# Variation Formula of APS $\eta$-Invariant and Chern-Simons Theory

In Perturbative Expansion of Chern-Simons Theory with Noncompact Gauge Group, the author proved the variation formula of the APS $$\eta$$-invariant. I have a few questions about their proof. I will explain in the following.

Let $$M$$ be a compact and closed three dimensional manifold. Let $$D$$ be a nilpotent differential operator acting on $$\mathfrak{g}$$-valued differential forms $$\Omega(M,\mathfrak{g})$$, where $$\mathfrak{g}$$ is a Lie algebra. Specifically, $$D$$ is the covariant derivative in some background gauge field $$B\in\Omega^{1}(M,\mathfrak{g})$$, i.e.

$$D\varphi=d\varphi+B\varphi$$

for a complex multiplet $$\varphi$$, and

$$Da=da+[B,a]$$

for a $$\mathfrak{g}$$-valued one-form $$a$$.

Let $$\ast$$ be a Hodge star operator such that $$\ast^{2}=1$$. One has the following elliptic operator

$$L=(D\ast+\ast D)J$$

where $$J$$ is $$+1$$ when acting on $$\Omega^{1}\oplus\Omega^{2}$$, and is $$-1$$ when acting on $$\Omega^{0}\oplus\Omega^{3}$$. In the following, I denote $$L_{-}$$ as $$L$$ restricted on odd forms $$\Omega_{-}=\Omega^{1}\oplus\Omega^{3}$$, i.e. $$L_{-}=L|_{\Omega_{-}}$$.

The APS $$\eta$$-invariant $$\eta(L_{-})$$ is, roughly speaking, the number of positive eigenmodes of $$L_{-}$$ minus the number of negative eigenmodes of $$L_{-}$$. It can be regularized in the following way

$$\eta(L_{-1},s)=\sum_{j}\frac{\mathrm{sign}(v_{j})}{|v_{j}|^{s}}$$

where $$s\in\mathbb{C}$$, and $$v_{j}$$ are non-zero eigenvalues of $$L_{-}$$.

Using a Laplace transform, one has the formula

$$\frac{\mathrm{sign}(x)}{|x|^{s}}=\frac{2}{\Gamma(\frac{s+1}{2})}\int_{0}^{\infty}dy\,y^{s}x\,e^{-x^{2}y^{2}}$$

Thus, one has

$$\eta(L_{-},s)=\frac{2}{\Gamma(\frac{s+1}{2})}\int_{0}^{\infty}dy\,y^{s}\,\mathrm{Tr}_{\Omega_{-}}(L_{-}e^{-y^{2}L_{-}^{2}})$$

where the trace is taken over odd forms. Next, one performs the variation with respect to the background gauge field $$B$$ and keep the metric fixed (i.e. $$\delta\ast=0$$). Integrating by parts and taking the limit $$s\rightarrow 0$$, one has the variation

$$\delta\eta(L_{-})=\frac{-2}{\sqrt{2\pi}}\lim_{y\rightarrow 0}(y\,\mathrm{Tr}_{\Omega_{-}}(\delta L_{-}e^{-y^{2}L_{-}^{2}}))$$

Then, the authors used the "fermionic interpretation" of differential form to proceed. For local coordinates $$x^{i}$$ on $$M$$, one defines the following two operations

$$\psi^{i}:\omega\rightarrow dx^{i}\wedge\omega$$

$$\chi_{j}:\omega\rightarrow \omega(\frac{\partial}{\partial x^{j}},\cdots)$$

where $$\omega\in\Omega^{q}(M)$$.

One finds that these two operations satisfy the Clifford algebra

$$\left\{\psi^{i},\chi_{j}\right\}=\delta^{i}_{j},\quad \left\{\psi^{i},\psi^{j}\right\}=\left\{\chi_{i},\chi_{j}\right\}=0$$

Define the Witten index $$(-1)^{F}$$ as

$$(-1)^{F}:\omega\rightarrow(-1)^{q}w,\,\,\,\,\mathrm{for}\,\,\,\forall\omega\in\Omega^{q}(M).$$

Then, one finds (equation 4.17)

## $$\ast\psi^{i}\ast=(-1)^{F}\chi^{i},\quad\ast\chi^{i}\ast=\psi^{i}(-1)^{F}$$

I tried this myself, but didn't find the correct Witten index factor. My own calculation is shown below.

In local coordinates, one has

$$\omega=\frac{1}{q!}\omega_{i_{1}\cdots i_{q}}dx^{i_{1}}\wedge\cdots\wedge dx^{i_{q}}$$

Hodge star operator is defined as

$$\ast:\Omega^{q}(M)\rightarrow\Omega^{n-q}(M)$$

such that $$\ast^{2}=1$$.

