I am reading the book Foundations of Differential Geometry Volume 1 by Kobayashi and Nomizu and in Chapter 2, Section 8, the holonomy theorem by Ambrose and Singer is introduced

Theorem: Let $P(M,G)$ be a principal fibre bundle, where $M$ is connected and paracompact. Let $\Gamma$ be a connection in $P$, $\Omega$ the curvature form, $\Phi(u)$ the holonomy group with reference point $u \in P$ and $P(u)$ the holonomy bundle through $u$ of $\Gamma$. Then the Lie algebra of $\Phi(u)$ is equal to the subspace of $\mathfrak{g}$, Lie algebra of G, spanned by all elements of the form $\Omega_v(X,Y)$ where $v\in P(u)$ and $X,Y$ are arbitrary horizontal vectors at $v$.

Let the subspace spanned by all elements of the form $\Omega_v(X,Y)$ where $X,Y$ are arbitrary horizontal vectors at $v$ be denoted by $\mathfrak{g}’$.

In the proof it is mentioned that $\mathfrak{g}’$ is actually an ideal of $\mathfrak{g}$ because $\Omega$ is tensorial form of type ad $G$.

I have the following questions

  1. Why is $\mathfrak{g}’$ as defined even a vector subspace of $\mathfrak{g}$?
  2. How is $\mathfrak{g}’$ an ideal of $\mathfrak{g}$?

$\mathfrak g'$ is a subspace by definition (it is given as the span of a collection of vectors).

Here's a hint to answer the ideal question. Remember that if $Z\in\mathfrak g$, then $Z=\dfrac d{dt}\Big|_{t=0} \exp(tZ)$, so $[Z,Y] = \dfrac d{dt}\Big|_{t=0} \text{ad}(\exp(tZ))(Y)$; moreover, if $Y\in\mathfrak g'$, then $\text{ad}(a)(Y)\in\mathfrak g'$ because $\Omega$ is tensorial of type $\text{ad}\ G$.

  • $\begingroup$ Thanks a lot!! That was extremely stupid of me to miss the fact that i needed to take the span of the image of the curvature form rather than the image itself $\endgroup$ – ravjotsk Aug 9 '19 at 17:25
  • $\begingroup$ You're most welcome. K-N. can be a difficult text to decipher at times :) $\endgroup$ – Ted Shifrin Aug 9 '19 at 17:31

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