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Dec 1, 2018 at 20:19 comment added Xenomorph @SebastianGoette Thank you for your comment. What you said is the version of APS theorem I first learned from physics books, but I couldn't get the $c_{2}(G)$. On the other hand, I obtained the factor $c_{2}(G)$ from a purely physical method, which is not mathematically rigorous.
Dec 1, 2018 at 19:15 comment added Sebastian Goette If both operators act on $\Omega^{\mathrm{odd}}(M;\mathfrak g)$, then we can consider the signature operator (not the spin Dirac operator, as you suggested) on $W=M\times[0,1]$, twisted by $W\times\mathfrak g$. On the twist bundle, we choose a connection that interpolates between $\nabla^0=d$ over $M\times\{0\}$ and $\nabla^1=d+A$ over $M\times\{1\}$. Then eqn 2.16 should follow from the APS theorem. From the integrand on $W$, only the part containing $c_2(\nabla)$ contributes. The number $c_2(G)$ is probably needed to get the precise second Chern form, but I have not checked the details yet.
Nov 30, 2018 at 17:50 comment added Xenomorph @SebastianGoette $L$ is acting on $\Omega^{\mathrm{odd}}(M,\mathfrak{g})$, where $\mathfrak{g}$ is the Lie algebra of $G$. I already gave the definition of $E$ in my post. No there is no $\mathrm{rank}(E)$ in $2.16$ in Witten's paper. I had already proven it myself last week but I haven't posted my proof.
Nov 30, 2018 at 9:37 comment added Sebastian Goette Could you please specify what space $L$ acts on? Is $E$ supposed to be a vector bundle over $M$ with structure group $G$, and $L$ acts on $\Omega^{\mathrm{odd}}(M;E)$? In that case, maybe there is a rk$(E)$ missing eqn 2.16? Otherwise, I have no idea yet how the $G$-connection $A$ acts.
Nov 22, 2018 at 3:31 history asked Xenomorph CC BY-SA 4.0