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Suppose $V$ and $W$ are vector spaces and $\bigwedge V$ and $\bigwedge W$ their exterior algebras. Then it is known that $\bigwedge (V \oplus W) \simeq \bigwedge V \otimes \bigwedge W$. Now my question is:


Is $\bigwedge V \otimes \bigwedge V \simeq \bigwedge V$ true?


The reason why I think that: The elements of $\bigwedge (V \oplus W)$ are linear combinations of wedges like $v_1 \wedge \ldots \wedge v_n \wedge w_1 \wedge \ldots \wedge w_m$ but on $\bigwedge (V \oplus V)$ the wedges $v_1 \wedge \ldots \wedge v_n \wedge v'_1 \wedge \ldots \wedge v'_{n'}$ can be reduced (using $v_i \wedge v_i =0 $) to something like $v''_1 \wedge \ldots \wedge v''_{n''} \in \bigwedge V$

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No, that is not true.

We can see this using two pieces of information:

  • If $V$ is of finite dimension $n$, then $\Lambda V$ has dimension $2^n$.

  • On the other hand, if $U$ and $W$ are vector spaces of finite dimension, we have $\dim U\otimes W=\dim U\cdot\dim W$.

If your isomorphism existed, we would then have that $2^n2^n=2^n$, which we don't for most values of $n$.

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If $V$ is infinite dimensional, though, there is an isomorphism $\Lambda V\otimes\Lambda V\cong\Lambda V$, coming from the fact that $V\oplus V\cong V$. –  Mariano Suárez-Alvarez Apr 11 '12 at 8:40

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