One has

$$(\ast\omega)_{j_{1}\cdots j_{n-q}}=\frac{1}{q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}$$

where the $$\epsilon$$ symbol is raised by the metric tensor. Therefore, one has

$$\ast\omega=\frac{1}{(n-q)!}\left(\frac{1}{q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}\right)dx^{j_{1}}\wedge\cdots\wedge dx^{j_{n-q}}$$

Then, one has

$$\psi^{i}\ast\omega=dx^{i}\wedge\ast\omega$$

$$=\frac{1}{(n-q+1)!}\left(\frac{(n-q+1)!}{(n-q)!q!}\epsilon^{i_{1}\cdots i_{q}}_{\qquad j_{1}\cdots j_{n-q}}\,\,\omega_{i_{1}\cdots i_{q}}\right)dx^{i}\wedge dx^{j_{1}}\wedge\cdots\wedge dx^{j_{n-q}}$$

Applying the Hodge star operator again, one has

$$(\ast\psi^{i}\ast\omega)_{k_{1}\cdots k_{q-1}}=\frac{1}{(n-q+1)!}\epsilon^{ij_{1}\cdots j_{n-q}}_{\qquad\quad\,k_{1}\cdots k_{q-1}}(\psi^{i}\ast\omega)_{ij_{1}\cdots j_{n-q}}$$

Thus, one has $$(\ast\psi^{i}\ast\omega)^{k_{1}\cdots k_{q-1}}=\frac{1}{(n-q)!q!}\epsilon^{ij_{1}\cdots j_{n-q}\,k_{1}\cdots k_{q-1}}\,\epsilon_{i_{1}\cdots i_{q}j_{1}\cdots j_{n-q}}\,\omega^{i_{1}\cdots i_{q}}$$

Rearranging indices of $$\epsilon$$ tensors, one has

$$\epsilon_{ij_{1}\cdots j_{n-q}\,k_{1}\cdots k_{q-1}}\epsilon^{i_{1}\cdots i_{q}j_{1}\cdots j_{n-q}}=(-1)^{(q-1)(n-q)}\epsilon_{ik_{1}\cdots k_{q-1}\,j_{1}\cdots j_{n-q}}\,\epsilon^{i_{1}\cdots i_{q}\,j_{1}\cdots j_{n-q}}$$

Using contraction rules of $$\epsilon$$ tensor, one has

$$\ast\psi^{i}\ast=(-1)^{(q-1)(n-q)}\chi^{i}$$

Question 1: I expect to have $$(-1)^{q}$$. Where did I make mistakes?

Next, the authors claims that

$$D=\sum_{i}\psi^{i}D_{i},\quad\ast D\ast\sum_{i}(-1)^{F}\chi^{i}D_{i}$$

Question 2: Can anybody explain why $$D$$ and $$\ast D\ast$$ can be expressed in the above ways?

Then, one has $$L_{-}^{2}=D\ast J\ast DJ+\ast DJD\ast J+(D\ast JD\ast J+\ast DJ\ast DJ)$$. The last term is a Lapacian operator. The first two terms can be written as

$$D\ast J\ast DJ=-\frac{1}{2}\psi^{i}\psi^{j}[D_{i},D_{j}](-1)^{F}$$

$$\ast DJD\ast J=-\frac{1}{2}\chi^{i}\chi^{j}[D_{i},D_{j}](-1)^{F}$$

Question 3: Can anybody tell me how to derive the above two expressions?

Then one has $$L_{-}^{2}=-(X-\Delta)$$, where $$\Delta=D\ast JD\ast J+\ast DJ\ast DJ$$, and $$X=\frac{1}{2}(\psi^{i}\psi^{j}[D_{i},D_{j}]+\chi^{i}\chi^{j}[D_{i},D_{j}])(-1)^{F}$$.

Next, using the expansion formula

$$e^{-y^{2}(\Delta-X)}=e^{-y^{2}\Delta}+\int_{0}^{y^{2}}dte^{-t\Delta}Xe^{-(y^{2}-t)\Delta}+\cdots$$

it is shown that one can obtain

$$\delta\eta(L_{-})=\frac{-1}{4\pi^{2}}\mathrm{Tr}_{\Omega_{-}}(\delta L_{-}X)=\frac{-1}{4\pi^{2}}\mathrm{Tr}_{\Omega_{-}}((\ast\delta D+\delta D\ast)JX).$$

Then, they author claims that since $$\ast$$ maps odd forms to even forms and maps even forms to odd forms, the above variation is equivalent to

$$\delta\eta(L_{-})=\frac{-1}{4\pi^{2}}\mathrm{Tr}((\ast\delta D)JX).$$

where the trace is taken over the full de-Rham complex

$$\Omega(M)=\Omega_{+}\oplus\Omega_{-}=\Omega^{0}\oplus\Omega^{1}\oplus\Omega^{2}\oplus\Omega^{3}.$$

Question 4: Can anybody explain to me the above statement that one can ignore $$\mathrm{Tr}(\delta D\ast JX)$$?

Since $$\psi^{i}$$ and $$\chi^{j}$$ are elements in the Clifford algebra, one can denote them as Dirac gamma matrices. In three dimensions, they are proportional to the Pauli matrices. In the end, the authors used the formulae

$$\mathrm{Tr}\ast\gamma^{i}J\gamma^{j}\gamma^{k}=\epsilon^{ijk}, \quad\mathrm{or}\quad\mathrm{Tr}(\sigma_{i}\sigma_{j}\sigma_{k})=2i\epsilon_{ijk}$$

with $$\delta D=\psi^{i}\delta B_{i}$$, and obtained

$$\delta(\eta_{-})=\frac{1}{8\pi^{2}}\int_{M}\epsilon^{ijk}\mathrm{Tr}_{\mathrm{ad}(\mathfrak{g})}\delta B_{i}[D_{j},D_{k}],$$

which can be integrated out and becomes a Chern-Simons action. Here the trace $$\mathrm{Tr}_{\mathrm{ad}(\mathfrak{g})}$$ is taken in the adjoint representation of the Lie algebra $$\mathfrak{g}$$.

Question 5: Where does the trace over the adjoint representation come from?

• Nov 10 '18 at 16:13