# Kernel of ${\rm GL}(n,F)$ on ${\rm PG}(n-1,F)$ over a division ring $F$

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?

• 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. – Derek Holt Mar 21 at 10:35
• @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. – Easy Mar 21 at 12:44
• 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 $(kv)A = (kV)\lambda= k(v\lambda) = k(vA)$. – Derek Holt Mar 21 at 21:28
• 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. – Easy Mar 22 at 4:22
• Yes I see what you mean! See my answer below. – 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 $$ac=da=ca$$.
• 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. – Easy Mar 23 at 3:08
• And from this point of view, we should define ${\rm PGL}(1,F)=1$? – Easy Mar 23 at 3:08