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I apologize for asking what may be a very basic question. But I have recently begun a more mathematical introduction to spinors and I am stuck at the formal definition of the Dirac operator. Specifically, if we have a bundle $S \longrightarrow X$, the Dirac operator $D$ is defined as a map $\Gamma (X,S) \longrightarrow \Gamma (X,S)$ via the sequence

$\Gamma (X,S) \overset{\nabla^S}{\longrightarrow} \Omega^1 (X) \otimes \Gamma (X,S) \overset{c}{\longrightarrow} \Gamma (X,S)$,

where $\nabla^S$ is the (spin) connection on $S$ and $c$ is "Clifford multiplication". Thus, I see often the Dirac operator written as $D= c \circ \nabla^S$ or

$D = \sum_i e_i \cdot \nabla^S_{e_i}$,

locally, where $e_i$ is an orthonormal frame for $TX$.

My confusion: The connection $\nabla^S$ acts as stated, but I don't understand the second map in the sequence. Clifford multiplication takes place in the Clifford algebra, yet there is no mention of the Clifford algebra in the definition (yet, it must of course be in there somewhere, since it is a Dirac operator!). Apparently, multiplication $c$ absorbs a one-form and spits out a section, so there must be some representation of $\Omega^1 (X)$ acting on $\Gamma (X,S)$, but the details are unclear to me or how this relates to Clifford multiplication?

An simple example, where a Dirac operator is constructed via the definition, would be wonderful!

Thank you.

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    $\begingroup$ $S$ is not just any bundle, it's the spinor bundle. This means that it carries an action of the Clifford algebra bundle, this seems to be the missing ingredient. $\endgroup$ – Peter Jul 3 '19 at 14:39
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As mentioned in the comments, the Clifford algebra is hiding in the definition of the spinor bundle. You will find a formal definition of the spinor bundle in whatever reference you are using for Dirac operators (or else it's time to find a better reference), so I will just say some informal things about it. Here's the story as I like to think of it: (I'll use Euclidean signature instead of Minkowski)

Dirac (working on flat $\mathbb R^n$) found that if he wanted a square root of $\Delta = - \sum \partial_i^2$ in the form of a first order operator of the form $D = \sum \gamma_i \partial_i$, then he needed the coefficients $\gamma_i$ to satisfy $\gamma_i\gamma_j + \gamma_j\gamma_i = -2\delta_{ij}$. (Square that operator $D$, compare to $\Delta$, and you'll come to the same conclusion). We call this the Clifford relation (and say that the $\gamma_i$ generate the Clifford algebra).

That relation cannot be satisfied by real or complex coefficients, hence the need for Dirac's gamma matrices. If you allow the coefficients $\gamma_i$ to be matrices then you'll be able to satisfy the Clifford relation. Of course now that you start using matrices as coefficients, it doesn't really make sense to try feeding this operator a real- or complex-valued function. You need to feed it a vector-valued function of the appropriate size for whatever size $\gamma$-matrices you're using.

We call such a vector-valued function a "spinor." Spinors are a bit more than just vector-valued functions, though: they're vector-valued functions equipped with an action of the Clifford algebra. That means that in the back of our minds, we've fixed some gamma matrices and we think of those as acting on our vector-valued functions.

Okay, that was the story on flat $\mathbb R^n$. It is our model. Now, when we think about generalizing this story to a manifold, we want to make sure that it always locally looks like this.

The manifold analogue of a vector-valued function is a section of a vector bundle. So when we construct the Dirac operator on a Riemannian manifold, our spinors will be sections of a vector bundle (the spinor bundle).

We need to decide what the analogue of $\gamma$-matrices should be. It turns out to be necessary to allow the Clifford relation to vary with the point $x \in M$. Any time you have a vector space $(V,g)$ equipped with an inner product you can define a Clifford algebra $Cl(V,g)$ by saying that its generators are the vectors of some basis $e_1, \ldots e_n$ of $V$ subject to the relation $e_ie_j + e_je_i = -2g(e_i,e_j)$. On a Riemannian manifold, each cotangent space is an inner product space. So we take advantage of that and define a Clifford algebra $Cl(T^*_xM,g_x)$ over each point $x \in M$. This defines a bundle of algebras over $M$, the "Clifford algebra bundle." I think of it as a generalization of $\gamma$-matrices.

Now we need something for "$\gamma$-matrices" to act on: That is, we need a vector bundle $S$ over $M$ that the Clifford algebra bundle can act on (a so-called Clifford module). That is, a bundle $S$ and an algebra bundle morphism $Cl(T^*M,g) \to \text{End}(S)$. To be a spinor bundle, we require that $S$ be no bigger than necessary: that in fact $Cl(T^*M,g) \cong \text{End}(S)$.

Existence and uniqueness questions for this bundle $S$ lead to the notion of `spin structures.' Some manifolds do not admit such a bundle for topological reasons. When a manifold does, it often admits many. (You'll find precise versions of those statements in the literature).

Okay, finally we can say what Clifford multiplication means and how it makes that definition of the Dirac operator make sense:

Whenever you form $Cl(V,g)$, the vector space is sitting inside of it as a subspace: $V \le Cl(V,g)$. Since we formed Clifford algebras from the cotangent spaces, each $T^*_xM \le Cl(T^*_xM,g_x)$. So that means that a 1-form, in addition to being a section of $T^*M$, is a section of $Cl(T^*M,g)$. The definition of $S$ was that $Cl(T^*M,g) \cong \text{End}(S)$, so that means that 1-forms act on spinors: $c: \Omega^1(M) \otimes \Gamma(S) \to \Gamma(S)$ is what we call "Clifford multiplication."

Now the abstract definition you wrote for the Dirac operator should make sense.

This is roughly always the story, but I left out some stuff: the story is a bit different for even- and odd-dimensional manifolds, the connection you use on $S$ needs to be tied to the geometry of $(M,g)$ in the sense of being a "lift of the Levi-Civita connection," and I didn't try to say anything about what a spin structure is.

I really like Thomas Friedrich's Dirac Operators in Riemannian Geometry, I think it's a great reference for this stuff.

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    $\begingroup$ This is an incredible answer! Thank you so much! $\endgroup$ – Henrymerrild Jul 10 '19 at 17:59

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