Unique $\mathbb{R}$-linear map between tensor products

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.

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 $$\text{im}\psi=\langle e_i^*\otimes e_i+e_j^*\otimes e_j, e_n^*\otimes e_m\mid i which shows that the rank is $$(\dim V)^2$$ for $$\dim V\geq 3$$.
• Why is $\varphi$ that you defined the same as the $\phi$ that the question asks for? – 2010 Kur Apr 26 at 0:57
• 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. – user43662 Apr 26 at 3:56