Noncommutative Synthetic Differential Geometry

If synthetic differential geometry is based on toposes related to commutative rings, what happens if we instead use noncommutative rings? How does "noncommutative synthetic differential geometry" behave?

There are many potential models for noncommutative synthetic differential geometry, just as there are in the commutative case. Some simple ones are the classifying toposes for rings, local rings, or \(\star\)-algebras. There should also be more complicated examples, analogous to the well-adapted toposes in synthetic differential geometry, but it's not obvious how these should be defined, or what they're adapted to.

I will not take any particular model as "correct". I would rather get a sense for what's likely to hold across models, to the extent possible.

Basics

Monoidal Connectives

Models of noncommutative SDG tend to be monoidal toposes. In addition to the standard \((A \times -) \dashv (A \to -)\) adjunction, we have a second \((A \star -)\dashv(A \wand -)\) adjunction, where \(A \star B \subseteq A \times B\). In the classifying topos for rings, \(\star\) is given by Day convolution over the tensor product of rings.

A suitable type theory for expressing these connectives is provided by Ulrich Schöpp's PhD thesis. As I discuss here, checking that the theory can be applied to a classifying topos reduces to picking out a relation in the underlying theory, and checking some simple properties. In our case, the relation is commutativity: \(a \sim b \iff ab = ba\). As a consequence, we have that \(R \star R = \{(a,b) \in R \times R \mid ab = ba\}\), the set of commuting pairs in \(R\).

In general, the monoidal connectives are all about commutativity. Intuitively, \(A \star B\) is the set of pairs \((a,b)\) for which everything in \(a\) commutes with everything in \(b\). \(A \wand B\) is the set of functions whose input is required to commute with the function itself!

For example, in "algebraic" models, both \(R \to R\) and \(R \wand R\) act as sets of polynomials. The difference is that for \(R \wand R\), the variable is assumed to commute with the coefficients, while in \(R \to R\), it isn't. The functions \(x \mapsto ax + b\) and \(x \mapsto xa + b\) are equal as elements of \(R \wand R\), but distinct in \(R \to R\).

Choice of Connective

As we've seen, we have two notions of product, and two notions of function. How do we choose between them?

It helps that the distinction is meaningless in global context. We can freely convert between \(\vdash p : A \times B\) and \(\vdash p : A \star B\), or between \(\vdash f : A \to B\) and \(\vdash f : A \wand B\).

So in a global definition, we need not worry about the outermost connective. Inner connectives require more care.

Consider the analogue of the Kock-Lawvere axiom. Writing \(D = \{d\in R\mid d^2 = 0\}\), the following map should be invertible. \[(a,b) \mapsto (d \mapsto a + bd) : (R \times R) \to (D \wand R)\] The map in question, and its axiomatized inverse, are globally-defined. So it doesn't matter whether we say \((R \times R) \to (D \wand R)\) or \((R \times R) \wand (D \wand R)\). But the inner connectives are crucial; it won't work if we use \(R \star R\) or \(D \to R\).

As a more complicated example, the most useful notion of ring seems to be one which is required to be commutative, but where multiplication is only defined on \(\star\)-pairs of inputs, rather than \(\times\)-pairs. \(R\) itself is such a ring, because \(R \star R\) is exactly the subset of \(R \times R\) on which multiplication commutes. And if \(R\) admits a "complex conjugation", the subset of "reals" is a commutative ring in this sense, but isn't even a ring in the ordinary sense; multiplication on \(\times\)-pairs of reals doesn't land back in the reals.

Synthetic Axioms

In ordinary synthetic algebraic geometry, whenever \(A\) is a finitely-presented commutative \(R\)-algebra, then \(\hom_{R-\text{alg}}(A,R) \to R\) is isomorphic to \(A\). In synthetic differential geometry, something similar holds, but for a smaller class of \(R\)-algebras.

By Blechschmidt's Generalized Nullstellensatz, this generalizes greatly, including to the "algebraic" models of noncommutative synthetic differential geometry. We have the exact same principle: if \(A\) is a finitely-presented \(R\)-algebra, then \(\hom_{R-\text{alg}}(A,R) \to R\) is isomorphic to \(A\).

But this isn't the whole story. This principle tells us how to understand types of the form \(X \to R\), but it's equally important to understand types of the form \(X \wand R\). For example, \(\{d\in R\mid d^2 = 0\} \wand R\) should consist exactly of the linear functions \(d \mapsto a + bd\). While I've checked this particular example directly, in the classifying topos of rings, there ought to be a more general principle at play.

