Boundaries, interfaces, and gluing

A boundary removes translations normal to it. Which supersymmetries can survive, what must happen to the fields, and how much can a protected count tell us about the answer? We will solve these questions for a free chiral multiplet, reproduce a known boundary identity, and then change one coupling.

Prepare with the research-entry method, the gauge-and-matter action, and the SQED–XYZ index calculation. The extra boundary language is developed here. The first project is a reproduction followed by a training extension, not a claim of a new duality.

Start with the boundary term

Take one free three-dimensional N=2 chiral, with four real supercharges, no dynamical gauge field, zero bulk superpotential, and vanishing masses and background sources. Let $x^\perp\leq0$, so the outward normal is $+\partial_\perp$. After continuing tangential time to Euclidean signature, the canonically normalized scalar action on the physical scalar contour is

$$S_\phi=\int_{x^\perp\leq0}d^3x\,\partial_\mu\bar\phi\,\partial_\mu\phi, \qquad \delta S_\phi\big|_\partial =\int_\partial d^2x\, \bigl(\partial_\perp\bar\phi\,\delta\phi +\delta\bar\phi\,\partial_\perp\phi\bigr). \tag{D06.1}$$

Here $\bar\phi=\phi^*$ on that contour. Dirichlet fixes $\phi|_\partial=0$ and hence $\delta\phi|_\partial=0$; Neumann leaves its boundary value free and imposes $\partial_\perp\phi|_\partial=0$. Each cancels (D06.1). Neither statement yet specifies the fermion or proves supersymmetry.

For that purpose use the Lorentzian boundary algebra of Dimofte–Gaiotto–Paquette, (2.1)–(2.2). The pair $Q_+,\bar Q_+$ obeys $\{Q_+,\bar Q_+\}=-2P_+$, where $P_+=P_0+P_1$ is tangential. It is a two-real-charge, two-dimensional N=(0,2) subalgebra. The other mixed anticommutators contain $P_\perp$ and cannot all remain symmetries of a fixed boundary. This count refers to the flat boundary problem, not a curved-background Killing-spinor count.

Introduce one odd coordinate $\theta$ and its conjugate $\bar\theta$. A useful local representation, with left derivatives and $P_+=i\partial_+$, is

$$Q=\partial_\theta+i\bar\theta\partial_+, \qquad \bar Q=-\partial_{\bar\theta}-i\theta\partial_+, \qquad Q^2=\bar Q^2=0,\quad\{Q,\bar Q\}=-2i\partial_+. \tag{D06.2}$$

A boundary chiral multiplet has an even lowest component, while a Fermi multiplet has an odd one. The bulk chiral splits into these two kinds of boundary data:

$$\Phi=\phi+\theta\psi_+-i\theta\bar\theta\partial_+\phi, \qquad \Psi=\eta+\theta f-i\theta\bar\theta\partial_+\eta. \tag{D06.3}$$

In the source’s boundary basis $\eta$ is $\bar\psi_-$, and the free equations identify $f$ with $\partial_\perp\bar\phi$ up to the fixed component normalization. Thus

Boundary condition	Whole multiplet set to zero	Component consequence
Neumann, N	$\Psi\rvert_\partial=0$	$\psi_-=0$, $\partial_\perp\phi=0$
Dirichlet, D	$\Phi\rvert_\partial=0$	$\phi=0$, $\psi_+=0$

Applying either differential in (D06.2) to a multiplet that vanishes identically along the boundary gives zero. This checks the completion, including tangential derivatives. These are boundary-adapted variables; $\eta$ is not an unexplained identification with the independent tilded column of the curved Euclidean action. The source itself suppresses some numerical factors in its bulk superspace shorthand. We use it for this multiplet structure, while (D06.1) fixes the scalar kinetic normalization directly.

There is also a short variational check of the fermionic data. Choose the normal, first-order superfield term to be $\int dx^\perp d\theta\,\Psi\partial_\perp\Phi$, with $\int d\theta\,\theta=1$. Its boundary variation is $\int d\theta\,\Psi\delta\Phi$. Subtracting $\int_\partial d\theta\,\Psi\Phi$ replaces it by $-\int d\theta\,\delta\Psi\,\Phi$. The two choices therefore select N and D respectively. In components the first expression is $f\delta\phi-\eta\delta\psi_+$: its minus sign follows from moving $\theta$ past the odd $\eta$. This is the boundary polarization in BC (2.13)–(2.17), not an additional field equation imposed on a fixed background auxiliary.

