05/16/2021, 11:39 PM
(05/16/2021, 10:00 AM)MphLee Wrote:JmsNxn Wrote:Oh, Mphlee. I guess I misunderstood what equivariant means. I think this is the perfect term. I'll cast my vote for that. I thought it was reserved for,
\(
\chi(f(t,x)) = f(t,\chi(x))\\
\)
I didn't realize it could be null in the t variable.
OFC! If that equation is valid for all t then, as a corollary, also for t=1... which is our initial function. \( \chi(f(1,x)) = g(1,\chi(x))\\ \).
I guess that is pretty obvious
I guess I was assuming that \( \chi \) would depend on \( t \); but now that I think about it; the point is that it doesn't.