As a Topos

How does noncommutative synthetic differential geometry fit into the wider landscape of topos logics?

Consider the most basic model: the classifying topos for rings. Compare it to the classifying topos of commutative rings, which models ordinary synthetic differential geometry. Naturally, we have a geometric morphism \(\mathbf{Set}[\mathbf{CRing}] \to \mathbf{Set}[\mathbf{Ring}]\), given by modeling the generic ring as the generic commutative ring. This has a left adjoint, modeling the generic commutative ring as a quotient of the generic ring. This adjunction of geometric morphisms expands into an adjoint triple \(\mathbf{Set}[\mathbf{CRing}] \mathrel{\substack{\overset{\times}{\leftarrow} \\ \hookrightarrow \\ \leftarrow}} \mathbf{Set}[\mathbf{Ring}]\), exhibiting \(\mathbf{Set}[\mathbf{Ring}]\) as a totally connected topos over \(\mathbf{Set}[\mathbf{CRing}]\).

Star-rings and Quantum Mechanics

A key motivation for noncommutative synthetic differential geometry is its connection to quantum mechanics.

Basics

Assume that \(R\) has a complex conjugation; it's not just a ring, but a \(\star\)-ring. We'll rename it to \(C\), since it behaves more like the complex numbers than the reals, and let \(R\) refer to the real subset, \(\{x \in C \mid x^\star = x\}\). We might as well also define \(I = \{x \in C \mid x^\star = -x\}\), for the imaginary numbers.

If \(-1\) has a square root, we have \(R \cong I\), nonuniquely. If \(2\) is invertible, we have \(C \cong R \times I\).

Both \(C\) and \(R\) are commutative rings if we require multiplication to take \(\star\)-pairs of inputs. If we don't, \(C\) is a noncommutative ring, and \(R\) isn't even a ring. In fact, the commutator of two real numbers is imaginary. \[a,b\in R \implies [ a , b ]^\star = (ab-ba)^\star = b^\star a^\star - a^\star b^\star = ba - ab = -[ a,b ] \implies [ a,b ] \in I\]

Presumably, we also want a set \(R^{\geq 0}\) of positive numbers. I haven't thought deeply about this, but analogy to C*-algebras suggests that positive numbers should be sums of terms of the form \(x^\star x\).

Physics

Set up a harmonic oscillator in standard fashion. \[\begin{align*} x, p &: R \wand R \\ H &= \frac{p^2}{2m} + \frac{m \omega^2 x^2}{2} \\ \dot{x} &= \frac{\partial H}{\partial p} = \frac{p}{m} \\ \dot{p} &= -\frac{\partial H}{\partial x} = -m \omega^2 x \end{align*}\]

Taking a cue from quantum mechanics, we simplify with a change of coordinates, equating \(R \times R\) with \(C\). \[\begin{align*} a &= \sqrt{\frac{m\omega}{2}}x + i\sqrt{\frac{1}{2m\omega}}p \\ H &= \omega \frac{a a^\star + a^\star a}{2} \\ \dot{a} &= -i\omega a \\ a &= \exp(-i\omega t) \end{align*}\]

That's all standard. What's new is that our "numbers" are noncommutative. It's not hard to check that the commutator \([ a,a^\star ]\) is conserved, so our phase space foliates into subspaces \(C_\hbar = \{a \in C \mid [a,a^\star] = \hbar\}\), as \(\hbar\) ranges over \(R\).

The choice of \(\hbar\) as a name is intended to be suggestive. \(C_0\) is the classical phase space; every other \(C_\hbar\) is quantum.

For example, none of these spaces are empty, yet \(C_0\) is the only one with elements definable in global context. After all, only in a classical system can position and momentum simultaneously take definite values.

What about indefinite values? Define \(\mathrm{Prob}(X) = \{f : (X \wand R^{\geq 0}) \overset{\text{linear}}{\wand} R^{\geq 0} \mid f(1) = 1\}\). On a classical space like \(C_0\), this describes probability distributions. On quantum \(C_\hbar\), it describes mixed states.

And indeed, I will show that if we minimize the expected Hamiltonian \(\mathrm{Prob}(C_1)\), and pass to the corresponding position distribution, the result is exactly the Gaussian you would expect from the ground state.

