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I'm studying this proof in the book cited in the title.

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The author say that any multiplicative function $\chi\colon\mathsf{GL}(k)\longrightarrow\mathbb{C}^\times$ is a power of the determinant. But what if I take $\chi(g)=\overline{\det g}$? Is this a mistake? If not and I'm missing something where can I find a proof of this fact?

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    $\begingroup$ The function $f$ is polynomial, and we can fix a generic $M$ and define $\chi(g)=f(gM)f(M)^{-1}$, so $\chi$ is also a polynomial function. Any character that is a polynomial function is a power of the determinant. $\endgroup$
    – Neil Strickland
    Commented Mar 30, 2018 at 20:15
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    $\begingroup$ A more pathological multiplicative function is $\chi(g) = \exp f(\log |\det g|)$, where $f$ is any solution to Cauchy's functional equation, and we further define $\chi(g) = 0$ whenever $g$ is singular. $\endgroup$
    – Adam P. Goucher
    Commented Mar 30, 2018 at 20:31
  • $\begingroup$ @NeilStrickland many thanks for the helpfull remark. Can you give me a rference where I can find your claim? I know almost nothing about Rep. Theory. $\endgroup$
    – Vincenzo Zaccaro
    Commented Mar 30, 2018 at 20:59
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    $\begingroup$ I'm sure it goes back much earlier than that, but a natural source for polynomial representations of $\mathrm{GL}(n)$ (the 1-dimensional ones being those @NeilStrickland describes) is Green - Polynomial representations of $\mathrm{GL}(n)$. $\endgroup$
    – LSpice
    Commented Mar 30, 2018 at 22:44
  • $\begingroup$ This is not a mistake, in this book they consider algebraic geometry, and hence consider only \textbf{polynomial} (i.e. regular) functions. The complex conjugate of a polynomial function is anti-holomorphic and hence not a "function" in Gelfand-Kapranov-Zelevinski context. $\endgroup$
    – S. carmeli
    Commented Mar 31, 2018 at 17:26

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