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I wanted to test my understanding of the Atiyah-Patodi-Singer theorem by studying flat bundles on $T^3$ explicitly, and miserably failed. Namely, I computed the eta invariant explicitly for flat bundles with known Chern-Simons invariants (taken from Borel-Friedman-Morgan) but they didn’t agree. Could you point out where I did mistakes in the following computation?

First, let's take an orthonormal metric on $T^3$, and consider the standard Dirac operator $D_\alpha$ on the spin bundle tensored by the flat line bundle $L_\alpha$, whose holonomies $(\alpha_1,\alpha_2,\alpha_3)$ around three generators of $T^3$ are all $\alpha_{i}\in \{\pm1\}$. The eigenvalues of $D_\alpha$ can be easily found by explicit computation: $$ \pm\sqrt{n_1^2+n_2^2+n_3^2}$$ where $n_i$ is in $\mathbb{Z}$ (in $\mathbb{Z}+1/2$) if $\alpha_i=+1$ (if $\alpha_i=-1$). From this we see that $\eta_\alpha(s)=0$ and in particular $\eta_\alpha(0)=0$, and $$ \xi_\alpha:=\frac{h_\alpha + \eta_\alpha(0)}2$$ is 1 or 0 iff $(\alpha_1,\alpha_2,\alpha_3)=(+1,+1,+1)$ or not.

Now, let’s consider two flat $Spin(n)$ bundles on $T^3$. The first one $P_1$ is the trivial one with trivial holonomy, and the second one $P_2$ has the holonomies (when projected to $SO(n)$) $$ g_1=diag(+,+,+,+,-,-,-,-,+,+,\ldots,+)$$ $$ g_2=diag(+,+,-,-,+,+,-,-,+,+,\ldots,+)$$ $$ g_3=diag(+,-,+,-,+,-,+,-,+,+,\ldots,+)$$ around $T^3$. In Borel-Firedman-Morgan it was shown that the difference of the CS invariants of $P_2$ and $P_1$ is $1/2$ (where the CS is defined mod 1). (For those who aren’t familiar with B-F-M, the moduli space of flat $Spin(n)$ connections on $T^3$ has two components, and these two are representatives from each.)

Now, take an irreducible representation $V$ of $Spin(n)$, and consider the Dirac operator $D_{V,i}$ on the spin bundle of $T^3$ tensored with $P_i \times_{Spin(n)} V$. The Atiyah-Patodi-Singer theorem says that $$ \frac{\xi_{V,1} - \xi_{V,2}}{2k(V)} = CS(P_1)-CS(P_2)$$ mod 1, where $k(V)$ is the quadratic Casimir of $V$ normalized so that $k(adj)$ is the dual Coxeter number.

So, I wanted to check this. When $V$ is the $n$-dimensional defining representation of $SO(n)$, the $\xi_{V,i}$ can be found by summing $\xi_\alpha$ computed above, by writing $$ P_2 \times_{Spin(n)} V = L_1\oplus L_2 \oplus \cdots \oplus L_{n}. $$ $L_2$ … $L_8$ give $\xi_\alpha=1$, while the others give zero. Furthermore, $k(V)=1$. Then we see $7/2 = 1/2$ mod 1, so far so good.

Now I have a problem when we take $W$ to be the adjoint representation. We have $$ P_2 \times_{Spin(n)} W = \bigoplus_{i<j} L_i \otimes L_j $$ and again I can easily compute the $\xi$ invariants. For definiteness, take $n=8$. Then every choice $i<j$ gives $\xi_\alpha=1$. And $k(W)=k(adj)=6$. So I seem to have $$ \frac{28}{12} = 1/2 $$ mod 1, which is not true.

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    $\begingroup$ Why can you divide both sides by $2k(V)$ in APS ? $\endgroup$ Nov 25, 2014 at 4:52
  • $\begingroup$ Well, Witten says so in p.47 of arxiv.org/abs/hep-th/0006010 . Yes I knew I shouldn't trust what's written in physicists' papers... $\endgroup$ Nov 25, 2014 at 5:58
  • $\begingroup$ APS says $\int_X ch(E) \hat A(X) - \xi$ is an index of a Fredholm operator, where $X$ is a manifold with boundary $Y$ ($=T^3$ in your case). Is it true that the index (more precisely the difference of indices for $i=1,2$) is divisible by $2k(V)$ ? $\endgroup$ Nov 25, 2014 at 8:18
  • $\begingroup$ If my computation is OK, I think the answer is no. I guess I need to ask Edward what he meant. $\endgroup$ Nov 25, 2014 at 8:44
  • $\begingroup$ @YujiTachikawa I am not sure what you are calling $\xi_a, h_a$ and the operation $\times_{Spin(n)}$, multiplying a $Spin(n)$ bundle with a representation. More importantly, I would like to know why you get : $$P_2 \times_{Spin(n)} W = \bigoplus_{i < j} L_i \otimes L_j$$ a direct sum of line bundles. $\endgroup$ Nov 25, 2014 at 12:39

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