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,
and
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.