# Spaces of Killing spinors for different orientation

Simply put, I want to understand how a change of orientation on a Riemannian spin manifold can change the space of Killing spinors.

To be more precise:

Let $M$ be a spin manifold (i.e. the first and second stiefel whitney class of $TM$ vanish).

Assume that $M$ is simply connected, so for every Riemannian metric $g$ on $M$ and every orientation $o$ on $M$ we have (up to equivalence) exactly on spin structure. Denote the associated complex spinor bundle by $\Sigma(M,g,o)$.

($\Sigma(M,g,o)=Spin(M,g,o)\times_\rho \Sigma_n$ where $\rho\colon Spin(n)\rightarrow Aut_{\mathbb C}(\Sigma_n)$ is the complex spinor representation and $n=dim(M)$.)

Denote the two orientations on $M$ by $o_1$ and $o_2$. Let $\alpha\in\mathbb{C}$. Are the spaces of $\alpha$-Killing spinors of $\Sigma(M,g,o_1)$ resp. $\Sigma(M,g,o_2)$ related? For example, do they have the same dimension?

Edit: We have an isomorphism $F\colon SO(M,g,o_1)\rightarrow SO(M,g,o_2)$ given by $(b_1,\ldots, b_n)\mapsto (-b_1,b_2,\ldots, b_n)$. (Assuming $M$ is connected.) If the spin structure on $(M,g,o_1)$ is $\eta\colon Spin(M,g,o_1)\rightarrow SO(M,g,o_1)$, then we get a spin structure on $(M,g,o_2)$ by $F\circ \eta\colon Spin(M,g,o_1)\rightarrow SO(M,g,o_2)$. The respective spinor bundles are the same as vector bundles, $\Sigma(M,g,o_1)=\Sigma(M,g,o_2)$. Clifford multiplication changes because of the isomorphism $F$. Namely, if $(b_1,\ldots,b_n)$ is an oriented ON basis of $(M,g,o_1)$ and $\varphi\in\Sigma(M,g,o_1)=\Sigma(M,g,o_2)$, we get $b_1\cdot_1\varphi=-b_1\cdot_2\varphi$ and $b_j\cdot_1\varphi=b_j\cdot_2\varphi$ for $j=2,\ldots, n$ where $\cdot_i$ is the Clifford multiplication in $\Sigma(M,g,o_i)$, $i=1,2$.

For the corresponding connections $\nabla^{1}$ and $\nabla^{2}$ I don't get a relation that seems to help and I can't relate the Killing spinors of $\Sigma(M,g,o_1)$ to those of $\Sigma(M,g,o_2)$.

• The map $F$ in the edit is a bundle diffeo, but it does not intertwine the $SO(n)$-actions. You can get that if $n$ is odd by reversing the sign of all basis vectors. You cannot get it in general if $n$ is even because the Euler class would change its sign. However, the union of the two principal bundles become an $O(n)$ bundle. If you twist a $Spin(n)$-principal bundle with $Pin(n)$, you get a $Pin(n)$ principal bundle with an induced connection. You can still associate a spinor bundle with connection, and the second connected component becomes a $Spin(n)$ principal bundle for $o_2$. – Sebastian Goette Dec 1 '15 at 9:17

If $M$ is even-dimensional, Clifford multiplication with the Clifford volume element is an isomorphism between the spaces of Killing spinors for $\alpha$ and for $-\alpha$. If you change the orientation, you can actually leave the Clifford multiplication $c\colon TM\to\operatorname{End}\Sigma(M,g,o)$ untouched, or you replace it by $-c$. Then either the $\alpha$-Killing spinors stay $\alpha$-Killing spinors or become $-\alpha$-Killing spinors.
If $M$ is odd-dimensional and you insist that the Clifford volume element acts by $1$, then you have to replace $c$ by $-c$, and the Killing constant $\alpha$ also changes sign.