construction of an invariant quantity

48 Views Asked by At

Let $G$ be a (matrix-)group acting on a manifold $M$ (e.g. the group of motions in the plane $\mathbb{R}^2$).

We consider the set of moving frames at each point $x \in M$, $(x, e_0, e_1, \dots,e_n)$, where the $e_i$ are independent vectors.

The Maurer-Cartan 1-forms associated to these frames are $$ \mathrm{d}x = e_i\omega^i \\ \mathrm{d}e_i = e_j\omega^j_i$$

They satisfy the Maurer-Cartan structure equations: $$ \mathrm{d}\omega^i = -\omega^i_j \wedge \omega^j \\ \mathrm{d}\omega^i_j = -\omega^i_k\wedge\omega^k_j $$

If we have a metric space, i.e. inner product relations between the $e_i$, we get additional relations between the Maurer-Cartan forms. (E.g. for the group of motions in the plane: $\omega^1_1 = \omega^2_2 = 0, \omega_1^2 = -\omega_2^1$.)

Integral geometry deals with the construction and integration of densities (a differential $k$-form) of geometric objects that are invariant under $G$. A necessary and sufficient condition for the density to be invariant is that its exterior differential vanish. As an example, the density of lines $L$ in the plane $\mathbb{R}^2$ invariant under the group of motions is $$ \mathrm{d}L = \omega_2^1 \wedge\omega^2. $$ If $G_L$ is the stabilizer of $L$ under $G$ then this is the volume form of the homogeneous space $G/G_L$.

This densitiy is invariant because $$ \mathrm{d} ( \mathrm{d} L ) = \mathrm{d} \omega_2^1 \wedge \omega^2 - \omega_2^1 \wedge \mathrm{d} \omega^2 \\ = (-\omega_1^1 \wedge \omega_2^1 - \omega_2^1 \wedge \omega_2^2) \wedge \omega^2 - \omega_2^1 \wedge (-\omega_1^2 \wedge \omega^1 - \omega_2^2 \wedge \omega^2) \\ =0 \quad (\mbox{recall that } \omega^1_1 = \omega^2_2 = 0, \omega_1^2 = -\omega_2^1)$$

It is clear that if we integrate the invariant density, we obtain an invariant measure $m(X) = \int_X \mathrm{d} L$. Further we can construct the ratio $P(X/Y) = \frac{m(X)}{m(Y)}$, which is also an invariant.

My question is if it is possible to construct invariant quantities, like the $P$ above, from non-invariant densities/measure.

Again, in the simple case of the group of motions in the plane for example, if I have as density $\mathrm{d} \tilde{L} = \omega^2$, which is not invariant since $\mathrm{d} ( \mathrm{d} \tilde{L} ) = \mathrm{d} \omega^2 = -\omega_1^2 \wedge \omega^1 \neq 0$. In a parameterisation of the group $\omega^2$ may have the form $\sin \theta\, \mathrm{d} b$ ($\theta$ being an angle of rotation and $b$ a shift in the direction perpendicular to $\tilde{L}$). So in a very naive thought one might be tempted to assume that in the construction $$ \tilde{P} (X/Y) = \frac{\int_X \omega^2}{\int_Y \omega^2}$$ the dependency of the angular parameter cancels out and the resulting $\tilde{P}$ is invariant.

Do quantities like $\tilde{P}$ exist and if so, how to prove (in general) that they are invariant?