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In a recent question Deane Yang mentioned the beautiful Riemannian geometry that comes up when looking at $G_2$. I am wondering if people could expand on the geometry related to the exceptional Lie Groups. I am not precisely sure what I am looking for, but ostensibly there should be answers forth coming from other who have promised such answers. I understand a bit about how the exceptional Lie groups come up historically, and please correct the following if it is incorrect, but when looking at the possible dynkin diagrams you see that there is no reason for $E_6$,$E_7$,$E_8$,$G_2$, and $F_4$ to not occur as root systems. While root systems are geometric, this is not what I am asking about.

Thanks

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    $\begingroup$ You might enjoy reading the entry on holonomy in wikipedia: en.wikipedia.org/wiki/Holonomy, and especially the section on Berger's classification of possible holonomy groups. $G_2$ is one of the two possible exceptional holonomy groups. (It can appear as the holonomy group of a 7-dim'l manifold, acting on the tangent bundle through its 7 dim'l irrep. If you google exceptional $G_2$ holonomy you will find a lot of literature on this subject.) $\endgroup$
    – Emerton
    Commented Dec 14, 2010 at 6:52
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    $\begingroup$ This survey of Joyce may also help: people.maths.ox.ac.uk/~joyce/cc.ps $\endgroup$
    – Emerton
    Commented Dec 14, 2010 at 6:53
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    $\begingroup$ The recent question mentioned above: mathoverflow.net/questions/49151/… $\endgroup$
    – Unknown
    Commented Dec 14, 2010 at 15:30

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I promised Sean a detailed answer, so here it is.

As José has already mentioned, it is only $G_2$ (of the five exceptional Lie groups) which can arise as the holonomy group of a Riemannian manifold. Berger's classification in the 1950's could not rule it out, and neither could he rule out the Lie group $\mathrm{Spin}(7)$, but these were generally believed to not possibly be able to exist. However, in the early 1980's Robert Bryant succeeded in proving the existence of local examples (on open balls in Euclidean spaces). Then in the late 1980's Bryant and Simon Salamon found the first complete, non-compact examples of such manifolds, on total spaces of certain vector bundles, using symmetry (cohomogeneity one) methods. (Since then there are many examples of non-compact cohomogeneity one $G_2$ manifolds found by physicists.) Finally, in 1994 Dominic Joyce stunned the mathematical community by proving the existence of hundreds of compact examples. His proof is non-constructive, using hard analysis involving the existence and uniqueness of solutions to a non-linear elliptic equation, much as Yau's solution of the Calabi conjecture gives a non-constructive proof of the existence and uniqueness of Calabi-Yau metrics (holonomy $\mathrm{SU}(n)$ metrics) on Kahler manifolds satisfying certain conditions. (In 2000 Alexei Kovalev found a new construction of compact $G_2$ manifolds that produced several hundred more non-explicit examples. These are the only two known compact constructions to date.) It is exactly this similarity to Calabi-Yau manifolds (and to Kahler manifolds in general) that I will explain.

When it comes to Riemannian holonomy, the aspect of the group $G_2$ which is important is not really that it is one of the five exceptional Lie groups, but rather that it is the automorphism group of the octonions $\mathbb O$, an $8$-dimensional non-associative real division algebra. The octonions come equipped with a positive definite inner product, and the span of the identity element $1$ is called the real octonions while its orthogonal complement is called the imaginary octonions $\mathrm{Im} \mathbb O \cong \mathbb R^7$. This is entirely analogous to the quaternions $\mathbb H$, except that the non-associativity introduces some new complications. In fact the analogy allows us to define a cross product on $\mathbb R^7$ in the same way, as follows. Let $u, v \in \mathbb R^7 \cong \mathrm{Im} \mathbb O$ and define $u \times v = \mathrm{Im}(uv)$, where $uv$ denotes the octonion product. (In fact the real part of $uv$ is equal to $-\langle u, v \rangle$, just as it is for quaternions.) This cross product satisfies the following relations: \begin{equation} u \times v = - v \times u, \qquad \qquad \langle u \times v , u \rangle = 0, \qquad \qquad {|| u\times v||}^2 = {|| u \wedge v ||}^2, \end{equation} exactly like the cross product on $\mathbb R^3 \cong \mathrm{Im} \mathbb H$. However, there is a difference, unlike the cross product in $\mathbb R^3$, the following expression is not zero: \begin{equation} u \times (v \times w) + \langle u, v \rangle w - \langle u, w \rangle v \end{equation} but is instead a measure of the failure of the associativity $(uv)w - u(vw)$, up to a factor. Note that on $\mathbb R^7$ there can be defined a $3$-form (totally skew-symmetric trilinear form) using the cross product as follows: $\varphi(u,v,w) = \langle u \times v, w \rangle$, which is called the associative $3$-form for reasons that we won't get into here.