Proof

I would like to be able to prove this entirely within the internal logic. What I am able to do, at the current time, is to minimize over globally defined elements of \(C_1 \wand C\), when modeled in the classifying topos for \(\star\)-algebras over the complex numbers, in a classical metatheory.

Let's begin.

By the synthetic axioms, maps \(C_1 \wand C\) are exactly polynomials, \(\sum_{m,n} c_{m,n} (a^\star)^m a^n\). Elements of \(\mathrm{Prob}(C_1)\), therefore, are characterized by their action on the monomials \((a^\star)^m a^n\).

Consider a globally-defined \(E \in \mathrm{Prob}(C_1)\). To minimize \(H\), we can minimize \(\frac{H}{\omega} - \frac{1}{2} = a^\star a\). Since the expectation of a positive function is required to be a positive number, we must have \(E(a^\star a) \geq 0\). I claim this can be uniquely achieved.

Uniqueness

I claim that if \(E \in \mathrm{Prob}(C_1)\) with \(E(a^\star a) = 0\), then \(E((a^\star)^m a^n) = \begin{cases} 1 & m=n=0 \\ 0 & \text{else} \end{cases}\).

Induct on \(\max(m,n)\). The base case, \(m=n=0\), follows immediately from the definition of \(\mathrm{Prob}(C_1)\).

For the inductive step, we consider first the case \(m = n\). If \(n = 1\), the result holds by assumption, so let \(n \geq 2\).

For arbitrary \(t \in R\), compute: \[\begin{align*} 0 &\leq E(((a^\star a+1)a^{n-1})^\star((a^\star a+t)a^{n-1})) \\ &= E((a^\star)^{n-1} (a^\star a+t)^2 a^{n-1}) \\ &= E((a^\star)^{n-1} ((a^\star)^2 a^2 + (2t+1) a^\star a + t^2) a^{n-1}) \\ &= E((a^\star)^{n+1} a^{n+1}) + (2t+1) E((a^\star)^n a^n) + t^2 E((a^\star)^{n-1} a^{n-1}) \\ &= E((a^\star)^{n+1} a^{n+1}) + (2t+1) E((a^\star)^n a^n) \end{align*}\]

Recall that \(E((a^\star)^{n+1} a^{n+1})\) and \(E((a^\star)^n a^n)\) are globally-defined elements of \(R\), so they're modeled as real numbers. If \(E((a^\star)^n a^n)\) is modeled by anything but zero, we get a contradiction by making \(t\) very positive or very negative. So \(E((a^\star)^n a^n) = 0\).

Now consider the case where \(m < n\). For arbitrary \(t \in C\), compute: \[\begin{align*} 0 &\leq E((a^m + ta^n)^\star(a^m + ta^n)) \\ &= E((a^\star)^m a^m) + t E((a^\star)^m a^n) + t^\star E((a^\star)^n a^m) + t^2 E((a^\star)^n a^n) \\ &\leq 1 + t E((a^\star)^m a^n) + t^\star E((a^\star)^n a^m) \end{align*}\]

The expressions \(E((a^\star)^m a^n)\) and \(E((a^\star)^n a^m)\) are globally-defined elements of \(C\), so they're modeled as complex numbers. By taking various values of \(t\) in the above equation, and observing that \(E((a^m + ta^n)^\star(a^m + ta^n))\) is real, we obtain enough information to prove that \(E((a^\star)^m a^n)\) and \(E((a^\star)^n a^m)\) are zero. This concludes the induction.

Existence

It remains to show that \(E((a^\star)^m a^n) = \begin{cases} 1 & m=n=0 \\ 0 & \text{else} \end{cases}\) is in fact an element of \(\mathrm{Prob}(C_1)\). The difficult condition is to show that \(E\) maps positive functions to positive numbers. How can we use the fact that \(f : C_1 \wand R^{\geq 0}\) only returns positive outputs, if we can't produce inputs to test it on?

I'm not sure. So for now, I work around the problem, by operating entirely within the model. This allows us to upgrade our knowledge that a function returns positive values to the knowledge that it is positive as a polynomial.

In the classifying topos for complex \(\star\)-algebras, \(C\) is interpreted as the forgetful functor from finitely presented \(\star\)-algebras over the complex numbers to sets. \(C_1\) and \(R^{\geq 0}\) are subspaces, interpreting as the functors \(A \mapsto \{a \in A \mid [a,a^\star] = 1\}\) and \(A \mapsto \{a \in A \mid a \text{ is positive}\}\), respectively.

