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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

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up vote 6 down vote accepted

$W_1 \wedge W_2$ 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.

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Actually I was thinking about this question when I was trying to prove the canonical isomorphism that you mentioned :$\wedge^n(W_1 \oplus W_2)=\bigoplus_{p+q=n} \wedge^p(W_1) \otimes \wedge^q(W_2)$ that I thought it should be $\wedge^n(W_1 \oplus W_2)=\bigoplus_{p+q=n} \wedge^p(W_1) \wedge \wedge^q(W_2)$ that when using the original post I see are equivalent. So do you know of any proof of this canonical isomorphism? – inquisitor Mar 28 '13 at 16:11
Also thanks for the clarification of $W_1\wedge W_2$, but I don't understand why the proof is not complete, what is missing? – inquisitor Mar 28 '13 at 16:17
Proofs can be found everywhere, e.g. Bourbaki's Algèbre (try to find the relevant chapter on exterior powers for yourself as an exercise in doing research). But it is best to prove it as an exercise, try to identify $\hom(\wedge^n (W_1 \oplus W_2),-)$ with $\hom(\oplus_{p+q=n} \dotsc,-)$ and then apply Yoneda's Lemma. – Martin Brandenburg Mar 28 '13 at 17:24

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