Classical geometric interpretation of spinors - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:20:27Z http://mathoverflow.net/feeds/question/66681 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors Classical geometric interpretation of spinors Benjamin 2011-06-01T19:22:17Z 2012-06-14T09:10:13Z <p>A lot of notions in differential geometry have direct meaning in Physics. For example:</p> <ul> <li>A Riemannian metric is a way to encode distances on a manifold and in Physics it is the gravitational field. The curvature of the Levi-Civita connection gives the strength of the gravitation in a certain sense,</li> <li>A principal $G$-connection is a object that allows us to do parallel transport conveniently with respect to an action of a certain Lie group $G$, and in Physics it is a gauge field, that is a field that is related to a fundamental interaction, for instance a principal $U(1)$-connection can be seen as the electromagnetic field. The curvature of the connection gives the field strength, in a way.</li> </ul> <p>I would like to have an interpretation of what is a spinor field (when the manifold on which we are working admits a spin structure) in classical differential geometry, that is a section of the spinor bundle. By classical differential geometry I mean typical manifolds, not supermanifolds. This is because, for me, spinors in the theory of supermanifolds, play a different role, since in a way they are "odd spacetime coordinates". I am interested in the geometry of classical fields: a spinor field represents "matter" (fermions) whereas gauge fields (that is, principal connections) represent "forces" (bosons). But this is Physics. I am interested in a mathematical interpretation like:</p> <ul> <li>Riemannian metric = gravitational field = a way to measure distances,</li> <li>Principal connection = gauge field = a way to do parallel transport,</li> <li>Spinor field = matter field = what in Mathematics?</li> </ul> <p>So my questions are:</p> <p><em>In classical differential geometry (that is, ordinary manifolds), how can we interpret geometrically spinor fields? How can we interpret the spin connection and its curvature?</em></p> <p>Thanks.</p> <p><strong>EDIT:</strong> In a comment below I was saying that spinor geometry is of fundamental importance to the Atiyah-Singer theorem. So perhaps this gives a lead to other people to help me with the interpretation of spinors in classical differential geometry.</p> http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors/66697#66697 Answer by robot for Classical geometric interpretation of spinors robot 2011-06-01T23:51:55Z 2011-06-01T23:51:55Z <p>A spinor field in classical differential geometry is just a section of a spinor bundle, which is by definition associated bundle constructed from the spinor representation and the orthonormal frame bundle. I am not aware of any truly geometrical interpretations but I am not an expert. Maybe it can be worthwhile to look at low dimension, especially at dimension 4. </p> <p>The use of spinor fields in classical differential geometry is however vast and very much influenced by physics. A prototype result is that existence of special solutions of certain `geometrical' equations implies restrictions on the geometry of the manifold.</p> <p>Also, if you want to define square root of the Laplace-Beltrami operator (i.e. Dirac operator) then you are led to spinor bundle and spinor fields. </p> http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors/66728#66728 Answer by Igor Khavkine for Classical geometric interpretation of spinors Igor Khavkine 2011-06-02T13:04:56Z 2011-06-02T19:43:13Z <p>The following is an elaboration of the point of view that spinors are "square roots" of vectors (or rather of <em>isotropic vectors</em>). I will restrict my attention to 3-dimensional Euclidean vectors, because I am much less familiar with higher dimensional cases and do not want to claim too much. Even then, I don't consider myself an expert and am likely to get some things wrong.</p> <p>Consider the vector space $\mathbb{C}^3$ with the bilinear (not Hermitian) form $v\cdot w=v_1 w_1 + v_2 w_2 + v_3 w_3$. The group $O(3)$ is the group of linear transformations preserving this bilinear form. Let $N\subset \mathbb{C}^3$ be the subset of <em>isotropic</em> (or <em>null</em>) vectors, those satisfying $v\cdot v=0$. Since we are in a complex vector space, there are plenty of such vectors. In fact, they form a subvariety of dimension 2, which is invariant under the action of $O(3)$.</p> <p>For fun, we can consider the question (or approximations to it) of whether the 2-dimensional space $N$ can be polynomially or rationally parametrized by $\mathbb{C}^2$. Now, I will not dwell on this question in its generality and will only notice that $N$ does in fact admit a local cover (so not quite a parametrization) by $\mathbb{C}^2$. Except at $v=0$ that is; I'm sure there is a more correct term than local cover that takes this into account, but I don't know it. This cover map is given by a homogeneous quadratic polynomial and is 2-to-1. This space $\mathbb{C}^2$ is the usual spinor space and this cover map is well known.</p> <p>At this point, we can start playing all the usual games with the cover map $\mathbb{C}^2\to N$. For example, paths in $N$ can be lifted to paths in $\mathbb{C}^2$. In particular, a closed path in $N$, produced by the action on a vector of a closed path in $O(3)$ that starts and ends at the identity, will lift to a path in $\mathbb{C}^2$, which will not necessarily be closed. This is the well known property that spinors are negated by a $2\pi$ rotation.</p> <p>It is perhaps not surprising then that the action of $O(3)$ on $N$ can be lifted to the action of a cover of $O(3)$ on $\mathbb{C}^2$. What <em>is</em> surprising to me in this construction is that the lifted action on $\mathbb{C}^2$ is <em>linear</em>. Perhaps this property of the construction is obvious for some reason, but I can't quite put my finger on it.