## Unsolved Mysteries of Fundamental Physics

2 January, 2019

In this century, progress in fundamental physics has been slow. The Large Hadron Collider hasn’t yet found any surprises, attempts to directly detect dark matter have been unsuccessful, string theory hasn’t made any successful predictions, and nobody really knows what to do about any of this. But there is no shortage of problems, and clues. Watch the talk I gave at the Cambridge University Physics Society for some ideas on this! Warning: this is for ordinary folks, not experts.

There are some squeaky sounds on the video at first, but they seem to go away pretty quick, so hang in there! You can also see my talk slides here:

and click on the links for extra information.

## Geometric Quantization (Part 6)

1 January, 2019

Now let’s do some more interesting examples of geometric quantization using the functor described in Part 4. Let’s look at the spin-j particle with j > 1/2.

To be specific, let’s consider the spin-3/2 particle. There’s nothing special about the number 3 here: everything I’ll say can be generalized. But the number 3 will give me a nice excuse to show you a picture of a curve called the ‘twisted cubic’.

We can build a spin-3/2 particle from three spin-1/2 particles, all having angular momenta pointing in the same direction. Classically this procedure amounts to tripling a vector, but quantum-mechanically it’s related to cubing. This is a bit mysterious to me, but let me explain.

Classically, if we take a vector of length 1/2 in $\mathbb{R}^3$ and triple it we get a vector of length 3/2. This gives a map from classical states of the spin-1/2 particle to classical states of the spin-3/2 particle. In other words: a map from the sphere of radius 1/2 to the sphere of radius 3/2. Simple! But this is not a symplectic map, since as we saw last time, we are giving the latter sphere a symplectic structure that’s 3 times as big. So, it’s not a valid classical process. Still, it’s a perfectly fine way to get our hands on lots of states of the classical spin-3/2 particle! All of them, in fact.

Quantum-mechanically the state space for the spin-1/2 particle is $\mathbb{C}^2.$ We can think of a guy in here as a linear functional on the dual ${\mathbb{C}^2}^\ast.$ We can cube this and get a homogeneous polynomial of degree 3 on ${\mathbb{C}^2}^\ast.$ The space of such polynomials is called $S^3(\mathbb{C}^2),$ and this is the space of states of the quantum spin-3/2 particle. So, we get a map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Simple! This describes the process of ‘triplicating’ the state of spin-1/2 particle and getting a spin-3/2 particle. But this is not a linear map, so it’s not a valid quantum process. As the saying goes, “you can’t clone a quantum”. Still, it’s a perfectly fine way to get our hands on lots of states of the quantum spin-3/2 particle! But not all of them, as we’ll soon see.

Geometric quantization should reconcile and combine classical and quantum mechanics. How can we do this here?

It’s pretty simple. We projectivize the map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Cubing sends any line through the origin in $\mathbb{C}^2$ into a unique line through the origin in $S^3(\mathbb{C}^2).$ So, we get a map from $\mathbb{C}\mathrm{P}^1$ to the projective space of $S^3(\mathbb{C}^2).$ And we define the image of this map to be the space of classical states of the spin-3/2 particle!

Let’s see what it looks like. I’m thinking of an element $(a,b) \in \mathbb{C}^2$ as a linear functional

$f(x,y) = ax + by$

If we cube it we get this homogeneous polynomial of degree 3:

$f(x,y)^3 = a^3 \, x^3 + 3a^2b \, x^2y + 3ab^2 \, xy^2 + b^3 \, y^3$

If we take $x^3, x^2y, xy^2$ and $y^3$ as our basis of the homogeneous polynomials of degree 3, we can identify $S^3(\mathbb{C}^2)$ with $\mathbb{C}^4.$ So we get

$\begin{array}{clll} \text{cubing} \colon & \mathbb{C}^2 &\to& \mathbb{C}^4 \\ & (a,b) & \mapsto & (a^3, 3a^2b, 3ab^2, b^3) \end{array}$

If we projectivize this map we get a map between projective varieties, which I’ll call

$P(\text{cubing}) \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^3$

The image of this map is the space of states of the classical spin 3/2-particle! It’s a copy of $\mathbb{C}\mathrm{P}^1$ sitting inside projective 3-space. But because cubing a nonlinear map, this copy will be twisted: not a ‘line’ but a ‘curve’. People call it the twisted cubic.

Just for kicks, let’s take a closer look at it. Let’s use homogeneous coordinates and write $[a,b]$ for the point in $\mathbb{C}\mathrm{P}^1$ corresponding to a nonzero vector $(a,b) \in \mathbb{C}^2.$ Similarly, any point in $\mathbb{C}\mathrm{P}^3$ can be written as a nonzero 4-tuple of complex numbers with a bracket around it. We get

$\begin{array}{cccl} P(\text{cubing}) \colon & \mathbb{C}\mathrm{P}^1 &\to& \mathbb{C}\mathrm{P}^3 \\ & [a,b] & \mapsto & [a^3, 3a^2b, 3ab^2, b^3] \end{array}$

so the twisted cubic is

$\{ [a^3, 3a^2b, 3ab^2, b^3] : \, (0,0) \ne (a,b) \in \mathbb{C}^2 \} \subset \mathbb{C}\mathrm{P}^3$

This is hard to visualize, so we can work with a copy of $\mathbb{C}^3$ that’s dense inside $\mathbb{C}\mathrm{P}^3,$ namely the set where $a = 1.$ The portion of the twisted cubic sitting in this set is

$\{ (b, 3b^2, b^3) : \, b \in \mathbb{C} \} \subset \mathbb{C}^3$

This is still hard to visualize, so we can restrict to the reals and think about the curve

$\{ (b, 3b^2, b^3) : \, b \in \mathbb{R} \} \subset \mathbb{R}^3$

This is also called the twisted cubic! People often rescale the $y$ axis to get rid of the number 3 here.

This version of the twisted cubic is still a bit hard to visualize, but it’s the intersection of two very nice surfaces: the surface $y = x^2$ and the surface $z = x^3.$ So, the twisted cubic is the black curve:

in this nice picture uploaded to ResearchGate by Alexander M. Kasprzyk.

Okay, back to serious business! Let’s call the twisted cubic $C_3 \subseteq \mathbb{C}\mathrm{P}^3.$ It’s exactly the sort of thing we can geometrically quantize using our functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

The reason is that it’s a projective variety and it’s linearly normal. Moreover when we quantize $C_3$ we get $\mathbb{C}^4,$ which is just what we want for the spin-3/2 particle!

