general | Ustun's Blog

Let ${M}$ be a smooth ${n}$ -manifold and ${p}$ be a point in ${M}$ . Consider the set ${ S_p}$ of all linear isomorphisms ${L_p:T_pM\rightarrow {\mathbb R}^n}$ between the tangent space at ${p}$ and ${{\mathbb R}^n}$ . Note that there is a natural left action of ${GL(n,{\mathbb R})}$ on ${S_p}$ . Since this action may be seen as a function composition, we denote the action by ${\circ}$ . Though, we will quite often drop the notation altogether hoping that it is clear. The disjoint union ${F = \sqcup_p S_p}$ is called the frame bundle of ${M}$ . (Of course, we need to have more conditions on ${F}$ but we will not go into details in this post.) The action of ${GL(n,{\mathbb R})}$ on ${S_p}$ induces a natural action on ${F}$ .

It is easy to define a bijection between any fiber of ${F}$ and ${GL(n,{\mathbb R})}$ . Using the bijection, we may define a group multiplication on ${F}$ which will, in turn, make the fiber isomorphic to ${GL(n,{\mathbb R})}$ , trivially. In general, this bijection is not canonical as we will see below. First, we fix an isomorphism ${L_p}$ in ${S_p}$ . Then, we send ${K_p \in S_p}$ to ${K_p \circ L_p^{-1} \in GL(n,{\mathbb R})}$ . Clearly, this map is injective and ${L_p}$ is sent to the identity matrix under this identification. Also, for any ${N\in GL(n,{\mathbb R})}$ , ${N \circ L_p \in S_p}$ and ${N\circ L_p \circ L_p^{-1} = N}$ i.e. the identification is onto. So, we have a bijection. As we can see, once an identity element ${L_p}$ is fixed, the fiber ${S_p}$ becomes a group isomorphic to ${GL(n,{\mathbb R})}$ . In other words, ${S_p}$ is a ${GL(n,{\mathbb R})}$ -torsor.

Let ${G}$ be a Lie subgroup of ${GL(n,{\mathbb R})}$ and ${P}$ be a subbundle of ${F}$ whose fibers (which we still denote by ${S_p}$ ) are isomorphic to ${G}$ in the above sense. Then, ${P}$ is called a ${G}$ -structure on ${M}$ . Clearly, the frame bundle ${F}$ is a ${GL(n,{\mathbb R})}$ -structure on ${M}$ .

Next, we discuss two examples of proper subbundles inducing various structures on ${M}$ . In our first example, we consider ${G}$ to be the orthogonal group ${O(n)}$ . Recall that the standard euclidean metric ${g_0}$ on ${{\mathbb R}^n}$ is fixed by ${O(n)}$ . In other words, for any ${N\in O(n)}$ , ${N^*g_0 =g_0}$ . We can use this property together with ${P}$ to define a Riemannian metric on ${M}$ . Let ${p\in M}$ , ${L_p\in S_p}$ and define the metric ${g_p}$ as the pullback ${L_p^*(g_0)}$ . We need to show that this definition is independent of the choice of ${L_p}$ . Let ${K_p\in S_p}$ , then ${K_p\circ L_p^{-1} \in O(n)}$ as ${P}$ is an ${O(n)}$ -structure. Hence, ${K_p^{*}(g_0) = (K_p\circ L_p^{-1}\circ L_p)^* (g_0)= L_p^*\circ (K_p\circ L_p^{-1})^* (g_0) = L_p^* (g_0)}$ . So, we can choose any isomorphism in the fiber in order to define ${g_p}$ . So, we see that an ${O(n)}$ -structure gives us a Riemannian metric. Next, we will go the other way around i.e. given a Riemannian metric ${g}$ , we construct an ${O(n)}$ -structure on ${M}$ . Each tangent space ${T_pM}$ is equiped with an inner product and we consider ${{\mathbb R}^n}$ equiped with the standard inner product. We define a fiber of ${P}$ to be the set of linear isometries between ${T_pM}$ and ${{\mathbb R}^n}$ . Next, we need to check that a fiber is isomorphic to ${O(n)}$ . Again, first, we fix an isomorphism ${L_p}$ . Then, given another isometry ${K_p}$ from ${T_pM}$ to ${{\mathbb R}^n}$ , ${K_p\circ L^{-1}_p}$ is an isometry from ${{\mathbb R}^n}$ to itself i.e. ${K_p \circ L^{-1}_p \in O(n)}$ . Also, for ${N\in O(n)}$ , ${N\circ L_p}$ is an isometry from ${T_pM}$ to ${{\mathbb R}^n}$ and ${N \circ L_p \circ L_p^{-1} = N\in O(n)}$ . Hence, as above, we have an isomorphism. It is easy to see that this correspondence between ${O(n)}$ -structures and Riemannian metrics is one to one.

