Let be a smooth -manifold and be a point in . Consider the set of all linear isomorphisms between the tangent space at and . Note that there is a natural left action of on . Since this action may be seen as a function composition, we denote the action by . Though, we will quite often drop the notation altogether hoping that it is clear. The disjoint union is called the frame bundle of . (Of course, we need to have more conditions on but we will not go into details in this post.) The action of on induces a natural action on .
It is easy to define a bijection between any fiber of and . Using the bijection, we may define a group multiplication on which will, in turn, make the fiber isomorphic to , trivially. In general, this bijection is not canonical as we will see below. First, we fix an isomorphism in . Then, we send to . Clearly, this map is injective and is sent to the identity matrix under this identification. Also, for any , and i.e. the identification is onto. So, we have a bijection. As we can see, once an identity element is fixed, the fiber becomes a group isomorphic to . In other words, is a -torsor.
Let be a Lie subgroup of and be a subbundle of whose fibers (which we still denote by ) are isomorphic to in the above sense. Then, is called a -structure on . Clearly, the frame bundle is a -structure on .
Next, we discuss two examples of proper subbundles inducing various structures on . In our first example, we consider to be the orthogonal group . Recall that the standard euclidean metric on is fixed by . In other words, for any , . We can use this property together with to define a Riemannian metric on . Let , and define the metric as the pullback . We need to show that this definition is independent of the choice of . Let , then as is an -structure. Hence, . So, we can choose any isomorphism in the fiber in order to define . So, we see that an -structure gives us a Riemannian metric. Next, we will go the other way around i.e. given a Riemannian metric , we construct an -structure on . Each tangent space is equiped with an inner product and we consider equiped with the standard inner product. We define a fiber of to be the set of linear isometries between and . Next, we need to check that a fiber is isomorphic to . Again, first, we fix an isomorphism . Then, given another isometry from to , is an isometry from to itself i.e. . Also, for , is an isometry from to and . Hence, as above, we have an isomorphism. It is easy to see that this correspondence between -structures and Riemannian metrics is one to one.
This examle can be generalized easily. Any structure on which is fixed by a Lie subgroup of can be carried to a manifold which admits a -structure. In fact, our second example will be of this type, again. We consider the correspondence between an almost complex structure and a -structure where . Before we discuss the correspondence, let us clearify a few things. We view as a subgroup of using the monomorphism
Let denote the action of on . In other words, where denotes the identity matrix. Or using, the monomorphism defined above
in matrix block form, as a real matrix. Of course, for any matrix , . Equivalently, we have . On the other hand, let . Then,
Therefore, if and only if and . Hence, we see that a real matrix can be identified with a complex matrix if and only if .
Now, we go back to the correspondence between a -structure and an almost complex structure. First, let us assume that we have a -structure and construct an almost complex structure . Let . Then, we define by . Next, we show that is well defined. Let . Since , we have . Hence,
Moreover, clearly, we have . Thus, we have constructed an almost complex structure.
Next, we go in the other direction. Given a complex structure , we form a subbundle of which consists of linear isomorphisms that satisfy . Take a basis of of the form and, then define with in the position and with in the position. It is easy to verify that . Hence, the subbundle is non-empty. Next, we want to show that the fibers are isomorphic to . Let be another isomorphism satisfying . Then, . Thus, , or equivalently, . Therefore, by above remarks. Also, given , that is is also an element of . So, the fibers are isomorphic to .
Our third example will be a -structure. However, I want to have a more detailed discussion of -structures before I present it in this context. So, I will include it in a future post.