$\texttt{Q}(C_3) = \mathbb{C}^4$

Why? Remember, $\texttt{Q}(C_3)$ is defined to be the smallest linear subspace $V \subseteq \mathbb{C}^4$ such that $C_3 \subseteq P(V).$ But $C_3$ twists around so much that the smallest $V$ that works is all of $\mathbb{C}^4.$

Now let’s take stock of where we are and draw some general conclusions from what we’ve seen in this example. We now have a firm grip on the space of quantum states of the spin-3/2 particle:

$S^3(\mathbb{C}^2) \cong \mathbb{C}^4$

and also the space of classical states of the spin-3/2 particle, the twisted cubic:

$C_3 = \{ [a^3, 3a^2b, 3ab^2, b^3] : \, (0,0) \ne (a,b) \in \mathbb{C}^2 \}$

Now for something cool: the latter sits inside the projectivization of the former! This is obvious, but it has a very nice physical meaning. While I’ve been calling $\mathbb{C}^4$ the space of quantum states of the spin-3/2 particle, it is very reasonable to argue that quantum states are actually points of its projectivization, $\mathbb{C}\mathrm{P}^3.$ I will skip the argument, which is old, famous and convincing:

• Wikipedia, Projective Hilbert space.

What matters for us here is that classical states of the spin-3/2 particle give some of these quantum states—far from all, but some of the nicest ones! They are the states obtained by ‘cubing’ a state of a spin-1/2 particle. In other words, they are the states where we can think of our spin-3/2 particle as made of three spin-1/2 particles with their spins perfectly aligned.

It’s nice to say this with a bit more physics jargon. These quantum states coming from classical ones are called ‘coherent states’ . A general quantum state of the spin-3/2 particle is a ‘quantum superposition’ of these coherent states. By this, I mean that the smallest linear subspace $V \subseteq \mathbb{C}^4$ for which $P(V)$ contains the twisted cubic is all of $\mathbb{C}^4.$

And while we’ve seen this in a particular example, it’s a completely general feature of our setup! Remember, projectivization

$\texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

is the left adjoint of quantization

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

This means that for any $M \in \texttt{Class}, V \in \texttt{Quant}$ we have

$\texttt{Q}(M) \subseteq V \quad \iff \quad M \subseteq \texttt{P}(V)$

We get something interesting if we take $V = \texttt{Q}(M)$ here. Since $\texttt{Q}(M) \subseteq \texttt{Q}(M),$ we get

$M \subseteq \texttt{P} (\texttt{Q}(M))$

Category theorists call this inclusion of $M$ in $\texttt{P} (\texttt{Q}(M))$ the ‘unit’ of our pair of adjoint functors. It says how classical states sit inside the projectivization of $\texttt{Q}(M).$ If we call points of $\texttt{P}(\texttt{Q}(M))$ quantum states, those in $M$ are called the coherent states.

Moreover every quantum state is a ‘quantum superposition’ of coherent states! In other words, the smallest linear subspace $V \subseteq \texttt{Q}(M)$ for which $M$ sits inside $\texttt{P}(V)$ is all of $\texttt{Q}(M).$

Why is this true? It’s just the definition of $\texttt{Q}(M)!$

So, I hope you see how much physics is packed into these adjoint functors $\texttt{Q}$ and $\texttt{P}.$ I’ll do more examples next time. In the meantime, you can read more about this approach to the spin-3/2 particle here:

• Dorje C. Brody and Lane P. Hughston, Geometric quantum mechanics.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

## Geometric Quantization (Part 5)

30 December, 2018

Now let’s start looking at some examples of the adjoint functors introduced in Part 4: quantization and projectivization. It’s really the examples that bring the subject to life. They give new insights into hoary old topics in physics, and also raise some puzzles about the relation between classical and quantum mechanics.

I’ll start with the classical spin-j particle and its quantization. I recently discovered through conversations on Twitter how few physicists have heard of the classical spin-j particle. They all know that the quantum spin-j particle has a Hilbert space $\mathbb{C}^{2j+1}$, an irreducible representation of $\mathrm{SU}(2).$ But the corresponding classical system whose quantization gives this Hilbert space seems remarkably little-known, especially given how simple it is. So, I’ll describe it and its geometric quantization slowly and carefully, before feeding it into our functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

which does this quantization quickly and automatically.

The space of states of the a classical spin-j particle is the sphere, $S^2.$ The radius of the sphere is the total angular momentum of the particle, namely j. A point on this sphere represents a possible angular momentum vector of the particle: a vector $\vec{J} \in \mathbb{R}^3$ with length j. There are 3 important observables in this system: the components of the angular momentum, $J_x, J_y$ and $J_z.$

To do classical mechanics with this system we want a symplectic structure on the sphere. A symplectic structure determines Poisson brackets of observables, and we want to make sure that

$\{J_x, J_y\} = J_z, \quad \{J_y, J_z\} = J_x, \quad \textrm{ and } \quad \{ J_z, J_x\} = J_z$

There’s a unique symplectic structure $\omega$ that does the job. A symplectic structure on the sphere is just a nowhere vanishing 2-form (since such a 2-form is automatically closed and nondegenerate), and you can almost guess $\omega$ just from the fact that it had better be rotation-invariant, so it has to be some multiple of the usual area 2-form on the unit sphere. If you figure out which multiple gives the above Poisson brackets, you see that $\omega$ must be $j$ times the area 2-form for the unit sphere. With painful explicitness:

$\omega = j \sin \theta d\theta \wedge d\phi$

This is a bit peculiar: you might expect $j^2$ here instead of $j,$ since the area of a sphere of radius $j$ in Euclidean space is proportional to $j^2.$ But that’s not what the calculation gives! If you want some more justification for the Poisson brackets we’ve chosen, which force this symplectic structure, you can think of the sphere as a coadjoint orbit of the group $\mathrm{SU}(2),$ and use the fact that any such coadjoint orbit gets a Poisson structure, which happens to give the above Poisson bracket formula in this example. Or, you can use dimensional analysis to see $\omega$ must be proportional to $j,$ since they both have units of action. But let’s not get into that here.

To geometrically quantize our symplectic manifold using Kähler quantization, we need to equip it with lots of extra structure, as explained in Part 2. For starters, we need to give it a complex structure. There’s a unique one invariant under rotations, namely the one that makes our sphere into a copy of the Riemann sphere $\mathbb{C}\mathrm{P}^1.$ There is then a unique Kähler structure on $\mathbb{C}\mathrm{P}^1$ whose imaginary part is our symplectic structure $\omega.$ The real part is a Riemannian metric: the usual metric on a round sphere of radius $\sqrt{j}$ in Euclidean space. (Again, that square root looks peculiar, but that’s what we get.)

