Bordism Categories and Symmetric Monoidal TQFTs

An $n$-dimensional Atiyah–Segal TQFT is a symmetric monoidal functor from a specified category of $(n-1)$-manifolds and $n$-dimensional bordisms to a specified algebraic target. Cutting becomes composition, disjoint union becomes tensor product, and orientation reversal supplies duals. In two dimensions these requirements are exactly the relations of a finite-dimensional commutative Frobenius algebra. They define an unextended topological theory; they do not assign data to points or assert that an arbitrary physical QFT is topological.

Required background. Bordism and tangential structures supply the geometric equivalence relation; state spaces, cobordisms, and gluing supply the physical interpretation; tensor products supply monoidality; and categories and functors supply composition and natural equivalence.

Helpful background. What is a topological field theory? contrasts the functorial axioms with metric-dependent dynamics.

The structured bordism category

Fix a dimension $n$ and, here, an orientation. The category $\operatorname{Bord}^{\mathrm{or}}_n$ has closed oriented $(n-1)$-manifolds $\Sigma$ as objects. A morphism

$$M:\Sigma_0\longrightarrow\Sigma_1$$

is a compact oriented $n$-manifold whose boundary is identified with

$$\partial M\cong \overline{\Sigma}_0\sqcup\Sigma_1,$$

together with collars that make gluing smooth. Morphisms are taken up to orientation-preserving diffeomorphism respecting those identifications. Composition glues outgoing to incoming collars; the identity is the cylinder $\Sigma\times[0,1]$. Disjoint union $\sqcup$ is the symmetric monoidal product, with the empty manifold as unit. The dual of $\Sigma$ is $\overline\Sigma$.

An Atiyah–Segal theory valued in finite-dimensional complex vector spaces is a strong symmetric monoidal functor

$$Z:\operatorname{Bord}^{\mathrm{or}}_n \longrightarrow \operatorname{Vect}^{\mathrm{fd}}_{\mathbb C}.$$

Thus $Z(\Sigma_0\sqcup\Sigma_1)\cong Z(\Sigma_0)\otimes Z(\Sigma_1)$, $Z(\varnothing)\cong\mathbb C$, and

$$Z(M_2\circ M_1)=Z(M_2)\circ Z(M_1).$$

Diffeomorphic bordisms give the same linear map. These are mathematical axioms, not a derivation from a path integral. Atiyah’s original formulation states the state-space, functoriality, disjoint-union, and duality conditions in Atiyah 1988, pp. 175–180.

The domain and target are part of the definition. Replacing oriented bordisms by framed, spin, or unoriented bordisms changes which manifolds are identified and which additional structures the functor must preserve. Replacing finite-dimensional vector spaces by super vector spaces, Hilbert spaces, or a higher category changes the available duals and the meaning of equivalence. In particular, a rule for closed manifolds alone is only a partition-function invariant; without compatible state spaces and bordism maps it is not an Atiyah–Segal TQFT.

Pair-of-pants generators in two dimensions

Set $n=2$ and $A=Z(S^1)$. The pair of pants with two incoming circles defines multiplication $\mu:A\otimes A\to A$; the incoming disk defines the unit $\eta:\mathbb C\to A$. Turning these bordisms around gives a coproduct $\Delta:A\to A\otimes A$ and counit $\varepsilon:A\to\mathbb C$. Diffeomorphisms and alternative decompositions imply

$$\mu(\mu\otimes\mathrm{id}) =\mu(\mathrm{id}\otimes\mu), \qquad \mu\circ\tau=\mu,$$

and the unit relations. The decisive compatibility is the Frobenius identity

$$\Delta\mu =(\mu\otimes\mathrm{id})(\mathrm{id}\otimes\Delta) =(\mathrm{id}\otimes\mu)(\Delta\otimes\mathrm{id}).$$

The bilinear form

$$\beta(a,b)=\varepsilon(ab)$$

is nondegenerate because the cap and cup bordisms exhibit $A$ as its own dual. Conversely, a finite-dimensional commutative algebra with such a nondegenerate invariant pairing determines $\Delta$ as the adjoint of $\mu$ and supplies all oriented surface maps. The equivalence between two-dimensional TQFTs and commutative Frobenius algebras is proved in Abrams 1996, pp. 569–587.

This is the exact first application developed with state spaces, cobordisms, and gluing: define the oriented bordism category, assign $\mu,\eta,\Delta,\varepsilon$ to its elementary surfaces, and recover the cylinder as the identity. The calculation is finite and exact. There is no convergence limit, metric, or Hamiltonian approximation.

Relations are the functoriality test

The construction mechanism is a generators-and-relations argument. A Morse function decomposes a surface into cylinders, caps, cups, and pairs of pants. Cerf moves relating two Morse functions translate into associativity, commutativity, unit, counit, and Frobenius relations. Once the algebra satisfies them, the composite map is independent of the chosen decomposition.

An independent check glues a cup and cap to the two legs of a pair of pants. Algebraically, the snake identities reduce the resulting cylinder to $\mathrm{id}_A$. A second check swaps the incoming circles; orientation-preserving diffeomorphism forces $\mu\tau=\mu$.

The adversarial failure assigns one map to $(\mu\otimes\mathrm{id})(\mathrm{id}\otimes\Delta)$ and an incompatible map to $(\mathrm{id}\otimes\mu)(\Delta\otimes\mathrm{id})$. Those composites represent diffeomorphic cuttings of the same bordism. If they differ, the assignment is not a functor on the bordism category. At most it is data on chosen decompositions; no decomposition-independent TQFT survives.

This also identifies the extension boundary. The circle algebra and surface maps determine the unextended oriented theory, but they do not specify a value on a point, modules on intervals, or intertwiners at corners. Supplying those lower-dimensional values requires a higher target and additional adjoint data; it cannot be obtained merely by reading $A$ as the point value.

Exercises

Show that the cylinder acts as the identity on $A$.

Solution

Decompose the cylinder using a coevaluation cup followed by an evaluation cap. The two possible contractions are the snake composites $(\mathrm{id}\otimes\mathrm{ev})(\mathrm{coev}\otimes\mathrm{id})$ and $(\mathrm{ev}\otimes\mathrm{id})(\mathrm{id}\otimes\mathrm{coev})$. Nondegeneracy of $\beta$ makes both identities. Functoriality therefore assigns $\mathrm{id}_A$ to the cylinder.

Why must $A$ be finite-dimensional when the target is $\operatorname{Vect}_{\mathbb C}$?

Solution

The cup and cap make $A$ dualizable. In ordinary complex vector spaces, dualizable objects are exactly finite-dimensional vector spaces: coevaluation would otherwise require an infinite sum not belonging to the algebraic tensor product.

References

• Abrams, Lowell. “Two-Dimensional Topological Quantum Field Theories and Frobenius Algebras.” Journal of Knot Theory and Its Ramifications 5 (1996): 569–587. DOI.
• Atiyah, Michael F. “Topological Quantum Field Theory.” Publications Mathématiques de l’IHÉS 68 (1988): 175–186. DOI; Open PDF.