The harmonic spinor equation states that
$D \psi=0$,
where $D$ is a Dirac operator and $\psi$ is a spinor. A spinor satisfying this equation is said to be harmonic. Is there some result that one always define a harmonic spinor on a spin manifold when the appropriate Dirac operator is defined? I make no other assumptions about the spinor.
Apologies if the question does not make sense, as I do not know anything about spin geometry.
It is not true that every spin manifold has a non-zero harmonic spinor.
The Lichnerowicz formula for the Dirac operator $D$ is $D^2 = \nabla^*\nabla + \frac{1}{4}s$ where $s$ is the scalar curvature. It follows that if $M$ is a compact spin manifold with positive scalar curvature, then $M$ does not admit a non-zero harmonic spinor. To see this, let $\sigma$ be a harmonic spinor. As $s > 0$ and $M$ is compact, we have $s \geq 4c > 0$ for some constant $c$. Therefore, using the $L^2$ norm, we have
$$0 = \langle D^2\sigma, \sigma\rangle = \langle\nabla^*\nabla\sigma, \sigma\rangle + \langle\frac{1}{4}s\sigma, \sigma\rangle \geq \langle\nabla\sigma, \nabla\sigma\rangle + \langle c\sigma, \sigma\rangle = \|\nabla\sigma\|^2 + c\|\sigma\|^2.$$
As $\|\nabla\sigma\| \geq 0$, $\|\sigma\| \geq 0$, and $c > 0$, we see that $\sigma = 0$. Taking a little more care, the same conclusion holds if $s \geq 0$ but $s \not\equiv 0$.
For more details, see Spin Geometry by Lawson and Michelsohn, chapter II, section 8.
In the case that $s \equiv 0$, every harmonic spinor is parallel. The existence of a non-zero parallel spinor on a closed spin manifold $M$ places a significant restriction on its geometry. For instance, if $M$ is simply connected, either $M$ or $M\times S^1$ admits a Ricci-flat Kähler metric, see Theorem 1.2 of Hitchin's Harmonic Spinors.
