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Let $V$ be an $n$-dimensional $\mathbb{K}$-vector space. By a simple calculus trick (*) on homogeneous functions of degree $n$ the determinant is a linear map on the $n$-th symmetric power of the vector space of endomorphisms: $$\det\,:\;S^n(End(V))\rightarrow\mathbb{K}\;.$$ Q: Is there an explicit formula? Reference?

It seems I have one in terms of the skew Hopf algebra convolution product of homomorphisms of the exterior algebra, which looks symmetric on $0$-degree homogeneous homomorphisms... (Being a left-right confuser I would have to check all the signs 3 times over. Plus, it looks suspiciously pretty and suggests benefits of skew Hopf algebra in the theory of determinants, so it should be in a good multilinear algebra textbook if my hunch is right. I'm not interested in quantum groups or braided monoidal category theory. Plain Hopf algebra is already interesting enough, as I once found in tensor calculus.)

(*) Serge Lang, Differential and Riemannian Manifolds (1995), p.7

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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
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    $\begingroup$ See mathoverflow.net/questions/507049 for discussion of what it might mean to give a "basis-free definition" $\endgroup$ Commented yesterday
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    $\begingroup$ 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

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Lang's observation is:

There is a unique linear functional $F\colon S^n(\operatorname{End}V)\to\mathbb K$ such that for any $A\in\operatorname{End}V$, $$F(A^n)=\det(A).$$

There was another functional $F'\colon S^n(\operatorname{End}V)\to\mathbb K$ that I constructed in the comments above:

An element $A_1\cdots A_n\in S^n(\operatorname{End}V)$ acts on $\wedge^nV$ by $$v_1\wedge\cdots\wedge v_n\mapsto \frac1{n!}\sum_{\sigma\in S_n}A_{\sigma1}v_1\wedge\cdots \wedge A_{\sigma n}v_n.$$ Since $\wedge^nV$ is one-dimensional, this linear map is multiplication by a scalar, which I will call $F'(A_1\cdots A_n)$.

I claim that the two linear functionals match. To check this, it suffices to check that $$F'(A^n)=\det(A).$$ But that's because $A^n$ acts on $\wedge^nV$ by $$v_1\wedge\cdots\wedge v_n\mapsto \frac1{n!}\sum_{\sigma\in S_n}Av_1\wedge\cdots \wedge Av_n=Av_1\wedge\cdots\wedge Av_n,$$ which is simply multiplication by $\det(A)$ by the definition of determinant.

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