Computing the basis of $\mathbb{T}_{2}\left(\mathcal{P}_{2}(\mathbb{R})\right)$ associated with the canonical basis in $\mathbb{R}^{3}$

Question

I am looking at the 2-covariant tensor space over the field of polynomials of degree equal or less than 2, $\mathbb{T}_{2}\left(\mathcal{P}_{2}(\mathbb{R})\right)$. (Sorry if this notation is not standard). If I start with the canonical basis of $\mathcal{P}_{2}$, $\mathcal{B}_{o}=\left\{1, t, t^{2}\right\}$. With this basis in mind, I can compute the basis of the dual space of $\mathcal{P}_{2}$, which would be such that each linear form would return the coefficient of the n-th degree monomial: $$\phi^1(ax^2+bx+c) = c$$ $$\phi^2(ax^2+bx+c) = b$$ $$\phi^3(ax^2+bx+c) = a$$ And therefore $\mathcal{B}^*_{o}=\left\{ \phi_1,\phi_2,\phi_3\right\}$. My problem is that I don't know how to go any further to calculate the basis asociated with the 2-covariant tensor space I mentioned. I know that, since $\dim(\mathcal{P}_{2})=3$, $\dim(\mathbb{T}_{2}\left(\mathcal{P}_{2}(\mathbb{R})\right)) = 3^2 = 9$. Am I missing some information? How could I proceed to calculate the basis? Thank you.

Answer 1

For a vector space $V$ with dual space $W$, the tensor space $T_kV$ is $T^kW$ (I guess by definition; I don't know where you're starting from). And $T^kW$ is the space spanned by all products of $k$ many vectors from $W$, where the product is associative ($\alpha\otimes(\beta\otimes\gamma)=(\alpha\otimes\beta)\otimes\gamma$) but not commutative ($\alpha\otimes\beta\neq\beta\otimes\alpha$). You have a basis $\{\phi^1,\phi^2,\phi^3\}$ for $W$. The corresponding basis for $T^2W$ is

$$\{\phi^1\otimes\phi^1,\;\phi^1\otimes\phi^2,\;\phi^1\otimes\phi^3,\;\phi^2\otimes\phi^1,\;\phi^2\otimes\phi^2,\;\phi^2\otimes\phi^3,\;\phi^3\otimes\phi^1,\;\phi^3\otimes\phi^2,\;\phi^3\otimes\phi^3\}.$$