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

In Riemannian geometry, there is a well-known notion of the scalar curvature on a Riemannian manifold $M$, which is a function on $M$ given by a suitable contraction the Riemannian curvature tensor. The scalar curvature gives a rough measure how much the geometry of $M$ differs locally from the flat Euclidean case, namely how much the volume (or surface) of a small ball with a given radius differs from the usual value in flat space.

I am interested in whether (or in which cases) this notion generalises to arbitrary Cartan geometries, given by a model Klein geometry $(G, H)$, an $H$-principal bundle $P$ and a one-form $\omega$ on $P$ with values in the Lie algebra $\mathfrak g$ of $G$, the Cartan connection. For what follows, you may assume that $H$ is the trivial subgroup.

The curvature is given by the $\mathfrak g$-valued two-form $\Omega = d \omega + \frac 1 2 [\omega, \omega]$. If and only if $\Omega$ vanishes, $P$ looks locally like $G$ in that there is a local diffeomorphism between $P$ and $G$ such that $\omega$ becomes the Maurer--Cartan form of $G$. In this sense, the curvature is completely analogous to the Riemannian curvature.

Now my question is whether one can form out of $\Omega$, $\omega$ and maybe some scalar product on $\mathfrak g$ (e.g. the Killing form) a scalar valued function (or maybe density) on $P$ which gives a rougher measure for how much the geometry differs from the Lie group $G$ (just as the scalar curvature does for Riemannian curvature)? For example, does it make sense to compare the volume of a small ball around the identity of $G$ with the "same" (in terms of the connection $\omega$) ball around a point in $P$, or is there some other well-established notion?

Of course, I would like to see whether one gets back the scalar curvature of Riemannian geometry (maybe up to some term depending on the torsion) in the case that $G$ is the group of Euclidean motions and $H$ the subgroup of rotations.

share|improve this question

1 Answer 1

up vote 12 down vote accepted

In the case that $H$ is trivial, $\omega$ is just a $\frak{g}$-valued coframing on the manifold $P$. The curvature $2$-form $\Omega = d\omega + \tfrac{1}{2}[\omega,\omega]$ can then be written in the form $$ \Omega = R(\omega\wedge\omega) $$ where $R:P\to {\frak{g}}\otimes\Lambda^2({\frak{g}}^\ast)$ is the curvature tensor. Since $H$ is trivial, there is no group action to take into account, so, technically, any $\lambda\in {\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ could be used to construct a scalar-valued curvature quantity $\lambda(R):P\to\mathbb{R}$.

Now, you seem to want to consider the scalar curvatures constructed from those $\lambda$ that are invariant under the adjoint action of $G$ on $\frak{g}$, and typically there are such elements. For example, the Lie algebra structure itself is an element $\mu\in {\frak{g}}\otimes\Lambda^2({\frak{g}}^\ast)=\mathrm{Hom}\bigl(\Lambda^2({\frak{g}}),{\frak{g}}\bigr)$. In the semi-simple case, one can use the Killing form to dualize this to an element $\mu^\ast\in {\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$, and then the resulting 'scalar curvature' $\mu^\ast(R)$ is a very natural invariant.

More generally, though, you might want to ask for the $G$-invariant elements in ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ and see if there is anything else interesting there. However, in the case that $\frak{g}$ is simple, it is not hard to show that the induced action of $G$ on ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$ only fixes multiples of $\mu^\ast$.

Remark: One possible reason for considering the $G$-invariant elements is this: There is a natural action of $G$ on Cartan connections for the pair $(G,\lbrace e\rbrace)$, namely, one can define $a\cdot \omega = \mathrm{Ad}(a)\bigl(\omega\bigr)$. Then one will have $\lambda\bigl(R(\omega)\bigr)=\lambda\bigl(R(a\cdot\omega)\bigr)$ when $\lambda$ has this kind of $G$-invariance.