</p> <p>Now, for a 3-dimensional Riemannian manifold, the above construction can be generalized, fibre-wise, by complexifying its tangent bundle. Since parallel transport of vectors leaves the sub-bundle of isotropic vectors invariant, it can also be lifted to parallel transport of spinors.</p> <p>So, finally getting to your question, in classical differential geometry of Riemannian 3-manifolds, I think one can think of spinors and various actions on them as (lifts of) isotropic vectors and corresponding actions on them. The spinor point of view reveals some nice properties that are not apparent otherwise: (a) the nonlinear subvariety isotropic vectors can be (2-to-1, quadratically) parametrized by a linear space, (b) rotations lift to the parametrizing space as linear transformations. I believe these ideas generalize to higher dimensions, but I'm afraid I wouldn't be able to give any of the details.</p> http://mathoverflow.net/questions/66681/classical-geometric-interpretation-of-spinors/66734#66734 Answer by Paul Siegel for Classical geometric interpretation of spinors Paul Siegel 2011-06-02T14:12:43Z 2011-06-02T14:12:43Z <p>As far as I know, this sort of structure was first invoked by Dirac in order to take a square root of the Laplacian, and this he was doing in order to write down Lorentz invariant Klein-Gordon equations. It is a useful exercise to try to solve the equation $D^2 = \Delta$ on a Euclidean space $V$ for a first order operator $D$; you will find that the coefficients have to satisfy certain relations that cannot be satisfied by ordinary real or complex numbers. The algebraic structure required to obtain these relations is provided by an algebra $A$ with $V$ as a linear subspace such that $v^2 = -||v||^2 1$ in the algebra. In other words, you need to take a "square root" of your quadratic form.</p> <p>In brief, a spinor bundle on a Riemannian manifold is a setting for taking a square root of the Riemannian metric. To be precise, it is a bundle $S$ on which tangent vectors act as bundle morphisms in such a way that $v^2 s = -||v||^2 s$. In Dirac's equation, the coefficients of $D$ were given by certain matrices (the "Pauli spin matrices"), and thus he was thinking of $D$ as taking values in a vector space which carries a representation of the algebra $A$. Thus the spinor bundle is a global version of that vector space.</p> <p>That tells you what properties the spinor bundle is supposed to have, but it doesn't tell you what the bundle actually is. If you look it up in a book, you will find that the spinor bundle is an associated bundle to a principal $Spin(n)$ (or $Spin^c(n)$) bundle via the spin representation, but to me that is only a little more helpful than defining a Riemannian metric to be a reduction of structure group from the principal $GL(n)$-bundle of frames to a $O(n)$-bundle.</p> <p>Here is what I would consider to be a more concrete and well-motivated description. Let us return to the algebra $A$ associated to a Euclidean space $V = \mathbb{R}^n$ as above. The universal example of such an algebra is the Clifford algebra $Cl(V)$, equipped with a natural left action of $V$. Choosing an orthonormal basis for $V$, one can describe $\mathbb{R}_n := Cl(V)$ as the universal algebra over $\mathbb{R}$ generated by symbols $e_1, \ldots, e_n$ subject to the relations $e_j^2 = -1$ and $e_j e_k = -e_k e_j$ for $i \neq j$. It is not hard to see that $Cl(V)$ is isomorphic as a vector space (but not as an algebra) to the exterior algebra of $V$, and thus $Cl(V)$ inherits a natural $\mathbb{Z}/2\mathbb{Z}$ grading, given by products of even / odd numbers of generators. Notice that right multiplication by the $j$th generator is an odd anti-involution, so a choice of orthonormal basis for $V$ gives $Cl(V)$ the structure of a $n$-multigraded super algebra.</p> <p>We can define a (real) spinor bundle of a $n$-manifold to be a bundle which is locally isomorphic to the trivial bundle whose fibers are given by $\mathbb{R}_n$ equipped with a left action of the tangent bundle and a $n$-multigrading structure coming from a choice of local orthonormal frame. There is an obvious notion of complex spinor bundle as well: just use the complex Clifford algebra $\mathbb{C}_n$. Note that the fiber dimension of this bundle will be twice that of the bundle obtained via the spin representation, but the multigrading operators can be used to "reduce" my version of the spinor bundle down to the usual version. There are lots of reasons why I believe it is more convenient to think of a spinor bundle as a bundle of Clifford algebras with extra supersymmetry data, but I will briefly focus on a topological reason that I think cuts to the heart of the matter.</p> <p>The existence of a real spinor bundle on a manifold $M$ (a "Spin structure") is a rather severe condition. The complexification of a real spinor bundle is a complex spinor bundle, but not all complex spinor bundles ("Spin$^c$ structures") arise in this way. For example, any complex manifold has a spin$^c$ structure, but even $\mathbb{C}P^2$ fails to have a spin structure. An orientation on $M$ can be recovered from a choice of spin$^c$ structure, and indeed "spin$^c$-able" is only a little bit stronger than orientable - most orientable manifolds that you can name are probably spin$^c$-able. My point in bringing this up is to relate spinor bundles to K-homology, the generalized homology theory dual to topological K-theory. In ordinary homology theory, a choice of orientation on an $n$-manifold $M$ is the same thing as a choice of fundamental class in $H_n(M)$. Similarly, a choice of real / complex spinor bundle on a $n$-manifold $M$ is the same thing as a choice of fundamental class in the $n$th degree real / complex K-homology of $M$ (the multigrading data are crucial here). This observation is the starting point for some of the more conceptual proofs of the Atiyah-Singer index theorem, but this answer has gone on long enough. I hope it helps!</p>