# Twisted K-theory

In mathematics, twisted K-theory (also called K-theory with local coefficients) is a variation on K-theory, a mathematical theory from the 1950s that spans algebraic topology, abstract algebra and operator theory.

More specifically, twisted K-theory with twist H is a particular variant of K-theory, in which the twist is given by an integral 3-dimensional cohomology class. It is special among the various twists that K-theory admits for two reasons. First, it admits a geometric formulation. This was provided in two steps; the first one was done in 1970 (Publ. Math. de l'IHÉS) by Peter Donovan and Max Karoubi; the second one in 1988 by Jonathan Rosenberg in Continuous-Trace Algebras from the Bundle Theoretic Point of View.

In physics, it has been conjectured to classify D-branes, Ramond-Ramond field strengths and in some cases even spinors in type II string theory. For more information on twisted K-theory in string theory, see K-theory (physics).

In the broader context of K-theory, in each subject it has numerous isomorphic formulations and, in many cases, isomorphisms relating definitions in various subjects have been proven. It also has numerous deformations, for example, in abstract algebra K-theory may be twisted by any integral cohomology class.

## The definition

To motivate Rosenberg's geometric formulation of twisted K-theory, start from the Atiyah-Jänich theorem, stating that

${\displaystyle Fred({\mathcal {H}}),}$

the Fredholm operators on Hilbert space ${\displaystyle {\mathcal {H}}}$, is a classifying space for ordinary, untwisted K-theory. This means that the K-theory of the space ${\displaystyle M}$ consists of the homotopy classes of maps

${\displaystyle [M\rightarrow Fred({\mathcal {H}})]}$

from ${\displaystyle M}$ to ${\displaystyle Fred({\mathcal {H}}).}$

A slightly more complicated way of saying the same thing is as follows. Consider the trivial bundle of ${\displaystyle Fred({\mathcal {H}})}$ over ${\displaystyle M}$, that is, the Cartesian product of ${\displaystyle M}$ and ${\displaystyle Fred({\mathcal {H}})}$. Then the K-theory of ${\displaystyle M}$ consists of the homotopy classes of sections of this bundle.

We can make this yet more complicated by introducing a trivial

${\displaystyle PU({\mathcal {H}})}$

bundle ${\displaystyle P}$ over ${\displaystyle M}$, where ${\displaystyle PU({\mathcal {H}})}$ is the group of projective unitary operators on the Hilbert space ${\displaystyle {\mathcal {H}}}$. Then the group of maps

${\displaystyle [P\rightarrow Fred({\mathcal {H}})]_{PU({\mathcal {H}})}}$

from ${\displaystyle P}$ to ${\displaystyle Fred({\mathcal {H}})}$ which are equivariant under an action of ${\displaystyle PU({\mathcal {H}})}$ is equivalent to the original groups of maps

${\displaystyle [M\rightarrow Fred({\mathcal {H}})].}$

This more complicated construction of ordinary K-theory is naturally generalized to the twisted case. To see this, note that ${\displaystyle PU({\mathcal {H}})}$ bundles on ${\displaystyle M}$ are classified by elements ${\displaystyle H}$ of the third integral cohomology group of ${\displaystyle M}$. This is a consequence of the fact that ${\displaystyle PU({\mathcal {H}})}$ topologically is a representative Eilenberg-MacLane space

${\displaystyle K(\mathbf {Z} ,2).}$

The generalization is then straightforward. Rosenberg has defined

${\displaystyle K_{H}(M)}$,

the twisted K-theory of ${\displaystyle M}$ with twist given by the 3-class ${\displaystyle H}$, to be the space of homotopy classes of sections of the trivial ${\displaystyle Fred({\mathcal {H}})}$ bundle over ${\displaystyle M}$ that are covariant with respect to a ${\displaystyle PU({\mathcal {H}})}$ bundle ${\displaystyle P_{H}}$ fibered over ${\displaystyle M}$ with 3-class ${\displaystyle H}$, that is

${\displaystyle K_{H}(M)=[P_{H}\rightarrow Fred({\mathcal {H}})]_{PU({\mathcal {H}})}.}$

Equivalently, it is the space of homotopy classes of sections of the ${\displaystyle Fred({\mathcal {H}})}$ bundles associated to a ${\displaystyle PU({\mathcal {H}})}$ bundle with class ${\displaystyle H}$.

## What is it?

When ${\displaystyle H}$ is the trivial class, twisted K-theory is just untwisted K-theory, which is a ring. However, when ${\displaystyle H}$ is nontrivial this theory is no longer a ring. It has an addition, but it is no longer closed under multiplication.

