Parity Anomaly and Gauge Invariance

To answer you Question 0, Yes you can use a real mass regulator. In that case after regularization the parity violating part of the partition function will be $$ \exp\left[\frac{1}{2}\lim_{M\to \infty}\sum_{i}\log \frac{\lambda_i+M}{\lambda_i-M}\right] = \exp\left[\lim_{M\to \infty}\sum_{i}\tanh^{-1}\left(\frac{M}{\lambda_i}\right)\right], $$ as opposed to $\exp[i\tan^{-1}(M/\lambda_i)]$ you would get from an imaginary mass. But you get the same result, since $$ \lim_{M\to \infty}\sum_{i}\tanh^{-1}\left(\frac{M}{\lambda_i}\right)=\frac{\pi i}2 \sum_{i}\mathrm{sgn}(\lambda_i)=\frac{\pi i} 2 \eta. $$ Although $\tanh^{-1}(M/\lambda)$ itself requires some regularization, which is nothing but adding a small imaginary part to $M$.

Basically parity anomaly says no matter how you regulate the massless Dirac fermion in 3d, you violate reflection symmetry.

I meant to answer this when it was first asked...

Witten is doing his path integral in Euclidean space. The point is that in Euclidean space you want the eigenvalues of the derivative part of the Dirac equation to have a factor of $i$ compared to the "$m$" part so that the denominator in the propagator never vanishes. Indeed the whole point of Wick rotation is to get rid of necessity of the $i\epsilon$'s that take care of the vanishing denominator that occurs in Minkowski signature. With $$ \gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu= 2\delta^{\mu\nu} $$ we can take the complete opertor to be $\gamma^\mu\partial_\mu+m$ so, with $\partial_\mu \to ip_\mu$ when acting on plane-waves, the free-field Euclidean signature propagator becomes $$ \frac{1}{i \gamma^\mu p_\mu+m}= \frac{-i\gamma^\mu p_\mu+m}{p^2+m^2}. $$ The denominator $p^2+m^2$ is now safely non-zero.

The matrix $i \gamma^\mu p_\mu$ is skew Hermitian and so has purely imaginary eigenvalues $ \pm i\sqrt{p^2}\equiv\pm i\sqrt{E^2+|{\bf p}|^2}$.

For fermions interacting with background gauge fields, we still have a skew-hermitian derivative part for the Dirac opertor, the eigenvalues $i\lambda_n$ are pure imaginary, and the Matthews-Salam determinant is $$ {\rm Det} (\gamma^\mu \nabla_\mu+m)=\prod_n (i\lambda_n+m). $$ Witten likes to multiply his whole action functional by $i$, so his derivative $i\gamma^\mu \nabla_\mu$ is Hermitian --- but then his mass must be pure imaginary. The overall factor of $i$ makes no difference to anything as it just gives a constant factor tha cancels out when you compute a physical quantity.