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}


\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.

Gronwall Inequality

There are a number of different statements of Gronwall’s inequality. In this post, we will consider only one of them, perhaps the weakest of all.

Proposition 1 Let {f(t)} be a non-negative continuous function on {\left[ a,b \right]} such that there are positive constants {C} and {K} satisfying

\displaystyle \begin{array}{rcl} f(t)\le C + K\int_{a}^{t}f(s)ds \end{array}

for all {t\in\left[ a,b \right]}. Then,

\displaystyle \begin{array}{rcl} f(t)\le Ce^{K(t-a)} \end{array}

for all {t \in \left[ a,b \right]}.

Proof: Define {U(t) = C + K\int_{a}^{t}f(s)ds}. Note that, by definition, {f(t)\le U(t)} and {U} is a strictly positive differentiable function. Also, we have {U'(t) = Kf(t)\le KU(t)}. In other words, {\frac{U'(t)}{U(t)}\le K} which means the relative rate of change of {U} is less than {K}. Hence, the growth of {U} is slower than an exponential function with relative rate of change {K}. That is {U(t) \le U(a) e^{K(t-a)}} (if you did not like this reasoning, you may integrate both sides of the previous inequality from {a} to {t}). So, we have the desired result {f(t) \le U(t) \le U(a)e^{K(t-a)}= Ce^{K(t-a)}}. \Box