Start with the mirror pair

$$ \operatorname{Coh}( ( \mathbb C^* )^2 ) \quad\leftrightarrow\quad \operatorname{Fuk}(T^*T^2). $$

On the B-side, quantization of $( \mathbb C^* )^2$ should replace the commutative Laurent polynomial algebra

$$ \mathbb C[x^{\pm 1},y^{\pm 1}] $$

by the quantum torus

$$ \mathbb C_q[x^{\pm 1},y^{\pm 1}] = \mathbb C\langle x^{\pm 1},y^{\pm 1}\rangle/(xy=qyx). $$

The question is:

What is the corresponding modification on the A-side?

The answer in this toy model is: twist the wrapped Fukaya category of $T^*T^2$ by a $B$-field whose flux through the base torus is nonzero.

Let

$$ M=T^*T^2. $$

Let $L=T^*_0T^2$ be a cotangent fiber. It is a basic generator of the wrapped Fukaya category.

One expects

$$ HW^0(L,L)\cong \mathbb C[ \pi_1(T^2) ]. $$

Since

$$ \pi_1(T^2)\cong \mathbb Z^2, $$

this gives

$$ HW^0(L,L) \cong \mathbb C[x^{\pm1},y^{\pm1}]. $$

Here $x$ and $y$ correspond to the two basic loops on $T^2$. Since $\mathbb Z^2$ is abelian, the product is commutative:

$$ xy=yx. $$

A quantum torus can be viewed as a twisted group algebra of $\mathbb Z^2$.

Let $e_m$ denote the basis element corresponding to

$$ m=(m_1,m_2)\in \mathbb Z^2. $$

Instead of the ordinary product

$$ e_m e_n=e_{m+n}, $$

define

$$ e_m\star e_n=\sigma(m,n)e_{m+n}, $$

where

$$ \sigma:\mathbb Z^2\times\mathbb Z^2\to \mathbb C^* $$

is a multiplicative 2-cocycle. The cocycle condition is

$$ \sigma(m,n)\sigma(m+n,k) = \sigma(n,k)\sigma(m,n+k). $$

This is exactly the condition that $\star$ be associative.

For example, take

$$ \sigma(m,n)=q^{m_1n_2}. $$

Then, if

$$ x=e_{(1,0)},\qquad y=e_{(0,1)}, $$

we get

$$ x\star y=q\,y\star x. $$

Thus the quantum torus is just a twisted group algebra:

$$ \mathbb C_q[x^{\pm1},y^{\pm1}] \cong \mathbb C_\sigma[ \mathbb Z^2 ]. $$

Now write

$$ T^*T^2\cong \mathbb R^2_{u,v}\times T^2_{\theta,\phi}. $$

Here $(\theta,\phi)$ are coordinates on the base torus and $(u,v)$ are cotangent fiber coordinates.

Take a $B$-field pulled back from the base:

$$ B=\beta\, d\theta\wedge d\phi. $$

In the wrapped Floer product, multiplying $e_m$ and $e_n$ is geometrically represented by a holomorphic triangle, or morally by a filling of two composable base loops. A $B$-field weights this triangle by

$$ \exp\left(2\pi i\int_{\Delta(m,n)}B\right). $$

Thus the product becomes

$$ e_m\star e_n = \exp\left(2\pi i\int_{\Delta(m,n)}B\right)e_{m+n}. $$

This is the desired multiplicative cocycle.

The commutator phase is the $B$-field holonomy around the parallelogram spanned by $m$ and $n$:

$$ \frac{e_m\star e_n}{e_n\star e_m} = \exp\left(2\pi i\beta(m_1n_2-m_2n_1)\right). $$

In particular,

$$ x\star y = \exp(2\pi i\beta)\, y\star x. $$

So if

$$ q=\exp(2\pi i\beta), $$

then

$$ x\star y=q\,y\star x. $$

The $A_\infty$-relations come from the boundary of one-dimensional moduli spaces of holomorphic polygons.

With a $B$-field, every polygon is weighted by

$$ \exp\left(2\pi i\int B\right). $$

When two polygons are glued, the integrals add. Therefore the weights multiply.

For three elements $e_m,e_n,e_k$, associativity compares the two decompositions of a quadrilateral into triangles. Since $B$ is closed, the total $B$-field holonomy of the quadrilateral is independent of the triangulation. Algebraically, this gives

$$ \sigma(m,n)\sigma(m+n,k) = \sigma(n,k)\sigma(m,n+k). $$

Thus the $B$-field holonomy automatically gives a multiplicative 2-cocycle.

On $( \mathbb C^* )^2$, write

$$ x=e^{u+i\theta},\qquad y=e^{v+i\phi}. $$

The logarithmic holomorphic symplectic form is

$$ \Omega=d\log x\wedge d\log y. $$

Expanding,

$$ \Omega = (du+i\,d\theta)\wedge(dv+i\,d\phi). $$

Therefore

$$ \operatorname{Re}\Omega = du\wedge dv-d\theta\wedge d\phi, $$

and

$$ \operatorname{Im}\Omega = du\wedge d\phi+d\theta\wedge dv. $$

The base part of $\operatorname{Re}\Omega$ is

$$ -d\theta\wedge d\phi. $$

This is exactly the kind of $B$-field term that produces the quantum torus phase.

