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Many definitions of the Dirac operator in the tradition of the Physics literature are hard to grasp for a mathematician. I would like to ask for a precise, general, definition of the Dirac operator in the setting of pure mathematics:

  • Which underlying structure (metrics, bundles, etc.) a manifold have in order to be able to define a Dirac?

  • What are the domain a range of the Dirac operator (perhaps the set of sections of a certain bundle?)

  • Is there an intuitive explanation of what the Dirac operator does? (say, along the lines of "the Laplacian represents diffusion, as in the heat equation $\partial_tu=\Delta u$).

I will be grateful for any suggestions.

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    $\begingroup$ perhaps it would help if you would indicate in what respect the Wikipedia entry is unclear/insufficient? en.wikipedia.org/wiki/Dirac_operator $\endgroup$ – Carlo Beenakker Jan 9 '15 at 17:06
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    $\begingroup$ concerning you last question: Bohr asked Dirac, and this was the answer: “What are you working on Mr. Dirac?” --- “I’m trying to take the square root of something” (that something being the Schrodinger equation) $\endgroup$ – Carlo Beenakker Jan 9 '15 at 17:08
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    $\begingroup$ Is this en.wikipedia.org/wiki/Clifford_analysis relevant to you? I suspect this might be what you see on the physics side. $\endgroup$ – Bombyx mori Jan 9 '15 at 17:28
  • $\begingroup$ There are a number of references that will answer this question, for example Lawson-Michelson, Berline-Getzler-Vergne... $\endgroup$ – Qiaochu Yuan Jan 9 '15 at 17:43
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    $\begingroup$ Have a look at Chapter 11 of these notes www3.nd.edu/~lnicolae/Lectures.pdf $\endgroup$ – Liviu Nicolaescu Jan 9 '15 at 18:16
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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 quaternionic representation, while Ettore Majorana used a real representation.

The physics implications were entirely different: Dirac had a complex field equation, and concluded that the field and its complex conjugate described two different particles. One was the electron, the complex conjugate was unknown at the time. Dirac hypothesised that it described a positively charged "antiparticle", with the same mass as the electron. This "positron" was discovered shortly afterwards, a triumph of mathematical physics.

Majorana, in contrast, had a real wave equation and hypothesised that it would describe charge-neutral particles that were their own antiparticle. We still do not know whether such particles exist in nature (the neutrino may or may not be of this type).

Historical note: the real representation of the Dirac equation is called the Majorana equation, but this was actually a rediscovery: Eddington had published it a decade before Majorana, so "Eddington-Majorana equation" would be a more appropriate name.

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  • $\begingroup$ Perhaps it is going too far from the initial question, but nevertheless: which is the point of working with modules over the Clifford Algebra? Which is the "first module" (since all this theory comes from Physics) that was used in this setting? $\endgroup$ – Jjm Jan 10 '15 at 8:20
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    $\begingroup$ I've added some perspectives from the physics context. $\endgroup$ – Carlo Beenakker Jan 10 '15 at 13:41
  • $\begingroup$ Thanks for the added material. But something is still puzzling me. Would not be the most natural choice to take $V$ as $\text{Cliff}(TX)$ itself? A connection in $TX$ leads to a connection in $\text{Cliff}(TX)$, and we would have a truly "root of $\Delta$", at least when looking at sections of $TX\subset\text{Cliff}(TX)$. $\endgroup$ – Jjm Jan 11 '15 at 8:00
  • $\begingroup$ @CarloBeenaker By the way, which would be the Dirac Equation for $V=\text{Cliff}(TX)$? $\endgroup$ – Jjm Jan 11 '15 at 8:11
  • $\begingroup$ @Jjm --- The case $V={\rm Cliff}(TX)$ is worked out on page 3 and 4 of the notes I linked to. This does not apply to the physics context because of the need to account for the spin degree of freedom of the electron. $\endgroup$ – Carlo Beenakker Jan 11 '15 at 11:28
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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) then it admits a vector bundle $\Sigma$ whose fibre at $x$ is an irreducible Clifford module of $Cl(T_xM)$. This means that we have a bundle map $$Cl(TM) \times \Sigma \to \Sigma$$ given fibrewise by the Clifford action of $Cl(T_xM)$ on $\Sigma_x$.

(Please allow me the notational abuse of confusing bundles with their sections, to save me some writing.)

Because we have a metric, we have musical isomorphisms $$\flat: TM \to T^*M \qquad\text{and}\qquad \sharp: T^*M \to TM~,$$ the latter of which can be composed with the above Clifford action to arrive at $$Cl(T^*M) \times \Sigma \xrightarrow{\operatorname{cl}} \Sigma~.$$

The Spin group $Spin(T_xM)$ is contained in $Cl(T_xM)$ and hence acts on $\Sigma_x$ by restriction. This representation need not be irreducible; although if $M$ is odd-dimensional it is. This allows us to view $\Sigma$ as a bundle of spinors, associated to a (not necessarily irreducible) spinor representation of the spin group. In fact, it is an associated vector bundle of the spin bundle.

