Abelian M5-brane on $S^6_q$

TL;DR

This paper calculates the conformal anomaly of the abelian M5 brane on a deformed six-sphere, confirming previous results and providing new insights into free energies and Renyi entropies.

Contribution

It presents the first computation of the conformal anomaly for the abelian M5 brane on $S^6_q$, including Casimir energy contributions and a proposed shift in Renyi entropy.

Findings

Results agree with previous $S^1 \times \mathbb{H}^5$ computations.

Identifies missing Casimir energy contributions in free energies.

Proposes a constant shift for the Renyi entropy of the selfdual two-form.

Abstract

We compute the conformal anomaly of the abelian M5 brane on a conical deformation $S^6_q$ of the round six-sphere. Our results agree with corresponding results on $S^1 \times \mathbb{H}^5$ that were obtained in arXiv:1511.00313. For the free energies we obtain missing Casimir energy contributions, inconsequental for the Renyi entropies, and we obtain the proposed constant shift for the Renyi entropy of the selfdual two-form.

Equations

$$Z(q,r)$$

$$a_6^S$$

$$a_6^B$$

$$a_6^F$$

$$a_6^S$$

$$a_6^B$$

$$a_6^F$$

$$a_6^{(2,0)}=5a_6^S+\frac{1}{2}{a}_6^B-2a_6^F=\frac{7}{12}$$

$$\mathcal{A}$$

$$c_3$$

$$c_3^S$$

$$c_3^B$$

$$c_3^F$$

$$a_6^S$$

$$a_6^B$$

$$a_6^F$$

$$a_6$$

$${\tilde{a}}_6^F$$

$$f_S(t)=\frac{e^{2t}(1+e^t)}{(e^t-1{)}^5}=\frac{2}{t^5}+...+\frac{31}{30240}t$$

$$f_{B^{+}}(t)=\frac{1-5e^t+10e^{2t}}{(e^t-1{)}^5}=\frac{6}{t^5}+...+\frac{191}{2016}t$$

$$f_F(t)=-\frac{4e^{2t}(1+e^t)}{(e^t-1{)}^5}=-\frac{8}{t^5}+...-\frac{31}{7560}t$$

$$f(t)$$

$$S_q$$

$$S_q^{S_q^6}$$

$$x^{\pm }$$

$$d{s}^2=d{x}^{I}d{x}^I=d{x}^{i}d{x}^i+2d{x}^{+}d{x}^{-}$$

$$x^{\pm }$$

$$\gamma$$

$$x^i{x}^i+x^{+}{x}^{-}$$

$$x^i$$

$$x^d$$

$$x^{d+1}$$

$$d{s}^2$$

Full text

1 Introduction

The $r$-dependence of the abelian partition function on $S^{6}$ was obtained in [3] from the spectrum of various differential operators on $S^{6}$. This $r$-dependence of the partition function is governed by the M5 brane conformal anomaly that for abelian gauge group was obtained in [6]. In this paper we will refine the computation in [3] to a conically deformed six-sphere $S^{6}_{q}$, also called the branched sphere [9]. For integer values of $q=1/\gamma$ we have $q$ different branches of $S^{6}$.

One motivation for considering such a deformation is to introduce another continuous parameter $\gamma$ in addition to the radius $r$. As we have one more parameter, we may compute the partition function that will be on the form

[TABLE]

In this paper we will only compute the conformal anomaly ${\cal{A}}(q)$.

Another motivation for considering $S^{6}_{q}$ is to study the gauge/gravity duality and in particular how entanglement entropy [8], [15], and more generally Renyi entropy [10], [18], is mapped from gauge theory to gravity theory. This has been studied in the past literature by using a conformal map from $S^{d}_{q}$ to $S^{1}_{q}\times{\mathbb{H}}^{d-1}$ for various dimensions, following [15]. Despite there is a conformal anomaly, we can compute the conformal anomaly itself on either manifold and get the same answer. This is because by the Wess-Zumino consistency condition, the anomaly has no anomaly itself, essentially because $d^{2}=0$. When there is no conformal anomaly and ${\cal{A}}(q)=0$, as is the case in odd dimensions, then we can still compute $c(q)$ and this will be conformally invariant simply because there is no conformal anomaly. This quantity was computed on the gauge theory side and for free field theories where an agreement was explicitly demonstrated for the free energies computed for both $S^{3}_{q}$ and $S^{1}_{q}\times{\mathbb{H}}^{2}$ [12].

In [17] both the nonsupersymmetric and the supersymmeric Renyi entropies are computed for the abelian M5 brane on $S^{1}_{q}\times{\mathbb{H}}^{5}$ and the generalization of the supersymmetric Renyi entropy to the nonabelian M5 brane was obtained in [18] and more generally to the nonabelian (1,0) SCFT’s in [19].

The quantity that we are computing goes by several different names in the literature: the conformal anomaly ${\cal{A}}(q)$, the heat-kernel coefficient $a_{6}(q)$, the Minakshisundaram–Pleijel zeta function $\zeta(0;q)$ evaluated at $s=0$. In our convention they are all equal to one another, ${\cal{A}}(q)=a_{6}(q)=\zeta(0;q)$ and they are functions of $q$. If we ignore the prefactor $c(q)$, then it is also related to the one-loop effective action $W$ and to the free energy $F$, as $Z=r^{a_{6}}=e^{-W}=e^{-\beta F}$ where $\beta=2\pi q$ is the inverse temperature.

Our results for the nonsupersymmetric case for the conformal anomalies are

[TABLE]

for the conformal scalar, two-form gauge field and fermions, respectively. When we put $q=1$, these results reproduce the results in [3] for the round $S^{6}$,

[TABLE]

The normalization of these conformal anomalies is such that for the 6d (2,0) tensor multiplet the conformal anomaly on $S^{6}$ is computed as

[TABLE]

Here $5$ is from the five conformal scalars on $S^{6}$, $1/2$ is because we work with a non-selfdual two-form, and the factor of $2$ corresponds to the $2$ in the (2,0) tensor multiplet. On a generic smooth six-manifold (without conical singularities), the conformal anomaly receives contributions from the Weyl tensor and the structure of the conformal anomaly is

[TABLE]

where the explicit expressions for $E_{6},I_{1},I_{2},I_{3}$ can be found in [6]. In [7] it was argued that the coefficient $c_{3}$ can be computed as

[TABLE]

We can confirm this by computing this number for each field in the tensor multiplet, from (1.1), (1.2) and (1.3) respectively. We then get

[TABLE]

The same coefficient $c_{3}^{S}$ was computed in [7] with the same result. We now see that the ratios between these three numbers are in precise agreement with the ratios between the coefficients $c_{3}$ that were obtained in [6] as $c_{3}^{S}:c_{3}^{B}:c_{3}^{F}=2:180:40$. For the agreement, one has to take into account that their convention is such that ${\cal{A}}=5a_{6}^{S}+\frac{1}{2}a_{6}^{B}+a_{6}^{F}$, which differs by a factor of $-2$ for $a_{6}^{F}$ from our convention. So we see that although these coefficients are computed in a case when supersymmetry is completely broken when $q$ is deformed away from $q=1$, they apparently still contain some very useful information.

