TITLE: SL(n) structures are the same as volume forms
QUESTION [0 upvotes]: I saw this fact stated in many places (as Wikipedia) but I could not find a proof of it anywhere. I don't really want a rigorous proof but I'd like to get the geometric intuition behind this, since right now it makes no sense to me.
Also any references on this are appreciated.

REPLY [2 votes]: Let's suppose we have that the tangent bundle $\pi:TX\to X$ is an $SL(n)$-bundle. This means, by definition, that we have an $SL(n)$-atlas $\{(U_i,\phi_i)\}_{i\in I}$, where $\phi_i: \pi^{-1}(U_i)\to U_i\times \mathbb{R}^n$ is a linear isomorphism and a homeomorphism, and moreover $\phi_i$ commutes with the projections to $U_i$, and such that the induced transition maps
$$
  \phi_{ji}: U_i\cap U_j \to GL(n),
$$
given by $x\mapsto (\phi_j)_x \circ (\phi_i)_x^{-1}$, actually land inside $SL(n)\subset GL(n)$. With this atlas in hand, we should be able to define a volume form on $X$. Indeed, we can take coordinates $x_1,\dots, x_n$ on each $U_i$ and let the volume form on $U_i$ be $v_i:=dx_1\wedge\dots\wedge dx_n$. The last thing we need to check is that these local volume forms glue into a global volume form on $X$. Well, you can check that on an overlap $U_i\cap U_j$, that the transition functions $\phi_{ji}$ induce transition functions $\tilde{\phi}_{ji}$ on $\bigwedge^n T^*X$ (imagine doing $(\phi_{ji}^{T})^{-1}$ to each vector in the wedge product). You can also check that $\tilde{\phi}_{ji} = \det(\phi_{ji})^{-1}$; in other words,
$$
  \tilde{\phi}_{ji}(v_i) = \det(\phi_{ji})^{-1} v_j.
$$
But since we have $\phi_{ji} \in SL(n)$, this determinant is $1$, and hence the $v_i$ glue into a volume form on $X$.

To get the rough idea of going backwards, take a volume form $v$ on $X$. In the overlap of two charts we can do a bit of linear algebra and choose transition matrices that preserve the volume form, which entails that the $\phi_{ji}$ will take values in $SL(n)$, hence giving an $SL(n)$-structure on $TX$.
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