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  • $\begingroup$ Could you repeat what the calculus trick is? I couldn't find what you were referring to in Lang's book. $\endgroup$ Commented yesterday
  • 1
    $\begingroup$ See mathoverflow.net/questions/507049 for discussion of what it might mean to give a "basis-free definition" $\endgroup$ Commented yesterday
  • 5
    $\begingroup$ Without knowing what Lang's definition is, my guess would be: An element $f_1\otimes\cdots\otimes f_n\in End(V)^{\otimes n}$ produces an endomorphism of $\wedge^n(V)$ as $v_1\wedge\cdots\wedge v_n\mapsto f_1(v_1)\wedge\cdots\wedge f_n(v_n)$. We symmetrize this construction, so $f_1\cdots f_n\in End(V)^{\otimes n}$ would act as $v_1\wedge\cdots\wedge v_n\mapsto\frac1{n!}\sum_{\sigma\in S_n}f_1(v_{\sigma1})\wedge\cdots\wedge f_n(v_{\sigma n})$. Now, note that since $\wedge^nV$ is one-dimensional, any endomorphism is multiplication by a scalar. This produces a map $S^n(End(V))\to\mathbb K$. $\endgroup$ Commented yesterday
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    $\begingroup$ @KentaSuzuki, oh, wow, yes, without looking, that is surely what it is! :) $\endgroup$ Commented yesterday
  • $\begingroup$ @Kenta, The trick is to take the $n$-th derivative at $0$ of a homogeneous function of degree $n$ on the vector space $End(V)$. It is a general remark on p.7 of Lang's last variant of his differential geometry book. It is not motivated by determinant function. $\endgroup$ Commented yesterday