Let $P\rightarrow M$ be a principal $G$ bundle and $\phi:H\rightarrow G$ be a morphism of Lie groups.
Can some one help me to understand the constrution of reducing the structrue group to $H$ i.e., I have to construct a principal $H$ bundle over $M$ from $P$.
How does one do this for trivial bundle $M\times G\rightarrow M$? I am sure reduction of the strucre is not the trivial bundle $M\times H\rightarrow M$.
Any help is appreciated.
A reduction of structure group usually is not something that happens canonically, but it expresses a choice of some additional structure. For example, consider the case that $\phi$ is the inclusion of $H:=O(n)$ into $G:=GL(n,\mathbb R)$ and that $p:P\to M$ is the frame bundle of $M:=\mathbb R^n$. Then you can view a point $u\in P$ as a basis of the tangent space $T_{p(u)}M$. Hence if you identify the tangent bundle $TM$ with the trivial bundle $M\times\mathbb R^n$, then you get a trivialization $M\times G\to P$, which associates to $(x,A)$ the basis of $T_xM=\mathbb R^n$ formed by the column vectors of $A$.
A reduction of $P$ to the structure group $H$ is equivalent to a Riemannian metric $g$ on $M$, with the subbundle in each $x\in M$ formed by the bases that are orthonormal for the inner product $g_x$ on $T_xM$. If you take the standard flat metric on $\mathbb R^n$, then this reduced bundle becomes the trivial bundle $M\times O(n)\subset M\times GL(n,\mathbb R)$F or a general metric it is much more complicated as a subbundle, although it still can be globally trivialized.