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First, I have to admit that I don't have much knowledge of Spin Geometry and Index Theory, the question could be too simple or naive and secondly there may be too many questions.

Let $D$ be the Dirac Operator for standard metric and $S$ be the Spin bundle on $S^{2}$.There is a unique Spin structure on $S^{2}$. How does $D$ look like, Can we write a general form for the harmonic spinors on $S^{2}$ ?

What is the general expression for equivariant index of $D$ ?

If $W \times S$ is the twisted spinor bundle and $D$ is the twisted Dirac operator, we can write the equivariant index as an integral over the fixed point manifold in terms of equivariant Chern Character of $W$ and $\hat{A}$ - genus of the fixed point manifold using the Aiyah-Segal-Singer theorem. What I am interested in is the final expression for $S^{2}$.

What can we say about the product of n $S^{2}$'s.


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up vote 3 down vote accepted

As Sebastian said, the Dirac operator can be identified with the $\bar{\partial}$-operator on the line bundle of degree $-1$ (the inverse of the Hopf bundle) and so its (equivariant) index is trivial. Now you ask for the equivariant index of the twisted Dirac. The reasonable group to study equivariance is $SU(2)$, because $SO(3)$ does not act on the spinor bundle. First note the the $K$-group of equivariant vector bundles on $S^2$ is $K_{SU(2)} (S^2) = KU_{SU(2)} (SU(2)/U(1)) \cong K_{U(1)} (*) = RU(1)$. Therefore, any $SU(2)$-equivariant vector bundle splits as a sum of line bundles, and the line bundles are precisely all powers of the Hopf bundle. So this reduces the problem to the study of the kernel of $\bar{\partial}$ on any power of the Hopf bundle, but as a $SU(2)$-representation. It is a pleasant exercise to prove that the space of holomorphic sections on the $k$th power of the Hopf bundle is canonically isomorphic to the space of degree $k$ homogeneous polynomials in two variables and the $SU(2)$-action on that is given by the action on the variables. So you get all irreducible representations of $SU(2)$ as the kernel of a twisted equivariant Dirac operator. This is of course not an accident, and you can read more about this close connection between index theory and representation theory in the papers (I gues that you now some of them):

G. Segal: ''Equivariant $K$-theory'', ''The representation ring of a compact Lie group'' R. Bott: ''Homogeneous vector bundles'' M. Atiyah, R. Bott: ''Lefschetz fixed point formula of elliptic complexes''

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thanks for your reply, very enlightening. – J Verma Apr 16 '11 at 20:23

There are many questions here. I can answer the first one quickly: there are no harmonic spinors on $S^2$ with the standard round metric. This follows from Lichnerowicz theorem. A good place to read about this is Nigel Hitchin's thesis Harmonic spinors published in Adv. Math. (1974) vol. 14 pp. 1-55. (MathSciNet link). More generally, on any compact spin manifold with positive scalar curvature the kernel of the Dirac operator is zero.

Concerning the other questions, you can try looking at this paper of Christian Bär, where the spectrum of the Dirac operator on space forms (particularly on round spheres) is determined: The Dirac operator on space forms of positive curvature., Journal of the Mathematical Society of Japan, vol. 48, No. 1, 1996. (link)

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@ Jose - Thanks for your reply and for the links. What can we say about the harmonic spinors if the metric on $S^{2}$ is not the standard one, say induced as a submanifold of $\mathbb{R}^{4}$. – J Verma Apr 15 '11 at 2:13
My guess is that if you cannot control the curvature (say, to be non-negative and positive at some point) then you cannot say much. Also computing the spectrum of operators such as Dirac, laplacian,... in the absence of symmetry is usually hopeless. – José Figueroa-O'Farrill Apr 15 '11 at 2:28
@ Jose - It seems that the kernel of Dirac operator is zero for any metric on $S^{2}$. Are there any cases if this is non-zero. Also in case the Spin bundle is twisted by a line bundle and then is it true that the kernel of twisted Dirac operator is also zero. Thanks for your time. – J Verma Apr 15 '11 at 2:48

No, the kernel of the (non-twisted) Dirac operator is always trivial: Every metric on $S^2$ is conformally equivalent to the metric of constant curvature $1.$ Moreover, the Dirac operator has a nice transformation formula with respect to conformal changes of the metric (see for example Lawson-Michelson) which implies that the kernel is invariant under conformal changes of the metric (at least in Dimension 2, but I think it is true generally). Another way to see this is the following: Every (Riemannian) spinor bundle on a Riemann surface (equipped with a compatible Riemannian metric) is of the form $S\oplus S^* ,$ where $S$ is a Riemann surface spin bundle. Using the Riemannian metric, one can identify $\bar K=K^* $, and $S^* =\bar{K} S ,... $ In that form the Dirac operator is just given by the $\bar\partial$ on $S$ and similarly the induced Riemannian $\partial$ on $S^*,$ which is of course the adjoint operator with respect to the Riemannian metric. But the degree of $S$ on $P^1$ is -1, so there cannot be holomorphic sections, which implies ther is no kernel.

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@ Sebastian - But a twisted Dirac Operator on a twisted spinor bundle can have non trivial kernel. – J Verma Apr 15 '11 at 18:43

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