# What does determinant bundle of a principal bundle say about the principal bundle

Let $$\pi:P\rightarrow M$$ be a principal $$Gl(n,\mathbb{R})$$ bundle.

Given $$x\in M$$ there is an open set $$U$$ containing $$x$$ and a local trivialization $$\pi^{-1}(U)\rightarrow U\times G$$. This gives a cover $$\{U_\alpha\}$$ of $$M$$ and local trivializations $$\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times G$$. These in turn give what are called as transition functions $$g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow Gl(n,\mathbb{R})$$. These transition functions determine the principal bundle.

Consider the determinant map $$det:Gl(n,\mathbb{R})\rightarrow \mathbb{R}^*=Gl(1,\mathbb{R})$$. This is smooth. So, is the composition $$h_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow Gl(n,\mathbb{R})\rightarrow \mathbb{R}^*$$. So, we have an open cover $$\{U_\alpha\}$$ of $$M$$ and maps $$h_{\alpha\beta}:U_{\alpha\beta}\rightarrow Gl(1,\mathbb{R})$$ satisfying the cocycle condition. These maps define $$G=Gl(1,\mathbb{R})$$ bundle on $$M$$. We call this the determinant bundle associated to $$P\rightarrow M$$.

I would like to understand what does this determinant bundle say about $$P\rightarrow M$$. In this question I came to know that if determinant bundle is trivial then the bundle is self dual. Are there any such properties?

The determinant bundle is a principal $$\mathbb{R}$$-bundle, it is trivial if and only if it has a $$(\mathbb{R}^+,\times)$$-reduction since $$\mathbb{R}^+$$ is contractible, this is equivalent that $$M$$ is orientable, in in fact the obstruction of the triviality of the determinant bundle is the 0-stiefel-Whithney class.
The answer of the question you link to refers to bundles of rank two, it is not true in general. Also, the staement in there refers to the associated vector bundle (corresponding to the obvious representation of $$GL(n,\mathbb R)$$ on $$\mathbb R^n$$ rather than to the principal bundle itself. Calling this associated bundle $$E$$, there is a simple characterization of the fact that the determinant bundle of $$P$$ is trivial, namely that $$E$$ is orientable. The point here is that the determinant of a matrix can be interpreted at the induced action on the top exterior power $$\Lambda ^n\mathbb R^{n*}$$. This easily implies that the determinant bundle of $$P$$ is the frame bundle of the line bundle $$\Lambda^n E^*$$ so triviality of these two bundles is equivalent. But triviality of $$\Lambda^n E^*$$ is well known to be equivalent to the fact that it has a global non-vanishng section and thus to orientability of $$E$$.