The corresponding nonsupersymmetric results that were obtained in [17] on $S^{1}_{q}\times{\mathbb{H}}^{5}$ are111We extract these results from [17] by putting ${\mbox{Vol}}({\mathbb{H}}^{5})=\pi^{2}\ln(r)$ and $\beta=2\pi q$ there.

[TABLE]

Also in [17] the supersymmetric Renyi entropies were obtained.

Firstly, we see that results on $S^{6}_{q}$ differ from results on $S^{1}_{q}\times{\mathbb{H}}^{5}$ by a term proportional to $q$. In [7] such a term was argued to correspond to the Casimir energy $E_{C}$ on $\mathbb{R}\times S^{d-1}$ and the following general structure was found for the conformal anomaly on $S^{d}_{q}$ for $d=4$ and $d=6$,

[TABLE]

Indeed we find this structure for the conformal scalar and the two-form gauge field. For the fermion we do not find this structure. The reason is simply because we break supersymmetry and to preserve supersymmetry we need to turn on a background R-gauge field. Doing so, we obtain a supersymmetric $(\gamma-1)$-deformation of $a_{6}^{F}$ that reads

[TABLE]

The R gauge field does not couple to the two-form and one of the five conformal scalars so those conformal anomalies remain unchanged in the supersymmetric setting. The supersymmetric Casimir energies for the Abelian M5 brane on $\mathbb{R}\times S^{5}$ can be extracted from the following single particle indices [4],

[TABLE]

for the conformal scalar, the two-form and the fermions respectively. The general structure of these single particle indices is

[TABLE]

where $\nu$ is the number of degrees of freedom and $E_{C}$ is the supersymmetric Casimir energy.

Secondly, there is a constant shift by $31/45$ for the two-form gauge field. In [17] a corresponding constant shift was proposed for the Renyi entropy. The Renyi entropy is [11]222In [11] the partition function is a function of temperatature, $\widetilde{Z}(T)$. Here it is a function of $\beta=1/T$. Thus $\widetilde{Z}(T/q)=Z(q/T)=Z(q\beta)$.

[TABLE]

In this paper, we will suppress the overall factor $\ln(r)$ and thus define

[TABLE]

We see from this expression that a Casimir energy term linear in $q$ does not contribute to $S_{q}$. A constant shift of $F(q)$ will contribute to $S_{q}$. More specifically, if $F(q)$ is shifted to $F(q)+C$, then $S_{q}$ gets shifted to $S_{q}+C$. We thus conclude that

[TABLE]

This is in exact agreement with the constant shift that was conjectured in [17].

In section 3 we describe the geometry of $S^{6}_{q}$. In section 4 we obtain the spectrum of two-form harmonics on $S^{6}_{q}$ after first reviewing the scalar and vector harmonics following [13]. In section 5 we compute $a_{6}$ for the two-form gauge potential. In section 6 we compute $a_{6}$ for a conformal scalar. In section 7 we present a representation theory method to compute $a_{6}$. First we reproduce the results for scalar and vector harmonics by this method, and then we use this method to compute $a_{6}$ for one fermion. In section 8 we consider the supersymmetric case. In section 9 we discuss results in the previous literature in relatation to our results.

There are five appendices. In section A we study the box operator on $S^{6}$ for higher rank differential forms. In section B we study the laplace operator. In section C we show how one can compute $a_{6}$ from a ‘half heat kernel’ if the eigenvalues factorize following [7]. In section D we obtain solutions of the Dirac operator on $S^{6}$, although we do not compute degeneracies. In section E we discuss the solution of the Killing spinor equation on $S^{1}_{q}$ and its generalization to $S^{6}_{q}$.

2 Note added

After the publication of this article, I was informed that conformal anomalies on branched spheres had been previously obtained by Dowker [1], [2] for the nonsupersymmetric case. In particular, my result for the conformal anomaly of the two-form gauge field in eq (1.2) had been previously obtained in [2].

In addition, Dowker informed me that it is not always true that a determinant factorizes even if the eigenvalues factorize. There may be a factorization anomaly and my derivation in the appendix C, which assumes the determinant factorizes, is then not correct. Nevertheless, Dowker assured me that this method works when used for the computation of anomalies in this paper due to some fortunate cancelation when one computes the average.

3 The geometry of $S^{6}_{q}$

The geometry of a conically deformed sphere is described in [13]. We begin by the conically deformed flat space $\mathbb{R}^{d+1}$ into which $S^{d}$ may be embedded. Let us denote the Euclidean coordinates on $\mathbb{R}^{d+1}$ by $x^{I}$ for $I=1,...,d,d+1$ and let

[TABLE]

such that $x^{I}=(x^{i},x^{+},x^{-})$ where $i=1,...,d-1$. The Euclidean flat metric on $\mathbb{R}^{d+1}$ is given by

[TABLE]

The cone $\mathbb{R}^{7}_{q}$ is now defined by making the identifications

[TABLE]

When $q=1,2,3,...$ this is a multi-cover of $\mathbb{R}^{7}$ and otherwise it is a cone. We also introduce the quantity

[TABLE]

The conically deformed sphere $S^{d}_{q}$ is the round sphere

[TABLE]

embedded in $\mathbb{R}^{d+1}_{q}$. There is a conical singularity in $\mathbb{R}^{d+1}_{q}$ at $x^{+}=x^{-}=0$. This conical singularity is a submanifold $\mathbb{R}^{d-1}$ with coordinates $x^{i}$. It induces conical singularities on $S^{d}_{q}$ at $x^{+}=x^{-}=0$ where $x^{i}x^{i}=r^{2}$. Hence this conical singularity is an equatorial sphere $S^{d-2}$. Local coordinates on $S^{d}$ may be chosen such that its embedding in $\mathbb{R}^{7}$ is given by

[TABLE]

where $n^{i}n^{i}=1$. Then the induced metric on $S^{d}$ is

[TABLE]

where $d\Omega_{d-2}=dn^{i}dn^{i}$ is the metric on unit $S^{d-2}$. Here $\tau\sim\tau+2\pi$ and $\theta\in[0,\pi/2]$. The conical deformation is obtained by just changing the global identification to $\tau\sim\tau+2\pi q$. This means that locally this is still $S^{d}$ away from $\theta=0$. For $d=2$ there are two distinct conical singularities at the north and south poles at $\theta=0$ for $n^{1}=\pm 1$ respectively. For $d>2$ the conical singularity is not a point but an equatorial sphere $S^{d-2}$. The local geometry in the vicinity of this conical singularity is $\mathbb{R}^{2}_{q}\times S^{d-2}$.