Next we need to choose a hermitian line bundle over the sphere, equipped with a hermitian connection whose curvature is $i \omega.$ By the miracle of algebraic topology, this exists precisely when the integral of $\omega$ over the sphere is $2 \pi$ times an integer. Since this area is $4 \pi j,$ this forces $j = 0,1/2,1,3/2,\dots.$ So, while we can work with the classical spin-$j$ particle for any real $j \ge 0,$ we can only quantize it when $j$ takes the usual integer or half-integer values!

Let’s focus on the spin-1/2 particle for a while. In this case we can easily describe the relevant line bundle over our sphere. It’s just the bundle I’ve been calling $L:$ the dual of the tautological line bundle on $\mathbb{C}\mathrm{P}^1.$ The holomorphic sections of this bundle are just linear functions $\psi \colon \mathbb{C}^2 \to \mathbb{C}.$ So, the space of sections is ${\mathbb{C}^2}^\ast,$ or just $\mathbb{C}^2$ if we identify this space with its dual using the usual inner product.

That’s good! We got the right answer! But as you can see, the process seems rather long and tortuous.

If we use our functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

everything goes a lot faster, mainly because all the choices we had to make are built into the definition of $\texttt{Class}.$ An object in this category, you’ll remember, is a linearly normal subvariety $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ for some arbitrary $n.$ When we quantize it we get $\texttt{Q}(M) = V$ where $V \subseteq \mathbb{C}^n$ is the smallest linear subspace with $M \subseteq PV.$

In other words, quantization ‘flattens out’ or ‘linearizes’ $M,$ replacing this possibly quite interesting projective variety by the smallest vector space whose projectivization contains this variety.

For the spin-1/2 particle, we take $M = \mathbb{C}\mathrm{P}^1.$ And we get $\texttt{Q}(M) = \mathbb{C}^2,$ since this is the smallest subspace of $\mathbb{C}^2$ whose projectivization contains $\mathbb{C}\mathrm{P}^1.$ Voilà!

That was pretty trivial. But it was trivial for an interesting reason. Remember from Part 4 that we have an adjoint functor going back from the quantum to the classical:

$\texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

This sends any subspace $V \subseteq \mathbb{C}^n$ to its projectivization $P V.$ Moreover, we have

$\texttt{Q} \circ \texttt{P} = 1_{\texttt{Quant}}$

In words: if we quantize the projectivization of a quantum system we get that quantum system back.

And that’s what we’re doing in the case of the spin-1/2 particle! The space of states of the classical spin-1/2 particle, $\mathbb{C}\mathrm{P}^1,$ was the projectivization of $\mathbb{C}^2.$ So when we quantize it, we just get $\mathbb{C}^2$ back.

That’s not how it will work for the spin-1 particle, or any higher-spin particle. The spin-j particle still has $\mathbb{C}\mathrm{P}^1$ as its classical state space, but when we quantize it we get $\mathbb{C}^{2j+1}.$ That’s what I’ll talk about next time.

Now, experts may be yawning at this point, because they already know how to handle such a higher-spin particle! We simply choose a different Kähler structure on $\mathbb{C}\mathrm{P}^1,$ and a different line bundle over it. Namely, we rescale the Kähler structure we’ve already been talking about to make the total area of the sphere be $4 \pi j.$ If j = k/2, this means we have to multiply our earlier Kähler structure by k. And to get a line bundle with a connection whose curvature is the imaginary part of this rescale Kähler structure, we just take the kth tensor power of our line bundle $L.$

The holomorphic sections of this new line bundle $L^{\otimes k}$ are homogeneous polynomials of degree k on $\mathbb{C}^2.$ The space of all these has dimension $2j+1,$ and this is the right space of states for quantum spin-j particle.

All this is great, and it’s an illustration of a theme I’ll eventually want to talk about much more: when you’re doing the Kähler quantization of some physical system, you can always multiply your Kähler structure by a natural number k, and take the kth tensor power of your line bundle, and get a new system!

What does this mean physically? I’ll give away part of the answer now: it corresponds to dividing Planck’s constant by k! You see, angular momentum is really measured in units of Planck’s constant, so using this procedure to go from the spin-1/2 particle to the spin-5/2 particle (for example) is the same as dividing Planck’s constant by 5.

However, in the setup described last time, we had no explicit choice over what Kähler structure or line bundle to use on our subvariety $M \subseteq \mathbb{C}\mathrm{P}^{n-1}.$ It inherited those structures from $\mathbb{C}\mathrm{P}^{n-1}.$ So to change those structures, we need to embed $M$ in a different way, perhaps into a different projective space.

This may seem a bit silly, and in a way it is—but not completely.

As I mentioned last time, we are describing the geometry of varieties extrinsically, through their embedding in projective space. This is a bit old-fashioned compared to the more modern intrinsic viewpoint. At some point I’d like to switch to an intrinsic approach. However, the interplay between the intrinsic and extrinsic approaches is a time-honored theme in algebraic geometry, and we should exploit it in geometric quantization!

In particular, even if we start with an abstract variety not embedded in any projective space, once we choose a line bundle $L$ over it we can often embed in a projective space built from $L.$ By definition, we can do this whenever $L$ is a ‘very ample’ line bundle. Sometimes $L$ isn’t very ample but $L^{\otimes k}$ is whenever k is large enough; then we call $L$ ‘ample’. The study of ample line bundles is a big deal in algebraic geometry.

So, it’s not so silly to start our research on geometric quantization by taking our varieties to come equipped with an embedding into a projective space, and then later think about how we get such an embedding if we don’t already have one.

So, next time I’ll geometrically quantize the spin-j particle by taking the space of states of a classical spin-j particle, $\mathbb{C}\mathrm{P}^1,$ and embedding it, not in itself, but in some higher-dimensional projective space. The math involved is well-known; the interesting part to me is its physical interpretation.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

## Geometric Quantization (Part 4)

28 December, 2018

Last time I showed that geometric quantization could be made into a functor—and that this functor has right adjoint, ‘projectivization’, going back from the quantum realm to the classical. This was just a preliminary version of something that deserves to be polished up a lot. I’d also like to look at a bunch of examples of how this functor works, because they raise a lot of interesting questions.

I’m a bit torn between what order to do all this stuff: polish and then give examples, or give examples and then polish? I think I’ll do a little easy polishing today—very superficial, nothing deep—and then look at examples in my next post.

One reason to do this is to clear away some clutter. In Part 2 I sketched the usual version of Kähler quantization, which is fairly complicated. Last time I switched to a stripped-down version where these complications are removed… but tried to explain how it was ‘backwards-compatible’ with the usual story, by showing how all the complications could easily be restored if you want them. Unfortunately this makes things look more complicated than they are!

