Euclidean continuation and independent spinors

In Lorentzian field theory, a barred fermion often means the Dirac adjoint of another field. In a Euclidean supersymmetry calculation, a second spinor is usually an independent variable. What changed, and how can we tell whether a familiar conjugation rule is still allowed?

We will continue a free fermion kinetic term and its vector bilinear, keeping every factor of $i$ visible. Then an explicit rotation will show why carrying the Lorentzian adjoint unchanged into Euclidean space fails. The necessary preparation is Spinors in three dimensions. The general distinction between independent fields and a reality prescription also applies in four dimensions; the explicit matrices below are three-dimensional and do not define the four-dimensional spinor conventions.

Fix the continuation before moving an index

Use units $\hbar=c=1$. Start with flat three-dimensional Lorentzian coordinates $(t,x,y)$, metric $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1)$, and a gauge-neutral Dirac fermion $\psi_L$. It has two complex components. This is a calculation of fermion kinematics, not a complete supersymmetric model; no supercharge algebra is being asserted for this fermion by itself.

Use the Euclidean matrices already introduced,

$$\gamma^1=\sigma_3,\qquad \gamma^2=-\sigma_1,\qquad \gamma^3=-\sigma_2,$$

which obey $\{\gamma^a,\gamma^b\}=2\delta^{ab}$. Our Lorentzian matrices are

$$\Gamma^0=\gamma^1,\qquad \Gamma^1=i\gamma^2,\qquad \Gamma^2=i\gamma^3.$$

The factors of $i$ make $(\Gamma^0)^2=1$ and $(\Gamma^1)^2=(\Gamma^2)^2=-1$. Distinct matrices anticommute, so $\{\Gamma^\mu,\Gamma^\nu\}=2\eta^{\mu\nu}$.

The Lorentzian Dirac adjoint is the row

$$\bar\psi_L=\psi_L^\dagger\Gamma^0.$$

Here the dagger includes transposition and complex conjugation; on products of odd fields it reverses their order. Take a real mass $m$ and the action

$$S_L=\int dt\,dx\,dy\; \bar\psi_L\bigl(i\Gamma^\mu\partial_\mu-m\bigr)\psi_L.$$

Both terms have mass dimension three: $[\psi_L]=1$ and $[m]=[\partial_\mu]=1$. The Grassmann parity of the fermion is independent of the signature.

Continue the kinetic bilinear

Choose $t=-i\tau$. Then

$$dt=-i\,d\tau,\qquad \partial_t=i\partial_\tau.$$

First regard $\psi_L$ and $\bar\psi_L$ as independent variables in the complexified action. Denote their continuations by the column $\psi_E$ and row $\chi_E$. Keeping a separate symbol for the row will prevent an accidental conjugation assumption.

Substitution into the differential operator gives

$$\begin{aligned} i\Gamma^\mu\partial_\mu-m &\longrightarrow i\Gamma^0(i\partial_\tau) +i\Gamma^1\partial_x+i\Gamma^2\partial_y-m\\ &=-\gamma^1\partial_\tau -\gamma^2\partial_x-\gamma^3\partial_y-m. \end{aligned}$$

There are two factors to track: a minus sign from this operator and a factor $-i$ from $dt$. Their product gives

$$S_L\big|_{\rm continued}=iS_E,\qquad S_E=\int d\tau\,dx\,dy\; \chi_E\bigl(\gamma^a\partial_a+m\bigr)\psi_E.$$

Consequently the weight changes from $e^{iS_L}$ to $e^{-S_E}$. The Euclidean coordinate order is $(x_E^1,x_E^2,x_E^3)=(\tau,x,y)$.

This local calculation assumes the usual vacuum continuation of the free theory, with its pole prescription. It is not a proof that every interacting theory or every curved Euclidean background can be obtained by rotating a global time coordinate.

A useful independent check is the free momentum-space kernel. On a mode $e^{ik\cdot x_E}$ it is $m+i\gamma^ak_a$. The Clifford algebra implies

$$(m+i\gamma^ak_a)(m-i\gamma^bk_b) =(m^2+k^2)\mathbf1, \qquad k^2=k_1^2+k_2^2+k_3^2.$$

Thus the inverse has the expected Euclidean denominator:

$$(m+i\gamma\cdot k)^{-1} =\frac{m-i\gamma\cdot k}{m^2+k^2}.$$

The formula applies where the denominator is nonzero. For $m=0$, the zero-momentum mode requires its own treatment.

