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In the three dimensions, the cross product $A \times B$ is said to be a $(1,2)$ tensor. But the definition I know about tensors is that $(1,2)$ tensor is a multilinear map $T : V^* \times V^2 \to \mathbb{R}$ where $V^*$ is the dual space of $V$. So how does the cross product fit in this definition? Could anyone explain?

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So recall that $V^*$ is the dual space of $V$, i.e. the space of linear maps from $V$ to the ground field $\mathbb{R}$.

So what do we mean by a $(1,2)$-tensor? Well, a tensor of type $(k,l)$ is defined as an element of the vector space $\otimes_k V \bigotimes \otimes_l V^*$. Now, I claim that we can identify tensors of type $(1,l)$ with $l$-multilinear maps taking values in $V$. You are perhaps already familiar with the case $l=1$: There is an identification $\mbox{Lin}(V,V) \cong V \otimes V^*$. The identification map we want to consider for general $l$ is rather similar, and I can elaborate if you want.

Clearly then, the cross product is a bilinear map from $\mathbb{R}^3 \times \mathbb{R}^3$ to $\mathbb{R}^3$ and hence falls in the category above, for $l=2$. I hope this helps!

You can also view a tensor of type $(1,2)$ as a map in the way you described it, but this may be a little confusing. Can you explain how you came to this point of view?

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  • $\begingroup$ Could you explain for higher $l$? What do you mean by $\otimes_k V \bigotimes \otimes_l V^*$? Also is $Lin(V,V)$ the vector space of linear maps from $V$ to $V$? $\endgroup$ – Keith Apr 10 '18 at 20:16
  • $\begingroup$ Can you tell me what your definition of a tensor / tensor product is? I mean the tensor product of $k$ copies of $V$ and $l$ copies of $V^*$. And yes, $\mbox{Lin}(V,V)$ is the spaces of linear maps from $V$ to $V$. There, the identification with tensors of type (1,1) is given by sending an elementary tensor $\alpha \otimes v$ to the linear map which sends $w \in V$ to the vector $\alpha(w)v$. For $l=2$, we have a map which sends $\alpha_1 \otimes \alpha_2 \otimes v$ to the bilinear map $(w_1,w_2) \mapsto \alpha_1(w_1)\alpha_2(w_2)v \in V$. Do you see the pattern? $\endgroup$ – Thomas Bakx Apr 10 '18 at 20:33

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