## Geometric Quantization (Part 7)

I’ve been falling in love with algebraic geometry these days, as I realize how many of its basic concepts and theorems have nice interpretations in terms of geometric quantization. I had trouble getting excited about them before. I’m talking about things like the Segre embedding, the Veronese embedding, the Kodaira embedding theorem, Chow’s theorem, projective normality, ample line bundles, and so on. In the old days, all these things used to make me nod and go “that’s nice”, without great enthusiasm. Now I see what they’re all good for!

Of course this is my own idiosyncratic take on the subject: obviously algebraic geometers have their own pefectly fine notion of what these things are good for. But I never got the hang of that.

Today I want to talk about how the Veronese embedding can be used to ‘clone’ a classical system. For any number k, you can take a classical system and build a new one; a state of this new system is k copies of the original system constrained to all be in the same state! This may not seem to do much, but it does something: for example, it multiplies the Kähler structure on the classical state space by k. And it has a quantum analogue, which has a much more notable effect!

Last time I looked at an example, where I built the spin-3/2 particle by cloning the spin-1/2 particle.

In brief, it went like this. The space of classical states of the spin-1/2 particle is the Riemann sphere, $\mathbb{C}\mathrm{P}^1.$ This just happens to also be the space of quantum states of the spin-1/2 particle, since it’s the projectivization of $\mathbb{C}^2.$ To get the 3/2 particle we look at the map

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

You can think of this as the map that ‘triplicates’ a spin-1/2 particle, creating 3 of them in the same state. This gives rise to a map between the corresponding projective spaces, which we should probably call

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

It’s an embedding.

Algebraic geometers call the image of this embedding the twisted cubic, since it’s a curve in 3d projective space described by homogeneous cubic equations. But for us, it’s the embedding of the space of classical states of the spin-3/2 particle into the space of quantum states. (The fact that classical states give specially nice quantum states is familiar in physics, where these specially nice quantum states are called ‘coherent states’, or sometimes ‘generalized coherent states’.)

Now, you’ll have noted that the numbers 2 and 3 show up a bunch in what I just said. But there’s nothing special about these numbers! They could be arbitrary natural numbers… well, > 1 if we don’t enjoy thinking about degenerate cases.

Here’s how the generalization works. Let’s think of guys in $\mathbb{C}^n$ as linear functions on the dual of this space. We can raise any one of them to the k power and get a homogeneous polynomial of degree k. The space of such polynomials is called $S^k(\mathbb{C}^n),$ so raising to the kth power defines a map

$\mathbb{C}^n \to S^k(\mathbb{C}^n)$

This in turn gives rise to a map between the corresponding projective spaces:

$P(\mathbb{C}^n) \to P(S^k(\mathbb{C}^n))$

This map is an embedding, since different linear functions give different polynomials when you raise them to the k power, at least if $k \ge 1.$ And this map is famous: it’s called the k Veronese embedding. I guess it’s often denoted

$v_k \colon P(\mathbb{C}^n) \to P(S^k(\mathbb{C}^n))$

An important special case occurs when we take $n = 2,$ as we’d been doing before. The space of homogeneous polynomials of degree k in two variables has dimension $k + 1,$ so we can think of the Veronese embedding as a map

$v_k \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^k$

embedding the projective line as a curve in $\mathbb{C}\mathrm{P}^k.$ This sort of curve is called a rational normal curve. When $d = 3$ it’s our friend from last time, the twisted cubic.

In general, we can think of $\mathbb{C}\mathrm{P}^k$ as the space of quantum states of the spin-k/2 particle, since we got it from projectivizing the spin-k/2 representation of $\mathrm{SU}(2),$ namely $S^k(\mathbb{C}^n).$ Sitting inside here, the rational normal curve is the space of classical states of the spin-k/2 particle—or in other words, ‘coherent states’.

