A Killing spinor on a Riemannian spin manifold is a section of the spinor bundle satisfying the equation:
\begin{align*}
\nabla_X\phi=\lambda X\cdot\phi
\end{align*}
Here $X$ is a vector field and $\phi$ is the spinor field.
Now I want to know if killing spinors have always constant norms, and what about the norms of the positive and negative parts, i.e., if $\phi=\phi_++\phi_-$, are the norms of $\phi_+$ and $\phi_-$ norms also constant?
What I can see is that if $\phi$ is a real killing spinor ($\lambda$ is a real constant number), then it has constant norm:
\begin{align*}
\nabla_X\|\phi\|^2&=\langle \nabla_X\phi,\phi\rangle+\langle\phi,\nabla_X\phi\rangle=\langle \lambda X\cdot\phi,\phi\rangle+\langle \phi, \lambda X\cdot\phi\rangle=0
\end{align*}
But what if $\lambda$ is not real, what about norms of the positive and negative parts?
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$\begingroup$ A good example to test your intuition is a Killing spinor on $S^{2n}\subset\mathbb R^{2n+1}$. As far as I remember, the spinor bundle is trivial, with fibre the spinor module of $\mathbb R^{2n+1}$. The grading at $x\in S^{2n}$ is induced by Clifford multiplication with $x$, so as a graded bundle, the spinor bundle is no longer trivial. Killing spinors $\phi$ are constant, but $\phi_\pm$ is not. $\endgroup$– Sebastian GoetteCommented Feb 13, 2023 at 16:08
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$\begingroup$ @SebastianGoette Killing spinors on $S^{2n}$ comes from parallel spinor of the cone of $S^{2n}$, i.e., $\mathbb{R}^{2n+1}$. Parallel spinors on $\mathbb{R}^{2n+1}$ are just constants. So, doesn't it imply that the norm of the pullbacks also are constant for indivisual positive and negative parts? $\endgroup$– ParthaCommented Feb 13, 2023 at 19:01
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$\begingroup$ No. $\Sigma^\pm$ are the $\pm i$ eigenspaces of Clifford multiplication with the radial vector of the cone. And the radial vector changes. Consider $S^2$. One spinor module can be identified with $\mathbb H$ where Clifford multiplication by the unit vectors in $\mathbb R^3$ acts as $I$, $J$, $K$ from the left and the complex unit $i$ acts on the right. The constant spinor $1$ lies in the $S^+$ at $e_1\in S^2$, and in $S^-$ at $e_2\in S^2$. $\endgroup$– Sebastian GoetteCommented Feb 14, 2023 at 7:33
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$\begingroup$ @SebastianGoette Thanks for the comment. I very much agree that the positive and negative spinors themselves need not be constant, my question is in your example are their norms constant? $\endgroup$– ParthaCommented Feb 14, 2023 at 15:04
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