20

No, this is not true: for each dimension $d \geq 4$, there is a closed, oriented $d$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.
The reason is simply that there are finitely presented groups which have no nontrivial finite quotient.
One example is Higman's group $H$, see
https://en.wikipedia....

19

Here's my favorite way to answer your question. Hopefully the answer to Robert Bryant's question is "yes".
Let $A$ be the ring of octonions (the "nonsplit" octonions over ${\mathbb R}$); it comes with an involution $\alpha \mapsto \bar \alpha$, from which there is a trace $Tr(\alpha) = \alpha + \bar \alpha$ and a norm $N(\alpha) = \alpha \cdot \bar \alpha$....

17

I can't give a comprehensive history (if you don't get that here, you might try [hsm.se]---a lot of mathematicians are active on that site), nor can I explain how or why the theory of spin manifolds first emerged. But I think I can say something about how and why spin manifolds became important.
The pre-history is an observation due to some combination of ...

answered Oct 14 '15 at 17:49

Paul Siegel

25.3k1414 gold badges104104 silver badges182182 bronze badges

17

I believe the bordism groups are nonzero in every dimension after some relatively small finite dimension, just by looking at the Poincaré polynomial in Anderson-Brown-Peterson's earlier paper "Spin Cobordism".
However, from their splitting result, all of these groups come from suspensions of Eilenberg-MacLane spectra. In particular, the image of the ...

17

Let $i$ denote an immersion $N \to M$. There is an exact sequence of vector bundles on $N$ given by
$$0 \to TN \to i^*TM \to \nu \to 0$$
where $\nu$ is the normal bundle. As total Stiefel-Whitney classes are multiplicative in short exact sequences (alternatively, $i^*TM \cong TN\oplus\nu$ smoothly), it follows that
\begin{align*}
i^*w_1(M) &= w_1(N) +...

answered Apr 30 '17 at 22:48

Michael Albanese

13.7k66 gold badges6464 silver badges113113 bronze badges

15

I'm a bit wary of resurrecting such an old question, but given that the precise content of the reconstruction theorem doesn't seem to be terribly well disseminated, please permit me to cross-post from math.SE and then make some extra comments:
"To be absolutely clear about the state of the art, Connes's theorem actually tells you the following:
A unital ...

15

Define the Wu manifold $W(n) = SU(n)/SO(n)$, the inclusion $SO \to SU$ given by thinking of $\Bbb C^n = \Bbb R^n \otimes \Bbb C$ (that is, including real matrices into complex matrices). Note that $W(1) = *$, $W(2) = S^2$, and $W(3)$ is what is usually called the Wu manifold.
There is a natural map $W(n) \to W(n+1)$, including $SU(n) \to SU(n+1)$ and then '...

14

The answer of course depends on the spin structure chosen. The paper
Johnson, Dennis,
Spin structures and quadratic forms on surfaces.
J. London Math. Soc. (2) 22 (1980), no. 2, 365–373.
proves that the set of spin structures on a closed surface $\Sigma_g$ of genus $g$ can be identified with the set of quadratic forms on $H_1(\Sigma_g;\mathbb{Z}/2)$. ...

14

A spin structure on a real vector space V equipped with a real quadratic form μ
is an invertible bimodule (i.e., a Morita equivalence)
from Cl(V,μ) to Cl(Rdim(V),ν).
Here ν is the direct sum of dim(V) copies of the canonical
quadratic form on R.
A spinc structure on a complex vector space V equipped with a complex quadratic form μ
is an invertible bimodule
...

answered Feb 24 '13 at 13:59

Dmitri Pavlov

21.2k44 gold badges4646 silver badges109109 bronze badges

14

The condition for having a $Pin^+$-structure is the vanishing of $w_2$, and for having a $Pin^-$-structure is the vanishing of $w_2 +w_1^2$, for manifolds of any dimension. This is because the Lie group $Pin^\pm(d)$ is a central extension
$$\mathbb{Z}/2 \longrightarrow Pin^\pm(d) \longrightarrow O(d)$$
so there are homotopy fibre sequences
$$BPin^\pm(d) \...

answered Dec 23 '16 at 7:54

Oscar Randal-Williams

15.9k22 gold badges5454 silver badges8282 bronze badges

13

The obstruction to spin$^c$ is often denoted by $W_3(M) \in H^3(M,\mathbb{Z})$. It is obtained as follows: Let
$$
\beta \colon H^2(M,\mathbb{Z}/2\mathbb{Z}) \to H^3(M,\mathbb{Z})
$$
be the Bockstein homomorphism obtained from the short exact sequence
$$
0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0
$$
Then we have $W_3(M) = \beta(w_2(M))$,...

