1
$\begingroup$

Hallo,

I am trying to read the paper "Calibrated embeddings in the special Lagrangian and coassociative cases" by Robert Bryant and I have a question. I hope that maybe one of you could give me some answers. The paper can be found here: http://arxiv.org/abs/math/9912246 . Here is my question: On page 12, in the middle, there are defined the spaces $\mathfrak{h}_{k}$ = { $x\in \mathbb{R}^{n\times n}$ | $\iota^{*}_{k}(x.\alpha)=0$, $\forall \alpha \in $ $\Lambda^{*}(\mathbb{R}^{n})^{G}$ }. I am not very familiar with the notation. I think $x.\alpha$ means the Lie-derivative of $\alpha$ in the direction $x$. Am I right? If so, using the formula for the Lie-derivative $\mathcal{L}_{X}= \iota d + d\iota$. I do not see how to plug in a matrix $x \in \mathbb{R}^{n\times n}$ in a form defined on $\mathbb{R}^{n}$. How can this be understood? I think, by using the $\mathbb{R}^{n}$-valued form $\nu$ defined by $\nu (v) = u(\pi ' (v))$ , for all $v \in T _{u} F$ (also on page 12 in the middle). But still, for $x \in \mathbb{R}^{n\times n} = Lie(GL(n,\mathbb{R}))=ker(\pi ')$ one gets $ \nu(x)=0 $. How is this to be understood? I would be very tankfull if somebody could help me with this.

greetings hapchiu

$\endgroup$
3
  • 1
    $\begingroup$ The notation $x.\alpha$ means the Lie algebra action of $x$ on the form $\alpha$, i.e., if $e^{tx}$ is the $1$-parameter subgroup of linear transformations of $\mathbb{R}^n$, then $x.\alpha$ is the velocity at $t=0$ of the curve of forms $(e^{tx})^\ast(\alpha)$. This is a standard notation for representations of Lie algebras. $\endgroup$ Feb 20, 2013 at 13:39
  • $\begingroup$ is there any interpretation of this using the Lie-derivative ? $\endgroup$
    – hapchiu
    Feb 20, 2013 at 14:00
  • $\begingroup$ @hapchiu: Yes, but you have to think of $\alpha$ as a constant coefficient form on $\mathbb{R}^n$ and think of $x$ as a vector field on $\mathbb{R}^n$, one whose flow is the $1$-parameter group of linear transformations $e^{tx}$ of $\mathbb{R}^n$. Then $x.\alpha$ is just the Lie derivative of $\alpha$ with respect to $x$ (thought of as a vector field). $\endgroup$ Feb 20, 2013 at 15:33

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.