The Levi-Civita connection on $TM$ defines an Ehresmann connection on the bundle of oriented orthonormal frames. That connection lifts to a connection on the spin bundle and hence induces a Koszul connection on $\Sigma$, called the spin connection. We thus get a bundle map $$ \Sigma \xrightarrow{\nabla} T^*M \otimes \Sigma $$

The Dirac operator is now simply the composition $$ \Sigma \xrightarrow{\nabla} T^*M \otimes \Sigma \xrightarrow{\operatorname{cl}} \Sigma~.$$

Thus the domain and range are the sections of $\Sigma$, the so-called spinor fields.

If $M$ is even-dimensional (and depending on signature perhaps after complexifying) then one can refine this. Under spin, $\Sigma$ splits into eigenbundles of the Clifford action by the volume form: $$ \Sigma = \Sigma_+ \oplus \Sigma_-$$ Then $\nabla : \Sigma_\pm \to \Sigma_\pm$, but $$ \operatorname{cl}: T^*M \otimes \Sigma_\pm \to \Sigma_\mp$$ whence now the Dirac operator, defined in the same way, breaks up into two maps $\Sigma_\pm \to \Sigma_\mp$.

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I'm adding more details to answer Igor's question. If we are in signature $(s,t)$, with $n=s+t$, being the dimension, then the answer depends on $s-t \mod 8$:

  • $s-t = 0 \mod 8$: $\Sigma$ is real and $\Sigma = \Sigma_+ \oplus \Sigma_-$, where $\Sigma_\pm$ are real.
  • $s-t = 1 \mod 8$: $\Sigma$ is complex but has a spin-invariant real structure: it does not split.
  • $s-t = 2 \mod 8$: $\Sigma$ is quaternionic and it splits if we complexify: $\Sigma_{\mathbb{C}} = \Sigma_+ \otimes \Sigma_-$, with $\Sigma_\pm$ complex conjugates.
  • $s-t = 3 \mod 8$: $\Sigma$ is quaternionic and remains so under spin: it does not split.
  • $s-t = 4 \mod 8$: $\Sigma$ is quaternionic and it splits $\Sigma = \Sigma_+ \oplus \Sigma_-$, where $\Sigma_\pm$ are also quaternionic.
  • $s-t = 5 \mod 8$: $\Sigma$ is complex but has a spin-invariant quaternionic structure: it does not split.
  • $s-t = 6 \mod 8$: $\Sigma$ is real and if we complexify then it splits $\Sigma_{\mathbb{C}} = \Sigma_+ \otimes \Sigma_-$, with $\Sigma_\pm$ complex conjugates.
  • $s-t = 7 \mod 8$: $\Sigma$ is real, does not split and remains real.
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  • $\begingroup$ Just wondering how you would make this more precise, when distinguishing complex and real spinor bundles $\Sigma$. In my understanding, if $\dim M$ is even, there is a unique irreducible representation of $Cl(T_xM)$, either over $\mathbb{R}$ or $\mathbb{C}$. But when $\dim M$ is odd, $Cl(T_xM)$ has two possible inequivalent irreducible representations over $\mathbb{C}$. Is there a common terminology that distinguishes between the two corresponding Dirac operators on complex spinors? Over $\mathbb{R}$ both representations are equivalent, so there is no ambiguity. $\endgroup$ – Igor Khavkine Jan 9 '15 at 19:43
  • $\begingroup$ @IgorKhavkine: I've added some details to the answer. $\endgroup$ – José Figueroa-O'Farrill Jan 9 '15 at 20:21
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    $\begingroup$ @Jjm: Clifford algebras were rediscovered in Physics by Dirac, who was looking for (and found!) a square root of the Klein-Gordon equation (basically a wave equation with a mass term). So this is $s=3, t=1$ (or perhaps $s=1, t=3$ since the Physics convention for clifford algebras differs from that of Clifford by a sign). In one case the Clifford module is real (the so-called Majorana spinors) and in the other case it's quaternionic (the so-called Dirac spinors, except that the quaternionic structure was not emphasised). $\endgroup$ – José Figueroa-O'Farrill Jan 10 '15 at 21:56
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    $\begingroup$ @Jjm Physically, (free) particles are interpreted as irreducible unitary representations of the Poincaré group and equations such as Dirac's,... can be interpreted as projections onto irreducible components of perhaps reducible representations. If you were to take a reducible Clifford module, you would not be describing a single particle. For example, take $X$ to be Minkowski spacetime and induce a representation of the Poincaré group taking the Clifford module to the Clifford algebra itself. (continued below) $\endgroup$ – José Figueroa-O'Farrill Jan 11 '15 at 16:29
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    $\begingroup$ (continued) This representation would not be irreducible and would decompose into a number of (non-spinorial) representations. None of these representations would include the one induced from the irreducible Clifford module. $\endgroup$ – José Figueroa-O'Farrill Jan 11 '15 at 16:31

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