A vector bilinear keeps track of the factors of i

Consider the Lorentzian vector $j_L^\mu=\bar\psi_L\Gamma^\mu\psi_L$. Its time component is $j_L^0=\psi_L^\dagger\psi_L$. Define the continued Euclidean vector by

$$J_E^a=\chi_E\gamma^a\psi_E.$$

The matrix dictionary immediately gives

$$j_L^0\longrightarrow J_E^1,\qquad j_L^1\longrightarrow iJ_E^2,\qquad j_L^2\longrightarrow iJ_E^3.$$

For a component calculation, write $\psi_E=(u,v)^T$ and $\chi_E=(a,b)$, with all four entries independent odd quantities. Keeping the row factors to the left gives

$$J_E^1=au-bv,\qquad J_E^2=-av-bu,\qquad J_E^3=iav-ibu.$$

In the Lorentzian theory the adjoint row would be $(u^\dagger,-v^\dagger)$, so its time component becomes $u^\dagger u+v^\dagger v$. In Euclidean calculations we do not impose that relation between $(a,b)$ and $(u,v)$.

As another check, the continued divergence is

$$\partial_\mu j_L^\mu \longrightarrow i\bigl(\partial_\tau J_E^1+\partial_xJ_E^2+\partial_yJ_E^3\bigr).$$

A vanishing Lorentzian divergence therefore becomes a vanishing Euclidean divergence. Omitting the spatial factors of $i$ would fail this check.

The row notation has a precise translation into the three-manifold paper’s lower-index spinors. With

$$C=(\epsilon_{\alpha\beta}) =\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad \chi_E=\widetilde\psi_E^{\,T}C,$$

we have $\chi_E\psi_E=\widetilde\psi_E^{\,T}C\psi_E=\widetilde\psi_E\psi_E$ in its contracted notation. This definition makes $\widetilde\psi_E$ an independent column; it does not define it as a conjugate. The contraction and gamma matrices agree with Appendix A.1, equations (A.1)–(A.3), of the paper’s version 3.

Why the old adjoint cannot simply be retained

A Euclidean spin rotation acts as $\psi_E\mapsto U\psi_E$, with $U\in SU(2)$. An independent row must transform as $\chi_E\mapsto\chi_EU^{-1}$ so that $\chi_E\psi_E$ is a scalar.

Suppose instead we imposed the unchanged Lorentzian rule $\chi_E=\psi_E^\dagger\gamma^1$. Reconstructing the row after a rotation would give $\psi_E^\dagger U^\dagger\gamma^1$, while the required transformed row would be $\psi_E^\dagger\gamma^1U^\dagger$. These agree only when $U^\dagger$ commutes with $\gamma^1$, which a general Euclidean rotation does not. The Lorentzian adjoint singled out a time direction; a Euclidean scalar cannot retain that preferred direction.

This does not mean Hermitian conjugation has ceased to exist. The row $\psi_E^\dagger$ is well defined and transforms appropriately under $SU(2)$. It means that a field denoted $\widetilde\psi_E$ or $\chi_E$ in the complexified theory is not automatically equal to that conjugate. Analytic continuation and complex conjugation also act differently on time: conjugating $t=-i\tau$ sends it to $+i\tau$. A Lorentzian adjoint relation is not automatically a pointwise relation at the same Euclidean time.

The representation itself gives a second test. Define the antilinear map

$$\mathcal J_E\psi=C\psi^*.$$

Here $*$ conjugates components. This map commutes with spin rotations: $CU^*=UC$. But

$$\mathcal J_E^2\psi=CC^*\psi=-\psi.$$

Consequently the proposed single-spinor reality condition $\psi=\mathcal J_E\psi$ would force $\psi=-\psi$, hence $\psi=0$. This is the pseudoreal structure of the $SU(2)$ doublet. In our Lorentzian basis the analogous map is $\mathcal J_L\psi=\sigma_1\psi^*$, which commutes with Lorentz spin transformations and has $\mathcal J_L^2=+1$; it admits nonzero Majorana spinors. Reality conditions involving additional paired spinors are a different possibility and must be specified explicitly.

The paper makes the distinction directly after equation (A.2): spinors related by Lorentzian conjugation are treated as independent in Euclidean signature. Its introduction also allows complex background auxiliary fields while retaining a real metric. Version 3, Appendix A.1 and introduction.

Independent variables do not choose an integration contour

Complexifying fields is an algebraic step. Defining a convergent functional integral requires additional choices.

For a finite-dimensional illustration, replace a complex scalar and its conjugate by independent variables $z,w$ in the polynomial $S=a\,wz$, with real $a>0$. The contour $z=x+iy$, $w=x-iy$, with real $x,y$, gives

$$S=a(x^2+y^2),\qquad \int_{\mathbb R^2}\frac{dx\,dy}{\pi}\,e^{-S}=\frac1a.$$

The different contour $w=-z^*$ makes the same polynomial equal $-a|z|^2$ and the integral divergent. The algebra alone did not select a contour or guarantee convergence. In a field theory, the action, boundary conditions, and desired observables determine which prescriptions are appropriate.

For fermions, “independent integration variables” means independent generators of a Berezin integral. There is no ordinary real-axis contour for an individual Grassmann generator. Bosonic contour choices, fermionic conjugation conventions, and symmetry of the resulting integral are related questions, but they are not interchangeable statements. Fixed background sources are another category: they are specified rather than integrated.