Let us now define a new coordinate $\theta_{D}=\pi/2-\theta\in[0,\pi/2]$. The metric (3.1) becomes

[TABLE]

Expanding this metric to first order in $\theta_{D}$ around $\theta_{D}=0$, we get

[TABLE]

where $d\Omega_{d-1}=d\theta_{D}^{2}+\sin^{2}\theta_{D}d\Omega_{d-2}$ and $t=r\tau$. In other words, near the equator $x^{i}=0$ we have locally the geometry of $S^{1}_{q}\times S^{d-1}$. In the limit $q\rightarrow\infty$ we have locally the geometry of $\mathbb{R}\times S^{d-1}$.

In this paper we find that the asymptotic behavior of $a_{6}$ as $q\rightarrow\infty$ is governed by the supersymmetric Casimir energy $E_{C}$ that one computes on $\mathbb{R}\times S^{d-1}$. That one shall take this limit to see the Casimir energy is natural since the infinite $\beta$ limit of the partition function behaves in this limit as $Z\sim e^{-\beta E_{C}}$. However, it is not immediately clear why it would be sufficient that only the local geometry near the equator is on the form $\mathbb{R}\times S^{d-1}$. On the other hand, $S^{d}_{q}$ in the infinite $q$ limit is conformally equivalent to $\mathbb{R}\times{\mathbb{H}}^{d-1}$. An explicit conformal transformation is presented between the two spaces for $d=3$ in [10], which also shows that the tip of the cone of $S^{d}_{q}$ is mapped to the boundary of ${\mathbb{H}}^{d-1}$. If the supersymmetric Casimir energy is conformally invariant, as suggested by the fact that it appears in the conformal anomaly on $S^{6}_{q}$, then we should be able to compute it on $\mathbb{R}\times{\mathbb{H}}^{d-1}$ and get the same result as one gets on $\mathbb{R}\times S^{d-1}$. As far as we aware, the computation of the Casimir energy on $\mathbb{R}\times{\mathbb{H}}^{d-1}$ has not yet been done in the literature, see for instance [14].

4 Spectrum on $S^{6}_{q}$

We will now obtain the eigenvalues and degeneracies for scalar harmonics and vector harmonics on $S^{6}_{q}$. This will reproduce the result in [13]. We will then follow the same strategy to obtain a new result, the eigenvalues and degeneracies for two-form harmonics on $S^{6}_{q}$.

Let us begin by the scalar harmonics on $S^{d}$. The scalar spherical harmonics are represented in $\mathbb{R}^{d+1}$ as a linear combination of elements of the form

[TABLE]

for $n=0,1,...$. They are constructed such that we have $\partial_{r}\widetilde{\phi}=0$. Let us put $\widetilde{\phi}=\frac{1}{r^{n}}\phi$. Then we find that for $\square=\partial_{I}\partial_{I}$,

[TABLE]

By using

[TABLE]

we get

[TABLE]

provided we also assume that $\square\phi=0$ which implies the tracelessness condition

[TABLE]

The eigenvalue of $\square$ is entirely determined by the power of the prefactor $1/r^{n}$.

4.1 Basis for harmonic differential forms

We introduce the following basis elements, which are independent of $r$, on $S^{6}_{q}$,

[TABLE]

where $\xi\in\{-1,0,1\}$ gives the charge $Q=\xi q$ under the $U(1)_{q}$ rotation under which

[TABLE]

Here $(\mathscr{P}^{\xi})^{m}_{i_{1}\cdots i_{n-2p-\xi}}$ are constant coefficients symmetric in the $i$ indices and subject to a tracelessness condition and some other regularity condition to be presented shortly. These basis elements provide a natural generalization of the scalar harmonics on $S^{6}$ to describe harmonic differential forms on $S^{6}_{q}$. A scalar field on $S^{6}_{q}$ carries charge $Q=0$ as it shall be periodic. A vector field has components $v^{I}=(v^{i},v^{+},v^{-})$ where $v^{i}$ carries charge $Q=0$ and $v^{\pm}$ carries charge $Q=\pm q$. The vector field components are expanded in the basis elements $(\mathscr{P}^{i})_{n,m},(\mathscr{P}^{\pm})_{n,m}$ and $(\mathscr{N}^{i})_{n,m},(\mathscr{N}^{\pm})_{n,m}$. Similarly, a two form has components $B^{IJ}=(B^{ij},B^{+-},B^{\pm i})$.

We want the basis elements to be regular everywhere. This means we need to avoid the presence of $(x^{+})^{-1}$ and $(x^{-})^{-1}$ respectively. Therefore we need to impose the following regularity constraints

[TABLE]

In addition, for vector and higher rank tensor harmonics, we need to assume that

[TABLE]

for $m=1,2,...$ as one can see by demanding regularity of $\mathscr{P}^{-1}_{0,m}$. This means we need to assume that $q\leq 1$. One may then analytically continue to get results valid for $q>1$.333One may embed $\mathbb{R}^{2}_{q}$ into $\mathbb{R}^{3}$ by cutting out a wedge with deficit angle $\alpha\in[0,2\pi]$ in a paper and gluing together the edges of the wedge. This produces a conical rice hat with $\beta=2\pi-\alpha\leq 2\pi$. Two get $\beta=2\pi+\alpha>2\pi$ one can instead insert a wedge with deficit angle $\alpha\in[0,2\pi)$ into a half-slit that one has cut in a paper. One then gets something that takes the shape of a saddle. When $\beta=4\pi$ one has a double cover of $\mathbb{R}^{2}$. One can keep inserting wedges indefinitely to create multicovered conical figures whose only singular point is the tip of the cone embedded into $\mathbb{R}^{3}$, but one will need to use several papers that one glues together along wedges.

One can show that444Here it is being understood that the coefficients of $\mathscr{P}$ and $\mathscr{N}$ are being identified. Alternatively we may consider the class of all possible coefficients and then the equalities really mean equalities between classes of elements.

[TABLE]

On the other hand, when $m=m^{\prime}\neq 0$, we have $\mathscr{P}^{\xi}_{n,m^{\prime}}\neq\mathscr{N}^{\xi}_{n,m^{\prime}}$.

