Fermionic loops, superconnections, and framing

The bosonic supersymmetric loop cancels the variation of a gauge connection against a scalar coupling. Can a loop preserve more supersymmetry by also coupling to matter fermions? Yes, but its connection becomes a matrix with Grassmann-odd off-diagonal entries. Its variation need only be a covariant derivative along the path. Closing that derivative around a circle then becomes part of the physics.

Keep compact $U(N)_k\times U(N)_{-k}$ with equal ranks, no additional center quotient, and ordinary fundamental traces normalized as in the preceding lesson. We discuss the undeformed conformal theory and take $k>0$. Its generic full algebra has twelve real Poincaré and twelve real conformal supercharges. The bosonic circle preserves four combinations; the fermionic circle below preserves twelve and is called one-half BPS. The manifest $\mathcal N=2$ description and the single charge used in localization are smaller choices inside that algebra. These fractions refer to the generic $\mathcal N=6$ theory; exceptional-level enhancement requires a separate count.

A connection that can change the node

It is useful to carry out the local component calculation in Drukker–Trancanelli’s explicit convention. In this section, $C_I$ and $\bar\psi^I$ map the second color space to the first, while $\bar C^I$ and $\psi_I$ map the first to the second. Here $I=1,2,3,4$ is a flavor index. Bars name the source’s fields; Euclidean variations do not require imposing conjugacy on independent integration variables.

Write $g=2\pi/k$, and use a unit circle $x^1=\cos\tau$, $x^2=\sin\tau$, $0\leq\tau\leq2\pi$. The source connection uses $M=\operatorname{diag}(-1,1,1,1)$:

$$L_F=\begin{pmatrix} \mathcal A&-i\sqrt g\,\eta^\alpha\bar\psi^1_\alpha\\ -i\sqrt g\,\psi_1^\alpha\bar\eta_\alpha&\widehat{\mathcal A} \end{pmatrix},\qquad \begin{aligned} \mathcal A&=A_\mu\dot x^\mu-i g\,M^I{}_J C_I\bar C^J,\\ \widehat{\mathcal A}&=\widehat A_\mu\dot x^\mu-i g\,M^I{}_J\bar C^J C_I, \end{aligned} . \tag{AB16.1}$$

The holonomy uses $\exp(i\int L_F d\tau)$. Thus both its scalar terms and its two fermionic terms have positive coefficients after multiplication by $i$. The spinor couplings $\eta,\bar\eta$ are commuting; the matter fermions are odd. Each off-diagonal block has the correct bifundamental color type. This is an even connection on a vector space with two graded summands, often called a $U(N|N)$ superconnection. It does not enlarge the microscopic gauge group or add a fermionic gauge field.

The three-dimensional dimensions are $[A]=1$, $[C]=1/2$, $[\psi]=1$ and $[k]=0$. For a general parametrization, multiply the scalar and fermionic terms by $v=|\dot x|$. Every term in the integrated connection is then dimensionless.

There are two important translations. DT uses $D C=\partial C+i(A C-C\widehat A)$ and orders earlier path factors on the left, as its (2.11) shows. The preceding lesson uses $\partial-i a$ and later factors on the left. Set $A=-a$, $\widehat A=-\widehat a$ and reverse the path parameter when converting the source holonomy: $x_D(u)=x(2\pi-u)$. The two reversals give the site’s $+i a\cdot\dot x$ and its ordering. Scalar length terms keep their sign. Finally the even flavor permutation

$$(C_1,C_2,C_3,C_4)=(Y^3,Y^4,Y^1,Y^2) \tag{AB16.2}$$

maps DT’s bosonic coupling $\operatorname{diag}(-1,-1,1,1)$ to the preceding lesson’s $\operatorname{diag}(1,1,-1,-1)$. These are changes of notation and path convention; we retain the already fixed positive-first-node matrix-integral phase. In particular, source spinor components below are not the three-dimensional lower-column gamma convention used in the bosonic lesson.

Check the off-diagonal variation

The needed source spinors are

$$\eta=(e^{i\tau/2},-i e^{-i\tau/2}),\qquad \bar\eta=\begin{pmatrix}i e^{-i\tau/2}\\-e^{i\tau/2}\end{pmatrix}. \tag{AB16.3}$$

They obey $\eta\bar\eta=2i$, not a unit-norm condition. Direct differentiation gives

$$\eta'=\frac i2\eta\sigma_3,\qquad \bar\eta'=-\frac i2\sigma_3\bar\eta,\qquad \eta(2\pi)=-\eta(0),\quad\bar\eta(2\pi)=-\bar\eta(0). \tag{AB16.4}$$

Here $\sigma_3=\operatorname{diag}(1,-1)$. The half-angle is fixed by supersymmetry, and its last minus sign will matter globally.

