# Image of Curvature form on a Principal Bundle

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 $$\text{ad}\ G$$.