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    $\begingroup$ The vector bundle $\operatorname{Sym}^2E$ is spanned by local sections which are symmetric products $u \cdot v$. Map to $(u\wedge v)^{\otimes 2}$. Extend by linearity. Is that not the sort of answer you are looking for? $\endgroup$ Commented Jan 26, 2019 at 14:48
  • $\begingroup$ Thanks for the answer but not quite. In local coordinates write $u = ai+bj$ and $v = ci+dj$. Then $u\otimes v = ac (i\otimes i) + ad (i\otimes j) + bc (j\otimes i) + bd (j\otimes j)$, and the corresponding matrix would be $M = (m_{i,j})$ with $m_{1,1} = ac, m_{1,2} = ad, m_{2,1} = bc, m_{2,2} = bd$. Since the map I am looking for should give the determinant of the matrix it should be zero on tensors of type $u\otimes v$. Here i considered the non symmetric case but the same question make sense replacing $E\otimes E$ with $Sym^2(E)$. $\endgroup$ Commented Jan 26, 2019 at 15:17
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    $\begingroup$ I don't even think that there is a natural map unless you're in characteristic $2$. Otherwise, these correspond to distinct irreducible representations of $GL_2$. $\endgroup$ Commented Jan 26, 2019 at 15:22
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    $\begingroup$ Why is there the tensor exponent "$\otimes 2$"? The usual determinant has target $\bigwedge^2 E$, not $(\bigwedge^2 E)^{\otimes 2}$. Are you asking about the square of the determinant? $\endgroup$ Commented Jan 26, 2019 at 16:41
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    $\begingroup$ The determinant of a matrix with a fixed frame depends skew-symmetrically (not symmetrically) on the frame, so the source of your map should be $\wedge^2E$. Also, as @JasonStarr explained, the target is $\wedge^2E$. So, in the end, I think you are asking about a natural map $\wedge^2E \to \wedge^2E$. And the answer to this question is, of course, the identity. $\endgroup$ Commented Jan 26, 2019 at 18:24