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For the usual $4$-dimensional Minkowski space $M$, the standard Dirac operator is given by $$ D: C^{\infty}(M) \to C^{\infty}(M), ~~~~~ f \mapsto \sum_{i=1}^4 \gamma_i\frac{\partial f}{\partial x_i}, $$ where the $\gamma_i$ are the usual gamma matrises. As we all know, $D$ was originally constructed as a square root of the Lapalcian $\sum_{i=1}^4 \frac{\partial^2 }{\partial x_i^2}$. Indeed, routine calculation will verify that $D$ does indeed square to give the Laplacian.

Moreover, there exists another square root of the Laplacian, namely $$ d + \ast d \ast: C^{\infty}(M) \to \Omega^1(M), $$ where $d$ is the usual exterior derivative, and $\ast$ is the Hodge $\ast$-operator.

How do these two square roots of the Laplacian relate to each other? Are they two separate objects, or just two ways of looking at the same thing?

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5 Answers 5

up vote 9 down vote accepted

As Sebastian and A.J. Tolland point out, the problem is that the two operators act on different bundles. However, the exterior algebra is canonically isomorphic to the Clifford algebra as vector spaces. Using this identification, you find two operators $D$ and $d+d^{\ast}$. Both operators have first order and they have the same principal symbol. Hence their difference is a tensor, i.e. linear over $C^{\infty}$-functions. Finally, both kill translation-invariant forms. Since invariant forms span all forms (over $C^{\infty}$-functions, they agree.

On a general manifold, there might be a problem due to curvature, but if I remeber correctly, they are equal on all Lorentz-manifolds. See Lawson-Michelson.

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I see that identifying the Clifford and exterior bundles allows us to view the two operators as operating on the same space. However, I don't see that they are equal: $D$ sends zero forms to zero forms, while $d+d^{\dagger}$ sends zero forms to one forms; how can they be equal? –  John McCarthy Mar 23 '11 at 12:26
    
The identification -- call it $i$ -- maps sections of one bundle to sections of the other. So the image $i(\psi)$ of a section $\psi$ of the Clifford bundle is a differential form. This form is, in general, of mixed degree, a section of $\Omega^* = \oplus_p \Omega^p$. –  userN Mar 23 '11 at 12:44
    
What's the explicit action of $i$ on a smooth function? –  John McCarthy Mar 23 '11 at 13:19
    
On a smooth function? i's domain is sections of the Clifford bundle. I think maybe you should have a look in Michelson-Lawson. –  userN Mar 23 '11 at 13:47
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First of all, your formulas looks odd, I guess you mean $d+\delta$ on $\Omega^*\to\Omega^*.$ Moreover, Dirac's Dirac operator does not map functions to functions, but sections in a bundle on which the gamma matrices act to itself. This is also the starting point for the whole story: to have some kind of geometric Dirac operator, on needs a bundle on which the tangent bundle acts by Clifford multiplication. This follows directly from the formula for the symbol of the square of an operator, together with the well-known formula for the symbol of an laplace. There is not a unqie choice of such a so Clifford module, for example, one can always tensor with some twisting bundle to obtain a new CLifford module. Nevertheless, under some topological assumptions (first and second Stiefel Whitney classes vanish), there exist the most simplest bundles on which the tangent bundle can act, the spinor bundels. Every other clifford modul is given by them through twisting. On these spinor bundle, there exist the raw dirac operator. Every other geometric Dirac operator is obtained form that one by the choice of an connection on the twisting bundle. FOr details, Lawson and Michaelson may be the best reference.

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Sebastian points out correctly that $D$ and $d+d^{\dagger}$ are not the same objects. $D$ acts on spinors; $d + d^{\dagger}$ acts on differential forms.

Physicists do sometimes make use of a trick relating spinors and differential forms: The Clifford algebra's underlying vector space is isomorphic to the fiber of the bundle of forms, and one can choose this isomorphism in such a way the Dirac operator becomes $d-d^{\dagger}$. (This is basically the A-twisting of spinors which is used in topological field theory.) In the literature on lattice QFT, the resulting differential forms are sometimes called Dirac-Kahler fermions.

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I second (third) the recommendations of Lawson-Michelsohn as a primary reference. For a book with less focus on the algebra, you could try "Elliptic operators, topology and asymptotic methods" by John Roe. There is, however, a couple of Wikipedia entries that you might find interesting.

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The links are broken. Here are the correct links: en.wikipedia.org/wiki/Lichnerowicz_formula en.wikipedia.org/wiki/Weitzenböck_formula –  Spiro Karigiannis Mar 23 '11 at 0:12
    
Fixed. Thanks for catching that. –  Raeder Mar 23 '11 at 8:11
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I am currently working through these ideas myself, trying to separate actual geometry from mere convention, so that I can consider various ways to dicretize Dirac operators.

As far as i understand so far, when Dirac wrote his 1928 paper, he was not thinking about manifolds, let alone bundles or sections. The paper uses a particular representation of a Clifford algebra which is appropriate to a Minkowski signature, hence spinors, etc. But even this was recognized in hindsight.

I personally would like to consider Johannes' ideas on the canonical isomorphism between exterior and Clifford algebras, but know that there is more to it than that. So, to me, another good reference is the book "Geometric algebra for physicists" by Doran and Lasenby.

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