Exterior product 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 
 A: 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.