Nor does complexification add physical supersymmetries. Three-dimensional $\mathcal N=2$ has four real Lorentzian Poincaré supercharges. Working with independent complex spinor parameters describes the complexified algebra; it does not establish eight real supercharges in a new unitary theory. On a curved background, the independent solutions of the relevant Killing-spinor equations must be counted in their stated convention.

In four dimensions, the same caution applies to the two chiral spinor sectors. Bilal’s equations (2.6)–(2.10) define their Lorentzian conjugation relation. A Euclidean calculation must state its own prescription; the paired equations in Appendix C, equation (C.1), of the three-manifold paper illustrate keeping the two parameters explicit. Four-dimensional $\mathcal N=1$ still refers to four real supercharges in its Lorentzian parent theory.

Exercises

1. Test the adjoint with a finite rotation

Take $U=(\mathbf1+i\gamma^2)/\sqrt2$. Verify that $U$ is unitary and calculate $U^\dagger\gamma^1-\gamma^1U^\dagger$. What does this say about imposing $\chi_E=\psi_E^\dagger\gamma^1$?

First hint

Use $(\gamma^2)^2=\mathbf1$ and $(\gamma^2)^\dagger=\gamma^2$.

Second hint

Insert $\gamma^1=\sigma_3$ and $\gamma^2=-\sigma_1$ after multiplying $U^\dagger U$.

Complete solution

The cross terms cancel:

$$U^\dagger U =\tfrac12(\mathbf1-i\gamma^2)(\mathbf1+i\gamma^2) =\mathbf1.$$

Using the explicit matrices,

$$U^\dagger\gamma^1-\gamma^1U^\dagger =\begin{pmatrix}0&-\sqrt2\,i\\\sqrt2\,i&0\end{pmatrix} =\sqrt2\,\sigma_2\ne0.$$

The proposed row relation is not preserved by this spin rotation. An independent row with transformation $\chi_E\mapsto\chi_EU^\dagger$ does preserve the scalar $\chi_E\psi_E$. The failure concerns the inserted $\gamma^1$, not the existence of a Hermitian adjoint.

2. Can a phase repair the Euclidean reality condition?

Replace $\mathcal J_E$ by $\mathcal J_\alpha\psi=e^{i\alpha}C\psi^*$, with real $\alpha$. Compute $\mathcal J_\alpha^2$. Compare with $\mathcal J_L\psi=\sigma_1\psi^*$, and exhibit a nonzero Lorentzian fixed spinor using ordinary complex test components.

First hint

An antilinear map conjugates the phase when applied a second time.

Second hint

The component equations for a fixed vector of $\mathcal J_L$ are $u=v^*$ and $v=u^*$.

Complete solution

The two phases cancel:

$$\mathcal J_\alpha^2\psi =e^{i\alpha}C\,e^{-i\alpha}C^*\psi =-\psi.$$

No phase changes the obstruction. In contrast, $\sigma_1\sigma_1^*=\mathbf1$, and $(u,v)=(z,z^*)$ is fixed by $\mathcal J_L$ for any complex $z$; $z=1$ gives a nonzero example. These commuting test components probe the spin representation. Physical fermion fields remain Grassmann odd. The check distinguishes a representation’s possible reality condition from its statistics.

3. Change a bosonic contour

For the complexified polynomial $S=a\,wz$, choose $z=x+iy$ and $w=e^{i\beta}(x-iy)$, with $a>0$ and real $\beta$. With the measure $dx\,dy/\pi$ held fixed as stated, determine when the integral converges absolutely and evaluate it there. Does the existence of the polynomial alone determine its value?

First hint

Absolute convergence is controlled by $\operatorname{Re}(ae^{i\beta})$.

Second hint

In polar coordinates, the integral is $2\int_0^\infty r\,e^{-ae^{i\beta}r^2}\,dr$.

Complete solution

Absolute convergence requires $\cos\beta>0$. In that region, putting $s=r^2$ gives

$$\int_0^\infty ds\,e^{-ae^{i\beta}s} =\frac{e^{-i\beta}}a.$$

At $\cos\beta<0$ the integrand grows exponentially; at $\cos\beta=0$ the integral is not absolutely convergent and would require an additional prescription. The result depends on the declared contour and measure. If a contour deformation were accompanied by a different pulled-back holomorphic measure, that factor would also have to be included. Merely naming the polynomial is insufficient. The check $\beta=0$ recovers $1/a$.

Checkpoint

Given a Euclidean fermion formula, identify the independent variables, their index contractions, and the stated conjugation prescription before simplifying it. Given an integral, also identify its bosonic contour or convergence prescription. These checks answer different questions.

For the next supersymmetry calculation, keep the two Euclidean spinor parameters independent until a background and any additional conditions have been specified. A tilde records a partner field or parameter; it is not permission to replace that object by a Hermitian conjugate.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Appendices A and B, conventions and geometry