4.2 The tracelessness constraint

The computation of the eigenvalues for the scalar harmonics on $S^{6}$ can be repeated verbatim for the basis elements on $S^{6}_{q}$. The eigenvalues of $\square$ are fully determined by the power of the prefactor $1/r^{n+\gamma m}$ so we get

[TABLE]

and we need to impose

[TABLE]

In particular, the eigenvalues are not sensitive to the charge $\xi$.

Explicitly (4.3) amount to the tracelessness conditions

[TABLE]

These conditions are nontrivial for $p=1,...,\left[\frac{n-\xi}{2}\right]$. By taking $p=0$ we get

[TABLE]

which are trivially realized since the highest nonvanishing components are $(\mathscr{P}^{\xi})^{m}_{i_{1}\cdots i_{n-\xi}}$. We will be mostly interested in the highest equation, that is $p=1$,

[TABLE]

since the lower tracelessness constraints (those with $p>1$) are relating the all the lower components to the highest components. Thus to obtain the degeneracies, we only need to study the most general solutions to the highest equations.

As an illustration of a general idea, let us consider the scalar harmonics. The highest components are $\mathscr{P}^{m}_{i_{1}\cdots i_{n}}$ and $\mathscr{N}^{m}_{i_{1}\cdots i_{n}}$ and all the lower components $\mathscr{P}^{m}_{i_{1}\cdots i_{n-2p}}$ and $\mathscr{N}^{m}_{i_{1}\cdots i_{n-2p}}$ are related to the highest components via the tracelessness constraints. The highest component $\mathscr{P}^{m}_{i_{1}\cdots i_{n}}$ is a rank $n$ symmetric tensor. As such it has

[TABLE]

independent components, where $d=6$ for $S^{6}_{q}$. When $m\neq 0$ there is an independent set of components $\mathscr{N}^{m}_{i_{1}\cdots i_{n}}$ so in total we have $d_{n,m\neq 0}=2d_{n,0}$, whereas when $m=0$ we have $d_{n,0}$ components. For the case of a round sphere, $\gamma=1$, we have $p=n+\gamma m=n+m$ is an integer, and we can define a degeneracy

[TABLE]

for scalar harmonics associated to the eigenvalue $p(p+d-1)$. It can be computed using the hockey-stick identity for binomial coefficients with the result

[TABLE]

which corresponds to the number of components of a rank-$n$ traceless symmetric tensor in $d+1$ dimensions. This is a nice consistency check for the dimensions $d_{n,m}$.

4.3 Transversality and coclosed constraints

To descend to one-form harmonics on $S^{d}$, we shall impose the constraints

[TABLE]

In polar coordinates, the first condition amounts to the transversality constraint555One way to see this is by using $\partial_{r}x^{I}=\frac{x^{I}}{r}$.

[TABLE]

while the second constraint becomes

[TABLE]

where $\theta^{m}$ denote local coordinates on $S^{6}$. This is just saying that $v_{m}$ is coexact.

The transversality and coexact constraints generalize to higher rank differential forms in a natural way,

[TABLE]

We now list the explicit form of the transversality and coexact constraints for the one-form and two-form harmonics.

4.3.1 One-form

[TABLE]

4.3.2 Two-form

[TABLE]

4.4 The highest equations

The highest equations are obtained by taking $p=0$. We shall also note that the highest nonvanishing components are $(\mathscr{P}^{\xi})^{m}_{i_{1}\cdots i_{n-\xi}}$ unless $m=0$ in which case we have the constraints (4.2), so that $(\mathscr{P}^{(-1)})^{0}_{i_{1}\cdots i_{n-1}}$ becomes the highest component with $\xi=-1$.

4.4.1 One-form

[TABLE]

When $m\neq 0$, all components are related to $(\mathscr{P}^{k})^{m}_{i_{1}\cdots i_{n}}$. When $m=0$, the equation (4.7) gives rise to the constraints

[TABLE]

and then we need to instead look at the next level, that is at $p=1$, where we find the constraints

[TABLE]

Let us now move on to (4.8). When $m=0$, this equation becomes

[TABLE]

and by using (4.10), we get

[TABLE]

We now notice that the tracelessness condition amounts to

[TABLE]

and so by combining these two conditions, we seem to get yet another constraint

[TABLE]

But this is really not a new constraint, since (4.11) follows beautifully from (4.9) upon contraction of two indices,

[TABLE]

Thus when $m=0$ the independent components are $(\mathscr{P}^{-})^{0}_{i_{1}\cdots i_{n-1}}$ and $(\mathscr{P}^{k})^{0}_{i_{1}\cdots i_{n}}$ subject to the constraints (4.9). This leads to the degeneracy

[TABLE]

When $m\neq 0$ the independent components are $(\mathscr{P}^{k})_{i_{1}\cdots i_{n}}$ and $(\mathscr{N}^{k})_{i_{1}\cdots i_{n}}$, leading to

[TABLE]

These results were first obtained in [13]. When $q=1$ we have the undeformed $S^{6}$ and we get

[TABLE]

Explicit computation gives

[TABLE]

which for $d=6$ reduces to

[TABLE]

4.4.2 Two-form

[TABLE]

We can eliminate $\mathscr{P}^{+-}$,

[TABLE]

Here (4.32) can be simplified to

[TABLE]

by applying the tracelessness condition (4.4).

When $m\neq 0$, we may use (4.28) to eliminate $\mathscr{P}^{-j}$,

[TABLE]

Using this, we can show that (4.31) gets automatically solved!

We also see that (4.29) gets automatically solved, essentially because $(\mathscr{P}_{(i_{1}k})^{m}_{i_{2}\cdots i_{n-1})ii}=0$ due to antisymmetry in $i_{1}k$.

So all that remains to check is (4.33) once we have solved for $\mathscr{P}^{+k}$ from (4.30),

[TABLE]

that is further simplified to

[TABLE]

by applying the tracelessness condition (4.4).

Then finally we need to see whether (4.33) is also satisfied. This is a hard problem. Let us simply assume this is the case.

Let us now summarize. For $m\neq 0$, we have found that the independent components are $(\mathscr{P}_{ij})^{m}_{i_{1}\cdots i_{n}}$. All the other tensors, as well as all the lower rank tensors are related to this one by the tracelessness conditions or by the conditions we have shown above.

Let us now assume that $m=0$. When $m=0$, the equation (4.28) implies the constraints

[TABLE]

We eliminate $\mathscr{P}^{+-}$,

[TABLE]

where (4.4.2) derives from (4.27). We notice that (4.4.2) is implied by (4.4.2) and (4.34). Equation (4.4.2) is implied by (4.2).

