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This is a follow up question to this question.

Let $\omega$ be an skew-symmetric bilinear form on $\mathbb{R}^{2n}$, which is unique up to change of basis. It is given by the formula $$\omega(\mathbf{x},\mathbf{y}) = \sum_{i=1}^n{x_iy_{i+n}-y_ix_{i+n}}$$

We can then write out the definition of $\mathrm{Sp}(n,\mathbb{R})$ as the group of linear operators $A: \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ such that they satisfy the condition:

$$\omega(A\mathbf{x},A\mathbf{y}) = \omega(\mathbf{x},\mathbf{y})$$ for all $\mathbf{x},\mathbf{y} \in \mathbb{R}^{2n}$.

Elements of $\mathrm{Sp}(n,\mathbb{R})$ are called symplectic transformations.

This is an analog of orthogonal transformations. If $\omega$ were chosen to be an inner product $\langle \cdot , \cdot \rangle$ instead, then the condition $$\langle A \mathbf{x} , A \mathbf{y} \rangle = \langle \mathbf{x,y} \rangle$$ is equivalent to preserving distances between points in $\mathbb{R}^n$, so a more down-to-earth point of view.

So my question is, is there a geometric meaning to the definition of symplectic transformations, like "distance-preserving transformation" in the orthogonal case? If not, is there maybe some other way to think about them?

mz71
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2 Answers2

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Write $x=\left(\begin{array}{c} a\\ b \end{array}\right),\,y=\left(\begin{array}{c} c\\ d \end{array}\right)$ with $a,\,b,\,c,\,d\in\mathbb{R}^{n}$ so $\omega\left(x,\,y\right)=a\cdot d-b\cdot c$. In the case $n=1$, this is a determinant, and $\left(\begin{array}{c} a\\ b\\ 0 \end{array}\right)\times\left(\begin{array}{c} c\\ d\\ 0 \end{array}\right)=\omega\left(x,\,y\right)\hat{k}$. More generally, $\omega\left(x,\,y\right)$ is a sum of $n$ such expressions.

Symplectic transformations are of especial interest in studying systems obeying Hamilton's equations. Any function of the phase space coordinates varies over time as documented here. One of the conservation laws that results is that Poisson brackets are preserved. It leads to an elegant phase space flow $q(t)=\exp(-t\{H,\,\cdot\})q(0),\,p(t)=\exp(-t\{H,\,\cdot\})p(0)$. Such a flow is probably the best way one can "visualize" the consequences.

J.G.
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  • That somewhat a good take on $\omega$, but that doesn't answer how symplectic transformations look like. – mz71 Apr 08 '20 at 10:47
  • @mzg147 Are you open to a discussion in terms of classical mechanics? If so, I can edit my answer later today to explain that symplectic transformations preserve Poisson brackets. – J.G. Apr 08 '20 at 11:35
  • I am open to bascially anything, so yes. I cannot guarantee that I will visually get it, though. – mz71 Apr 08 '20 at 13:11
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    @mzg147 I'll have a think about how best to explain visualizing Poisson brackets in an answer. In the mean time, this may be a valuable read. – J.G. Apr 08 '20 at 13:55
  • @mzg147 I've added a new paragraph. – J.G. Apr 08 '20 at 19:29
  • I don't think that I get it. Liouville's theorem says that Hamiltonian flow is a symplectomorphism. But symplectomorphisms are not linear transformations, so I don't see the connection. – mz71 Apr 09 '20 at 11:39
  • @mzg147 "symplectomorphisms are not linear transformations" Not in general, no. Try working out which ones are linear, though. – J.G. Apr 09 '20 at 11:46
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Here is a try: ${\Bbb R}^{2n}$ is the direct sum of the subspaces $E_i={\rm span}\{e_i,e_{i+n}\}$ for $i=1,...,n$. These subspaces come an area form and with projections which (deceptively) appears to be orthogonal projections in these coordinates. A symplectic transformation should preserve the sum of areas in these subspaces.

I say deceptively because it is simply our choice of coordinates which seem to imply some orthogonallity between the $E_i$'s. In fact, you could take any $n$ two-dimensional subspaces $E_1,...,E_n$ whose direct sum yields ${\Bbb R}^{2n}$. Define area forms $\omega_i$ in each $E_i$, evaluated on projections along the other spaces and $\sum_i \omega_i$ gives you a symplectic form. A symplectic transformation is then a transformation that preserves the sum or these 2-area forms. Any symplectic form can be transformed to the above by a suitable change of coordinates.

But this, of course, does not give you a good global geometric picture of what the symplectic transformation does. Also this picture is not independent of the choices made.

H. H. Rugh
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  • "symplectic if and only if it preserves area form in each E_i".. What is this area form exactly? This doesn't mean just area, because then $\mathrm{Sp}(2) = \mathrm{SL}(2)$ – mz71 Apr 08 '20 at 21:42
  • In each two dimensional space (i.e. equiv to ${\Bbb R}^2$) up to a multiplicative constant there is a unique area form, i.e. the determinant of a 2 by 2 matrix when expressing two vectors in a base. In dim 2 in fact Sp(2)=SL(2) (see e.g. wikipedia on Symplectic Group). I should have said (as written now) that the sympl trans preserves the sum of the areas not each individual. – H. H. Rugh Apr 08 '20 at 22:43
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    Perhaps I misunderstand the intention of this answer, but since a symplectic transformation need not preserve the "hyperbolic pairs" $e_i,e_{i+n}$, it is hard for me to understand what "preserves area" or "... sum of areas" would mean. Nor would "product of areas" quite work. It seems to me that there is simply no such alternative to the bare fact of "preserves an alternating two-form", or similar. – paul garrett Apr 08 '20 at 23:05
  • @paulgarrett Well there is in fact import products that are preserved, i.e. the exterior products, $\wedge^k \omega$, $1\leq k\leq n$. The top-one is a volume form called the Liouville measure which is thus preserved by symplectic transformations. This property form the basis for statistical mechanics (for Hamiltonian systems). – H. H. Rugh Apr 09 '20 at 20:46
  • Well, yes, various things expressible in terms of $\omega$ are preserved... But I don't think there's any simpler alternative. Yes, the top/volume form is preserved, but that's not to say much about individual hyperbolic pairs, and so on. The intrinsic symplectic structure "is what it is", I think. – paul garrett Apr 09 '20 at 21:49