Non-Abelian anomaly: why does non-Hermitian operator have complete basis of eigenvectors?

Question

In section 13.3 of his book [1], Nakahara computes the non-Abelian anomaly for a chiral Weyl fermion coupled to a gauge field by making use of an operator $$ \mathrm{i}\hat{D} = \mathrm{i}\gamma^\mu (\partial_\mu+\mathrm{i}\mathcal{A}_\mu \mathcal{P}_+) = \left(\begin{matrix}0 & \mathrm{i}\gamma^\mu\partial_\mu P_- \\ \mathrm{i}\gamma^\mu\nabla_\mu P_+ & 0\end{matrix}\right),\tag{13.43} $$ where $P_{\pm} = \frac{1}{2}(1+\gamma^5)$ are the chirality projection operators. Importantly, this operator is non-Hermitian, so its left and right eigenvectors may differ. These are defined as \begin{align} &\mathrm{i}\hat{D}\psi_i = \lambda_i\psi_i, \tag{13.44a}\newline\\ &\chi^\dagger_i(\mathrm{i}\overset{\leftarrow}{\hat{D}}) = \lambda_i \chi^\dagger_i, && (\mathrm{i}\hat{D})^\dagger\chi_i = \bar{\lambda}_i \chi_i. \tag{13.44b} \end{align} Nakahara then claims that these eigenvectors are complete, i.e. that $$ \sum_i \psi_i\chi_i^\dagger = 1. $$ How do we know this to be true? $\mathrm{i} \hat{D}$ is not Hermitian, so the usual theorems don't apply.

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[1] M. Nakahara, Geometry, topology, and physics, 2nd edition, Taylor & Francis, 2003

Answer 1

Consider a finite-dimensional non-Hermitian matrix ${\bf M}$ acting on a space $V$. Although such a matrix is not guaranteed to be diagonalizable, it will be in the generic case in which all its eigenvalues are distinct. It then possesses complete sets of left and right eigenvectors ${\bf u}_n$, ${\bf v}_n$ obeying $$ {\bf M}{\bf u}_n= \lambda_n {\bf u}_n,\nonumber\\ {\bf v}_n^\dagger {\bf M}= \lambda_n {\bf v}_n^\dagger.\nonumber $$ We have not distinguished between the left and right eigenvalues because the two sets of eigenvalues coincide: they are determined by the same characteristic equation. Although a right (left) eigenvector with eigenvalue $\lambda_n$ is no longer orthogonal to all right (left) eigenvectors with a different eigenvalue $\lambda_m$, it remains true that a {\it left/} eigenvector with eigenvalue $\lambda_n$ is orthogonal to every {\it right} eigenvector with a different eigenvalue $\lambda_m$. This is because $$ \lambda_n {\bf v}_n^\dagger {\bf u}_m= ({\bf v}_n^\dagger {\bf M}){\bf u}_m={\bf v}_n^\dagger ( {\bf M}{\bf u}_m) = \lambda_m {\bf v}_n^\dagger {\bf u}_m, $$ so $$ 0=(\lambda_n-\lambda_m) {\bf v}_n^\dagger {\bf u}_m. $$ This equation yields a proof of the equality of the left and right eigenvalues that does not require an appeal to the finite-dimensional notion of a characteristic polynomial: if ${\bf u}_m$ did not have a corresponding ${\bf v}_m$ with the same eigenvalue then ${\bf u}_m$ would be perpendicular to all vectors in $V$ --- in contradiction with the non-degeneracy of the inner product.

We may choose the phases and normalization of the two eigenfunction sets so that $$ {\bf v}_n^\dagger {\bf u}_m= \delta_{mn} $$ whence ${\bf v}_n^\dagger {\bf M} {\bf u}_m= \lambda_n \delta_{nm}$ which we can write as
$$ {\bf V}^\dagger {\bf M} {\bf U}= {\rm diag}(\lambda_1,\lambda_2,\ldots). $$ Although neither of the eigenvector families compose an orthonormal set, the matrices ${\bf U}$ and ${\bf V}$ obey ${\bf V}^\dagger {\bf U}= {\mathbb I}$ so we still have
$$ {\rm det\,}{\bf M}= {\rm det\,}({\bf V}^\dagger {\bf M} {\bf U}) = \prod_n\lambda_n. $$