Let $A, B, C$ be algebras. Suppose that the bounded derived category of $A$ $D^b(A)$ admits a recollement relative to $D^b(B)$ and $D^b(C)$.

Then, by a result of Alfred Wiedemann'paper 'on stratifications of derived module categories', the algebra $A$ has a finite global dimension if and only if so are $B$ and $C$.

Now, suppose that the bounded derived categories of $A$ and $B$ are equivalent. By Rickard's result, $D^b(A)$ admits a recollement relative to $D^b(B)$ and 0. Hence, the global dimension is a derived invariant.

But, BGS's Koszul duality give an equivalent between the bounded derived categories of the symmetric algebra and the exterior algebra. Obviousely, their global dimension are quite different. So, what's the reason?

Let $A, B, C$ be algebras. Suppose that $D^b(A)$ (the bounded derived category of $A$) admits a recollement relative to $D^b(B)$ and $D^b(C)$.

Then, by a result of Alfred Wiedemann's paper "On stratifications of derived module categories," the algebra $A$ has a finite global dimension if and only if so are $B$ and $C$.

Now, suppose that the bounded derived categories of $A$ and $B$ are equivalent. By Rickard's result, $D^b(A)$ admits a recollement relative to $D^b(B)$ and 0. Hence, the global dimension is a derived invariant.

But, BGS's Koszul duality give an equivalent between the bounded derived categories of the symmetric algebra and the exterior algebra. Obviously, their global dimension are quite different. So, what's the reason?