Given a vector bundle $V$ (on a scheme $X$, say), I can form $Sym(V[1])$, the symmetric algebra (in the derived/graded sense) on the shift of $V$; in other words this is the Koszul complex of the zero section of $V$, also known as the free exterior algebra on $V$. This is in fact a graded algebra, and its maximal non-vanishing graded component is $Sym^n(V[1]) = det(V)[n]$, where $n$ is the rank of $V$. Thus we naturally get the graded determinant of $V$.
Now let $E$ be a perfect complex. The expression $Sym(E[1])$ still makes sense. Similarly the graded determinant of $E$ makes sense (the morphism $Vect(X) \to Pic(D(X))$ sending $V$ to its graded determinant is additive and hence factors over $K$-theory; in more down to earth terms the graded determinant of $V_1 \to V_2 \to \dots \to V_n$ is $det(V_1) \otimes det(V_2)^{-1} \otimes \dots$, placed in degree $rank(V_1) - rank(V_2) + \dots$).
However there need not be a maximal degree component of $Sym(E[1])$, and even if there is, it need not have anything to do with the graded determinant of $E$ (as far as I can tell).
My question: Is there a good way of thinking about this? For example, is there an alternative generalization of $Sym(V[1])$ to perfect complexes which relates to graded determinants?