Flip the boundary condition

Add a two-dimensional Fermi multiplet $\Gamma$, with a positive physical kinetic term, to the N boundary. Couple it through the boundary superpotential

$$\int_\partial d^2x\,d\theta\,\Gamma J(\Phi)+\text{conjugate}, \qquad J(\Phi)=\lambda\Phi. \tag{D06.4}$$

The phase multiplying this term in BC (2.26) is absorbed into $\Gamma$ and $\lambda$. The additional Fermi constraint, usually called its $E$ term, is zero. Hence the condition $EJ=0$ for this zero-bulk-superpotential problem holds. The top component of a chiral superfield varies into a tangential derivative under (D06.2), so (D06.4) preserves both boundary charges. A nonzero bulk superpotential would require an additional cancellation; the same boundary rule cannot simply be transplanted to it.

Normalize $\lambda$ so that eliminating the boundary Fermi auxiliary gives $|J|^2$. Since $[\phi]=1/2$ and a two-dimensional Fermi lowest component has dimension $1/2$, $[\lambda]=1/2$. The scalar action gains $\kappa|\phi|^2$ at the boundary, where $\kappa=|\lambda|^2$ has dimension one. Its variation gives

$$(\partial_\perp+\kappa)\phi|_\partial=0. \tag{D06.5}$$

This statement uses the physical auxiliary elimination and its continued contour, not integration of a negative Euclidean auxiliary quadratic form over a positive real contour. At finite coupling (D06.5) is a Robin condition, accompanied by boundary fermion dynamics. At energies small compared with $\kappa$, the coupled combination approaches D. The superfield boundary equations identify the surviving $\Psi$ with $\Gamma$ and set $J=0$ in this low-energy limit. This is the known N-to-D flip of BC (2.26)–(2.27).

For a direct check, put $y=-x^\perp\geq0$ and send a scalar wave toward the boundary:

$$u(y)=e^{-ipy}+\mathcal R_\kappa(p)e^{ipy}, \qquad u'(0)=\kappa u(0) \quad\Longrightarrow\quad \mathcal R_\kappa(p)=\frac{ip+\kappa}{ip-\kappa}. \tag{D06.6}$$

For real $p>0$ its modulus is one. It tends to $+1$ when $p/\kappa\to\infty$ and $-1$ when $p/\kappa\to0$, the N and D reflection signs. Thus the flip is an infrared statement; it is not a replacement of the finite-coupling boundary action by Dirichlet at every frequency.

Check the anomaly before counting

A boundary fermion can have an anomalous response to background gauge fields. This is a global ‘t Hooft anomaly: it constrains coupling to background sources and matching under a flow. An uncancelled anomaly for a dynamical gauge transformation instead obstructs gauging. Our example has only a global flavor $U(1)$; we perform no gauge integral or flux sum.

Use the anomaly normalization of BC (2.41)–(2.46): a unit-charge left-moving complex fermion contributes $+\mathfrak f^2$, a right-moving one contributes $-\mathfrak f^2$. The symbols $\mathfrak f,\mathfrak r$ are formal flavor and R-curvature variables in that normalization; products mean the formal four-form polynomial, not a four-form integrated over the two-dimensional boundary. For scalar flavor charge $+1$ and R-charge $r$, define $b=\mathfrak f+(r-1)\mathfrak r$. Then

$$\mathcal A_{\rm N}=-\tfrac12 b^2, \qquad \mathcal A_{\rm D}=+\tfrac12 b^2, \qquad \mathcal A_\Gamma=[-\mathfrak f+(1-r)\mathfrak r]^2=b^2. \tag{D06.7}$$

The Fermi charges $(-1,1-r)$ make $\Gamma\Phi$ flavor neutral and R-charge one, as required for $d\theta$. Consequently $\mathcal A_{\rm N}+\mathcal A_\Gamma=\mathcal A_{\rm D}$. Use the same bulk background counterterms on both sides. Adding a common allowed contact term changes the bookkeeping but not this difference. Equation (D06.7) displays only flavor/R anomalies; it is not a calculation of gravitational anomalies or of every global anomaly.