So let me present the same construction again, more cleanly.

For any $n \ge 1,$ let $\texttt{Class}_n$ be the poset of linearly normal varieties in $\mathbb{C}\mathrm{P}^{n-1}$, ordered by inclusion. Let $\texttt{Quant}_n$ be the poset of linear subspaces of $\mathbb{C}^n,$ also ordered by inclusion. Define the order-preserving map

$\texttt{Q}_n \colon \texttt{Class}_n \to \texttt{Quant}_n$

as follows: for any $M \in \texttt{Class}_n$ let $\texttt{Q}_n(M)$ be the smallest linear subspace of $\mathbb{C}^n$ such that $M$ is contained in the projectivization of that subspace. And define the order-preserving map

$\texttt{P}_n \colon \texttt{Quant}_n \to \texttt{Class}_n$

as follows: for any $V \in \texttt{Quant}_n$ let $\texttt{P}_n(V)$ be the projectivization of $V.$

Now for some remarks. I usually make fun of mathematicians who number remarks in their paper—when they’re at a party, do they number their jokes? But I feel like numbering these, so you can easily skip them if you want:

1) Last time I said that varieties in $\texttt{Class}_n$ have to be smooth. This time I’ve dropped that condition! The reason is that last time I wanted to reassure skeptics by pointing out that our varieties could also be seen as Kähler manifolds, as traditional in geometric quantization. But when we do examples, it will also be fun to geometrically quantize non-smooth varieties, like those on top of these articles. (These pictures were drawn by Abdelaziz Nait Merzouk, and you can learn more by clicking on them.)

2) Everything I’m doing today would also work if we drop the condition that our varieties be linearly normal! However, when I geometrically quantize a projective variety $M,$ I want to get the vector space of all sections of the dual of the tautological line bundle on $M.$ The functors $\texttt{Q}_n$ only do this job if and only if $M$ is linearly normal. Otherwise they give a smaller vector space! So, I don’t think I want to drop linear normality.

It’s probably worth emphasizing that linear normality is an ‘extrinsic’ condition, depending not only on a variety but on its embedding in projective space. Later I will try a more ‘intrinsic’ approach to geometric quantization.

3) Everything I’m doing today would also work over an arbitrary field! I’m sticking to the complex numbers to keep physicists from thinking I’ve gone off the deep end. But it could be very interesting, at least for algebraists, to work more generally.

Okay, enough ‘remarks’—back to the main track. It’s a bit annoying to separately study geometric quantization for varieties in $\mathbb{C}\mathrm{P}^{n-1}$ for each different $n.$ We can get around this as follows.

For each $n$ we have an inclusion

$\mathbb{C}^n \to \mathbb{C}^{n+1}$

sending each vector to the same list of numbers with a zero tacked on at the end. This gives an inclusion of posets

$\texttt{Class}_{n-1} \hookrightarrow \texttt{Class}_n$

So, we can take the colimit of the diagram

$\cdots \hookrightarrow \texttt{Class}_{n-1} \hookrightarrow \texttt{Class}_n \hookrightarrow \texttt{Class}_{n+1} \hookrightarrow \cdots$

or in more lowbrow terms, the union of all the posets $\texttt{Class}_n,$ each included in the next, and get a big poset I’ll call $\texttt{Class}.$ (Sorry, this notation conflicts with the one I used last time. I should have put subscripts everywhere last time.)

Similarly, we have inclusions of posets

$\texttt{Quant}_{n-1} \hookrightarrow \texttt{Quant}_n$

so we can take the colimit of

$\cdots \hookrightarrow \texttt{Quant}_{n-1} \hookrightarrow \texttt{Quant}_n \hookrightarrow \texttt{Quant}_{n+1} \hookrightarrow \cdots$

and get a poset I’ll call $\texttt{Quant}.$

This poset $\texttt{Quant}$ seems familiar to me: it’s the poset of all finite-dimensional subspaces of a certain vector space called $\mathbb{C}^\infty,$ the colimit of all the spaces $\mathbb{C}^n.$ The poset $\texttt{Class}$ is less familiar to me. We can intuitively think of it as the poset of all linearly normal subvarieties of $\mathbb{C}\mathrm{P}^\infty$ that sit inside $\mathbb{C}\mathrm{P}^n$ for some finite $n$. However, while I’m familiar with $\mathbb{C}\mathrm{P}^\infty$ as a topological space, the colimit of all the manifolds $\mathbb{C}\mathrm{P}^n,$ I have never heard people talk about ‘subvarieties’ of $\mathbb{C}\mathrm{P}^\infty.$ This could easily be a hole in my education. (Is $\mathbb{C}\mathrm{P}^\infty$ just a non-noetherian scheme?)

Now, the nice thing is that our maps of posets

$\texttt{Q}_n \colon \texttt{Class}_n \to \texttt{Quant}_n, \qquad \texttt{P}_n \colon \texttt{Quant}_n \to \texttt{Class}_n$

get along with the inclusions

$\texttt{Class}_n \hookrightarrow \texttt{Class}_{n+1} , \qquad \texttt{Quant}_n \hookrightarrow \texttt{Quant}_{n+1}$

forming two commutative squares that I’m too lazy to draw here. So, we get well-defined maps of posets

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}, \qquad \texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

Last time we saw that the maps $\texttt{Q}_n$ and $\texttt{P}_n$ were adjoint for each $n.$ That implies the same result for $\texttt{Q}$ and $\texttt{P}.$ Don’t be scared; this just means that

$\texttt{Q}(M) \subseteq V \quad \iff \quad M \subseteq \texttt{P}(V)$

for all $M \in \texttt{Class}, V \in \texttt{Quant},$ and this is easy to check directly. (Or, if you’re a category theorist, dream up an abstract nonsense proof.)

Last time we also saw that $\texttt{Quant}_n$ is a reflective subcategory of $\texttt{Class}_n.$ That implies the same result for $\texttt{Quant}$ and $\texttt{Class}.$ This just means that

$\texttt{Q}( \texttt{P} (V)) = V$

for all $V \in \texttt{Quant}.$ And again, this is easy to check directly.

So, we are back where we started—but now we can stop worrying about which $\mathbb{C}\mathrm{P}^{n-1}$ our classical state spaces are subvarieties of, or which $\mathbb{C}^n$ our quantum state spaces are subspaces of.

This lets us do some new things!