Maybe I should expand on this, since it flew by so fast! Pick any direction you want the angular momentum of your spin-k/2 particle to point. Think of this as a point on the Riemann sphere and think of that as coming from some vector $\psi \in \mathbb{C}^2.$ That describes a quantum spin-1/2 particle whose angular momentum points in the desired direction. But now, form the tensor product

$\underbrace{\psi \otimes \cdots \otimes \psi}_{k}$

This is completely symmetric under permuting the factors, so we can think of it as a vector in $S^k(\mathbb{C}^2).$ And indeed, it’s just what I was calling

$v_k (\psi) \in S^k(\mathbb{C}^2)$

This vector describes a collection of k indistinguishable quantum spin-1/2 particles with angular momenta all pointing in the same direction. But it also describes a single quantum spin-k/2 particle whose angular momentum points in that direction! Not all vectors in $S^k(\mathbb{C}^2)$ are of this form, clearly. But those that are, are called ‘coherent states’.

Now, let’s do this all a bit more generally. We’ll work with $\mathbb{C}^n,$ not just $\mathbb{C}^2.$ And we’ll use a variety $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ as our space of classical states, not necessarily all of $\mathbb{C}\mathrm{P}^{n-1}.$

Remember, we’ve got:

• a category $\texttt{Class}$ where the objects are linearly normal subvarieties $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ for arbitrary $n,$

and

• a category $\texttt{Quant}$ where the objects are linear subspaces $V \subseteq \mathbb{C}^n$ for arbitrary $n.$

The morphisms in each case are just inclusions. We’ve got a ‘quantization’ functor

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

that maps $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ to the smallest $V \subseteq \mathbb{C}^n$ whose projectivization contains $M.$ And we’ve got what you might call a ‘classicization’ functor going back:

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

We actually call this ‘projectization’, since it sends any linear subspace $V \subseteq \mathbb{C}^n$ to its projective space sitting inside $\mathbb{C}\mathrm{P}^{n-1}$.

We would now like to get the Veronese embedding into the game, copying what we just did for the spin-k/2 particle. We’d like each Veronese embedding $v_k$ to define a functor from $\texttt{Class}$ to $\texttt{Class}$ and also a functor $\texttt{Quant}$ to $\texttt{Quant}.$ For example, the first of these should send the space of classical states of the spin-1/2 particle to the space of classical states of the spin-k/2 particle. The second should do the same for the space of quantum states.

The quantum version works just fine. Here’s how it goes. An object in $\texttt{Quant}$ is a linear subspace

$V \subseteq \mathbb{C}^n$

for some $n.$ Our functor should send this to

$S^k(V) \subseteq S^k(\mathbb{C}^n) \cong \mathbb{C}^{\left(\!\!{n\choose k}\!\!\right)}$

Here $\left(\!{n\choose k}\!\right)$, pronounced ‘n multichoose k’ , is the number of ways to choose k not-necessarily-distinct items from a set of n, since this is the dimension of the space of degree-k homogeneous polynomials on $\mathbb{C}^n.$ (We have to pick some sort of ordering on monomials to get the isomorphism above; this is one of the clunky aspects of our current framework, which I plan to fix someday.)

This process indeed defines functor, and the only reasonable name for it is

$S^k \colon \texttt{Quant} \to \texttt{Quant}$

Intuitively, it takes any state space of any quantum system and produces the state space for k indistinguishable copies that system. (If you’re a physicist, muttering the phrase ‘identical bosons’ may clarify things. There is also a fermionic version where we use exterior powers instead of symmetric powers, but let’s not go there now.)

The classical version of this functor suffers from a small glitch, which however is easy to fix. An object in $\texttt{Class}$ is a linearly normal subvariety

$M \subseteq \mathbb{C}\mathrm{P}^{n-1}$

for some $n.$ Applying the kth Veronese embedding we get a subvariety

$v_k(M) \subseteq \mathbb{C}\mathrm{P}^{\left(\!\!{n\choose k}\!\!\right)-1}$

However, I don’t think this is linearly normal, in general. I think it’s linearly normal iff $M$ is k-normal. You can take this as a definition of k-normality, if you like, though there are other equivalent ways to say it.

