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I'm trying to characterize the determinant in the following way. Consider a finite-dimensional vector space $V$ over $\mathbb{R}$ and the natural ring of its endomorphisms $\operatorname{End}(V)$. Given the determinant, we may define a function $d:\operatorname{End}(V)\rightarrow\mathbb{R}$ with the following properties:

  1. $d(1)=1$
  2. $d(fg)=d(f)d(g)$
  3. $d(f)\neq0\iff f$ has a two-sided inverse (i.e. $f$ is an automorphism)

Can we complete this set of conditions in such a way that the determinant could be defined. Is there some category theoretical point of view?

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    $\begingroup$ They don't. Take some power of the det instead. $\endgroup$
    – Abdelmalek Abdesselam
    Jan 10, 2021 at 19:52
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    $\begingroup$ Over $\mathbb{R}$ any positive real power of the absolute value of the determinant also satisfies these conditions. If you add the condition that $d$ is a polynomial then you get exactly the positive integer powers of the determinant. There are many different possible characterizations from here. $\endgroup$
    – Qiaochu Yuan
    Jan 10, 2021 at 21:06

1 Answer 1

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If you add the condition that $d$ is an alternating multilinear map of $f(e_1), \dots, f(e_n)$ where $\{e_1, \dots, e_n\}$ is the canonical basis of $V$ then you’ll get that $d$ is the determinant.

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    $\begingroup$ Worth noting though that this condition together with $d(1)=1$ is already sufficient to characterize the determinant, so conditions 2. and 3. from the question are not necessary anymore. $\endgroup$
    – Vercassivelaunos
    Jan 10, 2021 at 20:34
  • $\begingroup$ @Vercassivelaunos I agree with that. $\endgroup$
    – mathcounterexamples.net
    Jan 10, 2021 at 20:35

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