The more general case, when $H$ is nontrivial, can be handled similarly, and you'll want to look at the action of $H$ on ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g/h}})$ and consider the invariant elements $\lambda$. [NB: I have now modified this sentence to take into account the OP's comment below, as he pointed out that he wasn't restricting to torsion-free geometries.]

As to which of these invariants might be thought of as comparing some kind of local volume with some kind of 'geodesically parallel' volume (which is what the scalar curvature does in Riemannian geometry), I'd have to think about this some more.

Example: In the case that $G$ is the group of rigid motions of $\mathbb{R}^n$ and $H\simeq\mathrm{SO}(n)$ is the group of rotations about a point (in which the OP was particularly interested), one has $$ {\frak{g}}\simeq {\frak{so}}(n)\oplus \mathbb{R}^n \qquad\text{and}\qquad {\frak{h}}\simeq {\frak{so}}(n), $$ and the usual Cartan connection on the orthonormal frame bundle $P\to M$ of a Riemannian metric $g$ on an $n$-manifold $M$ is of the form $$ \omega = (\theta,\eta):TP\to {\frak{g}}\simeq {\frak{so}}(n)\oplus \mathbb{R}^n $$ where $\eta$ is the 'soldering form' and $\theta$ is the Levi-Civita connection form. The curvature of $\omega$ as a Cartan connection is then $$ \Omega = \bigl(d\theta+\theta\wedge\theta,\ d\eta + \theta\wedge\eta\bigr) = \bigl(R(\eta\wedge\eta),0\bigr), $$ where, now, $R:P\to \mathrm{Hom}\bigl(\Lambda^2(\mathbb{R}^n),{\frak{so}}(n)\bigr) \simeq {\frak{so}}(n)\otimes {\frak{so}}(n)$ is the usual Riemann curvature tensor. In this case, because the Cartan connection is torsion-free, the Cartan connection curvature takes values in $$ {\frak{h}}\otimes\Lambda^2\bigl(({\frak{g}/\frak{h}})^\ast\bigr)\simeq {\frak{h}}\otimes\Lambda^2\bigl(\mathbb{R}^n\bigr) \simeq {\frak{so}}(n)\otimes {\frak{so}}(n) $$ and, applying the above recipe to this curvature, the unique $H$-invariant element (up to multiples) in this space does indeed give the usual scalar curvature of the underlying Riemannian metric (up to a constant multiple).

NB: When $n=4$, the above space ${\frak{so}}(n)\otimes {\frak{so}}(n)$ actually has a $2$-dimensional space of vectors fixed under the action of $H$, but, of course, the first Bianchi identity says that the curvature takes values in a proper subspace of ${\frak{so}}(n)\otimes {\frak{so}}(n)$ that only intersects this $2$-dimensional space in a $1$-dimensional space after all. That brings up the point that, if one does consider only Cartan connections that satisfy some given torsion restrictions, then there can be Bianchi identities that say that the curvature of the connection must take values in some proper subspace of ${\frak{g}}^\ast\otimes\Lambda^2({\frak{g}})$, in which case, one will have to take this into account in considering which 'scalar curvatures' can be defined.

share|improve this answer
1  
Thanks for these thoughts so far; expressing all tensors using the natural coframe makes things much clearer. I have one question, though: In case of $H$ being non-trivial, why would I want to look at $\mathfrak h^* \otimes \Lambda^2 (\mathfrak g/\mathfrak h)$? I specifically do not want to restrict to torsion-free geometries, so I would need something like $\mathfrak g^* \otimes \Lambda^2 (\mathfrak g/\mathfrak h)$, wouldn't I? –  Marc Nieper-Wißkirchen Feb 14 '13 at 14:51
    
@Marc: Oh, yes, you are quite right. I lost sight of that for some reason in that final comment. –  Robert Bryant Feb 14 '13 at 15:55
    
@Robert: Please excuse my late silence — I have caught a serious cold and had no head for maths. Will think about your new input during the next days. –  Marc Nieper-Wißkirchen Feb 25 '13 at 14:36

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.