We can now summarize our result. The independent components are given by $(\mathscr{P}^{ij})^{0}_{i_{1}\cdots i_{n}}$ and $(\mathscr{P}^{+i})^{0}_{i_{1}\cdots i_{n-1}}$ subject to the constraints (4.34) and (4.4.2). The number of indepedent such components is given by

[TABLE]

This expression can be simplified to

[TABLE]

For $d=6$ we get

[TABLE]

As for the case when $m\neq 0$, we have assumed that

[TABLE]

So when $d=6$, we get

[TABLE]

As a consistency check, we then get

[TABLE]

where

[TABLE]

is the degeneracy of two-form harmonics on round $S^{6}$.

5 The two-form conformal anomaly

Let us now recall how we compute the heat kernel of a two-form gauge potential on round $S^{6}$. The spherical harmonics correspond to coexact forms. Therefore the heat kernel is computed as

[TABLE]

up to some zero mode. Here $T$, $V$ and $S_{0}$ refers to the two-form, the two one-form ghosts, and the three massless scalar ghosts. The reason we do not have the coefficients two and three multiplying these ghost contributions here, is because we are working with the coexact parts only. For a more detailed explanation we refer to [4]. To compute this, we thus need the refined dimensions for scalar, vector and two-form harmonics, which we obtained in the previous section. We also need the correponding eigenvalues. By using properties of the box operator $\square$ that we derive in appendix A, we find that when acting on massless scalar, vector and two-form harmonics respectively, we get

[TABLE]

Then by using the following relations that we derive in appendix B between the laplacians acting on a scalar, one-form and two-form and the box operator

[TABLE]

we get

[TABLE]

The determinant factorizes if the eigenvalues factorize. If we put $d=6$, then the factorized form of the eigenvalues is

[TABLE]

for $p=0,1,2$. As explained in [7] and in appendix C, each factor gives rise to its own ‘half heat kernel’,

[TABLE]

To extract the conformal anomaly $a_{6}$, we expand $(K_{p}+K_{5-p})/2$ around small $t$ and extract its constant term. We get

[TABLE]

and then

[TABLE]

We need to add $2$ by hand that accounts for two overcounted vector ghost zero modes by the heat kernels [3], [20], [21], [22].

6 The conformal scalar

For the conformal scalar, we have eigenvalues

[TABLE]

We then get

[TABLE]

This result was first obtained in [7].

7 Spectrum from representation theory

For the fermions we can not straightforwardly apply the same method as we did for they $p$-forms. We may try to solve the eigenvalue problem $\Gamma^{M}D_{M}\psi=\lambda\psi$ explicitly and find the eigenvalues and degeneracies by generalizing the approach in [12].

We will proceed in a different way and use $SO(7)$ representation theory. We obtain a universal method to compute the conformal anomaly on $S^{6}_{q}$ that seems to work for any field. We will use this universal method to reproduce the results that we got above for the scalar harmonics and vector harmonics. We then apply this method on the fermions.

We will obtained branching rules for irreducible representations corresponding to scalar harmonics, vector harmonics and fermion harmonics respectively, under $SO(7)\rightarrow SO(5)\times U(1)_{q}$. Here $SO(7)$ is the isometry group of $S^{6}$, and $SO(5)\times U(1)_{q}$ is the isometry of $S^{6}_{q}$. We then refine the dimension of the $SO(7)$ representation by turning on a chemical potential for the $U(1)_{q}$, but we do this in a novel way where instead of using the $U(1)_{q}$ charge for the refined dimension, we shall take the absolute value of the $U(1)_{q}$ charge. We explain why this prescription is needed for the scalar harmonic. We verify that this prescription also gives the right answer for the vector harmonics, which makes us rather confident that our method is universal and we then finally apply this method on the fermions.

The representations of $SO(7)$ that we will need for the tensor multiplet on $S^{6}$ can be found in the appendix of [3]. Let us here summarize the representations of $SO(5)$. We have the simple roots $\alpha_{1}$ and $\alpha_{2}$, the Cartan matrix

[TABLE]

We have the positive roots $\alpha_{1},\alpha_{2},\alpha_{1}+\alpha_{2},\alpha_{1}+2\alpha_{2}$. Their sum is

[TABLE]

The dimension of $SO(5)$ representation with Dynkin labels $(\Lambda_{1},\Lambda_{2})$ is

[TABLE]

7.1 Scalar harmonics

Scalar spherical harmonics on $S^{6}$ make up the irreducible representation $(n,0,0)$ of $SO(7)$. Under $SO(7)\rightarrow SO(5)\times U(1)$, we have the branching rule

[TABLE]

For $n=1$ this amounts to $(1,0,0)\rightarrow(1,0)_{0}\oplus(0,0)_{+1}\oplus(0,0)_{-1}$ which corresponds to the decomposition of a vector as $v^{I}=(v^{i},v^{+},v^{-})$. One may verify that the dimensions add up correctly,

[TABLE]

Here the dimension $s_{n}$ can be computed as the dimension of a rank $n$ symmetric tensor by removing traces,

[TABLE]

Let us now count the number of states in the representation $(n,0,0)$ with a given charge $Q=p-2q$. If $Q=0$, then these states are in the representation

[TABLE]

whose dimension is

[TABLE]

Let us next find the representation with charge $Q=+1$. It is

[TABLE]

There is a corresponding representation $R_{n,-1}$ with charge $Q=-1$. The dimensions of each of these are

[TABLE]

In general, states with charge $Q=0,\pm 1,\pm 2,...,\pm n$ are in a representation $R_{n,Q}$ of dimension $S_{n,Q}=\dim R_{0}=S_{n,0}$ if $Q=0$ and $S_{n,Q}+S_{n,-Q}=\dim\left(R_{+Q}\oplus R_{-Q}\right)=2S_{n-Q,0}$ if $Q=1,2,...,n$.

Let us now consider the following refined heat kernel,

[TABLE]

Let us now compare this with

[TABLE]

where we used $n+\gamma m=n+m+(\gamma-1)m$ and redefined $n+m$ as a new $n$ to go from the first to the second equality. By equating the two expressions, we find that

[TABLE]

Finally let us contrast with the naturally refined the degeneracy

[TABLE]

This is not the one that we shall use. Instead we shall use the following refined degeneracy

[TABLE]

in the refined heat kernel

[TABLE]

7.2 Vector harmonics

The vector harmonics appear in the product of a vector with scalar harmonics,

[TABLE]

The branching rule under $SO(7)\rightarrow SO(5)\times U(1)$ can be deduced from

[TABLE]

and

[TABLE]

From this, we can derive the branching rule

[TABLE]

One may check that the dimensions match,

[TABLE]

and that both sides amount to

[TABLE]

The refined dimension is defined as

[TABLE]

Let us introduce the heat kernel

[TABLE]

Explicitly this is

[TABLE]

where

[TABLE]

The sum (7.9) can be computed and series expanded for small values of $t$. We used Mathematica for this purpose. From this series expansion we can read off the constant term. It precisely agrees with what we obtained in (5.2).

