Invertible Field Theories and Generalized Cohomology

An extended field theory is invertible when every value and every morphism it assigns is tensor-invertible, so that stacking it with another theory gives the trivial theory. Its values therefore lie in the Picard part of the target: closed codimension-one manifolds receive lines rather than higher-dimensional state spaces, and closed spacetime manifolds receive nonzero phases or amplitudes. After the tangential structure, target, deformation relation, and positivity condition are fixed, the resulting Picard higher groupoid is encoded by a spectrum. Bordism spectra and generalized cohomology then organize deformation classes. This statement classifies functorial invertible theories under declared hypotheses; it does not prove microscopic realization by a gapped unitary system.

Required background. Dualizability and the cobordism hypothesis supply the fully extended classification; tangential structures fix the geometric domain; and background responses and invertible phases supply the physical response language.

Helpful background. Bordism and tangential structures supply the relevant bordism groups, while topological order and invertible phases distinguish invertibility from nontrivial topological degeneracy.

Tensor invertibility and Picard targets

Let

$$Z:\operatorname{Bord}^{G}_n\longrightarrow\mathcal C$$

be a fully extended symmetric monoidal field theory in dimension $n$, with tangential or symmetry type $G\to O(n)$. It is invertible if there is a theory $Z^{-1}$ and a monoidal equivalence

$$Z\otimes Z^{-1}\simeq\mathbb 1.$$

Equivalently, $Z$ factors through the maximal Picard $(\infty,n)$-groupoid $\mathcal C^\times$, obtained by retaining only tensor-invertible objects and invertible morphisms at every level. In an ordinary vector-space truncation, a tensor-invertible state space must be a one-dimensional line. A theory with a two-dimensional state space on some closed $(n-1)$-manifold cannot have a tensor inverse, because dimensions multiply:

$$\dim(V\otimes W)=\dim V\,\dim W\ne1.$$

Stacking is the abelian group operation on equivalence or deformation classes, the trivial theory is the identity, and complex conjugation often represents the inverse when the Hermitian structure is compatible. Freed and Hopkins give this factorization criterion and explain the passage from Picard higher groupoids to spectra in Freed and Hopkins 2021, §5.2, pp. 33–35.

Invertibility, full dualizability, and positivity are separate conditions. An invertible value is automatically dualizable, but a fully dualizable noninvertible algebra can still produce multidimensional state spaces. Conversely, an invertible functor may carry an indefinite Hermitian structure unless reflection positivity is imposed.

From bordism spectra to generalized cohomology

The group-completed structured bordism category has the homotopy type of a Madsen–Tillmann spectrum, customarily written $\Sigma^n M T G_n$ at finite dimension and $M T G$ after stabilization. Since a Picard target also determines a spectrum, an invertible theory becomes a spectrum map. The target spectrum matters:

$$\Sigma^n M T G_n\longrightarrow\Sigma^n I\mathbb C^\times$$

describes a discrete invertible theory whose partition function is character-valued, whereas maps to a shifted Anderson dual encode continuous deformation information. Under the stability and reflection-positive hypotheses of Freed–Hopkins, deformation classes are computed by

$$[M T G,\Sigma^{n+1}I\mathbb Z(1)].$$

For discrete reflection-positive theories, only the torsion subgroup is obtained. These distinctions, including the exact continuous and discrete targets, are proved in Freed and Hopkins 2021, Theorem 5.23, pp. 36–37, and Theorem 8.20 with Corollary 8.21, pp. 67–68. It is therefore unsafe to write “invertible theories equal a bordism group” without specifying the target spectrum and the equivalence relation.

In a finite purely torsion sector, the result often reduces to a character

$$\chi:\Omega_n^G\longrightarrow U(1), \qquad \chi([M\sqcup N])=\chi([M])\chi([N]).$$

Free bordism classes can instead support local or differential terms, so the character group alone need not retain all continuous data. Nor does a generalized-cohomology class by itself supply reflection positivity or a lattice realization; Freed and Hopkins explicitly separate their classification theorem from that realization assumption in Freed and Hopkins 2021, Introduction, pp. 1–4.

A spin-bordism character

As the exact first application, consider the nonbounding spin circle, which generates

$$\Omega^{\mathrm{Spin}}_1\cong\mathbb Z/2.$$

The nontrivial character sends the bounding spin circle to $+1$ and the nonbounding one to $-1$. Disjoint union adds bordism classes and multiplies signs, so stacking two copies gives the trivial character. Orientation alone cannot see this response: forgetting the spin lift identifies data on which the sign depends. This finite calculation is returned to topological order and invertible phases as a model of a low-dimensional invertible response. It establishes a bordism-valued functor, not a microscopic Hamiltonian.

An independent check evaluates the character on a spin boundary. Any boundary represents zero in $\Omega^{\mathrm{Spin}}_1$, so its phase must be $+1$. The test detects either a wrongly normalized generator or a rule that fails bordism invariance.

The adversarial counterexample is a two-dimensional semisimple TQFT with circle state space $A=\mathbb C\oplus\mathbb C$. It is fully dualizable in an appropriate Morita target, but $\dim A=2$. No finite-dimensional $W$ satisfies $A\otimes W\cong\mathbb C$, so the theory is not invertible and does not belong to this generalized-cohomology classification. The strongest surviving statement is a noninvertible extended TQFT classification in its declared target.

Exercises

Show that every state space of an invertible vector-space-valued TQFT is one-dimensional.

Solution

If $Z\otimes Z^{-1}\simeq\mathbb 1$, then on any closed hypersurface $\Sigma$ one has $Z(\Sigma)\otimes Z^{-1}(\Sigma)\cong\mathbb C$. Taking dimensions gives two positive integers whose product is one, so both are one.

Why does a character of $\Omega_n^G$ automatically respect gluing through a bordism?

Solution

If closed manifolds $M_0$ and $M_1$ are the boundary components of a structured bordism with the incoming orientation reversed, then $[M_1]-[M_0]=0$ in $\Omega_n^G$. A homomorphism therefore gives $\chi(M_1)=\chi(M_0)$. This is only the closed-manifold shadow of the extended functor.

References

• Freed, Daniel S. “Anomalies and Invertible Field Theories.” Proceedings of Symposia in Pure Mathematics 88 (2014): 25–45. DOI; Open PDF.
• Freed, Daniel S., and Michael J. Hopkins. “Reflection Positivity and Invertible Topological Phases.” Geometry & Topology 25 (2021): 1165–1330. DOI; Open PDF.
• Kapustin, Anton. “Symmetry Protected Topological Phases, Anomalies, and Cobordisms: Beyond Group Cohomology.” arXiv (2014). Open PDF.