Digression: In fact one can show that only on $\mathbb R^3$ and $\mathbb R^7$ can one construct such a cross product, and this is intimately related to the fact that only the spheres $S^2$ and $S^6$ can admit almost complex structures. But I digress...

Getting back to $G_2$ geometry: a $7$-dimensional smooth manifold $M$ is said to admit a $G_2$-structure if there is a reduction of the structure group of its frame bundle from $\mathrm{GL}(7, \mathbb R)$ to the group $G_2$ which can actually be viewed naturally as a subgroup of $\mathrm{SO}(7)$. For those familiar with $G$-structures, this tells you that a $G_2$-structure determines a Riemannian metric and an orientation. In fact, one can show on a manifold with $G_2$-structure, there exists a non-degenerate $3$-form $\varphi$ for which, given a point $p$ on $M$, there exists local coordinates near $p$ such that, in those coordinates, at the point $p$, the form $\varphi$ is exactly the associative $3$-form on $\mathbb R^7$ discussed above. Now one can show that there is a way to canonically determine both a metric and an orientation in a highly non-linear way from this $3$-form $\varphi$. Then one can define a cross product $\times$ by essentially using the metric to ``raise an index'' on $\varphi$. In summary, a manifold $(M, \varphi)$ with $G_2$-structure comes equipped with a metric, cross product, $3$-form, and orientation, which satisfy \begin{equation} \varphi(u,v,w) = \langle u \times v , w \rangle. \end{equation} This is exactly analogous to the data of an almost Hermitian manifold, which comes with a metric, an almost complex structure $J$, a $2$-form $\omega$, and an orientation, which satisfy \begin{equation} \omega(u,v) = \langle Ju , v \rangle. \end{equation} Essentially, a manifold admits a $G_2$-structure if one can identify each of its tangent spaces with the imaginary octonions $\mathrm{Im} \mathbb O \cong \mathbb R^7$ in a smoothly varying way, just as an almost Hermitian manifold is one in which we can identify each of its tangent spaces with $\mathbb C^m$ (together with its Euclidean inner product) in a smoothly varying way.

For a manifold to admit a $G_2$-structure, the necessary and sufficient conditions are that it be orientable and spin. (This is equivalent to the vanishing of the first two Stiefel-Whitney classes.) So there are lots of such $7$-manifolds, just as there are lots of almost Hermitian manifolds. But the story does not end there.

Let $(M, \varphi)$ be a manifold with $G_2$-structure. Since it determines a Riemannian metric $g_{\varphi}$, there is an induced Levi-Civita covariant derivative $\nabla$, and one can ask if $\nabla \varphi = 0$? If this is the case, $(M, \varphi)$ is called a $G_2$-manifold, and one can show that the Riemannian holonomy of $g_{\varphi}$ is contained in the group $G_2 \subset \mathrm{SO}(7)$. These are much harder to find, because it involves solving a fully non-linear partial differential equation for the unknown $3$-form $\varphi$. They are in some ways analogous to Kahler manifolds, which are exactly those almost Hermitian manifolds that satisfy $\nabla \omega = 0$, but those are much easier to find. One reason is because the metric $g$ and the almost complex structure $J$ on an almost Hermitian manifold are essentially independent (they just have to satisfy the mild condition of compatibility) whereas in the $G_2$ case, the metric and the cross product are determined non-linearly from $\varphi$. However, the analogy is not perfect, because one can show that when $\nabla \varphi = 0$, the Ricci curvature of $g_{\varphi}$ necessarily vanishes. So $G_2$-manifolds are always Ricci-flat! (This is one reason that physicists are interested in such manifolds---they play a role as ``compactifications'' in $11$-dimensional $M$-theory analogous to the role of Calabi-Yau $3$-folds in $10$-dimensional string theory.) So in some sense $G_2$-manifolds are more like Ricci-flat Kahler manifolds, which are the Calabi-Yau manifolds.