However, the direct sum of the twisted K-theories of ${\displaystyle M}$ with all possible twists is a ring. In particular, the product of an element of K-theory with twist ${\displaystyle H}$ with an element of K-theory with twist ${\displaystyle H'}$ is an element of K-theory twisted by ${\displaystyle H+H'}$. This element can be constructed directly from the above definition by using adjoints of Fredholm operators and construct a specific 2 x 2 matrix out of them (see the reference 1, where a more natural and general Z/2-graded version is also presented). In particular twisted K-theory is a module over classical K-theory.

## How to calculate it

Physicist typically want to calculate twisted K-theory using the Atiyah–Hirzebruch spectral sequence.[1] The idea is that one begins with all of the even or all of the odd integral cohomology, depending on whether one wishes to calculate the twisted ${\displaystyle K_{0}}$ or the twisted ${\displaystyle K^{0}}$, and then one takes the cohomology with respect to a series of differential operators. The first operator, ${\displaystyle d_{3}}$, for example, is the sum of the three-class ${\displaystyle H}$, which in string theory corresponds to the Neveu-Schwarz 3-form, and the third Steenrod square.[2] No elementary form for the next operator, ${\displaystyle d_{5}}$, has been found, although several conjectured forms exist. Higher operators do not contribute to the ${\displaystyle K}$-theory of a 10-manifold, which is the dimension of interest in critical superstring theory. Over the rationals Michael Atiyah and Graeme Segal have shown that all of the differentials reduce to Massey products of ${\displaystyle M}$.[3]

After taking the cohomology with respect to the full series of differentials one obtains twisted ${\displaystyle K}$-theory as a set, but to obtain the full group structure one in general needs to solve an extension problem.

### Example: the three-sphere

The three-sphere, ${\displaystyle S^{3}}$, has trivial cohomology except for ${\displaystyle H^{0}(S^{3})}$ and ${\displaystyle H^{3}(S^{3})}$ which are both isomorphic to the integers. Thus the even and odd cohomologies are both isomorphic to the integers. Because the three-sphere is of dimension three, which is less than five, the third Steenrod square is trivial on its cohomology and so the first nontrivial differential is just ${\displaystyle d_{5}=H}$. The later differentials increase the degree of a cohomology class by more than three and so are again trivial; thus the twisted ${\displaystyle K}$-theory is just the cohomology of the operator ${\displaystyle d_{3}}$ which acts on a class by cupping it with the 3-class ${\displaystyle H}$.

Imagine that ${\displaystyle H}$ is the trivial class, zero. Then ${\displaystyle d_{3}}$ is also trivial. Thus its entire domain is its kernel, and nothing is in its image. Thus ${\displaystyle K_{H}^{0}(S^{3})}$ is the kernel of ${\displaystyle d_{3}}$ in the even cohomology, which is the full even cohomology, which consists of the integers. Similarly ${\displaystyle K_{H}^{1}(S^{3})}$ consists of the odd cohomology quotiented by the image of ${\displaystyle d_{3}}$, in other words quotiented by the trivial group. This leaves the original odd cohomology, which is again the integers. In conclusion, ${\displaystyle K^{0}}$ and ${\displaystyle K^{1}}$ of the three-sphere with trivial twist are both isomorphic to the integers. As expected, this agrees with the untwisted ${\displaystyle K}$-theory.

Now consider the case in which ${\displaystyle H}$ is nontrivial. ${\displaystyle H}$ is defined to be an element of the third integral cohomology, which is isomorphic to the integers. Thus ${\displaystyle H}$ corresponds to a number, which we will call ${\displaystyle n}$. ${\displaystyle d_{3}}$ now takes an element ${\displaystyle m}$ of ${\displaystyle H^{0}}$ and yields the element ${\displaystyle nm}$ of ${\displaystyle H^{3}}$. As ${\displaystyle n}$ is not equal to zero by assumption, the only element of the kernel of ${\displaystyle d_{3}}$ is the zero element, and so ${\displaystyle K_{H=n}^{0}(S^{3})=0}$. The image of ${\displaystyle d_{3}}$ consists of all elements of the integers that are multiples of ${\displaystyle n}$. Therefore, the odd cohomology, ${\displaystyle \mathbb {Z} }$, quotiented by the image of ${\displaystyle d_{3}}$, ${\displaystyle n\mathbb {Z} }$, is the cyclic group of order ${\displaystyle n}$, ${\displaystyle \mathbb {Z} /n}$. In conclusion

${\displaystyle K_{H=n}^{1}(S^{3})=\mathbb {Z} /n}$

In string theory this result reproduces the classification of D-branes on the 3-sphere with ${\displaystyle n}$ units of ${\displaystyle H}$-flux, which corresponds to the set of symmetric boundary conditions in the supersymmetric ${\displaystyle SU(2)}$ WZW model at level ${\displaystyle n-2}$.

There is an extension of this calculation to the group manifold of SU(3).[4] In this case the Steenrod square term in ${\displaystyle d_{3}}$, the operator ${\displaystyle d_{5}}$, and the extension problem are nontrivial.