The Maurer-Cartan form on the Lie group $Gl(n,\mathbb{R})$ is a one-form taking values in $\mathfrak{gl}(n,\mathbb{R})$ as defined in the link. It has a rather concrete "extrinsic definition" as $\omega_{A}=A^{-1}dA$, which is apparently a slight abuse of notation.

Question: What precisely is the kernel of $\omega_{A}$, as referred to in the foliation literature? Is there a concrete way to see the kernel of $\omega_{A}$ using the above "extrinsic definition"?

I think the kernel of the lie-algebra-valued one form is locally the space of horizontal vectors as described on page 94 of this reference. The present question should be interpreted as asking how to most easily interpret this from the point of view of the above "extrinsic definition". For example, can we calculate this kernel by multiplying matrix of $\mathfrak{gl}(n,\mathbb{R})$ (perhaps by the natural identification with the tangent space) by the matrix of one-forms $\omega_{A}=A^{-1}dA$? (I don't think this is right.) How concrete can this calculation be made in the matrix case?

  • $\begingroup$ For 1) It is a vector space valued 1-form so it gives a linear map from vector fields to the Lie algebra. Do you have more context for 2)? I don't see it can have a kernel (viewing it as a map as in 1)) since it takes a tangent vector $v$ at a point $g$ to ${L_{g^{-1}}}_* v$ which is non-zero if $v$ is non-zero since left multiplicaiton by $g$ is a diffeomorphism. $\endgroup$ – Eric O. Korman Apr 11 '14 at 22:59
  • $\begingroup$ @Eric: This is precisely what is driving me nuts. Perhaps it is meant as a kernel of a flat connection. If I had a reference, I probably would not have had to ask this question. Sorry! $\endgroup$ – Jon Bannon Apr 11 '14 at 23:01
  • $\begingroup$ But do you have an example of where this appears in the foliation literature you mention? $\endgroup$ – Eric O. Korman Apr 11 '14 at 23:02
  • $\begingroup$ Ah! I found the reference as a pdf online. Start reading on page 81 of this: f3.tiera.ru/2/M_Mathematics/MD_Geometry%20and%20topology/… $\endgroup$ – Jon Bannon Apr 11 '14 at 23:05
  • 2
    $\begingroup$ Oh ok. It looks like they're using a much more general definition of Maurer-Cartan form (for example, their M.C. forms are not even on a Lie group necessarily). It seems like the prototypical example of such a form is a flat connection on a principal $G$-bundle. Note that the kernel of a connection defines a horizontal distribution and this distribution is integrable (and therefore gives rise to a foliation on the total space of the principal bundle) if and only if the connection is flat. $\endgroup$ – Eric O. Korman Apr 11 '14 at 23:11

For a general Lie group $G$ the Maurer-Cartan form $\omega$ is the unique left invariant $\mathfrak{g}$ valued 1-form so that $\omega_e \colon T_eG \to \mathfrak{g}$ is the identity map. To be clear, left invariance means that if $L_g\colon G \to G$ is the left multiplication map so that $L_g(h) = gh$ then $L_g^* \omega = \omega$. Equivalently, if you are more comfortable with left invariant vector fields, we have the formula $\omega(X) = X_e$ for any left invariant vector field $X$.

Morally you can think of the Maurer-Cartan form as a compact way to write the standard left-parallelization of the tangent bundle of $G$. Indeed, as a 1 form with values in $\mathfrak{g}$, for each $g\in G$ the restriction $\omega_g$ is a linear map from $T_gG$ to $\mathfrak{g}$. Notice that $T_gG$ and $\mathfrak{g}$ have the same dimension and also note that $\omega_g$ is surjective (for example, if $v \in \mathfrak{g}$, let $X$ be the left invariant vector field on $G$ so that $X_e = v$. Then $\omega_g(X) = v$). Thus we see that $\omega_g$ is an isomorphism for each $g\in G$ and also that $\omega$ gives a parallelization of $TG$ as a vector bundle.

Hopefully you see now why the Maurer-Cartan form does not have a kernel. Now, given a principal $G$-bundle (that is, a bundle over a manifold whose fiber is $G$ as well as a right action of $G$ on each fiber), there is a notion of a connection. This will be a $\mathfrak{g}$ valued 1-form on the total space which satisfies a couple properties, but the important point for us is that it is modeled on the Maurer-Cartan form. In particular, if you fix a point in the base manifold , the fiber is naturally a $G$-torsor, which is to say $G$ without any point singled out as the identity. If we (arbitrarily) fix some point as the identity then we can give the fiber the multiplicative structure of $G$. Then the restriction of the connection to this $G$ will be the Maurer-Cartan form. However, when considered on the total space the connection will have a kernel, called the horizontal space. This kernel will have the same dimension as the base manifold, and indeed allows us to lift vectors from the base manifold to the total space in a consistent manner.

Incidentally, considered from this perspective it is clear why the curvature of a connection $\omega$, defined as $\Omega = d\omega + \omega\wedge\omega$, only measures phenomena in the horizontal direction. Indeed, any Maurer-Cartan form satisfies the Maurer-Cartan equation $d\omega + \omega\wedge\omega = 0$, and so the curvature form will be zero when restricted to any fiber.

  • $\begingroup$ Thanks, Ben! This should do. $\endgroup$ – Jon Bannon May 20 '14 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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