7.3 Fermion harmonics

The fermion harmonics form the representation $(n,0,1)$ of $SO(7)$. They appear in the product of one fermion with the scalar harmonics,

[TABLE]

From the branching rule of the fermion,

[TABLE]

together with the branching rule (7.2) for the scalar harmonics and the product decomposition

[TABLE]

we can deduce the branching rule for the fermion harmonics,

[TABLE]

The corresponding refined dimension is

[TABLE]

More explicitly

[TABLE]

The heat kernel is

[TABLE]

This sum can be evaluated and then be series expanded for small values for $t$. The result is

[TABLE]

that is a refinement of

[TABLE]

that can be computed directly by using

[TABLE]

8 Supersymmetric conical deformation

The conformal Killing spinor on $S^{6}$ picks up a phase factor (E.6) on $S^{6}_{q}$ as we go around the circle once by letting $\tau\rightarrow\tau+2\pi q$. There is no way we can modify the Killing spinor solution since $S^{6}_{q}$ is locally the same as $S^{6}$ and hence the Killing spinor equations on $S^{6}_{q}$ and $S^{6}$ are locally the same, and so their solutions are locally the same. The only candidate solution is the Killing spinor on $S^{6}$, possibly up to a choice of spin structure, which is global data, but on $S^{6}$ there is only one spin structure, so there is no such freedom available. However, for the M5 brane, the supersymmetry parameter has an $SO(5)$ R-spinor index, so we can preserve supersymmetry by turning on the background R-gauge field so that it cancels this phase factor for some of the spinor components. In that way we can preserve some amount of supersymmetry on $S^{6}_{q}$.

For a field this amounts to modifying the covariant derivative as $D_{\tau}\rightarrow D_{\tau}+ik(q-1)$ where $k$ is the R-charge. The R-symmetry group is $SO(5)$ and has two Cartan generators $R_{1}$ and $R_{2}$. The five scalars form a vector of $SO(5)$. We may form two complex scalars with unit R-charges $(1,0)$ and $(0,1)$ under $(R_{1},R_{2})$. The complex conjegute scalars carry R-charges $(-1,0)$ and $(0,-1)$. All five scalars have conformal mass. For the complex scalars, the modified half heat kernels are therefore obtained by shifting the $U(1)_{q}$ quantum number $m\rightarrow m\mp(q-1)$, which amounts to shifting $\gamma m\rightarrow\gamma m\pm(\gamma-1)$,

[TABLE]

It turns out that with the shift, $K_{2}^{\pm}(t)$ is no longer equal to $K_{3}^{\pm}(t)$. We thus need to compute the sum of these and then divide by two, $\left(K_{2}^{\pm}(t)+K_{3}^{\pm}(t)\right)/2$ to get the full half heat kernel. The resulting conformal anomalies that we get this way are

[TABLE]

It is notable that the Casimir energy terms are identical for $a_{6}^{+}$ and $a_{6}^{-}$ while all the other terms are different. While this asymmetry between positive and negative R-charges for the scalars is mysterious to us, it was noted already in [17] in the context of Renyi entropies. From the above, we can compute the corresponding Renyi entropies with the result

[TABLE]

where

[TABLE]

is the Renyi entropy computed from an R-neutral conformal scalar (6.1). A general formula for $S^{\mu}-S=\Delta S^{s}(\mu)$ was presented in [17] for a conformal scalar (indicated by the superscipt $s$) with R-charge $\mu$. We find the following agreements with their general formula,

[TABLE]

We should expect the factor of $1/2$ since we have a complex scalar that corresponds to two real scalars. We get an extra minus signs due to the overall minus sign in the relation $q-1=-(\gamma-1)/\gamma$

Let us next consider the fermions. It is easy to see that eigenvalues of the Dirac operator on $S^{6}_{q}$ are $n+\gamma m+3$ from our computation in appendix D. Thus it shall be possible to recast the heat kernel in the form

[TABLE]

although we have not done it. Specifically, we have not obtained the degeneracies $d_{n,m}^{F}$ but used instead a representation theory method where these are implicit. Using this method, we obtained the following heat kernel

[TABLE]

We would now like to include the coupling to an R-gauge field. We do that by shifting the quantum number $m\rightarrow m\mp(q-1)/2$. But $m$ does not appear explicitly in our expression (8.2). It does appear in (8.3), so let us start there and make the replacement $n+\gamma m\pm(\gamma-1)/2$. That gives our deformed heat kernels,

[TABLE]

Now this deformation just amounts to multiplication by the overall factor $e^{\mp t(\gamma-1)/2}$ and so we can immediately conclude that in the representation theory expression we get the modified heat kernel as

[TABLE]

This deformed heat kernel gives the result in equation (1.7) for $\widetilde{a}_{6}^{F}$. We get the same result for $\widetilde{a}_{6}^{F}$ for both the plus and the minus signs. It has the structure that we expect of the supersymmetric conformal anomaly. It gives the supersymmetric Casimir energy, as the term linear in $q=1/\gamma$, and from the $1/q^{5}$ term we may read off $4$ degrees of freedom (so that to get the $8$ degrees of freedom of the 6d (2,0) tensor multiplet, we shall multiply $\widetilde{a}_{6}^{F}$ by $2$).

In [17] the supersymmetric Renyi entropy was obtained on $S^{1}_{q}\times{\mathbb{H}}^{5}$. It is given by

[TABLE]

where, if we insert $\mu=\pm(q-1)/2$,

[TABLE]

and $\Delta S^{f}(\mu)=\Delta S^{f}(-\mu)$. From our results for $a_{6}^{F}$ and $\widetilde{a}_{6}^{F}$ in (1.3) and (1.7), we can compute the corresponding Renyi entropies,

[TABLE]

We get

[TABLE]

Their difference is in agreement with [17],

[TABLE]

with $\Delta S^{f}_{q}$ given by (8.4).

