I am reading Peter Cameron's note on Classical Groups and I got confused with Proposition 2.1 on page 14.

I have no problem in proving that the elements in kernel are scalars. However, I don't understand the last paragraph of the proof, which shows that the kernel lies in ${\bf Z}(F)$. It seems that the argument tries to insist that an element $A$ in the kernel should induce a linear map. But I don't see any reason for so. (To be in the kernel, it just needs to stabilize all 1-spaces in $F^n$ but need not to stabilize any particular vectors, right?)

In fact, I believe that there is also a mistake in the paragraph before the proposition statement, which says that every $A\in{\rm GL}(n,F)$ induces a linear map on $F^n$. For a simple case, take $A=\lambda I$, then $$(kv)A=(\lambda k)v=(\lambda k\lambda^{-1})(\lambda v)=k^\lambda(vA)$$ so the induced map of $A$ should be semilinear instead?

  • $\begingroup$ There is no mistake. He is writing the $A$ on the right of its argument throughout, so why are you suddenly writing $A(kv)$ rather than $(kv)A$? It is important when dealing with spaces over non-commutative rings that scalars operate on one side and linear maps on the other. He has scalars on the left and linear maps on the right. $\endgroup$ – Derek Holt Mar 21 at 10:35
  • $\begingroup$ @DerekHolt Sorry, I was just used to putting $A$ on the left. I have modified it. But the same argument shows that $A$ is semilinear. And my main question is why the kernel cannot be all scalars? It seems that a semilinear scalar map fixes all 1-spaces as well. $\endgroup$ – Easy Mar 21 at 12:44
  • $\begingroup$ No it doesn't, just as in the commutative case, ${\rm GL}(n,F)$ corresponds to linear maps $V \to V$. Elements of ${\rm GL}(n,F)$ are matrices that act on the right. So when $A = \lambda I$, the entries of the row vector $v$ are multiplied on the right by $\lambda$, not the left. So we get $(kv)A = (kV)\lambda= k(v\lambda) = k(vA)$. $\endgroup$ – Derek Holt Mar 21 at 21:28
  • $\begingroup$ As you said, matrices act on the right. So for the noncommutative case, we have $\lambda v\neq v\lambda$ in general. Then reading on Peter's proof we should have $(ae_1)c=(ae_1)A=a(e_1A)=a(e_1c)$, and I see no necessity that $a,c$ commute. $\endgroup$ – Easy Mar 22 at 4:22
  • $\begingroup$ Yes I see what you mean! See my answer below. $\endgroup$ – Derek Holt Mar 22 at 4:37

I think the result about only central scalars being in the kernel only makes sense if $n > 1$.

We have shown so far that $A = cI$ for some scalar $c$. Now let $a \in F$ and consider $(ae_1+e_2)A = ace_1 + ce_2$. This must equal $d(ae_1+e_2)$ for some scalar $d$. From the $e_2$ coefficient, we get $c=d$, and hence $ac=da=ca$.

  • $\begingroup$ Em, I see. So we intentionally set coefficient and matrix on two sides of a vector so that ${\rm GL}(n,F)$ behaves linearly, otherwise ${\rm GL}(n,F)$ might behave semilinearly as I pointed out above. $\endgroup$ – Easy Mar 23 at 3:08
  • $\begingroup$ And from this point of view, we should define ${\rm PGL}(1,F)=1$? $\endgroup$ – Easy Mar 23 at 3:08

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