Paper guide: Section 3, geometry from one spinor

Section 3 of Supersymmetric Field Theories on Three-Manifolds moves between three kinds of statement: differential equations for a supersymmetry label, algebraic identities at one point, and global bundle conditions. Your reading task is to mark which kind is being used at each step.

Use arXiv:1212.3388v3, 22 October 2013. The entire assigned section is printed pages 10–12, or PDF pages 11–13. Prepare with geometry from one spinor and almost contact metric structures. Those lessons derive the bilinear identities; here we map them to the paper and reconstruct a consequential piece of information in the reverse direction.

The physical setting is three-dimensional $\mathcal N=2$, with four real Poincaré supercharges before Euclidean continuation. The two Euclidean charge sectors are independent complex solution spaces. This geometric argument chooses no dynamical gauge group or gauge rank.

Read the whole section in three stages

First read (3.1)–(3.2) and the paragraphs through footnote 5. The background is smooth, oriented, connected and Riemannian. The sources $A,V,H$ may be complex, and $\nabla_\mu V^\mu=0$. Fix them before asking for solutions. Since only $V$, rather than the graviphoton connection itself, enters these equations, flat graviphoton data are not determined by this test.

The equations specify every directional derivative of each two-component column. Along a curve they become a homogeneous linear ODE. Therefore a global solution that is zero at one point is identically zero. Evaluation at a point is injective on each solution space, so its complex dimension is at most two. This is a bound, not an existence theorem for arbitrary initial values. The transport argument explains the distinction.

Next read (3.3)–(3.6). The paper now explicitly stops using the Killing-spinor equations: a nonzero commuting column at a point suffices. Finally read the vector/covector definitions, footnote 6, and (3.7)–(3.8), including the closing paragraph for the opposite-charge label.

P v3 passage	What to establish	Canonical calculation and equation
(3.1), printed 10/PDF 11	The two charge equations use the same specified background	The Killing-spinor equations
(3.2), printed 10/PDF 11	Reversing $A,V$ exchanges the equations; $H$ stays fixed	The opposite-charge substitution
Bundle paragraphs, printed 10–11/PDF 11–12	Which transition functions multiply the spin lift?	Charged spinor sections
Footnote 5, printed 11/PDF 12	Why one zero forces a global solution to vanish	Local uniqueness and global existence
(3.3)–(3.4), printed 11/PDF 12	Normalize the Hermitian direction and check its unit length	M01.3–M01.4: normalized direction
(3.5)–(3.6), printed 11/PDF 12	Fix the sign of $\Phi$ and its transverse square	M01.6–M01.8: transverse tensor; M02.1–M02.2: metric compatibility
Text after (3.6) and footnote 6, printed 12/PDF 13	Compare the $+i$ vector eigenspace with Clifford annihilation	M01.10–M01.11: annihilator and index lowering
(3.7)–(3.8), printed 12/PDF 13	Track R-charge two and the antiholomorphic one-form	M01.12–M01.15: charged bilinear
Closing paragraph, printed 12/PDF 13	Repeat the algebra for the independent negative-charge label	The norm and bundle assumptions, with $L$ replaced by $L^{-1}$

The contact test $\eta\wedge d\eta\ne0$ and transverse integrability are not conclusions of (3.6). The former is a separate test in the prerequisite lesson; the latter is the next section’s differential argument.

Translate the notation before using reality

Paper notation or assumption	Meaning in the site conventions
$\eta_\mu$ and $\eta^\mu$	The one-form $\eta$ and its metric-dual vector $\xi=\eta^\sharp$; raising the index does not complex conjugate
$\zeta^\dagger$	Hermitian adjoint using the positive spin and R-line metrics; it is not the independent column $\widetilde\zeta$
$\zeta\gamma_\mu\zeta$	$\zeta^{\mathsf T}C\gamma_\mu\zeta$, with commuting column entries
$L\otimes S$, $L^{-1}\otimes S$	The charge-$+1,-1$ spinor bundles when a spin structure and R-line bundle are fixed separately
Conventional R-gauge transformations	$\zeta\mapsto e^{i\lambda}\zeta$, $A\mapsto A+d\lambda$ with real $\lambda$; $\operatorname{Im}A$ is a global, gauge-invariant one-form
$\Phi^\mu{}_\nu X^\nu=iX^\mu$	A holomorphic vector; the metric-dual row instead satisfies $X_\mu\Phi^\mu{}_\nu=-iX_\nu$
$P_\mu=\zeta\gamma_\mu\zeta$	A section of $L^2\otimes T^*M$, not generally a single uncharged one-form

The paper also permits a combined $\operatorname{Spin}^c$ structure. Then only the relevant product bundles need be specified globally; one must not assume that the two factors were separately chosen. Our worked example below does choose them separately.

In the section’s closing paragraph, $\widetilde\zeta$ defines its own normalized direction $\widetilde\eta$ and tensor $\widetilde\Phi$. Its bilinear $\widetilde P$ has R-charge $-2$ and is antiholomorphic relative to $\widetilde\Phi$. There is no assertion that $\widetilde\eta=\eta$, or that $\widetilde P=\overline P$.