9 Discussion

We have shown that conformal anomalies computed on $S^{6}_{q}$ agree with corresponding results on $S^{1}_{q}\times{\mathbb{H}}^{5}$, up to a constant term and to a linear term in $q$. The computation that has been done on $S^{1}_{q}\times{\mathbb{H}}^{5}$ is more subtle than our computation on $S^{6}_{q}$ since $S^{1}_{q}\times{\mathbb{H}}^{5}$ is noncompact and the spectrum is continuous. There appears a divergence for the zero mode along $S^{1}_{q}$ that one needs to regularize. The way this was done in [14] was to remove that zero mode by subtracting the infinite $q$ asymptotic part, which amounts to subtracting the infinite $q$ asymptotic from our conformal anomalies. Concretely that amounts to multiplying the conformal anomalies by the inverse leading power that generically appears to be $q$ and take the limit $q\rightarrow\infty$. The only term that survives this limit is the asymptotic term that was linear in $q$ before multiplying the whole anomaly by $q^{-1}$. So it is precisely this linear term that gets subtracted by this regularization. Thus we can explain the missing Casimir energies as being due to a careless regularization on the $S^{1}_{q}\times{\mathbb{H}}^{5}$ side and by using a better way to regularize we should expect to recover the linear term on the $S^{1}_{q}\times{\mathbb{H}}^{5}$ side as well. For the constant shift associated with the tensor gauge field [17], this may arise from missing to take into account boundary degrees of freedom that live on the asymptotic boundary $S^{1}\times S^{4}$ of $S^{1}\times{\mathbb{H}}^{5}$. A corresponding computation for a Maxwell field on $S^{1}\times{\mathbb{H}}^{3}$ [16] has shown that such a constant shift associated with the gauge field comes from a contribution from the boundary $S^{1}\times S^{2}$. Presumably this computation will have a straightforward generalization to a two-form gauge field in six dimensions on $S^{1}\times{\mathbb{H}}^{5}$.

In [19] the Renyi entropy for nonabelian (1,0) SCFT’s was obtained on $S^{1}_{q}\times{\mathbb{H}}^{5}$. One may ask if the nonabelian conformal anomaly can also be obtained on this space.666General results for the nonabelian 6d (2,0) conformal anomaly on smooth six-manifolds have been obtained in [23], [24]. The relation between the Renyi entropy and conformal anomaly as given in (1.8) can be at least partially inverted. If the conformal anomaly is on the form777The leading term $\sim q^{-5}$ may be lowered by supersymmetric cancelation.

[TABLE]

where we deleted the linear term in $q$, , then the corresponding Renyi entropy becomes

[TABLE]

Hence by knowing the coefficients in the nonabelian Renyi entropy of some (1,0) SCFT, as were obtained in [19], we can go back and obtain the conformal anomaly $a(q)$, at least up to the linear term in $q$. That is, up to the knowledge of the nonabelian Casimir energy.

In [19] it was argued that the supersymmetric Casimir energy (possibly rescaled by some factor) can be obtained from the leading term in the supersymmetric Renyi entropy for (1,0) supermultiplets as

[TABLE]

Here $\alpha,\beta,\gamma,\delta$ are the coefficients in the eight-form anomaly polynomial. For the abelian $(1,0)$ tensor multiplet and hypermultiplet these anomaly polynomials were obtained in [4]. From that we can read off the coefficients for the abelian $(1,0)$ tensor multiplet as $\alpha=1,\beta=1/2,\gamma=7/240,\delta=-1/60$ that gives us

[TABLE]

The Renyi entropy for the abelian tensor multiplet is computed from

[TABLE]

with the result

[TABLE]

Thus by looking at the leading term in the Renyi entropy, we confirm that the same number $1/360$ (up to a convention dependent minus sign) appears on both sides in the relation (9.1). On the other hand, the supersymmeric Casimir energy of the tensor multiplet, which may be extracted from the term linear in $q$ in $a_{6}^{tensor}$, is given by $E^{tensor}=-11/240$, which is in agreement with [4]. It does not seem obvious that the two numbers $-11/240$ and $1/360$ would be related.

For the hyper we look at the combination (the general case where two Cartans of $SO(5)$ are involved was considered in [17])

[TABLE]

where $\phi^{1}+i\phi^{2}$ carries R-charge $+1$ and $\phi^{3}$ and $\phi^{4}$ are R-neutral. It should be noted that there appears two copies of the holomorphic field $\phi^{1}+i\phi^{2}$ that leads to $2a^{+}_{6}$, rather than one holomorphic and one antiholomorphic field (that is, $\phi^{1}+i\phi^{2}$ and $\phi^{1}-i\phi^{2}$) that would lead to $a^{+}_{6}+a^{-}_{6}$. The holomorphic choice of $2a^{+}_{6}$ is necessary to have a supersymmetric cancelation of the $q^{-5}$ terms, but we think it is unlikely this can be understood at the classical level. The holomorphy is presumably related to having a chiral theory in euclidean signture where chiral spinors can not be also made real. More rigorously, this might be possible to trace back to holomorphic factorization of a nonchiral quantum theory [25]. From $a^{hyper}$ we read off the Casimir energy $E^{hyper}=1/240$ that corresponds to the Casimir energy at the points of enhanced $(2,0)$ supersymmetry, at the values of a mass parameter $m=\pm 1/2$ in [4].

Acknowledgments

I would like to thank Yang Zhou for explaining his work during the East Asia Joint Workshop on Fields and Strings 2018 at KIAS and Arkady Tseytlin for bringing his works to my attention by correspondence. This work was supported in part by NRF Grant 2017R1A2B4003095.

Appendix A The box operator on scalar, vector and two-form

We define the box operator on $\mathbb{R}^{d+1}$ as

[TABLE]

The metric in Euclidean and Polar coordinates is

[TABLE]

We have

[TABLE]

The nonvanishing Christoffel symbols are

[TABLE]

The box operator in Polar coordinates is

[TABLE]

Acting on a scalar, we have

[TABLE]

Let us define

[TABLE]

We now need the following covariant derivatives,

[TABLE]

where

[TABLE]

Then we get

[TABLE]

where we define

[TABLE]

We notice that the same result is obtained from

[TABLE]

Let us move on to the vector,

[TABLE]

We put

[TABLE]

just to avoid distraction. Then we have

[TABLE]

We expand out these covariant derivatives to get

[TABLE]

Now we insert

[TABLE]

We put $v_{r}=0$. We get

[TABLE]

This we can also express as

[TABLE]

if we let $\partial_{r}$ act on everything that stands to the right, so for instance $\partial_{r}\frac{1}{r}v_{m}$ shall be interpreted as $\partial_{r}\left(\frac{1}{r}v_{m}\right)$. Then this agrees with the equation (2.2) in [13]. We now notice that

[TABLE]

implies that

[TABLE]

Using this, we finally get

[TABLE]

Let us turn to the two-form. We find

[TABLE]

We now notice that

[TABLE]

implies that

[TABLE]

Using this, we finally get

[TABLE]

Appendix B Laplacian on scalar, vector and two-form

We have the following Laplacians on $S^{d}$,

[TABLE]

On $S^{6}$ we get

[TABLE]

This leads to

[TABLE]

