If we have a real vector space $V=W_1\oplus W_2$, is it true that $W_1 \otimes W_2 = W_1 \wedge W_2 $?
My guess is that this is true. The definition of the $k$-exterior power is the quotient of $V^{\otimes k}/I$ where $I$ is the subspace generated by the elements of the form $v_1\otimes \cdots \otimes v_k$ where $v_i=v_{i+1}$ for some $i$. Then $W_1 \otimes W_2$ is a subspace of $V^{\otimes 2}$ in which the only element of the form $v\otimes v$ is $0 \otimes 0 = 0$. Is this right? Is there another way to see this? thanks
Best Answer
$W_1 \wedge W_2$ isn't well-defined. Do you mean the subspace of $V \wedge V$ which is the image of the natural map $W_1 \otimes W_2 \hookrightarrow V \otimes V \twoheadrightarrow V \wedge V$? Then the answer is Yes (but your proof is not complete).
In general we have a canonical isomorphism $\wedge^n(W_1 \oplus W_2)=\bigoplus_{p+q=n} \wedge^p(W_1) \otimes \wedge^q(W_2)$. In this case, we have
$\wedge^2 V = (\wedge^2 W_1 \otimes \wedge^0 W_2) \oplus (\wedge^1 W_1 \otimes \wedge^1 W_2) \oplus (\wedge^0 W_1 \otimes \wedge^2 W_1) = \wedge^2 W_1 \oplus (W_1 \otimes W_2) \oplus \wedge^2 W_2$
and the claim follows.