It’s fun to combine quantum systems using the tensor product of their vector spaces, and now we can do it. If we have linear subspaces $V \subseteq \mathbb{C}^m$ and $W \subseteq \mathbb{C}^n,$ we can take their tensor product and get $V \otimes W \subseteq \mathbb{C}^m \otimes \mathbb{C}^n \cong \mathbb{C}^{mn}.$ So, we get a multiplication on $\texttt{Quant}.$ And this multiplication will be associative if we identify $\mathbb{C}^m \otimes \mathbb{C}^n$ with $\mathbb{C}^{mn}$ in the right way—namely, using the lexicographic ordering on the set of basis vectors $e_i \otimes e_j$ of $\mathbb{C}^m \otimes \mathbb{C}^n.$

So, let’s do this. $\texttt{Quant}$ then becomes a monoidal poset—that is, a monoid that’s also a poset, where the multiplication gets along with the ordering as follows:

$V \subseteq V', \; W \subseteq W' \quad \implies \quad V \otimes W \subseteq W \otimes W'$

(If you want to show off, you can equivalently say a monoidal poset is just a monoidal category that’s a poset.)

We can also combine classical systems, using the cartesian product. IF $M$ is a subvariety of $\mathbb{C}\mathrm{P}^{m-1}$ and $N$ is a subvariety of $\mathbb{C}\mathrm{P}^{n-1},$ their product $M \times N$ is naturally a subvariety of $\mathbb{C}\mathrm{P}^{mn-1},$ using a nice trick called the Segre embedding

$\mathbb{C}\mathrm{P}^{m-1} \times \mathbb{C}\mathrm{P}^{n-1} \hookrightarrow \mathbb{C}\mathrm{P}^{mn-1}$

I’m hoping that if $M$ is linearly normal in $\mathbb{C}\mathrm{P}^{m-1}$ and $N$ is linearly normal in $\mathbb{C}\mathrm{P}^{n-1},$ their product is linearly normal in $\mathbb{C}\mathrm{P}^{mn-1}.$ If this is true, we get a product on $\texttt{Class},$ which again will be associative if we carefully use the lexicographic ordering when defining the Segre embedding.

And now for the punchline: quantization and projectivization get along with these ways of combining classical and quantum systems!

$\texttt{Q}(M \times N) = \texttt{Q}(M) \otimes \texttt{Q}(N)$
$\texttt{P}(V \otimes W) = \texttt{P}(V) \times \texttt{P}(W)$

So, if we think of $\texttt{Class}$ and $\texttt{Quant}$ as monoidal categories, then $\texttt{Q}$ and $\texttt{P}$ are monoidal functors!

It would also be interesting to combine classical systems by disjoint union, and show that this corresponds to combining quantum ones by direct sum. But I think I should stop here.

Enough formalism! Let’s look at examples! Next time we will.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

## Geometric Quantization (Part 3)

27 December, 2018

Okay, I’ll stop warming up and actually do something. I’ve secretly been trying to convince you that it’s not utterly insane to start geometric quantization with something other than a symplectic manifold—since, after all, geometric quantization normally starts with much more than a mere symplectic manifold. In Parts 1 and 2 I explained that we often start with a Kähler manifold together with a suitable line bundle. This is a big pack of data that includes a manifold that’s simultaneously symplectic, complex and Riemannian, all in a compatible way—but also more, namely the line bundle.

Now I’ll do a stripped-down, bare-bones version of geometric quantization. It starts with something that sounds quire different: a smooth complex projective variety. This is a smooth submanifold of $\mathbb{C}\mathrm{P}^n$ that’s picked out by homogeneous polynomial equations.

In fact this is not as different as it sounds: if we give $\mathbb{C}\mathrm{P}^n$ its Fubini–Study metric, any smooth projective variety $M \subseteq \mathbb{C}\mathrm{P}^n$ becomes a Kähler manifold! We just restrict the Fubini–Study metric to $M.$

Furthermore, $M \subseteq \mathbb{C}\mathrm{P}^n$ has a god-given line bundle over it. You see, any point in $\mathbb{C}\mathrm{P}^n$ is really a line through the origin in $\mathbb{C}^{n+1}$, and these lines define a line bundle over $\mathbb{C}\mathrm{P}^n.$ This construction is so ridiculously tautological that people call the resulting bundle the tautological line bundle. This bundle has no holomorphic sections, but its dual has lots: every linear functional on $\mathbb{C}^{n+1}$ gives one, and that’s how they all arise! Let’s call this dual bundle $L_{\mathbb{C}\mathrm{P}^n}.$ If we restrict $L_{\mathbb{C}\mathrm{P}^n}$ to $M,$ we get a holomorphic line bundle with enough sections to be useful in geometric quantization. Let’s call this bundle $L_M.$

Even better, $M$ and this line bundle $L_M$ have the whole laundry list of extra structure and properties that I was bemoaning last time. As I already said, $M$ is Kähler. Furthermore, $L_M$ is holomorphic. Better yet, there’s a god-given hermitian metric on $L_M,$ coming from the standard inner product on $\mathbb{C}^{n+1}$ (and thus its dual). So, as I explained near the end last time, we get a connection on $L_M$ called the Chern connection. And if I’m not horribly mistaken, the curvature of this connection gives the symplectic structure on $M,$ at least up to those factors of $2\pi$ and $i$ that I keep losing track of.

In short, we get everything we want to use in Kähler quantization starting from something very simple: a smooth complex projective variety!

So let’s formalize this a little. Let’s define two categories, $\texttt{Class}$ and $\texttt{Quant},$ and make geometric quantization into a functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

I’m using a crude teletype font here because this is the first and crudest of a series of similar constructions. In fact I’ll restrain my desire to show off and do the simplest thing I can. My categories will be mere posets.

I’ll fix a number $n$ and let $\texttt{Class}$ be the poset of smooth linearly normal subvarieties of $\mathbb{C}\mathrm{P}^n,$ ordered by inclusion. Don’t worry—I’ll tell you what ‘linearly normal’ means in a minute. I’ll let $\texttt{Quant}$ be the poset of linear subspaces of $\mathbb{C}^{n+1},$ also ordered by inclusion.

You can think of any object of $\texttt{Class}$ as a ‘space of classical states’, since as we’ve seen it’s a symplectic manifold of a very nice kind, bedecked with all the bells and whistles we need for geometric quantization. Similarly you can think of any object of $\texttt{Quant}$ as a ‘space of quantum states’, since as a subspace of $\mathbb{C}^{n+1}$ it’s a Hilbert space.

