I am trying to prove the following.

Let $V_1, \ldots, V_k$ be finite dimensional vector spaces over a field $F$. There is a natural isomorphism between $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^k(V_1, \ldots, V_k;\ F)$.

Define a map $A:V_1^*\times\cdots\times V_k^*\to \mathcal L^k(V_1, \ldots, V_k;\ F)$ as \begin{equation*} A(\omega_1, \ldots, \omega_k)(v_1, \ldots, v_k)=\omega(v_1)\cdots\omega_k(v_k) \end{equation*} for all $(\omega_1, \ldots, \omega_k)\in V_1^*\times\cdots\times V_k^*$. It can be seen that $A$ is a multilinear map. By the universal property of tensor product, there exists a unique linear map $\tilde A: V_1^*\otimes\cdots\otimes V_k^*\to \mathcal L^{k}(V_1, \ldots, V_k; \ F)$ such that $\tilde A\circ \pi=A$.

We also know that \begin{equation*} \dim V_1^*\otimes\cdots\otimes V_k^*=\dim \mathcal L^k(V_1, \ldots, V_k; \ F) \end{equation*} Thus we just need to show that $\ker \tilde A=0$ to show that $V_1^*\otimes\cdots\otimes V_k^*$ and $\mathcal L^{k}(V_1, \ldots, V_k; \ F)$ are isomorphic.

My Problem: I want to show the triviality of the kernel in a basis free manner. But here I am stuck.

Can anybody help?

  • 1
    $\begingroup$ "I want to show the triviality of the kernel in a basis free manner." I don't think you should be able to do so -- this is a result that really requires the vector spaces to be finite-dimensional (or, more generally, the modules to be finite-free). $\endgroup$ – darij grinberg Mar 6 '15 at 5:40
  • $\begingroup$ Thanks. I can do it using bases. $\endgroup$ – caffeinemachine Mar 6 '15 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.