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I am reviewing materials in tensor products and I got stuck on this one, and I am never comfortable with "showing there exists a unique linear map" type of question.

Let $V$ be a real vector space of dimension of $n$ and let $V^*$ be the dual vector space of $V$.

  1. Show that there exists a unique $\mathbb{R}$ linear map $\phi:\Lambda^2(V^*) \otimes \Lambda^2(V) \rightarrow (V^*) \otimes (V)$ satisfying $\phi((f_1\wedge f_2)\otimes (v_1 \wedge v_2)) = (f_1(v_1)f_2-f_2(v_1)f_1)\otimes v_2 - (f_1(v_2)f_2-f_2(v_2)f_1)\otimes v_1$.

  2. What is the rank of $\phi$?

How should I start with this? I want to write that for $f_1,f_2 \in V^*$, there exists a homomorphism $\lambda_1:\Lambda^2(V^*) \rightarrow V^*\otimes V^*, \lambda_1(f_1\wedge f_2) = f_1\otimes f_2-f_2 \otimes f_1$, and for $v_1,v_2 \in V$, $\lambda_2:\Lambda^2(V) \rightarrow V\otimes V, \lambda_2(v_1\wedge v_2) = v_1\otimes v_2-v_2\otimes v_1$. So then $\lambda_1(f_1\wedge f_2)\lambda_2(v_1\wedge v_2)=f_1(v_1)f_2(v_2)-f_2(v_1)f_1(v_2)-f_1(v_2)f_2(v_1)-f_2(v_2)f_1(v_1) = (f_1(v_1)f_2-f_2(v_1)f_1)(v_2)-(f_1(v_2)f_2-f_2(v_2)f_1)(v_1)$.

How should I continue from here? I really have trouble with the tensor product type of argument. And I have no idea about the second part.

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Just unwind the tensors and wedges. I'll sketch this. Consider the map $\varphi:V^*\times V^*\times V\times V\rightarrow V^*\otimes_{\mathbb{R}} V$ via $(f_1(v_1)f_2-f_2(v_1)f_1)\otimes v_2 - (f_1(v_2)f_2-f_2(v_2)f_1)\otimes v_1$. You can check that this induces a map $\bigwedge^2 V^*\times\bigwedge^2 V\rightarrow V^*\otimes_{\mathbb{R}} V$ that is $\mathbb{R}-$bilinear (let me know if you need more explanation here - there are three universal properties nested here, and you need to check that two of the intermediate maps are alternating. Key hint: if either $f_1=f_2$ or $v_1=v_2$ then $\text{im}\varphi=0$). Then the universal property of the tensor product gives you the desired map.

Edit: I'll calculate the rank since you edited that in. If $\dim V=0$ or $1$ then the map clearly has rank $0$. Now consider $\dim V=2$ and let $\{e_1,e_2\}$ be a basis for $V$. Let $\psi$ be the induced map above. Then $\text{im}\psi=\langle e_1^*\otimes e_1+e_2^*\otimes e_2\rangle$ which shows that the rank is $1$ when $\dim V=2$. Finally, for $\dim V>2$, then $\text{im}\psi=\langle e_i^*\otimes e_i+e_j^*\otimes e_j, e_n^*\otimes e_m\mid i<j\leq\dim V, n\neq m\rangle$ which shows that the rank is $(\dim V)^2$ for $\dim V\geq 3$.

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  • $\begingroup$ Why is $\varphi$ that you defined the same as the $\phi$ that the question asks for? $\endgroup$ – 2010 Kur Apr 26 at 0:57
  • $\begingroup$ You have a map $\bigwedge^2 V^*\times \bigwedge^2 V\rightarrow \bigwedge^2 V^*\otimes_{\mathbb{R}} \bigwedge^2 V$ so the obvious thing to do is find a bilinear map $\bigwedge^2 V^*\times \bigwedge^2 V\rightarrow V^*\otimes_{\mathbb{R}} V$ with the same image as the desired map, and then use the universal property for tensor. This is precisely what you'll get if you do the procedure I outline, as each successive application of universal property (more specifically, of wedge, then wedge, then coproduct in $\mathbb{R}-$vect) gives the correct commutative triangle. $\endgroup$ – user43662 Apr 26 at 3:56

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