is called a 2-form. The wedge just says that the map is antisymmetric in its inputs. This follows from linearity + demanding that it vanishes if you put in the same input twice. It's worth checking that for yourself. The reason we want it to vanish if we input the same direction twice is that then we don't get a full plane, only the degenerate case of a single line.

Finally, a symplectic space is distinguished by the presence of a "symplectic form" $\omega$, which is just a special linear combination of planes in the sense I described above: $$\omega = \sum_i dx^i \wedge dp_i.$$

The reason we take this combination is that it is the one which "spans" the full space. It's essentially a math way of taking an even linear combination of all the special planes in the space.

But just how special are these planes? For a vector, I can decompose the same vector in many different bases. In the same spirit, I can decompose the same symplectic form $\omega$ into the sum of many different planes. For example, I could scale $x^i \to \lambda x^i$ and $p_i \to \lambda^{-1} p_i$. That leaves each term in the sum invariant, so $\omega$ is the same. I could also rotate $x^i \to R^i_j x^j$ and $p_i \to p_j R^j_i$. You should check that if $R^i_j$ is a rotation matrix, then this does indeed leave $\omega$ invariant. The full set of operations which leaves $\omega$ invariant is called the symplectic group.

Just like Euclidean geometry is essentially completely determined by the set of symmetry operations which leaves its preferred tensor (the metric) invariant, symplectic geometry is essentially completely determined by the symmetry operations which leaves $\omega$ invariant. There are more abstract ways to define this $\omega$, but there are theorems that say the one I have chosen is essentially already as general as possible (you can always pick a basis for phase space where it locally takes that form).

It's hard to go into the implications of this form on physics without going deeper into these special consequences, but that requires a lot of math that you explicitly mentioned you don't know very well yet. So I'll leave my explanation here for now, and just mention that the next steps for you should really be differential geometry, specifically tensor calculus and Lie groups. If you can fully understand the phase space of $\mathbb{R}^{2n}$ and understand what properties of $\omega$ make it really special, as well as the geometry of the symplectic group and how to use it to get Poisson brackets etc, that will take you really far. For completeness, I'll end with a hint about what properties of $\omega$ I'm talking about, with the significance of these properties left as an exercise for the reader:

  1. $\omega$ is a two-form.
  2. $\omega$ is closed, i.e., $d\omega=0$.
  3. $\omega$ is non-degenerate, i.e., it is invertible as a matrix in some basis.
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  • $\begingroup$ Thank you so much for this answer! I really appreciate your interest on answering my question! It's a very good and understandable answer. $\endgroup$ Commented Jul 8 at 18:47