In fact, if we allow the holonomy to be a proper subgroup of $G_2$, there are many examples of $G_2$-manifolds. For example, the flat torus $T^7$, or the product manifolds $T^3 \times CY2$ and $S^1 \times CY3$, where $CYn$ is a Calabi-Yau $n$-fold, have Riemannian holonomy groups properly contained in $G_2$. These are in some sense ``trivial'' examples because they reduce to lower-dimension constructions. The manifolds with full holonomy $G_2$ are also called irreducible $G_2$-manifolds and it is precisely these manifolds that Bryant, Bryant-Salamon, Joyce, and Kovalev constructed.

We are lacking a ``Calabi-type conjecture'' which would give necessary and sufficient conditions for a compact $7$-manifold which admits $G_2$-structures to admit a $G_2$-structure which is parallel ($\nabla \varphi = 0$.) Indeed, we don't even know what the conjecture should be. There are topological obstructions which are known, but we are far from knowing sufficient conditions. In fact, this question is more similar to the following: suppose $M^{2n}$ is a compact, smooth, $2n$-dimensional manifold that admits almost complex structures. What are necessary and sufficient conditions for $M$ to admit Kahler metrics? We certainly know many necessary topological conditions, but (as far as I know, and correct me if I am wrong) we are nowhere near knowing sufficient conditions.

What makes the Calabi conjecture tractable (I almost said easy, of course it is anything but easy) is the fact that we already start with a Kahler manifold (holonomy $\mathrm{U}(m)$ metric) and want to reduce the holonomy by only $1$ dimension, to $\mathrm{SU}(m)$. Then the $\partial \bar \partial$-lemma in Kahler geometry allows us to reduce the Calabi conjecture to a (albeit fully non-linear) elliptic PDE for a single scalar function. Any analogous ``conjecture'' in either the Kahler or the $G_2$ cases would have to involve a system of PDEs, which are much more difficult to deal with.

That's my not-so-short crash course in $G_2$-geometry. I hope some people read all the way to the end of this...

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    $\begingroup$ To echo emerton, many thanks. Great stuff. $\endgroup$
    – Deane Yang
    Commented Dec 14, 2010 at 16:33
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    $\begingroup$ +1!!! Thanks, Spiro. I knew this question was way the heck up your alley and was hoping (and expecting) you would answer, but this is beyond my wildest dreams. I will read this all the way through, and although it will take me some time and effort to do so, it will definitely be worth it. It seems to me that there is potential here to parlay this answer into a "What is...a $G_2$-manifold?" article. $\endgroup$ Commented Dec 14, 2010 at 17:30
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    $\begingroup$ "We certainly know many necessary topological conditions" -- are there any as simple as the "odd Betti numbers must be even" one for Kahler manifolds? $\endgroup$ Commented Dec 14, 2010 at 18:53
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    $\begingroup$ @Allen: We know $b_3$ (and by Poincaré duality also $b_4$) must be $\geq 1$, because the $3$-form (and its dual $4$-form) both represent non-trivial cohomology classes. This is for any $G_2$-manifold. If you insist that the holonomy be strictly $G_2$, and not a propers subgroup, there are 3 more conditions: the fundamental group $\pi_1(M)$ must be finite; there is a symmetric bilinear form on $H_2 (M, \mathbb R)$ (like the intersection form on $4$-manifolds) which must be definite (positive or negative depends on your sign convention), and the first Pontrjagin class $p_1(M)$ must be non-zero. $\endgroup$ Commented Dec 14, 2010 at 19:03
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    $\begingroup$ An expanded version of this answer has just appeared in the April 2011 issue of the Notices of the AMS as a "What is ...?" article. ams.org/notices/201104/rtx110400580p.pdf $\endgroup$ Commented Mar 30, 2011 at 17:14
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If you are willing to move away from Riemannian geometry, but still within differential geometry, then $G_2$ arises in a number of surprising and simple geometric circumstances, but it is not the compact $G_2$ of the answers above, but rather the `split' $G_2$. (Each simple Lie algebra has a compact real form and a ``maximally non-compact real form, called its split real form.) Consider the problem of rolling a sphere of radius 1 about on a sphere of radius R. The resulting configuration space is a 5-manifold $M^5$ which is a circle bundle over the product of the two spheres, the circle encoding the relative orientation of the two spheres at their points of tangency. This rolling problem defines a rank 2 distribution (linear sub-bundle of the tangent bundle) on $M^5$.

