24
$\begingroup$

My question essentially breaks down to

How do you, a working mathematician, think about (real) symplectic groups? How do you visualize symplectic (linear) transformations? What intuition do you have?

For example, we can think of rotations by thinking, "Okay, let's just restrict to a plane fixed by the transformation and see what it does there." This doesn't quite work for the symplectic group, though, since on the plane it is indistinguishable from the area-preserving linear transformations. The next smallest example is for a 4-dimensional vector space, which isn't particularly easy for me to visualize.

$\endgroup$
  • 4
    $\begingroup$ you might find John Baez's blog helpful [it certainly helped me] $\endgroup$ – Carlo Beenakker Aug 13 '17 at 14:04
  • 1
    $\begingroup$ arxiv.org/abs/math/9905052 $\endgroup$ – Vít Tuček Aug 14 '17 at 23:00
  • $\begingroup$ Robin: do I assume correctly that you understand that the lowest dimensional one is $SL(2, {\mathbb R})$ which forms a solid three-dimensional torus? $\endgroup$ – Richard Montgomery Aug 15 '17 at 18:40
  • $\begingroup$ @RichardMontgomery Indeed, I understand that $\mathrm{Sp}_2\mathbb{R}=\mathrm{SL}_2\mathbb{R}$ and that they are diffeomorphic to the interior of a solid torus (an open disk times a circle). $\endgroup$ – Robin Goodfellow Aug 15 '17 at 19:58
12
$\begingroup$

In a different direction from Victor Protsak's answer, I will focus on your comment

The next smallest example is for a 4-dimensional vector space, which isn't particularly easy for me to visualize.

It's true that the definition of the Lie group $\mathrm{Sp}(4,\mathbb{R})$ involves a $4$-dimensional vector space, but actually there are two nice $3$-dimensional homogeneous spaces on which this group acts.

The first, most obvious one, is the projective space $\mathbb{RP}^3$ of $1$-dimensional subspaces in $\mathbb{R}^4$.You can picture it as a compactification of $\mathbb{R}^3$ where you add a point at infinity for each possible direction of a line. The symplectic group acts on this space because it acts linearly on $\mathbb{R}^4$. The difference between the symplectic group and the full group of projective automorphisms $\mathrm{PGL}(4,\mathbb{R})$ is that $\mathrm{Sp}(4,\mathbb{R})$ does not act transitively on lines. Some lines are special, because they come from projectivizing Lagrangian planes. Conversely, any continuous transformation of $\mathbb{RP}^3$ which preserves lines and Lagrangian lines is an element of $\mathrm{Sp}(4,\mathbb{R})$ (this is a version of the fundamental theorem of projective geometry).

The second is the Lagrangian Grassmannian, the subset of the Grassmannian of $2$-planes consisting of planes where the symplectic form vanishes. It is a $3$-dimensional space again, which is a bit more topologically complicated. It's homeomorphic to $(\mathbb{S}^2\times \mathbb{S}^1)/\langle\iota\rangle$, where $\iota$ acts as the antipodal map on both spaces. This space admits an invariant conformal Lorentzian metric. This means that at each point (Lagrangian) $L$, there is a light cone, and in this case it consists of all Lagrangians intersecting $L$ in a line. Again, any continuous transformation of this space which preserves light cones comes from something in $\mathrm{Sp}(4,\mathbb{R})$. The Lorentzian structure is an accident coming from the isomorphism $\mathrm{Psp}(4,\mathbb{R})\cong \mathrm{SO}^0(3,2)$ and so does not exist for higher symplectic groups, but the incidence structure of Lagrangians is still present. You can imagine this space of Lagrangians as a compactification of $\mathbb{R}^{2,1}$, the $(2+1)$-dimensional Minkowski space, preserving its light cone structure.(see for instance https://arxiv.org/pdf/0706.3055.pdf for details)

If you get familiar with one or both of these spaces, it should help your intuition of how this particular group works.

Edit: Here is an additional description to answer R. van Dobben de Bruyn's comment

There is actually a more intuitive description due to Sophus Lie, the "Lie quadric". The space of Lagrangians in $\mathbb{R}^4$ is identified with the space of oriented circles on the $2$-sphere, where zero-radius circles ("point circles") are also allowed.