The substitution (3.2) is not Hermitian conjugation: conjugation also changes the complex sources and spinor representation. Likewise positivity of $\zeta^\dagger\zeta$ does not require $A,V,H$ to be real. These sources are fixed coefficients of an equation, not the metric that defines the pointwise Hermitian norm.

For footnote 6, the omitted positivity step is

$$\|X^a\gamma_a\zeta\|^2 =n\,\overline X^a(\delta_{ab}+i\Phi_{ab})X^b, \qquad n=\zeta^\dagger\zeta>0. \tag{PC02.1}$$

If $\Phi X=iX$, this norm vanishes, hence the column itself is zero. The converse follows by multiplying the annihilator equation by $\zeta^\dagger$ and by $\zeta^\dagger\gamma_a$. This uses the positive Riemannian Hermitian norm and commuting labels. It is not a division by a Grassmann-valued norm.

For $\zeta=(1,0)^{\mathsf T}$, the annihilating vector is $E_2+iE_3$, while $P=e^2+i e^3$ obeys $P\Phi=-iP$. Thus the same displayed components have opposite vector and covector eigenvalues. Calling $P$ holomorphic because $P^\sharp$ is holomorphic would reverse (3.8).

Restore a spinor from the charged bilinear

Here is an inverse calculation that helps interpret (3.7)–(3.8). It is a teaching reconstruction, not another use of the Killing-spinor equation.

In an oriented orthonormal frame use the prerequisite matrices $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$ and $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. For $\zeta=(a,b)^{\mathsf T}$, the established component formula is $P=(2ab,a^2-b^2,i(a^2+b^2))$. Instead of deriving it again, solve it for the symmetric spinor matrix:

$$Q(P)=\frac12 \begin{pmatrix} P_2-iP_3&P_1\\ P_1&-P_2-iP_3 \end{pmatrix} =\zeta\zeta^{\mathsf T},\qquad \det Q=-\frac14P_aP^a. \tag{PC02.2}$$

A nonzero complex null $P$ gives a nonzero rank-one symmetric matrix. On a patch where $Q_{11}\ne0$, choose $a=\sqrt{Q_{11}}$ and $b=Q_{12}/a$. The zero determinant then supplies $b^2=Q_{22}$. Where this pivot vanishes, use $Q_{22}$ instead. Locally the only freedom for a specified $P$ is the common sign of $(a,b)$.

This also shows what must be checked if both $\eta$ and $P$ are supplied independently. The bilinear determines

$$n=\sqrt{\frac{\overline P\cdot P}{2}},\qquad \rho=\frac{QQ^\dagger}{n^2},\qquad \eta_a=\operatorname{Tr}(\rho\gamma_a) =\frac{i(P\times\overline P)_a}{\overline P\cdot P}. \tag{PC02.3}$$

The positive root defines $n$; the cross product uses the fixed orientation. Indeed $QQ^\dagger=n\zeta\zeta^\dagger=n^2\rho$. The reconstructed direction must equal the proposed one. Nullness and transversality alone do not check its sign.

By contrast $\eta$ alone gives the projector $\rho=(1+\eta_a\gamma^a)/2$ and therefore a complex line of columns. It fixes neither their norm nor phase. The charged $P$ restores those up to sign in a chosen local line-bundle trivialization. It does not choose the spin and R transition functions.

Follow that sign around a loop

Take $M=S^1\times\mathbb R^2$, with angular coordinate $\theta\sim\theta+2\pi$ and $g=\ell^2d\theta^2+dx^2+dy^2$. Use the global coframe $(e^1,e^2,e^3)=(\ell d\theta,dx,dy)$. Specify the smooth bilinears

$$\eta=e^1,\qquad P=n_0 e^{im\theta}(e^2+i e^3),\qquad n_0>0,\quad m\in\mathbb Z. \tag{PC02.4}$$

They satisfy every pointwise identity in (3.3)–(3.8). The inverse matrix is $Q=\operatorname{diag}(n_0e^{im\theta},0)$, so a continuous local root is

$$\zeta(\theta)=\sqrt{n_0}\,e^{im\theta/2} \binom10,\qquad \zeta(\theta+2\pi)=(-1)^m\zeta(\theta). \tag{PC02.5}$$

For odd $m$, a single-valued periodic column does not exist in this fixed trivialization, although $\eta$ and $P$ are periodic. Patching the sign is part of specifying the bundle, not a discontinuity to be ignored.

Let the spin seam be $\varepsilon=\pm1$, and let the R-line seam be $h=\pm1$ in the chosen local trivializations. These define the endpoint law $\zeta(2\pi)=\varepsilon h\,\zeta(0)$. The charge-two bilinear has seam $h^2=1$ and cannot detect the distinction between the two choices of $h$. A global algebraic lift of the specified $P$ requires

$$(-1)^m=\varepsilon h. \tag{PC02.6}$$

The signs specify lifts and flat gluing data; a line bundle over a circle is topologically trivial, so this is not a claim of a nonzero first Chern class.

