Let $M_n$ be the space of all $n$ by $n$ matrices. Then a non-degenerated natural bilinear form on $M_n \otimes M_n$ is $tr(AB)$. The reason is as follows.

The natural linear form on $M_n$ is $f: M_n \to \mathbb{C}$, $f(A) = tr(A)$. The reason is as follows. Let $V$ be an $n$-dimensional vector space with a basis $v_1, \ldots, v_n$ and $V^*$ its dual vector space with the dual basis $v_1^*, \ldots, v_n^*$. We have an isomorphism $End(V) = M_n \to V^* \otimes V$ given by $(a_{ij})_{n \times n} \mapsto \sum_{i,j} a_{ij} v_i^* \otimes v_j$. There is a natural pairing $V^* \otimes V \to \mathbb{C}$ given by $(v_i^*, v_j) \mapsto v_i^*(v_j) = \delta_{ij}$. Therefore we have a natural map $M_n \to V^* \otimes V \mapsto \mathbb{C}$ given by $(a_{ij}) \mapsto \sum_{i,j} a_{ij} v_i^* \otimes v_j \mapsto \sum_{ij} a_{ij}\delta_{ij}= \sum_{i}a_{ii} = tr(A)$.

Now consider the natural map $V^* \otimes V \otimes V^* \otimes V \to \mathbb{C}$ given by $(f_1, v_1, f_2, v_2) \mapsto f_1(v_2)f_2(v_1)$. Then we have a map $M_n \otimes M_n \to V^* \otimes V \otimes V^* \otimes V \to \mathbb{C}$ given by $(A, B) \mapsto \sum_{ijkl} a_{ij} v_i^* \otimes b_{kl} v_j \otimes v_k^* \otimes v_l \mapsto \sum_{ij} a_{ij} b_{ji} = tr(AB)$.

Are there some other non-degenerated natural bilinear forms on $M_n \otimes M_n$? Thank you very much.

  • $\begingroup$ If $n$ is even, then there is the Darboux symplectic structure on $M_{n}$. Since $M_{n}$ is a vector space, the symplectic structure may be taken to be a (0,2)-tensor on $M_{n}$; that is, a skew-symmetric, nondegenerate, bilinear form. $\endgroup$ – Sinister Cutlass Nov 13 '15 at 4:04
  • $\begingroup$ A Riemannian metric, and even any pseudo-Riemannian metric on $M_{n}$ will also do the trick. $\endgroup$ – Sinister Cutlass Nov 13 '15 at 4:04

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