Let $(M,g)$ be a $d$-dimensional, oriented pseudo-Riemannian manifold, and $V$ the subbundle of $E=TM\oplus T^*M$ given by the graph of the musical linear isomorphism $g^\flat:TM\rightarrow T^*M$ associated to the metric $g$. The nondegeneracy of $g$ entails that $E$ decomposes as the Whitney sum $E=V\oplus V'$, where $V'$ is the graph of $-g^\flat=(-g)^\flat$. Indeed, the projections

$P_\pm(X\oplus\xi)=\frac{1}{2}(X\pm g^\sharp(\xi))\oplus(\xi\pm g^\flat(X))$

satisfy $P_-=\mathbb{1}-P_+$, $P_+(E)=V$ and $P_-(E)=V'$, where $g^\sharp:T^*M\rightarrow TM$ is the musical linear isomorphism associated to $g^{-1}$. Moreover, $E$ carries a canonical, pseudo-Riemannian metric $h$ of signature $0$


such that $(\mathbb{1}\oplus g^\flat)^*(h|_V)=g$. It can be shown that $E$ admits a spin structure associated to $h$ - for example, the (space of sections of the) exterior algebra bundle $\Lambda^*T^*M$ is a Clifford module, and its tensor product $\Lambda^*T^*M\otimes(\Lambda^dT^*M)^{1/2}$ with the real line bundle of half-densities over $M$ is a spinor bundle. Moreover, the space of spin structures on $(E,h)$ is an affine space modelled on the group $H^1(M,\mathbb{Z}_2)$ of real line bundles over $M$, which maps the corresponding spinor bundles onto each other by tensoring (the above preliminary results can be found on Chapter 2 of Marco Gualtieri's PhD thesis on generalized complex geometry, arXiv:math.DG/0401221).

Question(s): if $(M,g)$ admits a spin structure, does a choice of spin structure on $(TM\oplus T^*M,h)$ descend by restriction to $V$ to a choice of spin structure on $(M,g)$? Does this establish a one-to-one correspondence between both sets of spin structures? If so, how does this generalize to, say, generalized Riemannian metrics (i.e. rank-$d$ subbundles $W$ of a twisting of $TM\oplus T^*M$ by a Cech 1-cocycle $B$ with values at closed 2-forms, such that the restriction of $h$ to $W$ is positive definite)?

In other words, I want to know if, in the above convext, there is a specific converse to the well-known property that, given two pseudo-Riemannian vector bundles $V,V'$ and their Whitney sum $E=V\oplus V'$, a choice of spin structure for any two of these bundles uniquely determines a spin structure on the remaining one. In the case $V$ and $V'$ are Riemannian, this is Proposition 2.1.15, pp. 84-85 of H. B. Lawson and M.-L.Michelsohn, "Spin Geometry" (Princeton, 1989). See also M. Karoubi, "Algèbres de Clifford et K-Théorie". Ann. Sci. Éc. Norm. Sup. 1 (1968) 161-270.

More precisely, here we use the fact that there is a canonical isomorphism between $Spin(p,q)$ and $Spin(q,p)$ which covers the canonical isomorphism between $SO(p,q)$ and $SO(q,p)$ (see M. Karoubi, ibid.) to establish a canonical one-to-one correspondence between the set of spin structures on $(M,g)$ and the set of spin structures on $(M,-g)$. This correspondence, on its turn, is used to induce spin structures on $V$ and $V'$ from that of $(M,g)$ together with the orientation-preserving, isometric bundle isomorphisms $\mathbb{1}\oplus(\pm g^\flat)$. What I want to know is if, among the pairs of spin structures on $V$ and $V'$ which determine the given spin structure on $E$ in the above fashion, there is (only?) one pair which, once pulled back to $M$, is related by the above one-to-one correspondence between $Spin(p,q)$- and $Spin(q,p)$-structures.


1 Answer 1


OK, I am making an assumption: I can re-interpret the problem (using the musical isomorphism) as $V$ being the diagonal embedding of $TM$ inside $TM\oplus TM$ and studying spin structures on them.