So the quantum torus deformation can be thought of as coming from the base-torus component of the real part of the logarithmic holomorphic symplectic form.

The form

$$ du\wedge dv $$

is exact on the cotangent fibers:

$$ du\wedge dv=d(u\,dv). $$

So the fiber-fiber part of the $B$-field is gauge-trivial.

More explicitly, if

$$ B_{\mathrm{fib}}=c\,du\wedge dv, $$

then with

$$ \Lambda=c\,u\,dv $$

we have

$$ B_{\mathrm{fib}}=d\Lambda. $$

A $B$-field gauge transformation sends

$$ B\mapsto B-d\Lambda. $$

Thus $c\,du\wedge dv$ can be gauged away.

At the brane level, one must also change the connection on the Lagrangian. If $L=T^*_0T^2$, then $B_{\mathrm{fib}}|_L\neq 0$. One can choose a connection one-form

$$ A=-2\pi i\,c\,u\,dv $$

so that

$$ F_\nabla=dA=-2\pi i\,c\,du\wedge dv. $$

Then

$$ F_\nabla+2\pi i B_{\mathrm{fib}}|_L=0. $$

After the gauge transformation, both $B_{\mathrm{fib}}$ and this connection term disappear.

By contrast,

$$ d\theta\wedge d\phi $$

is not exact on the base torus. Its period is the invariant quantity, and this is what gives the quantum torus relation.

The ordinary wrapped Fukaya category gives

$$ HW^0(T^*_0T^2,T^*_0T^2) \cong \mathbb C[x^{\pm1},y^{\pm1}]. $$

Turning on a base $B$-field

$$ B=\beta\,d\theta\wedge d\phi $$

twists the wrapped product by a cocycle and gives

$$ HW^0_B(T^*_0T^2,T^*_0T^2) \cong \mathbb C_q[x^{\pm1},y^{\pm1}], $$

where

$$ q=\exp(2\pi i\beta). $$

So, in this toy model,

$$ \boxed{ \text{quantizing } \operatorname{Coh}( ( \mathbb C^* )^2 ) \text{ corresponds on the A-side to a }B\text{-field twist of } \operatorname{Fuk}(T^*T^2). } $$

Or more poetically:

The two wrapped products $xy$ and $yx$ differ by the $B$-field flux through one fundamental parallelogram of the base torus.

There is one important brane-level point. In the ordinary Fukaya category, an object is not just a Lagrangian submanifold $L$. One usually equips $L$ with extra data, for example a grading, spin or Pin structure, and a local system. Without a $B$-field, this local system is usually flat. In rank one, we can write it locally as a connection $\nabla=d+A$ on a $U(1)$ line bundle over $L$, with curvature $F_\nabla=dA$. Flatness means $F_\nabla=0$.

In the presence of a $B$-field, a disk contribution contains both an interior term and a boundary term. For a holomorphic disk or polygon $u:(\Sigma,\partial\Sigma)\to (M,L)$, the weight looks like $\exp(2\pi i\int_\Sigma u^*B)\cdot \exp(\int_{\partial\Sigma}A)$.

The reason a non-flat connection may be needed is that $B$-fields have gauge transformations. If $B\mapsto B+d\Lambda$, then $\int_\Sigma B$ changes by the boundary term $\int_\Sigma d\Lambda=\int_{\partial\Sigma}\Lambda$. To keep the total disk weight invariant, one must transform the connection $A$ on the brane at the same time. The gauge-invariant combination is morally $2\pi i\,B+F_\nabla$.

For a Lagrangian brane, the usual compatibility condition is $F_\nabla+2\pi i\,B|_L=0$, or equivalently $F_\nabla=-2\pi i\,B|_L$. Thus the connection on $L$ is flat only when $B|_L=0$ in the relevant sense. If $B|_L$ is nonzero, the brane must carry a possibly non-flat line bundle connection whose curvature cancels the restriction of the $B$-field.

For example, take $L=\mathbb R^2_{u,v}\subset T^*T^2$. Suppose the $B$-field has a fiber-fiber part $B_{\mathrm{fib}}=c\,du\wedge dv$. Then $B_{\mathrm{fib}}|_L=c\,du\wedge dv$. A compatible connection is $\nabla=d+A$, where $A=-2\pi i\,c\,u\,dv$. Indeed, $F_\nabla=dA=-2\pi i\,c\,du\wedge dv$, so $F_\nabla+2\pi i\,B_{\mathrm{fib}}|_L=0$.

This is the concrete meaning of saying that the curvature of the brane line bundle cancels the $B$-field restricted to $L$.

In contrast, if $L$ is one-dimensional, for example a line $L=\mathbb R\subset \mathbb C$, then every two-form restricts to zero on $L$. In that case $B|_L=0$, so there is nothing to cancel, and the brane local system remains flat.

In the example relevant to the quantum torus, the useful $B$-field is $B=\beta\,d\theta\wedge d\phi$. For the cotangent fiber $L=T^*_0T^2$, the coordinates $\theta,\phi$ are constant along $L$. Therefore $B|_L=0$. So the cotangent fiber itself does not need a non-flat connection for this base $B$-field. The effect of the $B$-field instead appears in the holonomy over the holomorphic triangles contributing to the wrapped product. This is what twists $xy=yx$ into $xy=qyx$.