Anti-Symmetry of Dirac Operator In his paper Fermion Path Integrals And Topological Phases, Witten states

“Whenever one has a theory of fermions, the quadratic part of the fermion action is always antisymmetric by virtue of fermi statistics and the corresponding fermion path integral is the Pfaffian of the antisymmetric bilinear form that appears in the action.”

I’m sure it is straightforward but how can we see this from the Euclidean Dirac action $\int d^dx \bar{\psi}D\psi$ that the Dirac operator is anti-symmetric? Does he mean anti-symmetric in all indices (gauge/spin etc)?
 A: Consider a  Grassman  linear operator "action"
$$
S=\bar\psi_i {L^i}_j\psi^j
$$
Here the index "$i$" include everything: spin label $\alpha$, spacetime $x$ colour etc: $i=(\alpha,x, c)$
for
$$
Z=\int d[\bar\psi_i] d[\psi^i] \exp\{S\}= {\rm det}[L]
$$
For Grassman $\chi^i$ and antisymmetic $Q_{ij}$ we have
$$
Z= \int d[\chi^i]\exp\left\{\frac 12  \chi^i Q_{ij}\chi^j\right\}
={\rm Pf}[Q].
$$
We can always write
$$
\bar \psi {L} \psi =\frac 12  [\bar\psi,\psi] \left[\matrix{0 & {L}\cr -{ L}^T &0}\right] \left[\matrix{ \bar\psi \cr \psi}\right].
$$
When  $L$ is $N$-by-$N$ we can now compute  the Pfaffian of the   skew symmetric matrix  and so find that
$$
{\rm Pf}\left[\matrix{0 & { L}\cr -{L}^T &0}\right]=(-1)^{N(N-1)/2} {\rm det}[ {L}], 
$$
The sign comes from the need to rearrange the terms in the measure $d\psi$ and $d\bar\psi$  so as  to put all the $d\bar\psi$'s before the $d\psi$'s  instead of in adjacent pairs. In my opinion the  dependence on $N$  makes this rewriting less transparent  in infinite dimensions, but Witten finds it useful.
There is discussion of some of this in the appendix to my paper here.