Actually, it turns out (see comments) that this "re-interpretation" is slightly different from the original construction. But the main bulk still goes through:

I will restrict to $\dim M=3$, in which case our (closed oriented) manifold is always spinnable. A spin structure $\mathfrak{s}$ on $M$ induces a canonical spin structure $\mathfrak{S}_0=\mathfrak{s}\oplus\mathfrak{s}$ on $TM\oplus TM$, and this is actually independent of the choice of $\mathfrak{s}$ (these appear in the notion of a 2-framing on 3-manifolds, which Atiyah and Witten have used for some of their QFT studies). As a result, the "restriction" $\mathfrak{S}_0|_V$ on $M$ is ill-defined.
[proof of claim of canonical spin structure (learned from conversation with Rob Kirby): the spin structure fixes a trivialization over the 1-skeleton, and over circles there are two trivializations, so changing a trivialization of $\mathfrak{s}$ is doubled in $\mathfrak{s}\oplus\mathfrak{s}$ which modulo-2 is no change.]

This also implies that the "restriction" to each $TM$-summand is ill-defined. (What I know that works: the collar-neighborhood theorem does allow an induced spin structure on $T(\partial X$) from a spin structure on $TX$ thanks to the splitting $TX|_\partial=T(\partial X)\oplus\underline{\mathbb{R}}$ near the boundary.)

  • $\begingroup$ Sorry, how does $V'$ equal the graph of $(−g)^b$, and why would a choice of spin structure for $V$ fix one for $V'$? For instance, if I take the subbundle $V$ to just be one of the two $TM$-summands, then I can talk about Whitney sums of spin structures $s_i\oplus s_j$ on $TM\oplus TM$ with the summands acting independently. $\endgroup$ Commented Dec 10, 2012 at 22:01
  • $\begingroup$ The first fact comes from the fact that we have a bundle projection $TM\oplus T^*M\ni(X,\xi)\mapsto P(X,\xi)=\frac{1}{2}(X+g^\sharp(\xi),\xi+g^\flat(X))\in V$ onto $V$ (here $g^\sharp:T^*M\rightarrow TM$ is the "musical" linear isomorphism induced by $g^{-1}$), which satisfies $(\mathbb{1}-P)(X,\xi)=\frac{1}{2}(X-g^\sharp(\xi),\xi-g^\flat(X))$, that is, $\mathbb{1}-P$ is a bundle projection onto $V'$. Hence we have $TM\oplus T^*M=P(TM\oplus T^*M)\oplus(\mathbb{1}-P)(TM\oplus T^*M)=V\oplus V'$. $\endgroup$ Commented Dec 11, 2012 at 0:29
  • $\begingroup$ As for the second fact, the reason is that the spin groups associated to the nondegenerate quadratic forms $Q$ and $-Q$ are canonically isomorphic, and the isomorphism descend to a canonical isomorphism between the corresponding special pseudo-orthogonal groups. One can use this isomorphism to build canonical pairs of principal bundle isomorphisms (one between spin principal bundles, other between orthonormal frame bundles) which commute with the corresponding spin covering maps and hence establish an one-to-one correspondence between the set of spin structures on $(M,g)$ and on $(M,-g)$. $\endgroup$ Commented Dec 11, 2012 at 0:30
  • $\begingroup$ Now I'm trying to understand your assumption a little better. There is no canonical (i.e. $g$-independent) metric $h$ on $TM\oplus TM$ as there is on $TM\oplus T^∗M$ above, but one can induce one from $g$ (with twice the signature of $g$) by setting $h(X_1\oplus X_2,Y_1\oplus Y_2)=\frac{1}{2}(g(X_1,Y_1)+g(X_2,Y_2))$. The factor $\frac{1}{2}$ guarantees that the diagonal map is isometric. There is a projection onto $V$ by setting $P(X,Y)=\frac{1}{2}(X+Y,X+Y)$, which satisfies $(\mathbb{1}−P)(X,Y)=\frac{1}{2}(X−Y,Y−X)$. That is, $\mathbb{1}−P$ is the projection onto the twisted diagonal $V'$. $\endgroup$ Commented Dec 11, 2012 at 1:37
  • $\begingroup$ Indeed, there seems to be some differences between your construction and mine. First, in your case the restriction of $h$ to $V'$ pulls back to $g$ under the twisted diagonal map $X\mapsto (X,-X)$, whereas in my case the pullback of $h$ yields $-g$. Moreover, the spin structures of $TM\oplus TM$ and $TM\oplus T^*M$ are completely different, due to the fact that the signature of $h$ is twice the signature of $g$ in the first case and zero in the second. In particular, the topological obstructions to the existence of spin structures and their classification also differ. $\endgroup$ Commented Dec 11, 2012 at 1:49

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