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A number of years ago I studied a preon model (Journal of Mathematical Physics 38:3414-3426, 1997) in which the preons interacted like group elements. I thought it curious that you could sometimes define continuous transformations over the group elements which preserved the group structure.

Specifically, find a continuous family of V such that [edit] if we define $g_i$ as the elements of the finite group,

${g_i}' = V g_i$

if $g_i g_j = g_k$ then ${g_i}' {g_j}' = {g_k}'$

[edit] V is continuous in the sense that the difference between two different transformations can be arbitrarily small. If represented as a matrix, the matrix elements can be arbitrarily small. I know there can be permutations, for which this isn't true, but I wasn't interested in those. I was representing the group elements as unit vectors in N-space where N is the group size, and the transformations as NxN complex matrices.

I was using complex numbers, and in that case at least, there aren't any such continuous transformations if the finite group is abelian. (That's why I don't think the model is physical.) Some non-abelian groups admit transformations isomorphic to SU(2) and SU(3), and of other things.

I'm confident that this is well-known, but I'm not deeply familiar with the research in group algebras, and I'm not sure where to look to find out more--e.g. if there's a pattern to this.

[edit] My apologies for a lack of precision in the initial question.

[edit2] Maybe a concrete example is in order The simplest non-abelian group has the Cayley table that I specify as $\begin{array}{c|cccccc} X&0&1&2&3&4&5 \\ \hline 0&0&1&2&3&4&5 \\ 1&1&2&0&4&5&3 \\ 2&2&0&1&5&3&4 \\ 3&3&5&4&0&2&1 \\ 4&4&3&5&1&0&2 \\ 5&5&4&3&2&1&0 \end{array}$

If an element in the group algebra is represented by a vector, with $g_0$ = (1,0,0,0,0,0), $g_1$=(0,1,0,0,0,0), etc, then if a, b, and c are infinitesimal small changes, $\bar{1}$ + $\delta V$ is a transformation which, when applied to the group elements, gives new elements whose products are isomorphic to those of the original group elements. $\bar{1}$ is the identity 6x6 matrix, and

$\delta V$ = $ \begin{array}{cccccc} 0&0&0 &0&0&0\\ 0&0&0 &a&-b&-a+b\\ 0&0&0 &-a&b&a-b\\ 0&a&-a& 0&c&-c \\ 0&-b&b&-c&0&c \\ 0&-a+b&a-b&c&-c&0 \\ \end{array}$

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    $\begingroup$ What is $V$, what is $g_i$, what's a "continuous transformation over the group elements", etc etc $\endgroup$
    – Qfwfq
    Commented Jul 29, 2019 at 22:56
  • $\begingroup$ I guess what you are looking is a continuous family of group automorphisms. Of course this depend on the group and the topology in the set of the automorphisms. If the group is finite then obviusly the set of automorphisms is finite. $\endgroup$ Commented Jul 29, 2019 at 23:16
  • $\begingroup$ msp.org/pjm/1978/76-1/pjm-v76-n1-p15-s.pdf this may help. $\endgroup$ Commented Jul 29, 2019 at 23:18
  • $\begingroup$ I'm starting to read that. Thanks. $\endgroup$ Commented Jul 30, 2019 at 4:10
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    $\begingroup$ This question is a good illustration of the kinds of problems that can arise when physicists and mathematicians try to talk to each other. In language comfortable for mathematicians, I think the question is: let $G$ be a finite group and let $A$ be its group algebra. Consider the automorphism group of $A$. What can the connected component of Aut$(A)$ containing the identity look like? $\endgroup$
    – Nik Weaver
    Commented Jul 30, 2019 at 5:11

1 Answer 1

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A nice example is in any algebra, by definition multiplying by a arbitary element is an automorphism of it group structure. So the algebra is the family you are looking for.

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    $\begingroup$ Yeah, except when it isn't. You need the element to be invertible in the algebra. $\endgroup$ Commented Jul 29, 2019 at 23:50
  • $\begingroup$ I meant just a morphism from $G$ to $G$ not an automorphism, which is what he is asking for. Sorry, this are just words, my idea is correct is a shame you couldn´t read this from what I wrote. $\endgroup$ Commented Jul 30, 2019 at 23:47
  • $\begingroup$ This is just the distributive law common!!! $\endgroup$ Commented Jul 30, 2019 at 23:51

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