# G-Structures 2

In this post, we briefly introduce the Lie group ${G_2}$, ${G_2}$-structures on a manifold and a ${G_2}$-manifold. Let us denote the three form ${dx^i\wedge dx^j \wedge dx^k}$ on ${{\mathbb R}^7}$ by ${dx^{ijk}}$. We set ${\phi_0 = dx^{123}+ dx^{145}-dx^{167}+dx^{246}+dx^{257}+dx^{347}-dx^{356}}$. This three form is non-degenerate in the sense that whenever we have two linearly independent vectors in ${{\mathbb R}^7}$, we can find a third vector such that the evaluation of ${\phi_0}$ on these vectors is non-zero. We define ${G_2 = \left\{ M\in GL(7,{\mathbb R}) \big| M^*\phi_0 = \phi_0 \right\}}$. One may prove that ${G_2}$ is a ${14}$-dimensional Lie subgroup of ${SO(7)}$.

Let us give a different descriptions of ${\phi_0}$. So, it does not look completely arbitrary. ${7}$ is the highest dimension that one may define a cross product. After we identify ${{\mathbb R}^8}$ with the octonions ${\mathbb O}$ equiped with some octonion product, for any two imaginary octonions ${x,y \in {\mathbb R}^7 \cong im(\mathbb O)}$ we define the cross product to be

$\displaystyle \begin{array}{rcl} x \times y = \frac{1}{2} [x,y] = \frac{1}{2}(xy-yx). \end{array}$

Then, we may define the ${3}$-form on ${{\mathbb R}^7}$ by ${\phi_0(x,y,z) = \left< x \times y, z\right>}$ where the inner product is the standard inner product. Of course, there is a choice on the octonion product and hence, ${\phi_0}$ may be different than the one we explicitly wrote above. However, we show that they are equivalent using the right octonion product. To show they are equivalent; first, we prove that ${\left}$ is indeed a ${3}$-form and then, evaluate it on the basis elements to see how the octonion product should be defined.

Using ${im(xy)=-im(yx)}$, we obtain ${x\times y = im(xy)}$ and thus, the above definition is equivalent to

$\displaystyle \begin{array}{rcl} \phi_0(x,y,z) &=& \left< xy, z\right>. \end{array}$

To prove that ${\phi_0}$ is alternating, it is enough to prove ${\phi_0(x,x,y)=0, \phi_0(x,y,x)=0}$ and ${\phi_0(y,x,x)=0}$ as we may replace ${x}$ by ${x+z}$ to get the desired equalities. However, also note that ${x \times y = - y\times x}$. Therefore, the first two equalities are enough. It is clear that ${x\times x = 0}$. Hence, we have the first equality. Furthermore,

$\displaystyle \begin{array}{rcl} \phi_0(x,y,x) &=& \left< xy, x\right> \\ &=& |x|^2\left< y,1\right> \\ &=& 0. \end{array}$

Thus, we have showed that ${\phi_0}$ is alternating.

Our next goal is to define the octonion product. Clearly, from the explicit definition, we want ${\phi_0(x_1,x_2,x_3)=1}$. In other words, ${\left = 1}$. So, a natural choice for the product ${x_1x_2}$ is ${x_3}$. Similarly, we can choose ${x_1x_4=x_5}$, ${x_1x_6=-x_7}$, ${x_2x_4=x_6}$, ${x_2x_5=x_7}$, ${x_3x_4=x_7}$ and ${x_3x_5=-x_6}$. Of course, as we are describing octonion multiplication, we should also define the multiplication with the ${8}$th generator but it is the generator of ${Re(\mathbb O)={\mathbb R}}$ part. So, it is just the trivial multiplication i.e. the multiplication coming from the vector space structure. We do not show that this indeed defines an octonion product.

Next, we need to show that they are equal and to do that, it is enough to evaluate on the basis elements. It is an easy computation which we omit.

Note that this definition makes an earlier claim more plausible, namely that ${\phi_0}$ is non-degenerate. Because ${\phi_0(x,y,x\times y) = \left}$ so, we only need to show that ${x\times y}$ is non-zero for linearly independent ${x}$ and ${y}$. However, that is a built-in property for a cross-product.

A ${G_2}$-structure on a manifold ${M}$ can be defined as a subbundle of the frame bundle of ${M}$ whose fibers are isomorphic to ${G_2}$. However, there is an equivalent, more convenient definition. In fact, this definition will follow the scheme of the previous post. More explicitly, since ${G_2}$ fixes ${\phi_0}$ on ${{\mathbb R}^7}$, we may pull it back to each space ${T_pM}$ to have a three form ${\phi}$ on the manifold and similarly, if we have such a three form on the manifold, then we may find a subbundle of the frame bundle whose fibers are ${G_2}$. So, having a three form ${\phi}$ on ${M}$ such that for any point ${p\in M}$, ${\phi_p}$ and ${\phi_0}$ can be identified by an isomorphism between ${{\mathbb R}^n}$ and ${T_pM}$, means that we can find a ${G_2}$-structure on ${M}$. By an abuse of notation, we call ${(M,\phi)}$ a manifold with a ${G_2}$-structure. Furthermore, since ${G_2}$ is a subgroup of ${SO(7)}$, it also fixes the standard metric and orientation on ${{\mathbb R}^7}$ giving rise to a Riemannian metric and orientation on the manifold. This immediately implies a non-orientable manifold does not admit a ${G_2}$-structure.

Next, we introduce ${G_2}$-manifolds. Given a manifold ${M}$ with a ${G_2}$-structure ${\phi}$ and the induced metric ${g}$, let ${\nabla}$ be the Levi Civita connection on ${(M,g)}$. If ${\nabla \phi =0}$, we call ${\phi}$ a torsion free ${G_2}$-structure. A manifold with a torsion free ${G_2}$-structure is called a ${G_2}$-manifold. In fact, there are a number of ways to define ${G_2}$-manifolds, as we can see in the following proposition.

Proposition 1 Let ${(M^7,\phi)}$ be a ${G_2}$-structure on ${M}$ with the induced metric ${g}$ and the Levi Civita connection ${\nabla}$. Then, the following are equivalent:

1. ${\nabla \phi = 0}$
2. ${Hol(g) \subseteq G_2}$
3. ${d\phi = 0}$ and ${d^*\phi = 0}$.

If any one of the conditions of the proposition holds (and hence, all), we call ${M}$ a ${G_2}$-manifold. The first example of a metric with ${G_2}$ holonomy is given by Bryant. The metric in his example is incomplete. Later, Bryant and Salamon constructed complete metrics with ${G_2}$ holonomy on non-compact manifolds. Then, Joyce constructed complete examples on compact manifolds.

# G-Structures

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.