13

As mentioned in Arun Debray's answer, a closed orientable smooth manifold $M$ is spin${}^h$ if and only if there is a principal $SO(3)$-bundle (or equivalently, an orientable real rank three bundle) $P$ such that $w_2(M) = w_2(P)$.
If $M$ is spin, then $w_2(M) = 0$. Taking $P$ to be the trivial bundle, we see that $w_2(M) = 0 = w_2(P)$, so $M$ is spin${}^h$....

answered Jul 7 '18 at 22:01

Michael Albanese

13.7k66 gold badges6464 silver badges113113 bronze badges

12

In addition to the above answers involving spinors and/or octonions, you might be interested in Cartan's original construction of the triality automorphisms, which is very explicit and takes just a couple of pages in his beautiful little paper Le principe de dualité et la théorie des groups simples et semi-simples (Bull. Sc. Math 49 (1925), 361–374). The ...

answered Dec 21 '12 at 15:28

Robert Bryant

84.2k77 gold badges255255 silver badges359359 bronze badges

12

Assume that $M$ is oriented throughout. Recall that $M$ has a $\text{Spin}^c$ structure iff the third integral Stiefel-Whitney class $\beta w_2 = W_3 \in H^3(M, \mathbb{Z})$ is trivial. Actually more is true: $\text{Spin}^c$ structures on $M$ are in bijection with trivializations of $W_3$, which are a torsor over $H^2(M, \mathbb{Z})$. So we get a functorial ...

answered Jun 5 '16 at 19:27

Qiaochu Yuan

98.8k2929 gold badges355355 silver badges643643 bronze badges

11

Hormander's approach to solving the $\bar \partial$ problem is basically this, and his paper is from 1965, predating Witten's work by a couple of decades! By varying the "weight" function $h$, you can get families of estimates on the solution of $\bar \partial$ problem. Check out Hormander's 1965 ACTA paper for more details. Really a fabulous paper. He ...

answered Jan 15 '14 at 22:31

Steven Gubkin

10.7k22 gold badges6767 silver badges101101 bronze badges

11

Let $(M,g)$ be an orientable pseudo-riemannian manifold. Each tangent space $T_xM$ is a pseudo-euclidean space and hence has an associated Clifford algebra $CL(T_xM)$, which is the fibre at $x\in M$ of the Clifford bundle $Cl(TM)$. If the manifold is spin (a topological condition which says that the oriented orthonormal frame bundle lifts to a spin bundle) ...

answered Jan 9 '15 at 18:27

José Figueroa-O'Farrill

30.2k44 gold badges8585 silver badges162162 bronze badges

11

Here's a standard explicit formula: Let $\mathbb{O}\simeq\mathbb{R}^8$ denote the algebra of octonions, and for $x\in\mathbb{O}$, let $L_x$ (respectively $R_x)$ denote the linear map from $\mathbb{O}$ to itself generated by left (respectively, right) multiplication by $x$ and let $C:\mathbb{O}\to\mathbb{O}$ be conjugation in the octonions. Now define $\rho(...

answered Jan 19 '15 at 19:05

Robert Bryant

84.2k77 gold badges255255 silver badges359359 bronze badges

11

Spin geometry is an active field and of course is not exhausted in the book of Lawson and Michelson.
In fact, nowadays, there are new books on the topic, including more recent results.
In the introduction of N.Ginoux's book, one finds,
''...one of the most famous achievements of spin geometry was the discovery of a topological obstruction to positive ...

10

For a non-Spin$^c$-manifold, Poincaré duality is best understood in the wrold of KK-theory. In G. Kasparov's paper on KK-theory and the Novikov conjecture, he shows that for any smooth compact manifold $M$, $C_0(M)$ is Poincaré dual in KK to $C_0(T^*M)$. In particular, the K-homology of $M$ is isomorphic to the K-theory of $T^*M$. The isomorphism is ...

10

Chapter 9 of Elements of Noncommutative Geometry, by Gracia-Bondia, Varilly, and Figueroa, has this perspective on spin$^c$ and spin structures.
The way to think about this algebraically is that the module of (continuous, say) sections of a spinor bundle over a (compact, Riemannian) manifold $M$ is Morita equivalence bimodule for the algebras $C(M)$ and $...

10

As it stands, the second definition is a concrete description of the spin group in dimension four. It defines an action of the simply connected group $SU(2)\times SU(2)$ on a vector space of real dimension four, which preserves a positive definite inner product, and this identifies $SU(2)\times SU(2)$ with the universal covering of the special orthogonal ...

10

From the perspective of the Gelfand–Naimark theorem, the heart of the reconstruction theorem is the following statement, Theorem 11.4 in Connes's paper:
Let $\mathcal{A}$ be a commutative unital complex $\ast$-algebra. Then $\mathcal{A} \cong C^\infty(X)$ for some compact oriented smooth $p$-manifold $X$ if and only there exist a faithful $\ast$-...