This examle can be generalized easily. Any structure on ${{\mathbb R}^n}$ which is fixed by a Lie subgroup ${G}$ of ${GL(n,{\mathbb R})}$ can be carried to a manifold which admits a ${G}$ -structure. In fact, our second example will be of this type, again. We consider the correspondence between an almost complex structure and a ${GL(m,{\mathbb C})}$ -structure where ${n=2m}$ . Before we discuss the correspondence, let us clearify a few things. We view ${GL(m,{\mathbb C})}$ as a subgroup of ${GL(2m,{\mathbb R})}$ using the monomorphism

$\displaystyle \begin{array}{rcl} N \mapsto \begin{pmatrix} Re(N) & -Im(N) \\ Im(N) & Re(N) \end{pmatrix} \end{array}$

Let ${J_0:{\mathbb C}^m\rightarrow {\mathbb C}^m}$ denote the action of ${i}$ on ${{\mathbb C}^m}$ . In other words, ${J_0 = iI}$ where ${I}$ denotes the ${m\times m}$ identity matrix. Or using, the monomorphism defined above

$\displaystyle \begin{array}{rcl} J_0 = \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix}, \end{array}$

in matrix block form, as a real ${n\times n}$ matrix. Of course, for any matrix ${N \in GL(m,{\mathbb C})}$ , ${J_0 N = iN=Ni=NJ_0}$ . Equivalently, we have ${N^{-1}J_0N=J_0}$ . On the other hand, let ${N = \begin{pmatrix} A & B \\ C & D \end{pmatrix} \in GL(n,{\mathbb R})}$ . Then,

$\displaystyle \begin{array}{rcl} J_0 N &=& \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} \begin{pmatrix} A & B \\ C & D \end{pmatrix} \\ &=& \begin{pmatrix} -C & -D \\ A & B \end{pmatrix} \end{array}$

and

$\displaystyle \begin{array}{rcl} N J_0 &=& \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} 0 & -I \\ I & 0 \end{pmatrix} \\ &=& \begin{pmatrix} B & -A \\ D & -C \end{pmatrix}. \end{array}$

Therefore, ${NJ_0 = J_0N}$ if and only if ${A=D}$ and ${C = -B}$ . Hence, we see that a real matrix ${N}$ can be identified with a complex matrix if and only if ${NJ_0=J_0N}$ .

Now, we go back to the correspondence between a ${GL(m,{\mathbb C})}$ -structure and an almost complex structure. First, let us assume that we have a ${GL(m,{\mathbb C})}$ -structure and construct an almost complex structure ${J:TM\rightarrow TM}$ . Let ${L_p\in S_p}$ . Then, we define ${J:T_pM\rightarrow T_pM}$ by ${J = L_p^{-1} J_0 L_p}$ . Next, we show that ${J}$ is well defined. Let ${K_p\in S_p}$ . Since ${L_p K_p^{-1} \in GL(m,{\mathbb C})}$ , we have ${K_pL_p^{-1} J_0 L_pK_p^{-1} = J_0}$ . Hence,

$\displaystyle \begin{array}{rcl} J &=& L_p^{-1}J_0L_p \\ &=& K_p^{-1}K_pL_p^{-1}J_0L_pK_p^{-1}K_p \\ &=& K_p^{-1}J_0K_p. \end{array}$

Moreover, clearly, we have ${J^2 = -I}$ . Thus, we have constructed an almost complex structure.

Next, we go in the other direction. Given a complex structure ${J}$ , we form a subbundle ${P}$ of ${F}$ which consists of linear isomorphisms ${L_p}$ that satisfy ${L_pJ=J_0L_p}$ . Take a basis of ${T_pM}$ of the form ${\left\{ e_1,\dots,e_m,Je_1,\dots,Je_m \right\}}$ and, then define ${L_p(e_i) = (0,\dots,0,1,0,\dots,0)}$ with ${1}$ in the ${i^{th}}$ position and ${L_p(Je_i)=(0,\dots,0,1,0,\dots,0)}$ with ${1}$ in the ${m+i^{th}}$ position. It is easy to verify that ${L_pJ=J_0L_p}$ . Hence, the subbundle is non-empty. Next, we want to show that the fibers are isomorphic to ${GL(m,{\mathbb C})}$ . Let ${K_p}$ be another isomorphism satisfying ${K_pJ = J_0 K_p}$ . Then, ${J = K_p^{-1}J_0K_p}$ . Thus, ${L_p K_p^{-1}J_0K_p = J_0L_p}$ , or equivalently, ${L_p K_p^{-1}J_0 = J_0 L_p K_p^{-1}}$ . Therefore, ${L_pK_p^{-1}\in GL(m,{\mathbb C})}$ by above remarks. Also, given ${N \in GL(m,{\mathbb C})}$ , ${NL_p J = NJ_0L_p = J_0NL_p}$ that is ${NL_p}$ is also an element of ${P}$ . So, the fibers are isomorphic to ${GL(m,{\mathbb C})}$ .

Our third example will be a ${G_2}$ -structure. However, I want to have a more detailed discussion of ${G_2}$ -structures before I present it in this context. So, I will include it in a future post.

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