For the family of preserved parameters $\bar\theta^{1I}$, the projected component variation in DT, (2.25) is

$$\delta(\eta\bar\psi^1) =4i\eta\bar\theta^{1I}\mathcal D_\tau C_I -2\eta\sigma_3\bar\theta^{1I}C_I, \qquad \mathcal D_\tau C=\partial_\tau C+i(\mathcal A C-C\widehat{\mathcal A}). \tag{AB16.5}$$

Repeated $I=2,3,4$ is summed. This is an ordinary even variation with an odd constant parameter $\bar\theta$, including its related conformal parameter. It is different bookkeeping from an odd differential with a commuting label. The second term arises from the conformal part of the circle’s preserved charge. The cubic matter terms have combined with the gauge derivative into the modified connection: explicitly, its scalar contribution is $g(M C\bar C)C-g C(M\bar C C)$, with the two color orders retained.

Now use (AB16.4), without dropping that conformal term:

$$\begin{aligned} 4i\mathcal D_\tau(\eta\bar\theta^{1I}C_I) &=4i\eta\bar\theta^{1I}\mathcal D_\tau C_I +4i\eta'\bar\theta^{1I}C_I\\ &=4i\eta\bar\theta^{1I}\mathcal D_\tau C_I -2\eta\sigma_3\bar\theta^{1I}C_I. \end{aligned} \tag{AB16.6}$$

Consequently the upper-right entry varies as $\delta(L_F)_{12}=4\sqrt g\,\mathcal D_\tau(\eta\bar\theta^{1I}C_I)$. A constant spinor coupling would leave the second term of (AB16.5) uncancelled. A naive condition $\delta L_F=0$ would also miss this loop: its variation is a covariant derivative, whose neighboring insertions cancel inside the ordered exponential. DT (2.29) supplies the matching diagonal and lower-left variations. When an odd parameter is moved through an odd matter insertion, its minus sign is essential to that cancellation.

Close the circle with the correct trace

Return to later-left transport notation $H'=K H$, $H(0)=1$. Let $G$ denote the off-diagonal supersymmetry gauge parameter, including the odd variation parameter. The component cancellations take the form

$$\delta K=G'+[G,K],\qquad \delta H(T)=G(T)H(T)-H(T)G(0). \tag{AB16.7}$$

To check the second identity, differentiate $G H-HG(0)$ and use the first. This works without commuting color matrices. Both $K$ and $G$ are even as graded matrices: their diagonal entries are even and their off-diagonal entries are odd. The bracket in (AB16.7) is the ordinary matrix commutator with Grassmann-valued entries.

Define the constant grading matrix and the supertrace by

$$\Gamma=\begin{pmatrix}1_N&0\\0&-1_N\end{pmatrix},\qquad \operatorname{sTr}\begin{pmatrix}A&B\\C&D\end{pmatrix} =\operatorname{Tr}A-\operatorname{Tr}D. \tag{AB16.8}$$

In the Cartesian frame used for the circle, the matter fields themselves return to their initial values. The anti-periodic couplings therefore give $K(T)=\Gamma K(0)\Gamma$ and $G(T)=\Gamma G(0)\Gamma$, where $T=2\pi$. Supertrace is cyclic for even graded matrices; ordinary trace is generally not cyclic when their odd entries are interchanged. Hence the single-winding fundamental observable is

$$W_F=\frac1{2N}\operatorname{sTr}(\Gamma H(T)) =\frac1{2N}\operatorname{Tr}H(T). \tag{AB16.9}$$

Indeed,

$$\begin{aligned} 2N\,\delta W_F &=\operatorname{sTr}\bigl(G(0)\Gamma H-\Gamma H G(0)\bigr)=0. \end{aligned} \tag{AB16.10}$$

The first equality uses $\Gamma^2=1$ and the endpoint condition. Omitting $\Gamma$ would instead subtract the two diagonal blocks and would not cancel the endpoint term. The twist is a transition in the graded description of this operator, not a change of spacetime spin structure. Equivalently, a basis change $R(\tau)=\operatorname{diag}(1_N,e^{i\tau/2}1_N)$ makes the connection periodic; its holonomy is $\Gamma H$ and its connection includes the compensating term $-R^{-1}R'$. Keeping that term gives the same observable. AGG, section 2.2 states this twist explicitly.