Here’s how the functor $\texttt{Q}$ works on objects. As we’ve seen, any smooth projective variety $M \subseteq \mathbb{C}\mathrm{P}^n$ comes with a line bundle $L_M \to M,$ namely the restriction of the dual of the tautological line bundle to $M.$ To quantize $M,$ we form the space of holomorphic sections of $L_M.$

Now, this doesn’t instantly look like a subspace of $\mathbb{C}^{n+1}.$ But we can identify it with one as follows. Every holomorphic section of $L_{\mathbb{C}\mathrm{P}^n}$ restricts to a holomorphic section of $L_M.$ And when $M$ is linearly normal, every holomorphic section of $L_M$ arises this way! This is just the definition of ‘linearly normal’.

In this case, the space of holomorphic sections of $L_M$ is a quotient of the space of holomorphic sections of $L_{\mathbb{C}\mathrm{P}^n}.$ But I’ve already mentioned that the latter space can be identified with ${\mathbb{C}^{n+1}}^\ast.$ So, we’ve got a quotient space of ${\mathbb{C}^{n+1}}^\ast.$ But by the magic of linear algebra, this corresponds to a subspace of $\mathbb{C}^{n+1}.$ And that’s $\texttt{Q}(M).$

This sounds complicated! Luckily, when you unravel it all you get a much simpler description, pointed out by Allen Knutson in a comment to Part 1. For any variety $M \subseteq \mathbb{C}\mathrm{P}^n$ there’s a smallest subspace $V \subseteq \mathbb{C}^{n+1}$ such that $M$ sits inside the projective space $\mathrm{P}V \subseteq \mathbb{C}\mathrm{P}^n.$ We have

$\texttt{Q}(M) = V$

We can take this as the definition of $\texttt{Q}$ if we want. This simpler description makes it blatantly obvious that

$M \subseteq M' \quad \implies \quad \texttt{Q}(M) \subseteq \texttt{Q}(M')$

So, $\texttt{Q}$ is a functor.

Now, I said in Part 1 that there’s also a reverse process going from the quantum realm back to the classical. This is even simpler: it’s ‘projectivization’, and it gives a functor

$\texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

It works like this. An object $V \in \texttt{Quant}$ is a linear subspace of $\mathbb{C}^{n+1}.$ This gives a projective variety $\mathrm{P}V \subseteq \mathbb{C}\mathrm{P}^n.$ Since this variety is linearly normal, we can define

$\texttt{P}(V) = \mathrm{P}V$

Sorry for the subtle difference in fonts… but it’s okay to use subtly different fonts for two things that are the same, right?

Again, it’s obvious that

$V \subseteq V' \quad \implies \quad \texttt{P}(V) \subseteq \texttt{P}(V')$

so $\texttt{P}$ is a functor.

Moreover, quantization and projectivization are adjoint functors! In fact quantization is the left adjoint of projectivization. This simply means that

$\texttt{Q}(M) \subseteq V \quad \iff \quad M \subseteq \texttt{P}(V)$

And this is obvious: a projective variety $M$ sits inside the projectivization of some linear subspace $V$ iff the smallest linear subspace whose projectivization contains $M$ is contained in $V.$

(Mathematicians have a tendency to say something is ‘obvious’ before reeling off a string of words no normal person would ever say.. All I mean is that no deep facts are required to see this.)

Even better, this adjunction has a special extra property:

$V = \texttt{Q}(\texttt{P}(V))$

for all $V \in \texttt{Quant}.$ Again this is obvious: the smallest linear subspace whose projectivization contains the projectivization of $V$ is $V.$

Thanks to this extra property we say $\texttt{Quant}$ is a reflective subcategory of $\texttt{Class}.$ A more familiar example of this situation is how $\textrm{AbGp}$ is a reflective subcategory of $\textrm{Gp}.$ The forgetful functor $\textrm{AbGp} \to \textrm{Gp}$ has a left adjoint, namely ‘abelianization’, with the special property that if we take an abelian group, forget that it’s abelian and think of it as a group, and then abelianize it, we get back the same abelian group we started with. (At least this is true up to natural isomorphism. But when we projectivize and then quantize we get back the same subspace exactly: that’s because the categories $\texttt{Quant}$ and $\texttt{Class}$ are mere posets, where all isomorphisms are identity morphisms.)

So, we’re seeing that just as abelian groups are especially nice groups, and we can ‘abelianize’ any group to make it nice, quantum state spaces are especially nice classical state spaces, and we can ‘quantize’ any classical state space to make it nice!

This is all very pretty, I think. However, it would be nice to remove some of the ridiculous restrictions I’ve imposed so far, like only considering projective varieties sitting inside a fixed projective space $\mathbb{C}\mathrm{P}^n,$ and only allowing inclusions as morphisms between these. That’s what I’d like to do next.

I should also note that while I’ve muttered words like ‘Hilbert space’, ‘symplectic struture’, ‘Kähler manifold’ and ‘hermitian line bundle’, my construction of the functors

$\texttt{P} \colon \texttt{Quant} \to \texttt{Class} , \qquad \texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

did not use the inner product on $\mathbb{C}^{n+1}$ at all! That’s why I was talking about the dual space ${\mathbb{C}^{n+1}}^\ast.$ So, all these structures involving inner products are not an intrinsic part of what’s going on—at least, not yet! But they can be slapped on at will, simply by endowing $\mathbb{C}^{n+1}$ with an inner product: any inner product, for example its standard one.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

## Geometric Quantization (Part 2)

26 December, 2018

Geometric quantization is often presented as a way to take a symplectic manifold and construct a Hilbert space, but in fact that’s a better description of ‘prequantization’, which is just the first step in geometric quantization. Even that’s not completely accurate: we need to equip our symplectic manifold with a bit of extra structure just to prequantize it. But more importantly, when we do this, the resulting ‘prequantum Hilbert space’ is too big: we need to chop it down significantly to get a Hilbert space suitable for the quantum description of our physical system. And to chop it down, we need to equip our symplectic manifold with a lot more extra structure. So much extra structure, in fact, that the whole idea of ‘quantizing a symplectic manifold’ starts sounding more like the wistful, nostalgic description of a naive hope than an honest account of what we’re really doing.

So, it’s probably better to admit that ‘quantizing a classical system’ is not a real thing. More precisely—because people will rightly object to that over-bold statement—there’s no systematic procedure that takes us all the way from the mathematical structures used to describe a large class of classical systems to those used to describe quantum systems, with no further inputs required. Sure, we can make quantization more systematic by limiting the class of systems it’s supposed to handle. We can also make it more systematic by having it do less: for example, just prequantization instead of full-fledged quantization. Both those options have been explored for many decades. Let’s instead try something new.

For starters, let’s accept the fact that the world was not created classically by God on the first day and quantized on the second.

The world is quantum; under some conditions it looks approximately classical. So instead of treating the lack of a purely functorial quantization procedure for symplectic manifolds as a failure, let’s accept it and try some new ways of thinking about the meaning of geometric quantization.