There remains a separate differential test. With the fixed sources $A=V=H=0$, the spin connection vanishes in this coframe and $\partial_\theta\zeta=im\zeta/2$. For $m\ne0$, this reconstructed section is not a solution of (3.1), even when its seam is allowed. If instead one chooses the different background $A=(m/2)d\theta$, $V=H=0$, the derivative cancels $iA_\theta\zeta$, and the other two equations vanish too. The corresponding charge-two equation is $(d-2iA)P=0$.

This example has three separate checks: pointwise tensor identities, an allowed global spin/R lift, and the fixed-source differential equation. Passing one does not imply the next. A local inverse square root is not a theorem reconstructing supersymmetry on an arbitrary specified background.

Exit calculations

1. Change the lift while keeping the tensors

For (PC02.4) with $m=1$, compare $(\varepsilon,h)=(+1,+1)$, $(-1,+1)$ and $(+1,-1)$. Which admit the algebraic lift? With the local connection $A=\tfrac12d\theta$ in each case, which admit the displayed Killing spinor? Can $\eta,P$ distinguish the last two choices?

Hint 1

The spinor and its square have different endpoint laws. Compare the spinor’s actual sign with the product $\varepsilon h$.

Hint 2

The local differential equation is the same in all three cases. Its solution still has to satisfy the prescribed seam; the connection’s return phase is not a replacement for that comparison.

Solution

The actual spinor changes sign. The first choice requires periodicity and fails; the second and third require a minus sign and pass. The local equation with $A=\tfrac12d\theta$ is solved in every patch, so precisely the same two cases give the displayed global solution.

Both tensor fields are identical in the last two descriptions: the spin sign squares to one and $P$ sees $h^2=1$. They do not identify the separately selected spin and R factors. Changing $h$ while fixing the local connection changes the specified flat R gluing, even though its square is invisible to $P$. A simultaneous change of local frames and connection components would instead be a description of the same bundle connection.

2. Conjugate a solution with a complex source

On Cartesian $\mathbb R^3$, set $H=V=0$ and $A=i\kappa\,dx^1$ with real nonzero $\kappa$. Verify that $\zeta=e^{-\kappa x^1}(1,0)^{\mathsf T}$ solves the positive-charge equation. Does the spinor column $C\overline\zeta$ solve the negative-charge equation on these same sources? Find a negative-charge solution, and explain whether a real R-gauge parameter can remove this $A$.

Hint 1

Along $x^1$, the two operators are $\partial_1+\kappa$ and $\partial_1-\kappa$.

Hint 2

The constant matrix $C$ changes the spinor representation but cannot change the derivative of its real exponential. A conventional gauge transformation adds a real one-form to $A$.

Solution

The positive-charge derivative is $(\partial_1+\kappa)e^{-\kappa x^1}=0$. All transverse derivatives vanish. In contrast

$$(\partial_1-\kappa)C\overline\zeta =-2\kappa e^{-\kappa x^1}C\binom10\ne0.$$

Every column $e^{+\kappa x^1}\widetilde c$, with constant $\widetilde c$, solves the negative-charge equation. It must be found using that equation, not assumed to be the conjugate of the first solution. The positive norm of $\zeta$ is $e^{-2\kappa x^1}$ and is nowhere zero; its nonconstant size causes no conflict with Section 3.

The imaginary one-form $\operatorname{Im}A=\kappa dx^1$ cannot be removed by the allowed real parameter $\lambda$. A complexified gauge transformation could formally do so, but it changes the geometric gauge convention that the paper explicitly chose.

3. Conjugate the bilinear but hold the direction fixed

In the reference orthonormal frame, keep $\eta=e^1$ but replace $P=e^2+i e^3$ by $P'=e^2-i e^3$. Both are transverse, null and have Hermitian norm squared two. Can $P'$ and the unchanged $\eta$ come from the same nonzero commuting spinor? Check by both the inverse matrix and (3.8).

Hint 1

Insert $P'=(0,1,-i)$ into (PC02.2). The nonzero diagonal entry has moved.

Hint 2

For $\eta=e^1$, the transverse block of $\Phi$ is $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$. Act on a covector from the right.

Solution

The inverse matrix is $Q(P')=\operatorname{diag}(0,-1)$. Its roots are $\zeta'=\pm(0,i)^{\mathsf T}$, whose normalized Hermitian direction is $-e^1$, not $e^1$. Independently, $(0,1,-i)\Phi=(0,i,1)=i(0,1,-i)$. Thus $P'$ is holomorphic for the original $\Phi$, whereas (3.8) requires the antiholomorphic eigenvalue.

Changing $\eta$ to $-e^1$ also changes $\Phi$ to $-\Phi$ and repairs the pair. The preliminary norm and transversality tests had missed this orientation-sensitive compatibility. For a separately supplied negative-charge spinor the paper constructs its own $\widetilde\eta$ and $\widetilde\Phi$; it does not assert that they equal the positive-charge tensors.

Before proceeding, be able to identify where Section 3 uses differential uniqueness, where it uses positive pointwise algebra, and where it retains bundle information. The next canonical step is integrability and transverse holomorphic geometry. The Section 4 guide then follows the source construction and tests how its local scalar spinor description patches globally.