Compare the two operators at a fixed framing

The fermionic connection changes a scalar coupling as well as adding fermions. In the temporary DT convention, let $L_B$ be the block-diagonal bosonic connection with $M_B=\operatorname{diag}(-1,-1,1,1)$. Then

$$L_F-L_B=B+F,\qquad B=-\frac{4\pi i}{k}\begin{pmatrix}C_2\bar C^2&0\\0&\bar C^2C_2\end{pmatrix}, \qquad \Lambda=i\sqrt{\frac{\pi}{2k}}e^{i\tau/2} \begin{pmatrix}0&C_2\\\bar C^2&0\end{pmatrix}. \tag{AB16.11}$$

Here $F$ means the off-diagonal part of the loop connection, not a matter auxiliary field. A shared odd charge $Q$, including Poincaré and conformal pieces, can be normalized so that $Q\Lambda=F$ and $QF=-8\mathcal D_\tau(e^{-i\tau}\Lambda)$. Two ingredients of the ordered-integral argument can be checked immediately:

$$8i e^{-i\tau}\Lambda^2=B,\qquad (C_2\bar C^2)C_2-C_2(\bar C^2C_2)=0. \tag{AB16.12}$$

The first identity recovers the precise changed scalar coupling; the second shows that the derivative on $\Lambda$ is the same whether computed with $L_B$ or $L_F$‘s diagonal blocks. Thus integrations by parts produce the required scalar insertions when two ordered insertion points meet. Fermions cannot simply be set to zero in the original operator to obtain the bosonic one.

The resulting shared-charge relation is $W_F-(W_L+W_R)/2=Q\mathcal V$, with normalization absorbed into $\mathcal V$. DT (3.9)–(3.15) displays the local identities and the first orders of its circle expansion; AGG (2.12) gives the cohomological relation used in localization. Using it for expectation values requires a common supersymmetric regulator and no uncancelled boundary contribution in the field integral. It is not an equality of the two operators on every field configuration.

We use the preceding lesson’s supersymmetric Hopf framing, linking number $+1$ in the stated sphere orientation. A framing displaces one copy of the loop along a chosen normal field before regulating coincident contractions. The integer linking number specifies geometric data; in a matter theory it does not, by itself, supply an exact formula converting all observables to another regulator. DT’s positive-first-node matrix integral (3.16), and the framing-one paragraph following (4.1), match our prescription. No pure Chern–Simons phase is being asserted as a universal ABJM framing conversion.

In this prescription, with $s_i=\ell\sigma_i$ and $t_i=\ell\widehat\sigma_i$,

$$\langle W_F\rangle_1 =\left\langle\frac1{2N}\left(\sum_{i=1}^N e^{2\pi s_i} +\sum_{i=1}^N e^{2\pi t_i}\right)\right\rangle =\frac{\langle W_L\rangle_1+\langle W_R\rangle_1}{2}. \tag{AB16.13}$$

The average retains the phase $\exp[i\pi k(\sum s_i^2-\sum t_i^2)]$ and both Weyl factors. The normalization makes the insertion equal one at $s=t=0$. Supergroup notation has reorganized the observable, not replaced the ABJM path integral by pure supergroup Chern–Simons theory.

A rank-one comparison with a convergence domain

For $N=1$ and integer $k>2$, the preceding lesson proved, using a positive Gaussian Abel regulator,

$$\langle W_L\rangle_1=e^{i\pi/k}\sec^2(\pi/k),\qquad \langle W_R\rangle_1=e^{-i\pi/k}\sec^2(\pi/k). \tag{AB16.14}$$

Their normalized sum therefore gives

$$\boxed{\langle W_F\rangle_1=\sec(\pi/k)} =1+\frac{\pi^2}{2k^2}+\frac{5\pi^4}{24k^4}+O(k^{-6}). \tag{AB16.15}$$

At $k=4$, the two node averages are $\sqrt2(1+i)$ and $\sqrt2(1-i)$, while the fermionic result is $\sqrt2$. This is not the modulus of a node average, which is $2$. It is also not a proof of framing independence: it is a real answer in one declared prescription. The $k^{-2}$ term agrees with the equal-rank-one specialization of DT (4.1).

Linearity is legitimate here at finite positive regulator, and the two limits were proved for $k>2$. The argument does not establish the same literal integral at $k=1,2$. Analytic continuation, if desired there, is an additional prescription rather than a consequence of the displayed Abel limit.