10

I would think that these these notes by Akhil Mathew provide the "exact definition" you are asking for:
In response to the follow-up question "which is the first Clifford module used in the physics context":
Two different representations (modules) of the Clifford algebra were studied in early work on the Dirac equation. Paul Dirac himself used a ...

answered Jan 9 '15 at 18:24

Carlo Beenakker

113k1111 gold badges272272 silver badges392392 bronze badges

10

I'm happy to be able to answer my own question! John Ratcliffe, Steven Tschantz and I showed that the Dirac operator on the Davis manifold (a closed hyperbolic 4-manifold constructed by Mike Davis) has non-zero kernel. This is in the preprint Harmonic spinors on the Davis hyperbolic 4-manifold that we posted on the arxiv today.
The method is to find a ...

10

For $d \geq 4$, let $M_d = (S^1)^{d-4}\times\mathbb{CP}^2$. As tori are parallelisable, $w(M_d) = w(\mathbb{CP}^2)$, in particular $w_1(M_d) = 0$ and $w_2(M_d) \neq 0$, so $M_d$ does not admit a Spin, Pin$^+$, or Pin$^-$ structure.
Therefore, there is no $d \geq 4$ such that every $d$-dimensional manifold admits a Pin$^+$/Pin$^-$ structure.

answered Dec 23 '16 at 15:07

Michael Albanese

13.7k66 gold badges6464 silver badges113113 bronze badges

9

For any finite-dimensional associative unital $\mathbb{R}$-algebra $A$, the set $S$ of noninvertible elements is the zero-set of a polynomial. Namely, let $f:A \to Hom (A,A)$ be the map $a \mapsto ( x \mapsto ax)$ to the linear endomorphisms of $a$. $f$ is an injective algebra homomorphism, since $f(a)=0$ means $ax =0$ for all $x \in A$, in particular $a= a1 ...

answered Dec 13 '14 at 15:47

Johannes Ebert

18.9k22 gold badges6262 silver badges105105 bronze badges

9

The Brauer-Picard 2-category of $SuperVect_{\mathbb R}$ (let's call it $sBrPic_\mathbb R$) is the homotopy fixed points of the Brauer-Picard 2-category of $SuperVect_{\mathbb C}$ w.r.t. the involution given by complex conjugation (let's call that involution $C$).
It is plausible that the Brauer-Picard 2-category of $SuperVect_{\mathbb H}$ (call it $sBrPic_\...

answered Jan 7 '16 at 23:49

André Henriques

37.8k55 gold badges101101 silver badges232232 bronze badges

9

The spin cobordism groups $\Omega^{spin}_n$ have been computed for $n \leq 127$; see section 10 of Secondary Invariants for String Bordism and tmf by Bunke and Naumann. They use MAPLE together with the decomposition of the 2-completion of $MSpin$ found by Anderson, Brown, and Peterson in their paper Spin Cobordism.
In particular, one sees that $\Omega^{spin}...

answered Mar 24 '19 at 19:46

Michael Albanese

13.7k66 gold badges6464 silver badges113113 bronze badges

9

SpinH-structures were studied by Shiozaki-Shapourian-Gomi-Ryu for applications to condensed-matter physics. They prove in Lemma D.9 that a closed manifold $M$ admits a spinH-structure iff it's orientable and there's a principal $\mathrm{SO}_3$-bundle $P\to M$ such that $w_2(P) = w_2(TM)$.
$\newcommand{\CP}{\mathbb{CP}}$This implies $\CP^2$ is spinH but not ...

9

You've got the right concepts, but they're presented in a way that makes me think some context could be helpful.
In #1, you're really talking about the special case where $V$ is one of the spinor representations. In this case, yes, the fermionic fields are sections of the associated spinor bundle $\mathcal{V}$. Sometimes, people write $\Pi \mathcal{V}$ to ...

Only top voted, non community-wiki answers of a minimum length are eligible

#### Related Tags

spin-geometry × 159dg.differential-geometry × 72

at.algebraic-topology × 24

riemannian-geometry × 21

lie-groups × 17

clifford-algebras × 17

gt.geometric-topology × 14

complex-geometry × 14

mp.mathematical-physics × 14

reference-request × 13

rt.representation-theory × 12

characteristic-classes × 11

smooth-manifolds × 9

vector-bundles × 9

index-theory × 9

ag.algebraic-geometry × 8

kt.k-theory-and-homology × 7

cobordism × 7

exceptional-groups × 7

spinor × 7

differential-topology × 6

kahler-manifolds × 6

sg.symplectic-geometry × 5

sp.spectral-theory × 5

noncommutative-geometry × 5