Reproduce the normalized half-index

A half-index counts boundary local operators in $\bar Q$ cohomology: $\bar Q$-closed operators modulo $\bar Q$-exact ones. Write its character as $\operatorname{Tr}(-1)^F q^{J+R_0/2}X^e$, where $J$ is boundary rotation, $e$ flavor charge, and $R_0(\phi)=0$ is an integral reference R-charge. It is not the superconformal value $1/2$ of the free bulk scalar. The source uses $(-1)^{R_0}$; it agrees with $(-1)^F$ on all the surviving letters below. For a general integral assignment the source’s footnote 12 requires a sign change of $q^{1/2}$. Do not insert fractional charges into $(-1)^{R_0}$ without specifying a convention.

We normalize the empty operator to one and include no extra background contact factor. The derivative $\partial_z$ has weight $q$. The closed-sphere index in the prerequisite used a variable $x$ in which this derivative had weight $x^2$; this grading comparison sets $q=x^2$, but does not identify the two geometries or their gluing prescriptions.

The boundary multiplets leave the following free letters, with $n\geq0$:

Boundary data	Statistics	Weight before the fermion sign
N: $\partial_z^n\phi$	bosonic	$q^nX$
D: $\partial_z^n\bar\psi_-$	fermionic	$q^{n+1}/X$
Added $\Gamma$: $\partial_z^n\bar\gamma$, $\partial_z^n\gamma$	fermionic	$q^nX$, $q^{n+1}/X$

A bosonic letter of weight $w$ contributes $\sum_{j\geq0}w^j=(1-w)^{-1}$; an odd letter contributes $1-w$. With $(a;q)_\infty=\prod_{n\geq0}(1-aq^n)$, we obtain

$$\mathbb I_{\rm N}=\frac1{(X;q)_\infty}, \qquad \mathbb I_{\rm D}=(q/X;q)_\infty, \qquad \mathbb F=(X;q)_\infty(q/X;q)_\infty. \tag{D06.8}$$

Therefore the fully normalized reproduction is

$$\boxed{\mathbb I_{\rm N}(X;q)\,\mathbb F(X;q) =\mathbb I_{\rm D}(X;q).} \tag{D06.9}$$

This is BC (3.17). There is no missing $(q;q)_\infty$, Weyl factor, or gauge volume in this ungauged example. Analytically we can work in $0<|q|<|X|<1$, where every geometric expansion used above converges. Elsewhere the products have meromorphic continuations, but a changed series expansion needs justification.

An independent way to compute the two bulk factors uses occupation number rather than mode products:

$$\mathbb I_{\rm N}=\sum_{j\geq0}\frac{X^j}{(q;q)_j}, \qquad \mathbb I_{\rm D}=\sum_{j\geq0} \frac{(-1)^j q^{j(j+1)/2}X^{-j}}{(q;q)_j}. \tag{D06.10}$$

For example, multiplying the first series by $1-X$ shifts $X$ to $qX$, reproducing the product recurrence and its value one at $X=0$. For the second, selecting $j$ distinct fermion modes explains the minimal exponent $1+\cdots+j$ and the remaining partition factor. These checks diagnose a misplaced fermion exponent or an extra normalization. Truncating $\log(a;q)_\infty$ after $L$ mode factors gives the explicit tail bound

$$\left|\sum_{n=L}^{\infty}\log(1-aq^n)\right| \leq\frac{|a||q|^L}{(1-|q|)(1-|a||q|^L)}. \tag{D06.11}$$

It follows by expanding $\log(1-z)$ and summing absolute values. Increasing arithmetic precision alone does not remove this truncation error. For a numerical starting point, $q=1/5$ and $X=1/\sqrt5$ give $\mathbb I_{\rm N}=2.031981381419196\ldots$ and $\mathbb I_{\rm D}=0.4921304934898421\ldots$. Ninety-seven product factors make the bound (D06.11) smaller than $10^{-68}$ for each product here; (D06.10) supplies a separate occupation-sum comparison.

