basis-independent isomorphism

Question

Could some one give me some help with this proof? Given the hint, I still don't have a clue about how to proceed. Thanks.

Answer 1

An element $A \in T_{l+1}^k(V)$ is a multilinear map $(V^*)^{l+1} \times V^k \rightarrow \mathbb{R}$ (assuming you're playing with real vector spaces; if you're not then use your favorite field, it doesn't matter. I assumed so because you tagged differential geometry and this resembles strongly the first book by Lee).

Fix $\omega_1, \cdots, \omega_l \in V^*$ and $v_1, \cdots v_k \in V$ and consider what you get by 'filling into' $A$:

$$A'(\cdot) := A(\omega_1, \omega_2, \cdots, \omega_l, \cdot, v_1, \cdots, v_k)$$

is something that when fed a functional $\omega \in V^*$ gives a real number in $\mathbb{R}$. Putting it differently, with $\omega$'s and $v$'s fixed, $A'$ can be thought of as a member of $V^{**}$, which is canonically (basis-independently) isomorphic to $V$. Moreover by the tensoriality of $A$ we know that the association of $A'(\cdot) \in V$ to the argument $\omega_1, \cdots, \omega_l, v_1, \cdots, v_k$ is multilinear.

Put another way, the process by which $A' : (V^*)^l \times V^k \rightarrow V$ is obtained from $A$ is basis independent and is given by $$ A \mapsto A' \\ V^{**} \ni A'(\omega_1, \cdots, \omega_l, v_1, \cdots, v_k) \\ A'(\omega_1, \cdots, \omega_l, v_1, \cdots, v_k)(\omega) := A(\omega_1, \cdots, \omega_l, \omega, v_1, \cdots, v_k) $$ (Notice that I wrote the range of $A'$ as $V$ and yet told you that $A'(\omega_1, \cdots, \omega_l, v_1, \cdots, v_k) \in V^{**}$. The reason is, as I stated in the comments, that $V^{**} \cong V$ when $V$ is finite-dimensional).

We'll now define an inverse to this. Assume that $B'$ is a multilinear map $(V^*)^l \times V^k \rightarrow V$. We'll define a tensor $B \in T^k_{l+1}(V)$ by $$ B(\omega_1, \omega_2, \cdots, \omega_l, \omega_{l+1}, v_1, v_2, \cdots, v_k) =\omega_{l+1}\big(B'(\omega_1, \omega_2, \cdots, \omega_l, v_1, \cdots, v_k)\big) $$

Now you have a correspondence in either direction, so to prove they're isomorphisms you should show they are mutual inverses.