Theorem: the local symmetry algebra of this distribution is that of the split $G_2$ if and only if $R =3$ or $R =1/3$. If you move to a double cover of $M^5$ then the split $G_2$ acts by symmetries of the lifted distribution, so that this double cover is a homogeneous space for $G_2$.

Gil Bor and I give some details of this theorem, and the action, and how it relates to Cartan's thesis construction of $G_2$ in the paper $G_2$ and the Rolling Distribution'', L'Enseignement Mathematique,
vol. 55, 157-196 (2009), or (arXiv:math/0612469). R. Bryant and L. Hsu detailed several other surprising realizations of $G_2$ including one on the space of all space curves having constant torsion 1 , in their Inventiones paper.

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The comments have so far addressed the issue of $G_2$ holonomy, which might be what was at the heart of Deane's answer to the question mentioned by the OP. Alas, of the exceptional Lie groups, only $G_2$ is part of the holonomy fellowship; namely, the possible holonomy groups of (simply-connected, complete) irreducible non-symmetric riemannian manifolds. To see the geometry associated to the other exceptional Lie groups, one needs to look elsewhere.

One kind of geometry associated to the exceptional Lie groups is the geometry of Riemannian symmetric spaces surrounding the Freudenthal-Tits magic square.

Closely related to this is a geometric realisation of the Lie algebras (not the groups, but then again more than just the root systems) of some of the exceptional Lie groups; namely $F_4$ and $E_8$ in some work of mine. (Sorry for the self-promotion.)

The idea was to find a geometrical interpretation to the spinorial construction of $F_4$ and $E_8$ which you can find described in Frank Adams's posthumous book Lectures on exceptional Lie groups by a procedure familiar to practitioners of supergravity: namely, the construction of a natural $\mathbb{Z}/2$-graded Lie (super)algebra from some geometric data, called the KIlling (super)algebra, as it is generated by Killing spinor fields.

The result is that $F_4$ and $E_8$ are the Killing Lie algebras of the round 8- and 15-sphere, respectively.

It is possible that the similar spinorial constructions of $E_6$ and $E_7$ can also be geometrised in this way, but I have not yet worked out the details.

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    $\begingroup$ @José: Very interesting. I did not know about this relation of $F_4$ to $S^8$. I will definitely read this paper. $\endgroup$ Commented Dec 14, 2010 at 13:56
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The compact group $F_4$ is the group of isometries of the octonionic projective plane $\mathbb{OP}^2$ endowed with an analog of Fubini-Study metric. I suspect the other real groups of type $F_4$ are the isometries groups of the octonionic hyperbolic plane and of the analogous objects built from split octonions. (Related question on mathoverflow.) One of the noncompact real forms of $E_6$ is the group of projective transformations (collineations) of $\mathbb{OP}^2$. The groups of type $F_4$ and $E_6$ arise in this context because of their close relationship to the exceptional Jordan algebras of hermitian three by three matrices over octonions. Indeed -- the group $E_6$ preserves the determinant of these matrices and $F_4$ preserves the determinant and the trace.

The most geometric approach to the exceptional groups that I am aware of (and which goes in this direction) is that of Rosenfeld. Unfortunately I don't have that book. He interprets groups of type $E_7$ and $E_8$ in a similar manner for $(\mathbb{C}\otimes\mathbb{O} ) \mathbb{P}^2$ and $(\mathbb{H}\otimes\mathbb{O} ) \mathbb{P}^2$. Some details and introduction to the subject is in Baez.

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  • $\begingroup$ The notes by Baez are certainly great. I had never heard of this book by Rosenfeld. It sounds fantastic! Thanks. $\endgroup$ Commented Jan 10, 2011 at 18:04
  • $\begingroup$ $E_7$ is related to $\mathbb H\otimes\mathbb OP^2$, $E_8$ is related to $\mathbb O \otimes \mathbb OP^2$. See e.g. Baez article on octonions. However there are some doubts whether these are projective spaces. For sure $F_{II}, E_{III}, E_{VI}, E_{VIII}$ of dimensions $16,32,64,128$ are Riemmanian symmetric spaces. $\endgroup$
    – user21230
    Commented Jan 30, 2019 at 11:49
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A review on the history of $G_2$ and its relation with the 7-dimensional geometry, is given in the following article of AMS Notices:

link text

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