Change the coupling, then ask what the index missed

Take two identical free bulk chirals with N conditions and one boundary Fermi of flavor charge $-1$, and change $J$ to $\lambda_1\Phi_1+\lambda_2\Phi_2$. Keep only the common flavor fugacity $X$ in the count. For $L=\sqrt{|\lambda_1|^2+|\lambda_2|^2}>0$, the unitary change of fields

$$\chi=\frac{\lambda_1\phi_1+\lambda_2\phi_2}{L}, \qquad \xi=\frac{-\bar\lambda_2\phi_1+\bar\lambda_1\phi_2}{L} \tag{D06.12}$$

preserves the canonical kinetic term. The coupled $\chi$ has $\kappa=L^2$, while the orthogonal scalar remains N. At low energy the conditions are D on $\chi$ and N on the orthogonal combination. Their flavor/R anomaly is zero, matching $2(-b^2/2)+b^2$ in the starting description, and their index is $\mathbb I_{\rm N}^2\mathbb F=\mathbb I_{\rm N}\mathbb I_{\rm D}$.

At $\lambda_1=\lambda_2=0$, however, both bulk fields remain N and the Fermi is decoupled. The same product still computes the graded count. To see what changed, restrict momentarily to the polynomial complex generated by the scalar zero-derivative letters and $\bar\gamma$:

$$d\bar\gamma=\lambda_1\phi_1+\lambda_2\phi_2, \qquad d\phi_i=0,\qquad d^2=0. \tag{D06.13}$$

At total degree $n\geq1$, assign degree one to each generator. There are $n+1$ even monomials and $n$ odd monomials. Multiplication by a nonzero linear polynomial is injective, so the cohomology has one even direction and no odd direction, represented by the orthogonal scalar to the power $n$. At zero coupling every one of those monomials survives. Their Euler difference is nevertheless $(n+1)-n=1$. This restricted complex isolates the cancellation; it does not discard the other fermion and derivative letters from the full index.

The reflection experiment detects the difference: taking $p\to0$ at fixed $L>0$ gives $-1$, whereas setting $L=0$ first gives $+1$. A matching index therefore supports the flip but cannot classify all boundary conditions. The loss of a scale at the rank-zero coupling is a controlled training example of this limitation.

From an interface to a gluing calculation

An interface separates two bulk regions. Folding one region reverses its normal and turns an interface problem into boundary data for a product theory. In the free example, the complementary N and D data can be joined by a quadratic boundary superpotential so that both chiral and conjugate boundary data match. This is the identity-interface construction of BC (5.1)–(5.2), not a new interface duality.

Gluing partition functions also requires a pairing. If boundary states $|B_\alpha\rangle$ have a nonsingular Gram matrix $G_{\alpha\beta}=\langle B_\alpha|B_\beta\rangle$ on a chosen complete subspace, its identity operator is

$$\mathbf1=\sum_{\alpha,\beta}|B_\alpha\rangle (G^{-1})_{\alpha\beta}\langle B_\beta|. \tag{D06.14}$$

Replacing $G^{-1}$ by a Kronecker delta assumes an orthonormal basis and can change an answer. Protected field-theory pairings can be complex bilinear forms rather than positive Hermitian norms; their precise contraction must come from the chosen path integral. They may involve continuous holonomies, flux sectors, a contour, and anomaly-cancelling boundary matter. None of these data follows from merely multiplying (D06.8).

Literature checked 20 October 2026. BC v2 supplies the boundary flip, anomaly matching, and identity interface reproduced above. Its section 6 is a next reading for the already familiar SQED–XYZ pair, with the boundary choices retained. Zhao–Betzios–Roehl, IF v2, (1.1)–(1.3), proves factorizations in specified SQED and Chern–Simons–Yang–Mills examples. Its SQED construction uses topologically twisted amplitudes. Its section 3 instead uses a rigid hemisphere background and explicitly different boundary spin data. Neither prescription can be substituted for our half-index without a dictionary. Wilson-loop-labelled gluing factors are additional data; section 4 separates the demonstrated cases from suggested extensions.

A bounded next project is to reproduce one of IF’s specified finite-rank pairings, first recording its twist, boundary states, line labels, and integration measure. Start by recovering its stated result before modifying a boundary coupling. We make no claim that such a modification is unexplored. A disagreement in this first reproduction calls for checking orientation, anomaly inflow, normalization, and convergence before interpreting it as new physics.

Exercises

1. A scalar condition is not a full boundary condition

Suppose someone sets only $\phi|_\partial=0$ and leaves $\psi_+|_\partial$ arbitrary. Test the preserved algebra. Then show that choosing a negative Robin coefficient $\kappa=-a<0$ produces a normalizable negative eigenvalue of $-\partial_y^2$ on $y\geq0$. Explain why this does not arise from the positive-kinetic Fermi coupling above.

