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Generalization of De Rham cohomology for spinor fields

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Is there a generalization of De Rham cohomology for spinors fields?

I can see that one can construct p form fields out of spinor field by contraction of the type $\bar{\psi} \gamma^{a_1} \gamma^{a_2}...\gamma^{a_p}\chi$.

Now we can consider the integral of the p form fields on p-cycles. There is a natural derivative like operation acting on spinor fields $\gamma \cdot \partial$. Does this map have the the desired properties to make a co-homology in some way. If I cannot use this map can I use some other operator to suitably generalize the exterior derivative operator.

Are there any connections to objects such as topological solitons?

I have a background in physics and not in mathematics, please keep that in view when you write your answer.


This post imported from StackExchange Mathematics at 2015-06-17 17:28 (UTC), posted by SE-user Prathyush

asked Jun 17, 2015 in Mathematics by Prathyush (695 points) [ revision history ]
edited Jun 21, 2015 by Prathyush

See also some interesting comments in the corresponding discussion of this question on MO.

Related paper.

Harmonic Spinors

2 Answers

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A cohomology theory can be defined in which the Dirac operator $\not{D}$ plays the role of the exterior derivative. I'll try to describe what I know about the subject in some detail, and give you some references. For a specific choice of the Dirac operator, this theory is customarily called: "Dirac cohomology".

Just for your intuition, a theory involving gamma matrices (Clifford algebra) is likely to behave like a quantum theory rather than a geometrical theory (unless you consider noncommutative geometry). Actually the Clifford algebra is a quantization of the Grassmann algebra of differntial forms.

In contrast to the De-Rham case, the square of a Dirac operator ($H_D \doteq \not{D}^2 $) is not zero in general, so this theory is not a cohomology theory in the usual sense. However on the subspace of the zero modes of $H_D$, the Dirac operator is nilpotent $\not{D}^2 = 0$, therefore an appropriate choice for a cohomology operator. I don't know if there is any application of a dual to the Dirac cohomology, thus my answer will not refer to the dual "Dirac homology".

The interest in the subject started from remarkable relationships observed by Pengpan and Ramond between massless multiplets of D=11, N=1 supergravity which is believed to be the low energy local limit of M-theory. Kostant invented Dirac cohomology to provide an explanation of this phenomenon.

Dirac cohomology is mainly applied as a tool in the study of group representations. From the physical point of view, these types of applications treat a quantum mechanical toy model of a superparticle moving on a Riemannian manifold with torsion. (please see Frohlich and Gawedzki for an explicit construction from a superparticle Lagrangian.) Upon quantization, the supersymmetry generator becomes a Dirac operator. On the zero modes subspace, the Dirac operator acts as a BRST operator, thus it is instructive to think about its kernel as a space physical states.

There are some generalizations of the Dirac cohomology to 1+1 dimensional theories. In these cases the relevant Dirac operators are operators on infinite dimensional spaces such as loop spaces. Applications in more realistic cases e.g. 3+1 dimensional physics are still open problems.

When, the configuration manifold on which the superparticle moves is homogeneous, (such as a Lie group or a cost space), there are cases where the solution of the Kostant-Dirac equation (the kernel of the Dirac operator):

$$\not{D} \psi = 0$$

can be expressed purely in algebraic terms. This case happens when there is a frame (vielbein) in which both the metric and the gauge fields to which the Dirac operator is coupled have constant coefficients. (Gauge fields satisfying this property are called homogeneous). The Killing metric on the coset space certainly satisfies this property. When the Dirac operator is also coupled to torsion this requirement extends to the torsion field. Assuming that these conditions are satisfied, we can work with an algebraic version of the Dirac operator. The Dirac operator constructed by Kostant is of this type; it is based not on a Levi-Civita type of connection but rather contains torsion. This choice has the property that its square is given by a very simple formula expressed in terms of Casimirs granting it its remarkable properties. This operator is called the Kostant-Dirac operator or the cubic Dirac operator.

