Let $M$ be a closed $4$d Riemannian manifold and $Z$ be its twistor space of $M$, i.e., the bundle of almost complex structures on $M$. Let $V$ be a Spin$^{\mathbb{C}}$ bundle, $V_+$ denote the positive spin bundle. We know $Z$ admits more than one almost complex structure. So it can have canonical Spin$^{\mathbb{C}}$ bundle once we fix an almost complex structure and we can have other Spin$^{\mathbb{C}}$ bundles by twisting it with a complex line bundle. Now let $\pi:Z\rightarrow M$ be the projection. Can we realize $\pi^*(V_+)$ as a subbundle of some Spin$^{\mathbb{C}}$ bundle of positive spinors on $Z$ and if yes how?

$\begingroup$ What do you mean by a Spinᶜ bundle? If it is a vector bundle with a reduction of structure group from $GL(n)$ to $\operatorname{Spin}^c(n)$, this will be impossible for $n$ large enough by a simple dimension count. If it is the (positive and negative) spinor bundle defined by a Spinᶜstructure, this should follow from the splitting of $TZ = \pi^* TM\oplus L$, where $L$ is the second power of the complex line bundle whose fiber over $Z$ are (positive) spinors annihilated by the Lagrangian in $TX$ defined by the almost complex structure $Z$. $\endgroup$– Bertram ArnoldSep 14, 2021 at 8:01

$\begingroup$ @BertramArnold I meant the positive spinor bundle defined by a Spin$^\mathbb{C}$ structure. Can you please explain a bit about your comment as an answer? Also, I think you meant $M$ instead of $X$. Notice $\pi^*(V_+)$ is a complex line bundle on $Z$ of dimension $2$. $\endgroup$– ParthaSep 14, 2021 at 10:40
1 Answer
$\newcommand{\spinors}{\mathbb{S}}$The tangent bundle $TZ$ fits into an exact sequence $$ 0\to T_{/M}Z\to TZ\xrightarrow{\pi_*}\pi^*TM\to 0 $$ where the fiberwise tangent bundle $T_{/M}Z$ is twodimensional and equipped with a canonical complex structure. Explicitly, the fiber of $Z\to M$ over a point $m$ are the (oriented) complex structures on $T_mM$, which are isomorphic to the projective line of the twodimensional complex bundle $\spinors_+(M)\to M$ of positive spinors (which you denote by $V_+$). Note that this projective line is defined without the choice of a spin structure, as the action of $\operatorname{Spin}(4)\cong SU(2)\times SU(2)$ on $\mathbb{CP}^1$ factors through projection of the first factor to $PSU(2)$ and therefore descends to $SO(4)$. The isomorphism sends a complex structure to the line of spinors which are annihilated by Clifford multiplication with vectors $v\in T_mM\otimes\mathbb C$ which are antiholomorphic with respect to the chosen complex structure.
The exact sequence splits using the Euclidean metric on $TZ$, whose existence you take for granted when you talk about the spinor bundle of $Z$; to be precise, the splitting can be obtained via the LeviCivita connection, with the metric on $TZ$ defined as the orthogonal sum of the induced metric on $T_{/M}Z$ and the pulled back metric on $M$. The fiberwise tangent bundle has a complex structure from the description of the fiber as the projective line of a complex vector bundle, and the pullback $\pi^*TM$ has a canonical complex structure by definition of $Z$. It follows that the sum $TZ$ has a (almost) complex structure as well.
The spinor bundle of the sum of two evendimensional vector spaces is the tensor product of the respective spinor bundles. This shows that $\spinors_+(TZ)\cong \spinors_+(T_{/M}Z)\otimes \pi^*\spinors_+(TM)\oplus\spinors_(T_{/M}Z)\otimes \pi^*\spinors_(TM)$. Thus we can realize $\pi^*\spinors_+(TM)$ as a subbundle of the positive spinors on $Z$ if $\spinors_+(T_{/M}Z)$ is trivial, which we can achieve by a judicious choice of Spinᶜstructure on the vertical tangent bundle, coming from a $PU(2)$equivariant Spinᶜstructure on $\mathbb{CP}^1$; it is an easy exercise that such Spinᶜstructures are in bijection with the integers, with the positive spinor bundle given by the line bundle $\mathcal O(k)$ and the negative spinor bundle given by the line bundle $\mathcal O(k+2)$, which can be made equivariant iff $k$ is even, so that we may take $k = 0$.

$\begingroup$ Can you clarify "different choices of spin structure are related by an isomorphism which acts by a scalar multiple of the identity on...", because $\mathbb S_+(M)$ is attached to a given spin structure, so when you have two different spin structures I don't know what map you're talking about and how their projective lines are isomorphic. $\endgroup$ Sep 15, 2021 at 17:30

$\begingroup$ Right, sorry  locally, there's a unique spin structure which has a nontrivial automorphism, which acts by $\operatorname{id}$ on the spinor bundle and therefore trivially on its projective line. I updated the answer to clarify this point. $\endgroup$ Sep 15, 2021 at 19:42

$\begingroup$ @BertramArnold +1 for the answer but your description slightly disagrees with my intuition. So, at a point $m\in M$ and say $J$ being an almost complex structure at $m,$ $V_+=L_J\otimes(\Lambda^{0,0}\oplus \Lambda^{0,2}_J),$ where $L_J$ is a line depending on $J$ now as we pull it back to $Z,$ there should be a twisting somewhere as we are kind of varying the pull back as $J$ varies. Your description straightaway nullifies the varying over the $J$ part, which kind of contradicts with my understanding. $\endgroup$– ParthaSep 16, 2021 at 21:09

$\begingroup$ I guess the intuitive picture is that the line $L_J$ depends on the choice of Spinᶜstructure on the fibers; as I explained, it is exactly the positive spinor bundle of the fiber, which can be taken to be trivial. Note however that it is not possible to simultaneously realize the (pullback of the) positive and negative spinors as subbundles of the positive spinors on $Z$, as the positive and negative spinor bundle of the fiber are line bundles whose degrees differ by $2$. $\endgroup$ Sep 17, 2021 at 7:39