Exercises

1. Change the half-angle

Replace the first component of $\eta$ by $e^{i a\tau}$ and the second by $-i e^{-i a\tau}$, with real $a$. Keep the projected field variation (AB16.5). Determine which $a$ turns it into $4i\mathcal D_\tau(\eta\bar\theta C)$. Then explain why choosing an integer $a$ just to make the coupling periodic fails.

Hint 1

Differentiate the proposed row before considering its endpoint value.

Hint 2

The coefficient of $\eta\sigma_3\bar\theta C$ is fixed independently by the conformal supersymmetry transformation.

Solution

The changed row obeys $\eta'=i a\eta\sigma_3$, so its derivative contribution is $4i\eta'\bar\theta C=-4a\eta\sigma_3\bar\theta C$. Matching the coefficient $-2$ in (AB16.5) gives $a=1/2$. An integer $a$ would remove the anti-periodicity but leave a residual coefficient $4a-2$ when subtracting the proposed total derivative from the actual variation. The endpoint problem cannot be solved by changing a coupling that the local supersymmetry equation has already fixed. At $a=1/2$ the row returns with a minus sign, and (AB16.9) supplies the needed twist.

2. Find the missing boundary sign

Work with one-dimensional color blocks. Let $\rho$ and $\chi$ be independent Grassmann generators, and take $G(0)=\left(\begin{smallmatrix}0&\rho\\0&0\end{smallmatrix}\right)$ and $H=\left(\begin{smallmatrix}1&0\\\chi&1\end{smallmatrix}\right)$. Use $G(T)=\Gamma G(0)\Gamma$ to calculate $\delta H$, $\delta\operatorname{Tr}H$, and $\delta\operatorname{sTr}H$. What goes wrong if the entries are treated as commuting numbers?

Hint 1

The upper-right block of $G$ changes sign across the seam.

Hint 2

Keep the product order until after subtracting $HG(0)$, then use $\chi\rho=-\rho\chi$.

Solution

Since $G(T)=-G(0)$,

$$\delta H=-G(0)H-HG(0) =\begin{pmatrix}-\rho\chi&-2\rho\\0&-\chi\rho\end{pmatrix} =\begin{pmatrix}-\rho\chi&-2\rho\\0&\rho\chi\end{pmatrix}.$$

Thus $\delta\operatorname{Tr}H=0$, whereas $\delta\operatorname{sTr}H=-2\rho\chi\ne0$. Equivalently $\delta\operatorname{sTr}(\Gamma H)=0$. Commuting the two odd entries would incorrectly give a nonzero ordinary-trace variation. This small example tests both the seam and the Grassmann sign; it is a diagnostic of the holonomy argument, not a replacement for the ABJM field transformations.

3. Traverse the circle twice

For this exercise define the winding-$p$ fermionic operator with fixed normalization $1/(2N)$ and twist $\Gamma^p$, rather than renormalizing its zero-field value. Derive its localized insertion. At $N=1$, use the inherited Abel-limit node formula $\langle W_L^{(p)}\rangle_1=e^{i\pi p^2/k}\sec^2(\pi p/k)$ and its complex conjugate, valid for positive integer $k>2p$. Evaluate the fermionic result at $p=2,k=8$. Why should it not be forced to equal one at zero fields?

Hint 1

Each traversal contributes one endpoint sign. Apply the definition of supertrace after multiplying by $\Gamma^p$.

Hint 2

For even winding the two node terms are subtracted. The gauge phase depends on $p^2$, while the matter factor depends on $p$.

Solution

The insertion is

$$W_F^{(p)}\longmapsto\frac1{2N} \left(\sum_i e^{2\pi p s_i}-(-1)^p\sum_i e^{2\pi p t_i}\right).$$

This is the winding sign in AGG (2.13), with our explicitly fixed normalization. For rank one it gives $\cos(\pi p^2/k)\sec^2(\pi p/k)$ at odd $p$, and $i\sin(\pi p^2/k)\sec^2(\pi p/k)$ at even $p$. At $p=2,k=8$, the result is $2i$ and the convergence condition $8>4$ holds. At zero fields the even-winding insertion is zero: it is a graded difference of equal-dimensional blocks. Dividing by that zero is not a normalization convention. Nor is this operator the square of the single-winding traced loop; the path ordering and trace are performed before taking its expectation value.

The practical distinction is now visible in the operator itself: a fermionic loop uses odd off-diagonal couplings, a supersymmetry variation that is a covariant derivative, and a trace compatible with the circle’s seam. Its computable expectation value still requires the specified framing and integration prescription.

The defect-observable project follows a real mass deformation of this framed loop and extracts its prescribed Bremsstrahlung response from the complex phase.