Atiyah–Segal Functorial TQFT and Gluing

Atiyah–Segal functoriality turns a geometric cut along a closed hypersurface into composition of linear maps, and disjoint union into tensor product. For a two-dimensional oriented theory, every closed-surface amplitude is therefore a contraction of the multiplication, coproduct, unit, and trace of one commutative Frobenius algebra. The answer is independent of a pants decomposition precisely because the Frobenius relations hold. This is an unextended functorial statement; projective theories and theories with boundary anomalies require additional data.

Required background. Bordism categories and symmetric monoidal TQFTs define the functor and Frobenius operations. State spaces, cobordisms, and gluing provide the physical sewing interpretation.

Helpful background. Gluing, reduction, and composition theorems explain why gauge-theoretic path integrals need corrections beyond the axiomatic rule, while topological field theory supplies representative physical models.

Cutting is categorical composition

Let a bordism $M:\Sigma_0\to\Sigma_2$ contain a separating closed hypersurface $\Sigma_1$ and write

$$M=M_2\circ_{\Sigma_1}M_1, \qquad M_1:\Sigma_0\to\Sigma_1, \quad M_2:\Sigma_1\to\Sigma_2.$$

For an anomaly-free Atiyah–Segal functor,

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

If $\Sigma_0=\Sigma_2=\varnothing$, the result is a scalar. Cutting a closed $M$ open along $\Sigma$ produces a vector in $Z(\overline\Sigma)\otimes Z(\Sigma)$, and regluing contracts it with the evaluation pairing. The mapping cylinder of a diffeomorphism $f:\Sigma\to\Sigma$ gives an automorphism $Z(f)$; closing that cylinder yields its categorical trace. Atiyah derives these duality and trace consequences from the axioms in Atiyah 1988, pp. 176–181.

Orientation is what selects the dual pairing: an incoming copy of $\Sigma$ is $\overline\Sigma$, so its state space is identified with $Z(\Sigma)^\vee$, not with $Z(\Sigma)$ by an unqualified equality. In a Hermitian theory a further conjugate-linear structure may identify these spaces, but that is additional to the bilinear Atiyah–Segal axioms. Keeping the two steps separate prevents a gluing contraction from being mistaken for a positive inner product.

The statement assumes genuine functoriality. If gluing is only projective,

$$Z(M_2\circ M_1)=\alpha(M_2,M_1)\, Z(M_2)Z(M_1),$$

then the multiplier $\alpha$ is extra anomaly data. Ignoring it can make two decompositions disagree by a phase.

A genus-two Frobenius contraction

Let $A=Z(S^1)$ be a finite-dimensional commutative Frobenius algebra with multiplication $\mu$, coproduct $\Delta$, unit $1=\eta(1)$, and trace $\varepsilon$. Define the handle operator and Euler element by

$$H=\mu\circ\Delta:A\to A, \qquad e=H(1)=\mu\Delta(1).$$

A closed genus-$g$ surface is obtained from a disk by attaching $g$ handles and then capping the last circle. Hence

$$Z(\Sigma_g)=\varepsilon\!\left(H^g(1)\right) =\varepsilon(e^g).$$

For genus two,

$$Z(\Sigma_2)=\varepsilon\!\left[ \mu\Delta\bigl(\mu\Delta(1)\bigr) \right].$$

This formula is independent of which separating curves define the pants decomposition. One elementary move changes the bracketing of three multiplications and is controlled by associativity. The other essential move slides a multiplication past a coproduct and is exactly the Frobenius identity. A Morse-theoretic presentation reduces general changes of decomposition to such moves; the two-dimensional equivalence with Frobenius algebras makes this independence precise Abrams 1996, pp. 579–587.

For a concrete check, take the semisimple algebra

$$A=\bigoplus_{i=1}^{r}\mathbb C p_i, \qquad p_ip_j=\delta_{ij}p_i, \qquad \varepsilon(p_i)=\theta_i\ne0.$$

Adjointness gives

$$\Delta(p_i)=\theta_i^{-1}p_i\otimes p_i, \qquad e=\sum_i\theta_i^{-1}p_i.$$

Therefore

$$Z(\Sigma_g)=\sum_{i=1}^{r}\theta_i^{\,1-g}, \qquad Z(\Sigma_2)=\sum_i\theta_i^{-1}.$$

The answer depends only on the genus and Frobenius weights, not on the chosen cut system.

This is the exact first application returned to state spaces, cobordisms, and gluing: cut a genus-two surface into pairs of pants, perform the Frobenius contraction, and verify it against a second decomposition.

Pairing and anomaly failure tests

Nondegeneracy of

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

is not optional. It identifies the state space of the oppositely oriented circle with $A^\vee$ and supplies the cup/cap snake identities. If $\beta$ is degenerate, choose $0\ne x$ with $\beta(x,a)=0$ for all $a$. The cylinder composite built from coevaluation and evaluation annihilates $x$, contradicting the identity-cylinder axiom.

An independent algebraic check evaluates the same genus-two surface by the handle formula and by an explicit contraction of four trivalent tensors in a basis. The inverse matrix $\beta^{ij}$ must appear on every sewn circle. In the idempotent basis both computations reduce to $\sum_i\theta_i^{-1}$.

The adversarial alternative retains a nondegenerate Frobenius algebra but omits a projective mapping-class multiplier. Then local pants moves can work while a loop in decomposition space returns the state with a phase. The strongest surviving object is a projective or relative theory, not an absolute symmetric monoidal functor to vector spaces. One must either trivialize the multiplier coherently or enlarge the target to remember the anomaly.

The axiomatic gluing equation also has no hidden integration measure: the inverse pairing already performs the finite contraction. In a gauge-theory path integral, residual fields, determinants, and boundary polarizations must first be controlled before its result can be shown to realize this functorial rule. That analytic construction belongs to the BV–BFV treatment rather than following from the topology alone.

Exercises

Compute the torus amplitude in the idempotent example.

Solution

For $g=1$, $Z(T^2)=\varepsilon(e)=\sum_i\theta_i^{-1}\varepsilon(p_i)=\sum_i1=r$. It equals the dimension of $A$, as expected from the trace of the identity on $Z(S^1)$.

Show explicitly how degeneracy breaks the cylinder.

Solution

If $x$ lies in the radical of $\beta$, every coefficient obtained by pairing $x$ with one leg of coevaluation vanishes. Thus the cup–cap composite sends $x$ to zero. Since $x\ne0$, that composite cannot equal $\mathrm{id}_A$.

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.