I have a little doubt about tensor product acting on vectors. I was reading Spivak's Calculus on Manifolds, and Spivak introduces the tensor product of multilinear functionals. Latter he introduces the idea of the wedge product built based on the tensor product to produce differential forms.
Everything's fine here, but I've heard that it's possible to consider both the tensor and wedge product acting on vectors and that this will bring some object called "blade" into play.
I've serched a little, but I still didn't understand how those notions can be extended to vectors, I mean, if I let $V$ be a vector space over $\mathbb{R}$ and if we define $T_1,T_2:V\times V\to\mathbb{R}$ are both bilinear maps and if $v_1,v_2,w_1,w_2\in V$ it's perctly fine that $T_1\otimes T_2(v_1,v_2,w_1,w_2)=T_1(v_1,v_2)T_2(w_1,w_2)$ but I can't understand what will hapen if instead of applying to multilinear functions I apply this product on vectors.
The only idea I had was as follows: identifying $V$ with the dual to the dual and consider the elements of $V$ as linear functions on the elements of the dual, in other words, if $v\in V$ I think of $v$ as the function $v: V^\ast\to\mathbb{R}$ such that if $\omega \in V^\ast$ then $v(\omega)=\omega(v)$. Is this the way to construct the notion of tensor and wedge product for vectors?
Sorry if any of those things are silly. And thanks in advance.
I guess it works as an interpretation, but for vectors, $v\otimes w$ is basically just the ordered pair $\langle v,w\rangle$, with the additional criteria that $$\langle \lambda\cdot v,w\rangle=\langle v,\lambda\cdot w\rangle \\ \langle v_1,w\rangle+\langle v_2,w\rangle = \langle v_1+v_2,w\rangle \\ \langle v,w_1\rangle+\langle v,w_2\rangle = \langle v,w_1+w_2\rangle $$ I think the notation $v\otimes w$ is misleading. With that we have
I guess somethng similar must go on in case of wedge product..