Principal $\mathrm{SL}_n$-bundles

Question

It seems to be well-known that a principal $\mathrm{SL}_n$-bundle on a scheme or manifold $X$ is the same as a vector bundle of rank $n$ whose determinant is a trivial line bundle. One direction is clear to me: A principal $\mathrm{SL}_n$-bundle corresponds to a principal $\mathrm{GL}_n$-bundle whose cocycles can be chosen to live in $\mathrm{SL}_n$, which means that we have a vector bundle of rank $n$ whose determinant has trivial cocycles. Conversely, if we are given a vector bundle of rank $n$ whose determinant is a trivial line bundle, this corresponds to a principal $\mathrm{GL}_n$-bundle whose determinant bundle has cocycles which are just coboundaries (not trivial). So how do we get cocycles in $\mathrm{SL}_n$?

Edit. Here is an attempt to spell out the details in MarianoSuárez-Alvarez's comment. Consider the short exact sequence of groups $1 \to \mathrm{SL}_n \to \mathrm{GL}_n \to \mathrm{GL}_1 \to 0$. It is split in the sense that $\mathrm{GL}_n \to \mathrm{GL}_1$ has a section, for example $\lambda \mapsto \mathrm{diag}(\lambda,1,\dotsc,1)$. We obtain an exact sequence of pointed sets $$0 \to H^1(X,\mathrm{SL}_n) \to H^1(X,\mathrm{GL}_n) \to H^1(X,\mathrm{GL}_1) \to 0$$ because $H^i(X,\mathrm{GL}_n) \to H^i(X,\mathrm{GL}_1)$ is surjective for $i=0,1$. But this means that $0 \to \mathrm{Bun}_{\mathrm{SL}_n}(X) \to \mathrm{Vect}_n(X) \to \mathrm{Pic}(X) \to 0$ is exact, proving the claim. Essentially the same argument is given in archipelago's answer without using the language of cohomology.

Answer 1

If $$GL_n\rightarrow E\stackrel{\pi}{\rightarrow} B$$ is a principal $GL_n$-bundle with a trivializing open cover ${U_i}$, trivializations $$\psi_i\colon U_i\times GL_n\rightarrow \pi^{-1}(U_i)$$ and induced cocylces $$\psi_{ij}\colon U_i\cap U_j\rightarrow GL_n,$$ the determinant of this bundle is the principal $GL_1$ bundle given by the coclyes $$det(\psi_{ij})\colon U_i\cap U_j\rightarrow GL_n,p\mapsto det(\psi_{i,j}(p)).$$

So suppose the determinant bundle is trivial, then the cocyles $det(\psi_{ij})$ are just coboundaries, i.e. we have maps $\eta_i\colon U_i\rightarrow GL_1$, such that $$det(\psi_{ij})(p)=\eta_i(p)^{-1}\eta_j(p).$$ Now one can define new trivializations $\psi_i'$ of the original bundle by using $\psi_i$ and $\eta_i$, such that the induced cocyles have determinant one.

If I guess correctly $\psi_i'(p,A)=\psi_i(p,A\eta_i(p)^{-1/n})$ does the job since it should induce cocylces given by $\psi' _{ij}(p)=\psi_{ij}(p)\eta_j(p)^{-1/n}\eta_i(p)^{1/n}$, which have determinant one.

Answer 2

Disclaimer: this answer does not address the question of cocycles specifically, and is only about topological vector bundles (i.e., it might not be useful at all).

You can think of the equivalence this way: the short exact sequence $1 \to SL_n \to GL_n \xrightarrow{\mathrm{det}} GL_1 \to 1$ gives rise to a fibration sequence of classifying spaces, $BSL_n \to BGL_n \to BGL_1$. Now fix a vector bundle $E$ with classifying map $f : X \to BGL_n$. The universal property of fiber sequences says that the space of lifts $\tilde{f} : X \to BSL_n$ is equivalent to the space of null-homotopies of the composite $X \xrightarrow{f} BGL_n \xrightarrow{B\mathrm{det}} BGL_1$. Those null-homotopies are trivializations of the determinant bundle of $E$.