Luckily, a projectively normal subvariety of projective space is k-normal for all $k \ge 1.$ And even better, projectively normal varieties are fairly common! In particular, any projective space is a projectively normal subvariety of itself.

So, we can redefine the category $\texttt{Class}$ by letting objects be projectively normal subvarieties $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ for arbitrary $n \ge 1.$ I’m using the same notation for this new category, which is ordinarily a very dangerous thing to do, because all our results about the original version are still true for this one! In particular, we still have adjoint functors

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

defined exactly as before. But now the kth Veronese embedding gives a functor

$v_k \colon \texttt{Class} \to \texttt{Class}$

Intuitively, this takes any state space of any classical system and produces the state space for k indistinguishable copies that system that are all in the same state. It has no effect on the classical state space $M$ as an abstract variety, just its embedding into projective space—which in turn affects its Kähler structure and the line bundle it inherits from projective space. In particular, its symplectic structure gets multiplied by k, and the line bundle over it gets replaced by its kth tensor power. (These are well-known facts about the Veronese embedding.)

I believe that this functor obeys

$\texttt{Q} \circ v_k = S^k \circ \texttt{Q}$

and it’s just a matter of unraveling the definitions to see that

$\texttt{P} \circ S^k = v_k \circ \texttt{P}$

So, very loosely, the functors

$v_k \colon \texttt{Class} \to \texttt{Class}, \qquad S^k \colon \texttt{Quant} \to \texttt{Quant}$

should be thought of as replacing a classical or quantum system by a new ‘cloned’ version of that system. And they get along perfectly with quantization and its adjoint, projectivization!

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 the symmetric tensor powers of a Hilbert space as the corresponding way to clone a quantum system.

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

### 10 Responses to Geometric Quantization (Part 7)

1. Avi Levy says:

I gather you meant to write “$n$ multichoose $k$” instead of $\binom{n}{k}$ for the dimension of $S^k(\mathbb{C}^n)$?

• John Baez says:

Yes, you are right. You can choose the same variable repeatedly when you’re forming a monomial!

2. Or $n+k-1\choose k$.

3. John Baez says:

The standard notation for $n$ multichoose $k$ is this:

$\left(\!{n\choose k}\!\right)$

and Allen’s formula for it is correct.

• John Baez says:

Here’s how we check Allens’s formula. Suppose we want to know the dimension of the space of degree-9 homogeneous polynomials in 4 variables a,b,c,d. There’s a basis of monomials, and here’s a typical one:

aabbbdddd

I’ve shown one where c doesn’t appear at all, just to make sure I understand degenerate cases.

We can draw a circle for each variable, and a line each time we change from one variable to the next one in alphabetical order. In this style, the above monomial can be depicted like this:

oo|ooo||oooo

We have 4-1 lines and 9 circles. So, this is also a picture of how to choose 9 distinct things out of 9+4-1. So, the dimension is

$\displaystyle{ \binom{9+4-1}{9} }$

In general, we get

$\displaystyle{ \left(\!{n\choose k}\!\right) = \binom{n+k-1}{k} }$

• I think you mean draw degree # of circles and number of variables minus 1 lines….

4. Ishi Crew says:

i think a book by wm feller went through that—for me, very difficult.bose-einstein statistics.

5. John Huerta says:

Great stuff! It’s intriguing how you’re treating k copies of the spin-1/2 particle as a bunch of “identical bosons”, when by the spin-statistics theorem, half-integer spin particles are fermions.

• John Baez says:

This has bothered me for decades: back when I did loop quantum gravity and spin foams, I used to think about building spacetime from spin. But it’s just a mathematical fact that the spin-k rep of SU(2) is the symmetrized kth tensor power of the spin-1/2 rep. There’s no way around it!

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