Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

Thanks.

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
1  
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
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.