Here I'll be following mainly the following article by Brink and Ramond in the follwing example. The simplest example of the Kostant-Dirac operator is the case of a group manifold, with Lie algebra commutation relations:

$$ [T_a, T_b ] = i f_{ab}^c T_c$$

In this case, the Kostant-Dirac operator is given by:

$$\not{D} = T_a \gamma^a - \frac{i}{12} f_{abc}\gamma^a\gamma^b\gamma^c $$

Where the gamma matrices satisfy the anti-commutation relations:

$$ \{\gamma_a, \gamma_b \} = 2 g_{a b}$$

and $ g_{a b}$ is the Lie algebra Cartan-Killing tensor.

Here, the Kostant-Dirac operators acts on the space colored spinors $V \otimes S$, where $V$ is the vector space on which the symmetry generators $T_a$ act and $S$ is the spinor space on which the gamma matrices act ($S$ is an irreducible representation space of a Clifford algebra of the group manifold dimension).

Given an explicit matrix representation of the symmetry operators, one can construct this operator a s a (big) matrix.

The square of the cubic Kostant-Dirac operator is given by:

$$\not{D}^2 = 2 C_2 + \frac{D}{12}C_2^{adj}$$,

Where $C_2$ is the second Casimir of the representation $T_a$ and $ C_2^{adj}$ is the second Casimir of the adjoint representation and $D$ is the dimension of the group manifold. This case is not very interesting because the square of the cubic Dirac operator is positive and its kernel is zero.

A more interesting case, which is the one treated by Kostant, is when the manifold is taken as a coset space $G/H$ where $G$ and $H$ are of equal rank. In this case the Kostant-Dirac operator is defind by:

$$\not{D} = T_m \gamma^m - \frac{i}{12} f_{mnp}\gamma^m\gamma^m\gamma^p $$ In this case the indices run over the coset generators only (i.e. without the subgroup generators).
In this case the square of the Kostant-Dirac operator is given by:

$$\not{D}^2 = C_2(G) + \frac{D}{24}C_2(G)^{adj} - C_2(H) - \frac{D}{24}C_2(H)^{adj} $$,

Where $C_2(G)$, $ C_2(H)$ are the quadratic Casimirs of $G$ and $H$ etc. The Kostant Dirac operator commutes with the operators:

$$ L^i = T^i+f_{mn}^i\gamma^{mn}$$

Where $i$ runs over the sugroup $H$ indices. These operators generate an algebra isomorphic to the subgroup $H$ algebra. Therefore the restriction of the solution of the Kostant-Dirac equation to this subalgebra splits into $H$ representations. The main result of Gross, Kostant, Ramond, and Sternberg is that the representation splits into a number of representations equal to the Euler number of $G/H$. These are called Euler multiplets. Pengpan and Ramond noticed from their study of N=1 D=11 supergravity that the massless spectrum corresponding to the little group $SO(9)$ are arranged in a tower of Euler multiplets corresponding to the cost space $F_4/SO(9)$. A full understanding of this phenomenon is still lacking.

In the field theoretical infinite dimensional case, Juoko Mickelsson constructed a cubic Dirac operator corresponding to the supersymmetric Wess-Zumino-Witten model.

This post imported from StackExchange Physics at 2015-06-28 22:32 (UTC), posted by SE-user David Bar Moshe
answered Jun 27, 2015 by David Bar Moshe (3,845 points) [ no revision ]
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This is a well-known vector space isomorphism between the Clifford algebra (algebra of gamma matrices) and the exterior algebra (algebra of forms), described eg. on the Wikipedia page for Clifford algebras. *It is not an algebra homomorphism!* It is of course possible to describe the exterior differential in the gamma matrix basis and one gets the ordinary de Rham cohomology. It's not a good way to do things, though, since the cohomology ring structure comes from the exterior multiplication, not the Clifford multiplication.

One modification of de Rham cohomology that can be done, however, is de Rham cohomology valued in some vector bundle. One could consider "spinor-valued de Rham cohomology". A spinor valued p-form takes a p-tuple of tangent vectors and produces a spinor. It looks like $\psi dx_1 \wedge ... dx_p$. The exterior derivative is defined by the ordinary formula, extended by linearity.

answered Jun 21, 2015 by Ryan Thorngren (1,605 points) [ revision history ]

Interesting points, Yes I don't see any obvious way to generalize the cup product.

I can't see why a spinor valued p form will be useful however

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