0
$\begingroup$

I am trying to understand the following automorphism of general linear group in terms of matrices.

(0) Let $V$ be a finite dimensional vector space over a field $k$.

(1) Given $u\in {\rm GL}(V)$, let $\check{u}$ denote contragradient of $u$.

(2) Let $t:V\rightarrow V^*$ denote a correlation map onto dual space.

Then the map $u\mapsto t^{-1}\check{u}t$ is an automorphism of ${\rm GL}(V)$.

Q.1 If we consider ${\rm GL}_n(k)$ instead of ${\rm GL}(V)$, how this map is intrepreted? I didn't understand properly the maps in (2).

Q.2 The map in (1) in terms of matrices is transpose map, am I right?


The automorphism mentioned above appears on first page of the paper A new type of automorphism of general linear group over a ring by Reiner I. (this link)

$\endgroup$
  • 1
    $\begingroup$ From a google search, it appears that the matrix for the contragradient is the transpose of the inverse. I can't seem to find a reasonable definition of the "correlation" map except in the context of a bilinear form --- are you talking about correlation for the standard inner product on $k^n$? $\endgroup$ – John Hughes Nov 28 '17 at 12:35
  • $\begingroup$ I am following a paper "A new type of automorphism of general linear group over a ring" by I. Reiner. The mentioned things before Q.1 are in front page of the paper. $\endgroup$ – Beginner Nov 28 '17 at 12:37
  • $\begingroup$ OK, that's fine. But I'm not following that paper, and if you want my help, you're going to have to give more details. At the very least, a link might be nice... $\endgroup$ – John Hughes Nov 28 '17 at 12:39
  • $\begingroup$ Ah... you need to read more carefully: it says that $t$ is a correlation mapping $V$ to its dual, not the correlation. Perhaps to save us time, you might mention what reference 7 in the paper is (or even link to it), since that's apparently where we (and you) can learn more about such correlation maps. $\endgroup$ – John Hughes Nov 28 '17 at 12:51
  • $\begingroup$ OK; thanks for the help, suggestion and noticing error; I will do it as you said. $\endgroup$ – Beginner Nov 28 '17 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.