To make things specific, let’s consider Kähler quantization. Here we start not with a raw symplectic manifold, but much more. For starters, we take a Kähler manifold, which is a smooth manifold $M$ equipped with a beautiful trio of structures:

1) a symplectic structure $\omega$

2) a complex structure $J$

3) a Riemannian metric $g$

fitting together in a way that obeys this equation:

$\omega(v, J w) = g(v, w)$

whenever we have two tangent vectors $v,w$ at any point of $M.$ This equation says how any two of $\omega, g, J$ are enough to determine the third.

Each tangent space of a Kähler manifold is a complex Hilbert space—which is downright suspicious, given that in quantization we’re aiming to get a complex Hilbert space. Here’s how it works:

1) the symplectic structure $\omega$ gives the imaginary part of the inner product on each tangent space of $M,$

2) the complex structure $J$ gives the operation of multiplication by $i$ on each tangent space, and

3) the Riemannian metric $g$ gives the real part of the inner product on each tangent space.

The equation

$\omega(v, J w) = g(v, w)$

says that the imaginary part of the inner product of $v$ and $i w$ equals the real part of the inner product of $v$ and $w.$

In particular, each finite-dimensional Hilbert space is a Kähler manifold. Similarly, an infinite-dimensional Hilbert space is a kind of infinite-dimensional Kähler manifold.

So, you can think of a complex Hilbert space as a special kind of Kähler manifold: a flat and simply-connected one. Or—less precisely but quite evocatively—you can think of a Kähler manifold as a curved generalization of a complex Hilbert space!

If Kähler quantization were a systematic procedure for extracting a Hilbert space from a Kähler manifold, these facts would lead us to a clear view of what this procedure is doing. Namely, it’s taking a general Kähler manifold and ‘flattening it out’ in a very clever way, producing a Hilbert space. Perhaps a better word than ‘flattening’ would be ‘linearizing’, since this emphasizes the all-important linearity built into quantum mechanics.

Something like this is true—but it can’t be the whole story, because Kähler quantization requires more input than a mere Kähler manifold to produce a Hilbert space! We also need a complex line bundle $L$ over our manifold $M.$ Vectors in our Hilbert space will not be just complex-valued functions on $M;$ rather, they will be (nice) sections of this line bundle.

It would be nice to understand this in a deeply physical way. For example, we might try to insist that the ‘true’ space of classical states is not the manifold $M$ but something with one extra dimension built from this line bundle $L.$ This thing is not a symplectic manifold: it’s called a contact manifold. Fans of contact geometry will argue, quite convincingly, that the ‘phases’ so important in quantum mechanics are already lurking in classical mechanics, invisible when we use symplectic manifolds, but evident when we use contact manifolds! If we take this seriously, maybe we shouldn’t be trying to geometrically quantize symplectic manifolds in the first place: maybe we should be using contact manifolds.

However, even taking our Kähler manifold $M$ and this complex line bundle $L$ over it as input is not enough to systematically construct the long-sought Hilbert space. We need more!

1) We need to equip $L$ with a connection whose curvature is $i \omega.$ This is our way of making $L$ compatible with the symplectic structure on $M.$ We need this even for prequantization: it lets us turn smooth functions on $M$ (‘classical observables’) into operators on the space of sections of $L$ (‘quantum observables’).

2) We need to equip $L$ with the structure of a holomorphic line bundle. This is our way of making $L$ compatible with the complex structure on $M.$ We need this to cut down the big vector space obtained in prequantization to a smaller one: the space of holomorphic sections of $L.$

3) We need to equip $L$ with the structure of a hermitian line bundle. I would like to say this is our way of making $L$ compatible with the Riemannian structure on $M,$ since that would make everything very symmetrical—but it doesn’t seem correct! Instead, we need this to put an inner product on the space of holomorphic sections of $L,$ so we can get a Hilbert space.

All this must sound like a real slag-heap of mathematics, if you’re not already in love with it! It’s beautiful when you get to know it, especially when you look at examples. But I find it difficult to motivate everything on physical grounds, and I also find it a bit difficult to keep track of all these interacting structures.

So, next time I’ll present a simplified version of geometric quantization, where we throw out most of these structures and keep only enough to get a complex vector space, not a complex Hilbert space. In fact I’ll throw out the symplectic and Riemannian structures on our manifold $M$ and only keep the complex structure! This is pretty heretical from the viewpoint of physics, but in fact it’s quite standard in mathematics. Indeed, I don’t want to get your hopes up too much: if you know some algebraic geometry, you should find most of what I say quite familiar. But I think this is a good way to get started on setting up a pair of adjoint functors: ‘quantization’ and the reverse process I mentioned last time: ‘projectivization’.

### A question

Since I’ve got your ear, let me ask a question. Suppose we have a holomorphic hermitian line bundle $L$ on a Kähler manifold $M$. I can think of three ways in which a connection $D$ on $L$ can be compatible with all this structure:

1) We can demand that the curvature of the connection $D$ equal $i \omega.$

2) We can demand that the connection $D$ be compatible with the holomorphic structure of $L.$ Let me spell this out a little. Since $L$ is a holomorphic line bundle there’s a $\overline{\partial}$ operator that we can apply to any section of $L$ and get an $L$-valued 1-form. This is an $L$-valued (0,1)-form, meaning that written out in holomorphic complex coordinates $z^i$ it has $d \overline{z}^i$ terms but no $dz^i$ terms. On the other had, we can use the connection $D$ to take the covariant derivative of $s,$ which is a $L$-valued 1-form $D s,$ and take its (0,1) part, denoted $\pi_{0,1} D s.$ Then, we can demand that these agree:

$\overline{\partial} s = \pi_{0,1} D s$

3) We can demand that the connection $D$ be hermitian. This means that the directional derivative of the inner product of two sections of $L$ can be computed using a kind of product rule where we differentiate each section using $D.$ Namely:

$v \langle s, t \rangle = \langle D_v s, t \rangle + \langle s, D_v t \rangle$

for any vector field $v$ on $M$ and any sections $s,t$ of $L.$

I’m confused by when we can find a connection obeying all three of these conditions. Here are some things I know.

First, any holomorphic hermitian line bundle has a unique connection obeying 2) and 3). This is called the Chern connection and its construction is actually given here:

• Wikipedia, Hermitian metrics on a holomorphic vector bundle.

Does this connection also obey 1)? I see no reason why, in general: condition 1) involves the symplectic structure on $M,$ while conditions 2) and 3) don’t mention the symplectic or Riemannian structure on $M,$ just its complex structure.