The way to set it up is to use $\mathbb{C}^2$ as your real 4-dimensional vector space, fix a complex-valued symplectic form $\omega$ and use $Im(\omega)$ as your real-valued symplectic form. Then, the projection $(\mathbb{C}^2\backslash\{0\})\rightarrow\mathbb{CP}^1$ sends Lagrangian planes to circles 2-to-1 (and so Lagrangians to oriented circles 1-to-1). Two Lagrangians intersect if and only if the corresponding circles are tangent. The group $\mathrm{Sp}(4,\mathbb{R})$ acts on oriented circles and points as tangency-preserving transformations. These transformations take a bit of time to get used to, since they can send a point to a circle... You just have to remember that, unlike in projective geometry, the transformations on the "space of circles" do not preserve points.

The transformations which do preserve the "point circles" are the same as the Moebius transformations acting on $\mathbb{CP}^1$. This is a $6$-dimensional subgroup of the $10$-dimensional $\mathrm{Sp}(4,\mathbb{R})$. There is also a "increase radius by $r$" operation which takes every circle in $\mathbb{C}\subset\mathbb{CP}^1$ to a circle centered at the same point, of radius $r$ bigger. Circles with negative orientation are thought of as having "negative radius". If two circles are externally tangent, they have opposite orientations, and if you add the same amount of "signed radius" to them, they stay tangent.

These two types of transformations (Moebius and radius addition) together generate the full group of tangency-preserving transformations. (for more details see https://en.wikipedia.org/wiki/Lie_sphere_geometry)

$\endgroup$
  • 1
    $\begingroup$ Nice answer, but it also raises further questions. What does the configuration of Lagrangian lines look like? Does it have some combinatorial description? Without that, it's still hard to visualise $\operatorname{Sp}(4,\mathbb R)$. $\endgroup$ – R. van Dobben de Bruyn Aug 15 '17 at 22:57
  • $\begingroup$ This is a lovely answer. I feel quite comfortable with symplectic groups, but just from the "reductive groups with certain root datum" perspective, and this gives me a new perspective from which to regard them. $\endgroup$ – LSpice Aug 18 '17 at 15:44
12
$\begingroup$

This is an outgrowth of an extended comment dealing with the superficial difference between symplectic and orthogonal rotations embedded in the question. I posit that the theory for symplectic groups is completely parallel to the split orthogonal case.

The main distinction is that symplectic groups are always split, whereas orthogonal groups come in many different "flavors" depending on the Witt index of the underlying form, from anistropic groups, such as the "usual" compact real orthogonal groups ${\rm O}(n)$, to split groups, e.g. complex orthogonal groups and real ${\rm O}(n,n)$. Witt's theorem on isomorphisms of orthogonal spaces remains true in the symplectic setting. Once you stop thinking of just compact real orthogonal groups and consider general orthogonal forms that have (many) isotropic subspaces, most differences recede. In fact, orthogonal story is richer in some ways, and if you have a way of thinking of orthogonal transformations in the presence of isotropic subspaces, it should guide you in creating the corresponding mental picture in the symplectic case.

To illustrate the point made in the previous paragraph, consider a symplectic vector space $W$ and a nondegenerate $2$-plane $V$. Then any symplectic rotation of $V$ extends to a symplectic rotation of $W$ that is the identity transformation on the (skew-)orthogonal complement of $V$ in $W$.

More generally, the same remarks apply to $\varepsilon$-hermitian forms on vector spaces over a division algebra with an involution and the corresponding isometry groups (when the involution is trivial, one gets the orthogonal and symplectic cases depending on whether $\varepsilon$ is $1$ or $-1$). For details, you may consult books on geometric algebra or geometry of classical groups.

$\endgroup$
  • 3
    $\begingroup$ It may be worth noting that "symplectic groups are always split" doesn't mean what an algebraic-group theorist might (or at least I did) think. Rather than "symplectic groups have no non-split forms" (which is false, at least $p$-adically), I think it means "the group of similitudes of any non-degenerate alternating form is split." $\endgroup$ – LSpice Aug 14 '17 at 3:40
  • 2
    $\begingroup$ Good point! There is certain incongruency in my answer in that I was using algebraic groups lingo in a geometric algebra context, so the terms "orthogonal group" and "symplectic group" are to be understood in this narrow sense, in line with the hints in the final paragraph. $\endgroup$ – Victor Protsak Aug 14 '17 at 3:48
  • 1
    $\begingroup$ (By the way, isometries rather than similitudes.) $\endgroup$ – Victor Protsak Aug 14 '17 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.