If $V$ is a vector bundle of rank $n$, the corresponding universal algebra $A$ which makes $V$ trivial (i.e. $V \otimes A \cong A^n$), or equivalently the algebra of the corresponding $\mathrm{GL}_n$-torsor, is given by $$A = \mathrm{Sym}(V^n) \otimes_{\mathrm{Sym}(\Lambda^n V)} \mathrm{Sym}^{\mathbb{Z}}(\Lambda^n V).$$ Here, we define $\mathrm{Sym}^{\mathbb{Z}}(\mathcal{L})=\bigoplus_{z \in \mathbb{Z}} \mathcal{L}^{\otimes z}$ for the line bundle $\mathcal{L}=\Lambda^n V$, and the tensor product is taken with respect to the morphism $\delta : \mathcal{L} \to \mathrm{Sym}^n(V^n)$ which maps $v_1 \wedge \dotsc \wedge v_n$ to $\sum_{\sigma \in \Sigma_n} \mathrm{sgn}(\sigma) \prod_{i=1}^{n} \iota_i(v_{\sigma(i)})$. This description is global in nature and actually generalizes to arbitrary cocomplete linear tensor categories. Details can be found in my thesis, Section 4.9.

If $V$ is a vector bundle of rank $n$, the corresponding universal algebra $A$ which makes $V$ trivial (i.e. $V \otimes A \cong A^n$), or equivalently the algebra of the corresponding $\mathrm{GL}_n$-torsor, is given by $$A = \mathrm{Sym}(V^n) \otimes_{\mathrm{Sym}(\Lambda^n V)} \mathrm{Sym}^{\mathbb{Z}}(\Lambda^n V).$$ Here, we define $\mathrm{Sym}^{\mathbb{Z}}(\mathcal{L})=\bigoplus_{z \in \mathbb{Z}} \mathcal{L}^{\otimes z}$ for the line bundle $\mathcal{L}=\Lambda^n V$, and the tensor product is taken with respect to the morphism $\delta : \mathcal{L} \to \mathrm{Sym}^n(V^n)$ which maps $v_1 \wedge \dotsc \wedge v_n$ to $\sum_{\sigma \in \Sigma_n} \mathrm{sgn}(\sigma) \prod_{i=1}^{n} \iota_i(v_{\sigma(i)})$. This description is global in nature and actually generalizes to arbitrary cocomplete linear tensor categories. Some details can be found in my thesis, Section 4.9.

The idea of the construction of $A$ is the following: $\mathrm{Sym}(V^n)$ is the universal algebra $B$ with a morphism of $B$-modules $V \otimes B \to B^n$. Then we construct $B \to A$ so that the determinant of this morphism becomes invertible over $A$, so that $V \otimes A \cong A^n$.

We could also construct $A$ as a quotient of $\mathrm{Sym}(V^n) \otimes \mathrm{Sym}((V^*)^n)$, which introduces morphisms $V \otimes A \to V^n$ and $A^n \to V \otimes A$, and the quotient should be made in such a way that these morphisms become inverse to each other.