# Question in the paper of Robert Bryant “Calibrated embeddings in the special Lagrangian and coassociative cases”

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

• 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. – Robert Bryant Feb 20 '13 at 13:39
• i have edited my question once again so one actually can read it :). – hapchiu Feb 20 '13 at 13:43
• Great. I delete my two earlier comments as they are obsolete. A general remark on formatting here: it is in general not necessary to split up formulas in the way you did for the first. It is true there is an issue with displaying braces { , however there is another way around this: namely use two backslashes instead of one so \\{ or use the commands \lbrace and \rbrace . Moreover, the symbols _ and * cause problems as the have an additional meaning here. The latter can be avoided using \ast instead. If issues persist include the problematic formula in backticks ` (both at end and start). – user9072 Feb 20 '13 at 13:47
• is there any interpretation of this using the Lie-derivative ? – hapchiu Feb 20 '13 at 14:00
• @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 $x\ldot\alpha$ is just the Lie derivative of $\alpha$ with respect to $x$ (thought of as a vector field). – Robert Bryant Feb 20 '13 at 15:33