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For a $\mathbb K$-vector space $V$ let us write $\Lambda^i V^*:=\mathsf{Hom}(\Lambda^i V, \mathbb K)$.

Given two $\mathbb K$-vector spaces $V$ and $W$ and a linear map $f:V\longrightarrow W$ there is an induced linear map $f^*: \Lambda^i W^*\longrightarrow \Lambda^i V^*$ given by $$(f^* \varepsilon)(v_1\wedge\ldots \wedge v_i):=\varepsilon(f(v_1)\wedge \ldots \wedge f(v_i)).$$ In particular, the projections $\pi_V:V\oplus W\longrightarrow V$ and $\pi_W:V\oplus W\longrightarrow W$ induce linear maps $$\pi_V^*:\Lambda^iV^*\longrightarrow \Lambda^i (V\oplus W)^*\quad \textrm{and}\quad \pi_W^*:\Lambda^i W^*\longrightarrow \Lambda^i (V\oplus W)^*.$$ There is also a graded product $$\wedge:\Lambda^i V^*\times \Lambda^j V^*\longrightarrow \Lambda^{i+j} V^*.$$ Is it true that any $\omega\in \Lambda^k (V\oplus W)^*$ is of the form $$\omega=\sum_{i+j=k} \pi_V^* \alpha_i\wedge \pi_W^* \beta_j,$$ for unique $\alpha_i\in \Lambda^i V^*$ and $\beta_j\in \Lambda^j W^*$?

Thanks.

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2 Answers 2

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Yes. This is due to the isomorphism $$\bigwedge(V\oplus W)\cong \left(\bigwedge V\right)\hat\otimes\left(\bigwedge W\right) $$ where $\hat\otimes$ is the skew-symmetric graded product of algebras. You can prove this by abstract nonsense: prove the RHS is universal for maps $\phi$ from $V\oplus W$ into algebras with the property that $\phi(u+v)\phi(u+v)=0$, $u\in U$, $v\in V$.

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We'll, you have to be a bit careful, since not every element of a tensor product is décomposable. There may be several terms in each component corresponding to $\wedge^i V^{\star}\otimes \wedge^j W^{\star}$

You can see that for yourself: take $V$ and $W$ of dimension $2$ and the element $d x_1 \wedge dx_3 + dx_2 \wedge d x_4\in \wedge^1 V^{\star}\otimes \wedge^1 W^{\star} \subset \wedge^2(V\oplus W)^{\star}$.

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