Hint 1

Read the lowest-component $Q$ transformation from (D06.3).

Hint 2

Try $u(y)=e^{-ay}$ in the Robin condition and normalize its squared norm.

Solution

$Q\phi=\psi_+$, so the proposed scalar-only condition is not stable unless $\psi_+=0$ as well. Then $\bar Q\psi_+=-2i\partial_+\phi$ also vanishes along the boundary. The full $\Phi=0$ condition is closed under both charges.

For $a>0$, $u=\sqrt{2a}\,e^{-ay}$ has unit norm and $u'(0)=-a u(0)=\kappa u(0)$. It obeys $-u''=-a^2u$. The scalar quadratic form is $\int_0^\infty|u'|^2dy+\kappa|u(0)|^2=-a^2$, so the negative sign is a genuine instability of this trial scalar boundary action. The supersymmetric coupling used here gives $\kappa=|\lambda|^2\geq0$; no such mode exists. This check does not classify more general supersymmetric boundary theories.

2. Keep the symmetries that the interaction actually preserves

Assign separate flavor rotations to $\Phi_1$ and $\Phi_2$. Can fixed nonzero $\lambda_1,\lambda_2$ and a single Fermi preserve both? For $\lambda_1=1$, $\lambda_2=i$ in chosen units, find the coupled and uncoupled fields, their boundary scales, and the degree-two cohomology dimensions of (D06.13). Compare with zero coupling.

Hint 1

Each term in $\Gamma(\lambda_1\Phi_1+\lambda_2\Phi_2)$ must have the same zero flavor charge.

Hint 2

At degree two, the differential maps two odd monomials into three even ones. A nonzero linear polynomial has no annihilator in a polynomial ring.

Solution

The first term would require Fermi charge $(-1,0)$ and the second $(0,-1)$ under independent rotations. A single field cannot have both charges. With fixed couplings the common diagonal $U(1)$ remains; independent fugacities would require transforming the couplings as spurions, a different count.

Here $L=\sqrt2$, $\chi=(\phi_1+i\phi_2)/\sqrt2$ and the orthogonal scalar is $(i\phi_1+\phi_2)/\sqrt2$. Their Robin coefficients are $2$ and $0$. At degree two, the map sends $\bar\gamma\phi_1$ and $\bar\gamma\phi_2$ to $\phi_1^2+i\phi_1\phi_2$ and $\phi_1\phi_2+i\phi_2^2$. These are independent, so the cohomology dimensions are $(1,0)$ in even/odd parity. At zero coupling they are $(3,2)$. Both have Euler difference one, while their state spaces and reflection data differ.

3. Glue in a nonorthogonal basis

In an ordinary two-dimensional Hilbert space take $B_1=e_1$ and $B_2=\tfrac12e_1+\tfrac{\sqrt3}{2}e_2$, with orthonormal $e_1,e_2$. Compute the Gram matrix and use it to recover the squared norm of $v=e_1+e_2$ from $h_\alpha=\langle B_\alpha|v\rangle$. What does the naive expression $\sum_\alpha|h_\alpha|^2$ give? Explain the problem as $B_2$ approaches $B_1$.

Hint 1

The appropriate scalar is $h^\dagger G^{-1}h$.

Hint 2

Replace $1/2$ temporarily by $s$, with $B_2=s e_1+\sqrt{1-s^2}e_2$, and examine $\det G$.

Solution

The matrices and overlaps are

$$G=\begin{pmatrix}1&1/2\\1/2&1\end{pmatrix}, \qquad G^{-1}=\frac43\begin{pmatrix}1&-1/2\\-1/2&1\end{pmatrix}, \qquad h=\begin{pmatrix}1\\(1+\sqrt3)/2\end{pmatrix}.$$

Substitution gives $h^\dagger G^{-1}h=2$, the correct norm. The naive sum gives $2+\sqrt3/2$. In the varying family, $\det G=1-s^2$. At $s=1$ the two vectors no longer span the Hilbert space and the inverse does not exist; discarding the missing direction is not the identity on the original space. An actual field-theory gluing formula needs an appropriate complete basis and a controlled inverse pairing. This finite-dimensional exercise verifies that algebraic requirement, not a universal factorization theorem.