Second, we can find a connection obeying 1) iff $\omega /2 \pi$ represents, in de Rham cohomology, the Chern class of the line bundle $L.$ If $\omega / 2 \pi$ is an integral 2-form we can always find some line bundle $L$ for which this is true.

Can we find a holomorphic line bundle for which this is true? Can we then choose a holomorphic connection on this line bundle obeying 1)? If I understood Picard groups better, I might know.

But as you can see, my understanding of how conditions 1)-3) interact is rather weak. I feel I should have seen this spelled out carefully in a book on geometric quantization—but I don’t think I have.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

## Geometric Quantization (Part 1)

1 December, 2018

I can’t help thinking about geometric quantization. I feel it holds some lessons about the relation between classical and quantum mechanics that we haven’t fully absorbed yet. I want to play my cards fairly close to my chest, because there are some interesting ideas I haven’t fully explored yet… but still, there are also plenty of ‘well-known’ clues that I can afford to explain.

The first one is this. As beginners, we start by thinking of geometric quantization as a procedure for taking a symplectic manifold and constructing a Hilbert space: that is, taking a space of classical states and contructing the corresponding space of quantum states. We soon learn that this procedure requires additional data as its input: a symplectic manifold is not enough. We learn that it works much better to start with a Kähler manifold equipped with a holomorphic hermitian line bundle with a connection whose curvature is the imaginary part of the Kähler structure. Then the space of holomorphic sections of that line bundle gives the Hilbert space we seek.

That’s quite a mouthful—but it makes for such a nice story that I’d love to write a bunch of blog articles explaining it with lots of examples. Unfortunately I don’t have time, so try these:

• Matthias Blau, Symplectic geometry and geometric quantization.

• A. Echeverria-Enriquez, M.C. Munoz-Lecanda, N. Roman-Roy, C. Victoria-Monge, Mathematical foundations of geometric quantization.

But there’s a flip side to this story which indicates that something big and mysterious is going on. Geometric quantization is not just a procedure for converting a space of classical states into a space of quantum states. It also reveals that a space of quantum states can be seen as a space of classical states!

To reach this realization, we must admit that quantum states are not really vectors in a Hilbert space $H$; from a certain point of view they are really 1-dimensonal subspaces of a Hilbert space, so the set of quantum states I’m talking about is the projective space $PH.$ But this projective space, at least when it’s finite-dimensional, turns out to be the simplest example of that complicated thing I mentioned: a Kähler manifold equipped with a holomorphic hermitian line bundle whose curvature is the imaginary part of the Kähler structure!

So a space of quantum states is an example of a space of classical states—equipped with precisely all the complicated extra structure that lets us geometrically quantize it!

At this point, if you don’t already know the answer, you should be asking: and what do we get when we geometrically quantize it?

The answer is exciting only in that it’s surprisingly dull: when we geometrically quantize $PH,$ we get back the Hilbert space $H.$

You may have heard of ‘second quantization’, where we take a quantum system, treat it as classical, and quantize it again. In the usual story of second quantization, the new quantum system we get is more complicated than the original one… and we can repeat this procedure again and again, and keep getting more interesting things:

• John Baez, Nth quantization.

The story I’m telling now is different. I’m saying that when we take a quantum system with Hilbert space $H,$ we can think of it as a classical system whose symplectic manifold of states is $PH,$ but then we can geometrically quantize this and get $H$ back.

The two stories are not in contradiction, because they rely on two different notions of what it means to ‘think of a quantum system as classical’. In today’s story that means getting a symplectic manifold $PH$ from a Hilbert space $H.$ In the other story we use the fact that $H$ itself is a symplectic manifold!

I should explain the relation of these two stories, but that would be a big digression from today’s intended blog article: indeed I’m already regretting having drifted off course. I only brought up this other story to heighten the mystery I’m talking about now: namely, that when we geometrically quantize the space $PH,$ we get $H$ back.

The math is not mysterious here; it’s the physical meaning of the math that’s mysterious. The math seems to be telling us that contrary to what they say in school, quantum systems are special classical systems, with the special property that when you quantize them nothing new happens!

This idea is not mine; it goes back at least to Kibble, the guy who with Higgs invented the method whereby the Higgs boson does its work:

• Tom W. B. Kibble, Geometrization of quantum mechanics, Comm. Math. Phys. 65 (1979), 189–201.

This led to a slow, quiet line of research that continues to this day. I find this particular paper especially clear and helpful:

• Abhay Ashtekar, Troy A. Schilling, Geometrical formulation of quantum mechanics, in On Einstein’s Path, Springer, Berlin, 1999, pp. 23–65.

so if you’re wondering what the hell I’m talking about, this is probably the best place to start. To whet your appetite, here’s the abstract:

Abstract. States of a quantum mechanical system are represented by rays in a complex Hilbert space. The space of rays has, naturally, the structure of a Kähler manifold. This leads to a geometrical formulation of the postulates of quantum mechanics which, although equivalent to the standard algebraic formulation, has a very different appearance. In particular, states are now represented by points of a symplectic manifold (which happens to have, in addition, a compatible Riemannian metric), observables are represented by certain real-valued functions on this space and the Schrödinger evolution is captured by the symplectic flow generated by a Hamiltonian function. There is thus a remarkable similarity with the standard symplectic formulation of classical mechanics. Features—such as uncertainties and state vector reductions—which are specific to quantum mechanics can also be formulated geometrically but now refer to the Riemannian metric—a structure which is absent in classical mechanics. The geometrical formulation sheds considerable light on a number of issues such as the second quantization procedure, the role of coherent states in semi-classical considerations and the WKB approximation. More importantly, it suggests generalizations of quantum mechanics. The simplest among these are equivalent to the dynamical generalizations that have appeared in the literature. The geometrical reformulation provides a unified framework to discuss these and to correct a misconception. Finally, it also suggests directions in which more radical generalizations may be found.

Personally I’m not interested in the generalizations of quantum mechanics: I’m more interested in what this circle of ideas means for quantum mechanics.

One rather cynical thought is this: when we start our studies with geometric quantization, we naively hope to extract a space of quantum states from a space of classical states, e.g. a symplectic manifold. But we then discover that to do this in a systematic way, we need to equip our symplectic manifold with lots of bells and whistles. Should it really be a surprise that when we’re done, the bells and whistles we need are exactly what a space of quantum states has?

I think this indeed dissolves some of the mystery. It’s a bit like the parable of ‘stone soup’: you can make a tasty soup out of just a stone… if you season it with some vegetables, some herbs, some salt and such.

However, perhaps because by nature I’m an optimist, I also think there are interesting things to be learned from the tight relation between quantum and classical mechanics that appears in geometric quantization. And I hope to talk more about those in future articles.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.