I have asked this question in the Mathematics forum but I received no answer.
Let $E$ be an algebraic vector bundle of rank $r$ and degree $d$, 
Then $\Lambda^2 E$ is of rank $r'=r(r-1)/2$, but is of determinant 0, because $$\Lambda^{r'}(\Lambda^2 E)\subset \Lambda^{r(r-1)}E=0$$
which I can't understand! because its determinate has to be a line bundle? Is there any mistake in that calculus? 
Thanks 



Exterior product
Let me expand my comments into an answer.


Before thinking about vector bundles, it makes sense to see what happens with vector spaces. If $(a,b,c)$ is a basis for a $3$-dimensional vector space $V$, then $\bigwedge^{2}V$ has basis $(a \wedge b,b \wedge c,c \wedge a)$. Then the determinant $\bigwedge^{3}(\bigwedge^{2}V)$ has basis $(a \wedge b) \wedge (b \wedge c) \wedge (c \wedge a)$. But you can't just remove the parentheses and say that this is zero. It is nonzero. The wedges live in different vector spaces.


One interesting statement is that if $W$ is $n$-dimensional, then there is a canonical isomorphism
$$ \bigwedge^{k}W^{\*} \otimes \bigwedge^{n}W \to \bigwedge^{n-k}W $$


First of all, the special case $k = n$ shows that $\det W$ is dual to $\det W^{\*}$.


Returning to the example $V$, if we plug in $W = V^{\*}$, we obtain $\bigwedge^{2}V \otimes \bigwedge^{3} V^{\*} \cong \bigwedge^{1} V^{\*}$. In other words $\bigwedge^{2} V$ is isomorphic to $V^{\*} \otimes \det V$. Finally
$$ \bigwedge^{3}(\bigwedge^{2} V) \cong \bigwedge^{3}(V^{\*} \otimes \det V) \cong (\bigwedge^{3} V^{\*}) \otimes (\det V)^{\otimes 3} \cong (\det V)^{\otimes 2}.$$


I do realise that this is a pretty special example, since $3 = 2 + 1$, whilst in general $n \ne 2 + 1$. Still, I hope this helps a bit in answering your question.


