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A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure.

For example:
• $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} (n)^{(1)} = 0$.
• But $Sp(n)$-structures (symplectic structures) are not of finite type because the group of symplectomorphisms is infinite dimensional.
• It can also be shown that $CO(n)$-structures (conformal structures) are of finite type and order $2$ (except if the dimension is $2$, in which case it is not of finite type).

Are there any finite type $G$-structures of order greater than $2$?

More generally, are there $G$-structures of any order?


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could you explain the meaning of $\mathfrak{g}^{(k)}$? and its relation to $\pi$? – Olivier Bégassat May 5 '11 at 23:28
Actually, while the definition involves the bundle $B_G$, it depends only on the group $G$. You may think of $\mathfrak{g}^{(k)}$ as the space of all $k+1$-multilinear functions $T : (\mathbb{R}^n)^{k+1} \rightarrow \mathbb{R}^n$ which are completely symmetric and such that, given any $v_1, \ldots, v_k \in \mathbb{R}^n$, the linear map $\phi : \mathbb{R}^n \rightarrow \mathbb{R}^n$ given by $\phi(u) = T(v_1, \ldots, v_k, u)$ belongs to $\mathfrak{g}$. – Leandro May 5 '11 at 23:35
when you write $\mathbb{R}^n$ do you mean $T_m M$, the tangent space at $m\in M$? Is your notion of $G$ structure the same as that where you ask for an atlas of $M$ such that the transition maps induced on the local trivialisations of $TM$ act like $G$ (for some representation $\roh:G\rightarrow \mathrm{GL}(n)$? – Olivier Bégassat May 5 '11 at 23:54
@Olivier: I think that you're correct. And I think that having a structure of order 3 would mean the following: there should exist a $G$-manifold $M$, and a non-trivial (locally-defined) automorphism around some point $m\in M$ that fixes the 2-jet of that point. – André Henriques May 6 '11 at 0:00
up vote 13 down vote accepted

Yes, $G$-structures exist of each finite order. In other words, for every $k\ge1$, there is an $n\ge1$ and a subgroup $G\subset GL(n,\mathbb{R})$ such that its Lie algebra $\frak{g}$ satisfies ${\frak{g}}^{(k-1)}\not=0$ while ${\frak{g}}^{(k)}=0$.

There is no known classification of such algebras, but here is a simple example of an algebra ${\frak{g}}_k\subset {\frak{gl}}(k{+}3,\mathbb{R})$ such that ${\frak{g}}_k$ has order $k$: Let $e_1,\ldots, e_{k+3}$ be the standard basis of $\mathbb{R}^{k+3}$, with dual basis $x^1,\ldots, x^{k+3}$. Let ${\frak{g}}_k$ be the (abelian, nilpotent) subalgebra of ${\frak{gl}}(k{+}3,\mathbb{R})$ with basis $l_1,\ldots,l_k$, where $$ l_i = e_{i+3}\otimes x^1 + e_{i+2}\otimes x^2. $$ One computes that ${\frak{g}}_1^{(1)}=0$ and that, for $k>1$, the space ${\frak{g}}_k^{(1)}$ has dimension $k{-}1$, with basis $q_2,\ldots,q_k$, where $$ q_i = e_{i+3}\otimes (x^1)^2 + 2e_{i+2}\otimes x^1x^2 + e_{i+1}\otimes (x^2)^2. $$ Continuing on in this way, one finds that the dimension of ${\frak{g}}_k^{(j)}$ is $k{-}j$ for $0\le j\le k$.

For each $n$, there is an upper bound on the order of the subalgebras of ${\frak{gl}}(n,\mathbb{R})$ of finite type, but I do not know what that is. There are estimates for this upper bound, but I don't think they are very tight.

Meanwhile, a theorem of Cartan (originally proved over $\mathbb{C}$ by a classification (but with some omissions), and later completed by others and worked out over $\mathbb{R}$ as well) says that, if $G\subset GL(n,\mathbb{R})$ acts irreducibly on $\mathbb{R}^n$, then $\frak{g}$ has order $1$, $2$, or $\infty$. The list of the irreducibly acting $G\subset GL(n,\mathbb{R})$ that have order $2$ or $\infty$ is known and can be found in my 1996 survey paper, Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), Sémin. Congr., vol. 1 (1996), pp. 93–165. See the tables in Appendix A.

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Robert: This is interesting. What are those subgroups $G\subset GL(n,\mathbb R)$? – André Henriques May 6 '11 at 0:02
@André: Which ones? The ones of finite order $k>2$ or the irreducibly acting ones of order $2$ or $\infty$? – Robert Bryant May 6 '11 at 4:55
I was asking about the ones of finite order $> 2$. – André Henriques May 6 '11 at 9:22
@Robert: And which are the irreducibles of order 2? – José Figueroa-O'Farrill May 6 '11 at 9:48
Tank you Robert for the added explanations. – André Henriques May 7 '11 at 15:07

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