We compute: \[\begin{align*} & \llbracket C_1 \wand R^{\geq 0} \rrbracket \\ &= B \mapsto \int_{A} \llbracket C_1 \rrbracket(A) \to \llbracket R^{\geq 0} \rrbracket(B \otimes A) \\ &= B \mapsto \int_{A} \{a \in A \mid [a , a^\star] = 1\} \to \{t \in B \otimes A \mid t \text{ is positive}\} \end{align*}\]

\(E\) defines a natural transformation from this to \(\llbracket C \rrbracket\). \[E_B(f) = (\mathrm{id}_ B \otimes \text{extract constant component})(f_{\mathbb{C}[ a ]/([a , a^\star] = 1)}(a))\]

We are to show that this always outputs a positive element of \(B\). In other words:

In an arbitrary star-algebra \(B\), consider an arbitrary positive element of the polynomial algebra \(B[ a ]/([a , a^\star] = 1)\). When expanded as a polynomial so that the \(a^\star\)s in each term precede the \(a\)s, must the constant term be positive?

The answer is yes. Given any polynomial \(f\), directly expanding the constant component of \(f^\star f\) yields a sum consisting entirely of positive terms.

Results

We have found the unique global element of \(\mathrm{Prob}(C_1)\) of minimal expected Hamiltonian. We now extract the position distribution.

Recall: \[a = \sqrt{\frac{m\omega}{2}}x + i\sqrt{\frac{1}{2m\omega}}p \in C\] \[\sqrt{\frac{1}{2m\omega}}(a + a^\star) = x\]

We lift this map \(C_1 \to R\) to a map \(\mathrm{Prob}(C_1) \to \mathrm{Prob}(R)\), and apply it to \(E\). \[f \mapsto E\left(a \mapsto f\left(\sqrt{\frac{1}{2m\omega}}(a + a^\star)\right)\right) \in \mathrm{Prob}(R)\]

To fully determine this map, it suffices to evaluate at the powers of \(x\). \[\begin{align*} x^n &\mapsto E\left(a \mapsto \left(\sqrt{\frac{1}{2m\omega}}(a + a^\star)\right)^n\right) \\ &= (2m\omega)^{-n/2} E\left(a \mapsto (a + a^\star)^n\right) \end{align*}\]

We need the constant term of \((a+a^\star)^n\), when it is reexpressed with the \(a^\star\)s preceding the \(a\)s. But this can be rephrased as a combinatorial problem! It equals the number of ways of putting an \(a\) or an \(a^\star\) in each of \(n\) slots, then matching each \(a\) to a later \(a^\star\). This is simply the number of matchings of \(n\) items, which is \((n-1)!!\) when \(n\) is even, and zero when \(n\) is odd.

We conclude: \[\begin{align*} x^{2n} &\mapsto (2m\omega)^{-n} (2n-1)!! \\ x^{2n + 1} &\mapsto 0 \end{align*}\]

What function does this distribution describe? The Gaussian \(\sqrt{\frac{m\omega}{\pi}} e^{-m\omega x^2}\). \[\begin{align*} \int_{-\infty}^\infty \sqrt{\frac{m\omega}{\pi}} e^{-m\omega x^2} x^{2n} &= (2m\omega)^{-n} (2n-1)!! \\ \int_{-\infty}^\infty \sqrt{\frac{m\omega}{\pi}} e^{-m\omega x^2} x^{2n+1} &= 0 \end{align*}\]

This Gaussian is exactly the correct expression for the amplitude of the quantum harmonic oscillator in its ground state, in units where \(\hbar = 1\).

Conclusion

I have shown that noncommutative synthetic differential geometry supports non-classical phase spaces, over which physics is naturally quantum. But the argument, as it stands, is unsatisfying. It requires me to drop into the model, rather than working synthetically, and it's longer than it ought to be. I am hopeful that these flaws can be repaired, once more of the theory of noncommutative synthetic differential geometry has been developed.

I would also like to develop more of quantum theory within this framework. How do we discuss pure states, or spectra? I've solved for the ground state, but how would I ask for the excited states?

I'm especially curious what noncommutative synthetic differential geometry has to say about quantum field theory. Ordinary synthetic differential geometry treats infinite-dimensional spaces differently from classical geometry; the same should be true in the noncommutative case.