# 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.

• Thank you for your answer. Can you explain in more detail, how you see in which way the Clifford multiplication changes if you change the orientation? Does the connection on the spinor bundles stay the same? I have edited my starting post in order to explain my thoughts on the topic in more detail. Maybe you can read through it and help me understand how this is related to your answer. – uro Nov 29 '15 at 9:53