Eigenvalues are

[TABLE]

for Laplace on vector harmonics. For two-form harmonics we get

[TABLE]

Appendix C Factorization of the determinant

Here we explain the heat kernel method that was used in [7] when the eigenvalues factorize. If the eigenvalues of some quadratic differential operator $\triangle$ take the form of a product

[TABLE]

then the corresponding determinant should factorize. When the product is a divergent infinite product, the determinant is defined through a zeta function

[TABLE]

as

[TABLE]

and the factorization means that

[TABLE]

Let us assume the eigenvalues are on the form

[TABLE]

where $\widehat{\lambda}_{n}$ and $\widehat{\lambda}_{n}^{a,b}$ are dimensionless888Subscript $a,b$ is a short notation that means either $a$ or $b$.. Then this derivative becomes

[TABLE]

The relation (C.1) now amounts to the relation

[TABLE]

This means that we can compute $\zeta(0)$ from the half heat kernels

[TABLE]

instead of using the original heat kernel

[TABLE]

which is much harder to compute in our application where the eigenvalues $\lambda$ are quadratic, while $\lambda_{a,b}$ are linear in $n$.

There seems to be no simple relation for the other heat kernel coefficients. For example, for the massless scalar on $S^{6}$ with eigenvalues $\widetilde{\lambda}=n(n+5)$ we get the heat kernel

[TABLE]

and the half heat kernels

[TABLE]

Appendix D Explicit solutions for fermion harmonics

Let us comment on explicit solutions for the fermion harmonics. We will restrict ourselves to $S^{6}$ and only briefly mention $S^{6}_{q}$ at the end. Let us begin by considering the conformal Killing spinor equation The conformal Killing spinor equation is

[TABLE]

from which follows that

[TABLE]

where

[TABLE]

We also find that

[TABLE]

The most general solution is most easily found using stereographic coordinates on $S^{6}$ where the metric is

[TABLE]

Then the solutions are simply obtained from the flat space solution ${\cal{\varepsilon}}={\cal{\varepsilon}}_{0}+x^{M}\Gamma^{\widehat{M}}{\cal{\varepsilon}}_{1}$ by a conformal map. We get

[TABLE]

where

[TABLE]

and $\Gamma^{\widehat{M}}$ are the flat space gamma matrices obeying $\{\Gamma^{\widehat{M}},\Gamma^{\widehat{N}}\}=2\delta^{\widehat{M}\widehat{N}}$.

We now define two scalars out of one spinor as

[TABLE]

where the bar means hermitian conjugate. We then get

[TABLE]

Let us now assume that

[TABLE]

Then also we get

[TABLE]

and so we get

[TABLE]

By demanding that $\phi_{\pm}$ are scalar harmonics with eigenvalues

[TABLE]

we get the equations

[TABLE]

which have the solutions

[TABLE]

Thus we see that $\phi_{+}$ is composed of fermions with eigenvalues $-(n+3)$ and $n+2=(n-1)+3$. But as both $\phi_{+}$ and $\phi_{-}$ have the same eigenvalue, they must be joined together into one spherical harmonic multiplet with that eigenvalue ($n^{2}+5n$), which means that we have a decomposition of one scalar harmonic into two fermionic harmonics, thus realizing the group theory decomposition

[TABLE]

We would now like to solve

[TABLE]

We make the ansatz

[TABLE]

Then we get

[TABLE]

and so the eigenvalue problem reduces to

[TABLE]

The secular equation becomes

[TABLE]

We take

[TABLE]

to get $\square=n^{2}+5n$.

Appendix E Killing spinor on $S^{1}_{q}$

Let us begin with $\mathbb{R}^{2}$ where we may use polar coordinates

[TABLE]

The metric is

[TABLE]

The vielbein is

[TABLE]

in Cartesian coordinates, and

[TABLE]

in polar coordinates. We have the following $SO(2)$ tangent space transition when going between these vielbeins

[TABLE]

Thus the $SO(2)$ transition matrix in the vector representation is

[TABLE]

The corresponding transitition matrix in the spinor representation is

[TABLE]

and a constant spinor ${\cal{\varepsilon}}_{0}$ with respect to $E^{1},E^{2}$ corresponds to a spinor

[TABLE]

with respect to $e^{\widehat{R}},e^{\widehat{\tau}}$. This is a Killing spinor on $S^{1}$ that is antiperiodic,

[TABLE]

But we would like to have a periodic Killing spinor. Indeed we can get a periodic Killing spinor if we cover $S^{1}$ by two overlapping coordinate patches. We may take $0<t<2\pi$ in one patch $U_{\alpha}$, and $-\pi<s<\pi$ in the other patch $U_{\beta}$. On the overlap, these coordinates are related as $s=t-\pi$ and the transition matrices on $U_{\alpha}$ and $U_{\beta}$ are related as $g_{\alpha}=-g_{\beta}$. The Killing spinor is now defined on each patch as

[TABLE]

and on the overlap, there is a transition

[TABLE]

that gives us the periodic Killing spinor if we alternate once from $U_{\alpha}$ to $U_{\beta}$ as we go around $S^{1}$ once, and it gives us the anti-periodic Killing spinor if we alternate twice from $U_{\alpha}$ to $U_{\beta}$ and then back again to $U_{\alpha}$ as we go around $S^{1}$ once. However, it is only the periodic Killing spinor that generalizes to $S^{d}$ for $d>1$.

Now with the periodic Killing spinor on $S^{1}$, it is easy to see that this will pick up a phase factor as we go around $S^{1}_{q}$ by letting $\tau\rightarrow\tau+2\pi q$,

[TABLE]

This is easily generalized to $S^{6}_{q}$. We introduce the following coordinates on $mb{R}^{7}$,

[TABLE]

where $\theta^{i}$ are polar coordinates on $S^{4}$ and $n^{m}n^{m}=1$ for $m=1,...,5$. The metric on $\mathbb{R}^{7}$ is

[TABLE]

and $S^{6}$ is characterized by the constraint $R^{2}+r^{2}=1$. The vielbein is

[TABLE]

and is related to the Cartesian vielbein $dx^{M}$ on $\mathbb{R}^{7}$ by the transition matrix

[TABLE]

where $g_{2}$ is the transition matrix we computed above on $\mathbb{R}^{2}$ and $g_{4}$ is the transition matrix for $\mathbb{R}^{4}$ in polar coordinates, which commutes with $g_{2}$. One class of conformal Killing spinors on $S^{6}$ is now obtained from the constant spinor on $\mathbb{R}^{7}$ by transformig it by a local Lorentz transformation to the above coordintes,

[TABLE]

and again we see that we pick up the phase factor (E.6) as we go around $S^{6}_{q}$ once, along the $\tau$ direction.